Chapter 1 Relations And Functions

Given:

Let A be the set of all students in a boys' school.

Relation R is defined as R = {(a, b) : a is sister of b}, where a, b ∈ A.

Relation R' is defined as R' = {(a, b) : the difference between heights of a and b is less than 3 meters}, where a, b ∈ A.

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To Show:

1. The relation R is an empty relation.

2. The relation R' is a universal relation.

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Proof for Relation R (Empty Relation):

The set A consists of all students of a boys' school, which means every student in set A is a boy.

For an ordered pair (a, b) to be in the relation R, the element 'a' must be the sister of element 'b'.

Since 'a' is a student from the boys' school, 'a' must be a boy. A boy cannot be a sister.

Therefore, it is impossible to find any pair (a, b) in A × A that satisfies the condition "a is sister of b".

Since no element of A × A is in the relation R, the relation R contains no ordered pairs.

Hence, R is the empty relation (R = ∅).

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Proof for Relation R' (Universal Relation):

The relation R' is defined by the condition that the difference between the heights of any two students 'a' and 'b' from the school is less than 3 meters.

Let the height of student 'a' be $h_a$ and the height of student 'b' be $h_b$. The condition is $|h_a - h_b| < 3$ meters.

The height of a human being is a finite positive value. A difference of 3 meters (approximately 9 feet 10 inches) between the heights of two school students is a physical impossibility.

The difference in height between any two students in the school will realistically be much smaller than 3 meters.

Therefore, for any pair of students (a, b) chosen from the school, the condition $|h_a - h_b| < 3$ meters will always be true.

This means that every possible ordered pair (a, b) from A × A is an element of the relation R'.

Hence, R' is the universal relation (R' = A × A).

Given:

Let T be the set of all triangles in a plane.

A relation R is defined on T as R = {(T₁, T₂) : T₁ is congruent to T₂}.

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To Show:

The relation R is an equivalence relation.

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Proof:

To prove that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.

1. Reflexivity:

For a relation to be reflexive, (a, a) must belong to R for every a in the set. In our case, for any triangle T₁ ∈ T, we must check if (T₁, T₁) ∈ R.

The condition is that T₁ must be congruent to T₁.

By the properties of geometry, every triangle is congruent to itself (T₁ ≅ T₁).

Thus, (T₁, T₁) ∈ R for all T₁ ∈ T.

Therefore, R is reflexive.

2. Symmetry:

For a relation to be symmetric, if (a, b) ∈ R, then (b, a) must also ∈ R. In our case, if (T₁, T₂) ∈ R, we must check if (T₂, T₁) ∈ R.

Let (T₁, T₂) ∈ R. This means that T₁ is congruent to T₂ (T₁ ≅ T₂).

If T₁ is congruent to T₂, then it follows from the properties of congruence that T₂ is also congruent to T₁ (T₂ ≅ T₁).

This implies that (T₂, T₁) ∈ R.

Therefore, R is symmetric.

3. Transitivity:

For a relation to be transitive, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also ∈ R. In our case, if (T₁, T₂) ∈ R and (T₂, T₃) ∈ R, we must check if (T₁, T₃) ∈ R.

Let (T₁, T₂) ∈ R and (T₂, T₃) ∈ R. This means:

T₁ is congruent to T₂ (T₁ ≅ T₂)

T₂ is congruent to T₃ (T₂ ≅ T₃)

From the transitive property of congruence, if T₁ is congruent to T₂ and T₂ is congruent to T₃, then T₁ is congruent to T₃ (T₁ ≅ T₃).

This implies that (T₁, T₃) ∈ R.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it satisfies all the conditions for an equivalence relation.

Hence, R is an equivalence relation.

Given:

Let L be the set of all lines in a plane.

A relation R is defined on L as R = {(L₁, L₂) : L₁ is perpendicular to L₂}.

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To Show:

The relation R is symmetric, but not reflexive and not transitive.

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Proof:

We check the three properties for the relation R.

1. Reflexivity:

For R to be reflexive, (L₁, L₁) must ∈ R for every line L₁ ∈ L. This would mean L₁ is perpendicular to L₁.

A line cannot be perpendicular to itself. A line is parallel to itself.

Since L₁ is not perpendicular to L₁ for any line L₁, (L₁, L₁) ∉ R.

Therefore, R is not reflexive.

2. Symmetry:

For R to be symmetric, if (L₁, L₂) ∈ R, then (L₂, L₁) must also ∈ R.

Let (L₁, L₂) ∈ R. This means that line L₁ is perpendicular to line L₂ (L₁ ⊥ L₂).

If L₁ is perpendicular to L₂, then L₂ is also perpendicular to L₁. The relationship of being perpendicular is mutual.

Thus, L₂ ⊥ L₁, which means (L₂, L₁) ∈ R.

Therefore, R is symmetric.

3. Transitivity:

For R to be transitive, if (L₁, L₂) ∈ R and (L₂, L₃) ∈ R, then (L₁, L₃) must also ∈ R.

Let (L₁, L₂) ∈ R and (L₂, L₃) ∈ R. This means:

L₁ is perpendicular to L₂ (L₁ ⊥ L₂)

L₂ is perpendicular to L₃ (L₂ ⊥ L₃)

In a two-dimensional plane, if two lines (L₁ and L₃) are perpendicular to the same line (L₂), then they must be parallel to each other (L₁ || L₃).

For the relation to be transitive, we would need L₁ to be perpendicular to L₃. But L₁ is parallel to L₃, not perpendicular.

Since (L₁, L₃) ∉ R, the condition for transitivity is not met.

Therefore, R is not transitive.

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In conclusion, the relation R is symmetric but neither reflexive nor transitive.

Given:

Let A = {1, 2, 3}.

A relation R on A is given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

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To Show:

The relation R is reflexive, but not symmetric and not transitive.

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Proof:

We check the three properties for the relation R.

1. Reflexivity:

A relation is reflexive if (a, a) ∈ R for every element a ∈ A.

The elements of A are 1, 2, and 3. We need to check if (1, 1), (2, 2), and (3, 3) are in R.

Looking at the set R, we can see that (1, 1) ∈ R, (2, 2) ∈ R, and (3, 3) ∈ R.

Since the condition is met for all elements of A, R is reflexive.

2. Symmetry:

A relation is symmetric if for every (a, b) ∈ R, the pair (b, a) is also in R.

Let's check the pairs in R:

• (1, 1) is in R, and its reverse (1, 1) is in R. (OK)
• (2, 2) is in R, and its reverse (2, 2) is in R. (OK)
• (3, 3) is in R, and its reverse (3, 3) is in R. (OK)
• (1, 2) is in R. We check for its reverse, (2, 1). The pair (2, 1) is not in the set R.

Since we found a pair (1, 2) whose reverse (2, 1) is not in R, the relation is not symmetric.

Therefore, R is not symmetric.

3. Transitivity:

A relation is transitive if for every pair of the form (a, b) ∈ R and (b, c) ∈ R, the pair (a, c) is also in R.

We look for a chain of pairs (a, b) and (b, c) in R.

We can see that (1, 2) ∈ R and (2, 3) ∈ R.

According to the transitivity rule, the pair (1, 3) must also be in R for the relation to be transitive.

However, looking at the set R, the pair (1, 3) is not present.

Since we found a case where (a, b) ∈ R and (b, c) ∈ R, but (a, c) ∉ R, the relation is not transitive.

Therefore, R is not transitive.

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In conclusion, the relation R is reflexive but neither symmetric nor transitive.

Given:

Let Z be the set of all integers.

A relation R is defined on Z as R = {(a, b) : 2 divides (a - b)}.

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To Show:

The relation R is an equivalence relation.

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Proof:

To prove that R is an equivalence relation, we must show that it is reflexive, symmetric, and transitive.

1. Reflexivity:

For any integer a ∈ Z, we must check if (a, a) ∈ R. The condition is that 2 divides (a - a).

The expression $(a - a) = 0$.

Since 2 divides 0 (as $0 = 2 \times 0$), the condition is satisfied for any integer a.

Therefore, R is reflexive.

2. Symmetry:

Let (a, b) ∈ R for any two integers a, b ∈ Z. We need to show that (b, a) ∈ R.

If (a, b) ∈ R, then 2 divides (a - b). This means that $(a - b)$ is an even integer, so we can write:

Now, consider the expression (b - a):

Since k is an integer, -k is also an integer. Thus, (b - a) is also a multiple of 2.

This means 2 divides (b - a), so (b, a) ∈ R.

Therefore, R is symmetric.

3. Transitivity:

Let (a, b) ∈ R and (b, c) ∈ R for any three integers a, b, c ∈ Z. We need to show that (a, c) ∈ R.

If (a, b) ∈ R, then 2 divides (a - b). So, $a - b = 2k_1$ for some integer $k_1$.

If (b, c) ∈ R, then 2 divides (b - c). So, $b - c = 2k_2$ for some integer $k_2$.

Now, let's consider the expression (a - c). We can write it as:

Substituting the expressions from above:

Since $k_1$ and $k_2$ are integers, their sum $(k_1 + k_2)$ is also an integer. Thus, (a - c) is a multiple of 2.

This means 2 divides (a - c), so (a, c) ∈ R.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, R is an equivalence relation.

Given:

Let A = {1, 2, 3, 4, 5, 6, 7}.

A relation R on A is defined as R = {(a, b) : both a and b have the same parity (i.e., both are odd or both are even)}.

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Part 1: Show that R is an equivalence relation.

1. Reflexivity:

For any element a ∈ A, the pair (a, a) must be in R. The condition is that 'a' and 'a' are both odd or both even. This is always true, as any number has the same parity as itself. Therefore, R is reflexive.

2. Symmetry:

If (a, b) ∈ R, then (b, a) must be in R. If (a, b) ∈ R, it means 'a' and 'b' have the same parity. The statement "a and b have the same parity" is the same as "b and a have the same parity". Thus, if (a, b) ∈ R, then (b, a) ∈ R. Therefore, R is symmetric.

3. Transitivity:

If (a, b) ∈ R and (b, c) ∈ R, then (a, c) must be in R. If (a, b) ∈ R, 'a' and 'b' have the same parity. If (b, c) ∈ R, 'b' and 'c' have the same parity. If 'a' has the same parity as 'b', and 'b' has the same parity as 'c', then 'a' must have the same parity as 'c'. Therefore, (a, c) ∈ R. Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation.

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Part 2: Show properties of subsets.

Let's define two subsets:

$O = \{1, 3, 5, 7\}$ (the set of odd numbers in A)

$E = \{2, 4, 6\}$ (the set of even numbers in A)

1. All elements of O = {1, 3, 5, 7} are related to each other.

Take any two elements 'a' and 'b' from the set O. Both 'a' and 'b' are odd. According to the definition of R, if both elements are odd, they are related. For example, (1, 3) ∈ R, (5, 7) ∈ R, (1, 7) ∈ R, etc. Therefore, all elements of {1, 3, 5, 7} are related to each other.

2. All elements of E = {2, 4, 6} are related to each other.

Take any two elements 'a' and 'b' from the set E. Both 'a' and 'b' are even. According to the definition of R, if both elements are even, they are related. For example, (2, 4) ∈ R, (2, 6) ∈ R, (4, 6) ∈ R, etc. Therefore, all elements of {2, 4, 6} are related to each other.

3. No element of O = {1, 3, 5, 7} is related to any element of E = {2, 4, 6}.

Take any element 'a' from set O (so 'a' is odd) and any element 'b' from set E (so 'b' is even). For 'a' and 'b' to be related, they must both be odd or both be even. This is not the case, as one is odd and the other is even. Therefore, no element from O is related to any element from E. For example, (1, 2) ∉ R, (3, 4) ∉ R, (5, 6) ∉ R, etc.

This shows that the equivalence relation R partitions the set A into two disjoint equivalence classes: the set of odd numbers and the set of even numbers.

Question 1 (i):

Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}.

The relation can be written as R = {(x, y) : y = 3x}.

Let's list the pairs in R for x, y $\in$ A:

• If x = 1, y = 3(1) = 3. So, (1, 3) $\in$ R.
• If x = 2, y = 3(2) = 6. So, (2, 6) $\in$ R.
• If x = 3, y = 3(3) = 9. So, (3, 9) $\in$ R.
• If x = 4, y = 3(4) = 12. So, (4, 12) $\in$ R.
• If x = 5, y = 3(5) = 15. Since 15 $\notin$ A, no pairs with x $\geq$ 5 are in R.

So, R = {(1, 3), (2, 6), (3, 9), (4, 12)}.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ A.

This requires y = x, which means $3x = x$. This implies $2x = 0$, so $x = 0$.

Since 0 is not in the set A = {1, 2, ..., 14}, there is no element x in A such that (x, x) $\in$ R.

For example, for x = 1, (1, 1) $\notin$ R because $3(1) \neq 1$.

Therefore, R is not reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

Consider the pair (1, 3) $\in$ R (since $3 = 3 \times 1$).

For R to be symmetric, (3, 1) must also be in R.

The condition for (3, 1) $\in$ R is $1 = 3(3)$, which is $1 = 9$. This is false.

Since (1, 3) $\in$ R but (3, 1) $\notin$ R, R is not symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Consider the pair (1, 3) $\in$ R.

We look for a pair starting with 3 in R. (3, 9) $\in$ R.

For R to be transitive, (1, 9) must also be in R.

The condition for (1, 9) $\in$ R is $9 = 3(1)$, which is $9 = 3$. This is false.

Since (1, 3) $\in$ R and (3, 9) $\in$ R, but (1, 9) $\notin$ R, R is not transitive.

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Conclusion for (i): The relation R is neither reflexive, nor symmetric, nor transitive.

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Question 1 (ii):

Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}.

The set N = {1, 2, 3, ...}. The condition on x is $x \in N$ and $x < 4$. So possible values for x are 1, 2, 3.

Let's list the pairs in R:

• If x = 1, y = 1 + 5 = 6. Since 6 $\in$ N, (1, 6) $\in$ R.
• If x = 2, y = 2 + 5 = 7. Since 7 $\in$ N, (2, 7) $\in$ R.
• If x = 3, y = 3 + 5 = 8. Since 8 $\in$ N, (3, 8) $\in$ R.

So, R = {(1, 6), (2, 7), (3, 8)}.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ N.

This requires y = x, which means $x = x + 5$. This implies $0 = 5$, which is false.

So, there is no element x in N such that (x, x) $\in$ R.

For example, for x = 1, (1, 1) $\notin$ R because $1 \neq 1 + 5$.

Therefore, R is not reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ N.

Consider the pair (1, 6) $\in$ R.

For R to be symmetric, (6, 1) must also be in R.

The condition for (6, 1) $\in$ R is $1 = 6 + 5$ and $6 < 4$. This requires $1 = 11$ (false) and $6 < 4$ (false).

Since (1, 6) $\in$ R but (6, 1) $\notin$ R, R is not symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ N.

We need to find if there exist pairs (x, y) $\in$ R and (y, z) $\in$ R.

From the pairs in R: (1, 6), (2, 7), (3, 8).

Consider (1, 6) $\in$ R. For transitivity, we need to check if there is a pair (6, z) in R.

A pair (y, z) is in R if $z = y + 5$ and $y < 4$. In this case, y = 6. Since $6 \nless 4$, there is no pair (6, z) in R.

Similarly, there are no pairs (7, z) or (8, z) in R because 7 and 8 are not less than 4.

Since the condition "(x, y) $\in$ R and (y, z) $\in$ R" is never satisfied for any x, y, z in N, the implication "implies (x, z) $\in$ R" is vacuously true.

Therefore, R is transitive.

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Conclusion for (ii): The relation R is transitive but neither reflexive nor symmetric.

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Question 1 (iii):

Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}.

The condition "y is divisible by x" means that $y = kx$ for some integer k.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ A.

This requires x to be divisible by x.

For any x $\in$ A = {1, 2, 3, 4, 5, 6}, x is divisible by x (since x = 1 * x, and 1 is an integer).

Thus, (x, x) $\in$ R for all x $\in$ A.

Therefore, R is reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

Assume (x, y) $\in$ R. This means y is divisible by x, so $y = kx$ for some integer k.

For R to be symmetric, (y, x) must also be in R. This means x must be divisible by y, so $x = my$ for some integer m.

Substitute y from the first equation into the second: $x = m(kx) = (mk)x$.

If $x \neq 0$, then $mk = 1$. Since x, y $\in$ A, x and y are positive integers. For mk to be 1, both m and k must be either 1 or -1. As x, y are positive, k must be positive, so k=1. This means y=x. If y=x, then m=1. So symmetry only holds for pairs where x=y.

Consider the pair (2, 4) $\in$ R (since 4 is divisible by 2, $4 = 2 \times 2$).

For R to be symmetric, (4, 2) must also be in R.

The condition for (4, 2) $\in$ R is that 2 is divisible by 4. This is false (2 = $k \times 4$ has no integer solution for k).

Since (2, 4) $\in$ R but (4, 2) $\notin$ R, R is not symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Assume (x, y) $\in$ R and (y, z) $\in$ R.

(x, y) $\in$ R means y is divisible by x, so $y = kx$ for some integer $k$.

(y, z) $\in$ R means z is divisible by y, so $z = my$ for some integer $m$.

We need to check if (x, z) $\in$ R, which means z is divisible by x.

Substitute y from the first equation into the second: $z = m(kx) = (mk)x$.

Since k and m are integers, the product mk is also an integer.

This shows that z is an integer multiple of x, so z is divisible by x.

Thus, (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Therefore, R is transitive.

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Conclusion for (iii): The relation R is reflexive and transitive but not symmetric.

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Question 1 (iv):

Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}.

The set Z is the set of all integers: {... -2, -1, 0, 1, 2, ...}.

The condition "x – y is an integer" is always true for any two integers x and y, since the difference of any two integers is always an integer.

This means R = Z $\times$ Z, which is the universal relation on Z.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ Z.

This requires x – x to be an integer.

For any integer x, $x - x = 0$. Since 0 is an integer, the condition is satisfied.

Thus, (x, x) $\in$ R for all x $\in$ Z.

Therefore, R is reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ Z.

Assume (x, y) $\in$ R. This means x – y is an integer.

Let $x - y = k$, where k is an integer.

Consider y – x. We have $y - x = -(x - y) = -k$.

Since k is an integer, -k is also an integer.

Thus, y – x is an integer, which means (y, x) $\in$ R.

So, (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ Z.

Therefore, R is symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ Z.

Assume (x, y) $\in$ R and (y, z) $\in$ R.

(x, y) $\in$ R means x – y is an integer. Let $x - y = k_1$, where $k_1$ is an integer.

(y, z) $\in$ R means y – z is an integer. Let $y - z = k_2$, where $k_2$ is an integer.

Consider x – z. We have $x - z = (x - y) + (y - z) = k_1 + k_2$.

Since $k_1$ and $k_2$ are integers, their sum $k_1 + k_2$ is also an integer.

Thus, x – z is an integer, which means (x, z) $\in$ R.

So, (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ Z.

Therefore, R is transitive.

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Conclusion for (iv): The relation R is reflexive, symmetric, and transitive. It is an equivalence relation.

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Question 1 (v):

Relation R in the set A of human beings in a town at a particular time.

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(a) R = {(x, y) : x and y work at the same place}

Reflexivity: For any person x $\in$ A, does x work at the same place as x? Yes, a person works at the same place as themselves. R is reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x and y work at the same place. If x and y work at the same place, then y and x also work at the same place. So (y, x) $\in$ R. R is symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. This means x and y work at the same place, and y and z work at the same place. If x works at the same place as y, and y works at the same place as z, then x must work at the same place as z. So (x, z) $\in$ R. R is transitive.

Conclusion for (v) (a): The relation is reflexive, symmetric, and transitive (an equivalence relation).

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(b) R = {(x, y) : x and y live in the same locality}

Reflexivity: For any person x $\in$ A, does x live in the same locality as x? Yes. R is reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x and y live in the same locality. If x and y live in the same locality, then y and x also live in the same locality. So (y, x) $\in$ R. R is symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. This means x and y live in the same locality, and y and z live in the same locality. If x lives in the same locality as y, and y lives in the same locality as z, then x must live in the same locality as z. So (x, z) $\in$ R. R is transitive.

Conclusion for (v) (b): The relation is reflexive, symmetric, and transitive (an equivalence relation).

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(c) R = {(x, y) : x is exactly 7 cm taller than y}

Reflexivity: For any person x $\in$ A, is x exactly 7 cm taller than x? No, the height difference between x and x is 0 cm. R is not reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x is exactly 7 cm taller than y. Let height(x) = $h_x$ and height(y) = $h_y$. The condition is $h_x = h_y + 7$. For R to be symmetric, (y, x) must be in R, meaning y is exactly 7 cm taller than x ($h_y = h_x + 7$). If $h_x = h_y + 7$, then $h_y = h_x - 7$. So y is exactly 7 cm shorter than x, not 7 cm taller. R is not symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. This means x is exactly 7 cm taller than y ($h_x = h_y + 7$) and y is exactly 7 cm taller than z ($h_y = h_z + 7$). For R to be transitive, (x, z) must be in R, meaning x is exactly 7 cm taller than z ($h_x = h_z + 7$). Substitute the second equation into the first: $h_x = (h_z + 7) + 7 = h_z + 14$. This means x is exactly 14 cm taller than z, not 7 cm. R is not transitive.

Conclusion for (v) (c): The relation is neither reflexive, nor symmetric, nor transitive.

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(d) R = {(x, y) : x is wife of y}

Reflexivity: For any person x $\in$ A, is x the wife of x? No, a person cannot be their own spouse. R is not reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x is the wife of y. If x is the wife of y, then y must be the husband of x (assuming a traditional marriage). Can y be the wife of x? No, the wife is x. So (y, x) $\notin$ R. R is not symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. If (x, y) $\in$ R, then x is the wife of y, which implies x is female and y is male. If (y, z) $\in$ R, then y is the wife of z, which implies y is female and z is male. For the premise "(x, y) $\in$ R and (y, z) $\in$ R" to be true, y must be both male (from the first pair) and female (from the second pair). This is impossible for a human being. Since the premise of the transitivity condition is never met, the implication "implies (x, z) $\in$ R" is vacuously true.

Therefore, R is transitive.

Conclusion for (v) (d): The relation is transitive but neither reflexive nor symmetric.

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(e) R = {(x, y) : x is father of y}

Reflexivity: For any person x $\in$ A, is x the father of x? No, a person cannot be their own parent. R is not reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x is the father of y. If x is the father of y, is y the father of x? No, y is the child of x. So (y, x) $\notin$ R. R is not symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. If (x, y) $\in$ R, then x is the father of y. If (y, z) $\in$ R, then y is the father of z. For R to be transitive, (x, z) must be in R, meaning x is the father of z. If x is the father of y, and y is the father of z, then x is the grandfather of z, not the father of z. R is not transitive.

Conclusion for (v) (e): The relation is neither reflexive, nor symmetric, nor transitive.

Given:

The set is R (real numbers).

The relation R is defined as R = {(a, b) : a $\le$ b$^2$}, where a, b $\in$ R.

---

To Show:

R is neither reflexive nor symmetric nor transitive.

---

Proof:

We examine the three properties of relations.

---

1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

For the relation R on the set of real numbers, we need to check if (a, a) $\in$ R for every real number a.

The condition for (a, a) $\in$ R is that a $\le$ a$^2$.

This inequality $a \le a^2$ is not true for all real numbers. For example, consider a real number between 0 and 1, such as $a = \frac{1}{2}$.

If $a = \frac{1}{2}$, then $a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

Is $\frac{1}{2} \le \frac{1}{4}$? No, $\frac{1}{2} > \frac{1}{4}$.

So, $(\frac{1}{2}, \frac{1}{2}) \notin$ R because $\frac{1}{2} \not\le (\frac{1}{2})^2$.

Since there exists at least one real number 'a' for which $a \not\le a^2$, R is not reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R for some real numbers a and b, and we need to check if (b, a) $\in$ R.

If (a, b) $\in$ R, it means a $\le$ b$^2$.

For R to be symmetric, (b, a) must be in R, which means b $\le$ a$^2$.

Let's consider a counterexample. Choose a pair (a, b) such that a $\le$ b$^2$ is true, but b $\le$ a$^2$ is false.

Consider $a = 1$ and $b = 2$.

Is (1, 2) $\in$ R? Check the condition: $1 \le 2^2 \implies 1 \le 4$. This is true. So (1, 2) $\in$ R.

Is (2, 1) $\in$ R? Check the condition: $2 \le 1^2 \implies 2 \le 1$. This is false. So (2, 1) $\notin$ R.

Since (1, 2) $\in$ R but (2, 1) $\notin$ R, R is not symmetric.

---

3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R and (b, c) $\in$ R for some real numbers a, b, and c, and we need to check if (a, c) $\in$ R.

If (a, b) $\in$ R, then a $\le$ b$^2$.

If (b, c) $\in$ R, then b $\le$ c$^2$.

For R to be transitive, if $a \le b^2$ and $b \le c^2$, it must imply $a \le c^2$ for all a, b, c.

Let's consider a counterexample. We need values a, b, c such that $a \le b^2$ and $b \le c^2$ are true, but $a \le c^2$ is false.

Consider $a = 3$, $b = -2$, and $c = 1.5$.

Is (3, -2) $\in$ R? Check the condition: $3 \le (-2)^2 \implies 3 \le 4$. This is true. So (3, -2) $\in$ R.

Is (-2, 1.5) $\in$ R? Check the condition: $-2 \le (1.5)^2 \implies -2 \le 2.25$. This is true. So (-2, 1.5) $\in$ R.

Now check if (a, c) = (3, 1.5) $\in$ R. The condition is $3 \le (1.5)^2$.

Is $3 \le 2.25$? This is false.

Since (3, -2) $\in$ R and (-2, 1.5) $\in$ R, but (3, 1.5) $\notin$ R, R is not transitive.

---

In conclusion, the relation R = {(a, b) : a $\le$ b$^2$} on the set of real numbers is neither reflexive, nor symmetric, nor transitive.

Given:

Set A = {1, 2, 3, 4, 5, 6}

Relation R in A is defined as R = {(a, b) : b = a + 1}, where a, b $\in$ A.

---

To Check:

Whether R is reflexive, symmetric, or transitive.

---

Let's list the pairs in R:

• If a = 1, b = 1 + 1 = 2. So, (1, 2) $\in$ R.
• If a = 2, b = 2 + 1 = 3. So, (2, 3) $\in$ R.
• If a = 3, b = 3 + 1 = 4. So, (3, 4) $\in$ R.
• If a = 4, b = 4 + 1 = 5. So, (4, 5) $\in$ R.
• If a = 5, b = 5 + 1 = 6. So, (5, 6) $\in$ R.
• If a = 6, b = 6 + 1 = 7. Since 7 $\notin$ A, no pairs with a = 6 are in R.

So, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.

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Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

This requires b = a, which means $a = a + 1$. This implies $0 = 1$, which is false.

So, there is no element a in A such that (a, a) $\in$ R.

For example, for a = 1, (1, 1) $\notin$ R because $1 \neq 1 + 1$.

Therefore, R is not reflexive.

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Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R. This means b = a + 1.

For R to be symmetric, (b, a) must also be in R. This means a = b + 1.

Substitute b from the first equation into the condition for the second: $a = (a + 1) + 1 = a + 2$.

This implies $0 = 2$, which is false.

Consider the pair (1, 2) $\in$ R (since $2 = 1 + 1$).

For R to be symmetric, (2, 1) must also be in R.

The condition for (2, 1) $\in$ R is $1 = 2 + 1$, which is $1 = 3$. This is false.

Since (1, 2) $\in$ R but (2, 1) $\notin$ R, R is not symmetric.

---

Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means b = a + 1.

(b, c) $\in$ R means c = b + 1.

We need to check if (a, c) $\in$ R, which means c = a + 1.

Substitute b from the first equation into the second: $c = (a + 1) + 1 = a + 2$.

For the relation to be transitive, we need $c = a + 1$. We found $c = a + 2$. These are not the same.

Consider the pairs (1, 2) $\in$ R and (2, 3) $\in$ R.

For R to be transitive, (1, 3) must also be in R.

The condition for (1, 3) $\in$ R is $3 = 1 + 1$, which is $3 = 2$. This is false.

Since (1, 2) $\in$ R and (2, 3) $\in$ R, but (1, 3) $\notin$ R, R is not transitive.

---

Conclusion: The relation R is neither reflexive, nor symmetric, nor transitive.

Given:

The set is R (real numbers).

The relation R is defined as R = {(a, b) : a $\le$ b}, where a, b $\in$ R.

---

To Show:

R is reflexive and transitive but not symmetric.

---

Proof:

We examine the three properties of relations.

---

1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

For the relation R on the set of real numbers, we need to check if (a, a) $\in$ R for every real number a.

The condition for (a, a) $\in$ R is that a $\le$ a.

This inequality is always true for any real number a.

Thus, (a, a) $\in$ R for all a $\in$ R.

Therefore, R is reflexive.

---

2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R for some real numbers a and b, and we need to check if (b, a) $\in$ R.

If (a, b) $\in$ R, it means a $\le$ b.

For R to be symmetric, (b, a) must be in R, which means b $\le$ a.

If $a \le b$, it does not necessarily imply $b \le a$. This only happens when $a = b$.

Consider a counterexample. Choose distinct real numbers a and b such that a $\le$ b is true.

Let $a = 1$ and $b = 2$.

Is (1, 2) $\in$ R? Check the condition: $1 \le 2$. This is true. So (1, 2) $\in$ R.

Is (2, 1) $\in$ R? Check the condition: $2 \le 1$. This is false. So (2, 1) $\notin$ R.

Since (1, 2) $\in$ R but (2, 1) $\notin$ R, R is not symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R and (b, c) $\in$ R for some real numbers a, b, and c, and we need to check if (a, c) $\in$ R.

If (a, b) $\in$ R, then a $\le$ b.

If (b, c) $\in$ R, then b $\le$ c.

By the property of real numbers, if $a \le b$ and $b \le c$, then $a \le c$.

This means that (a, c) $\in$ R.

Thus, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ R.

Therefore, R is transitive.

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In conclusion, the relation R = {(a, b) : a $\le$ b} on the set of real numbers is reflexive and transitive but not symmetric.

Given:

The set is the set of all real numbers, denoted by ℝ.

The relation R is defined as R = {(a, b) : $a \le b^3$}, where a, b ∈ ℝ.

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To Check:

If the relation R is reflexive, symmetric, or transitive.

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Checking the Properties:

1. Reflexivity:

A relation is reflexive if (a, a) ∈ R for every a ∈ ℝ. The condition for this is $a \le a^3$.

Let's test this condition. For integer values like a = 2, the condition $2 \le 2^3$ (i.e., $2 \le 8$) is true.

However, the relation must hold for all real numbers. Let's consider a fractional value between 0 and 1, for example, a = $\frac{1}{2}$.

The condition becomes $\frac{1}{2} \le (\frac{1}{2})^3$, which simplifies to $\frac{1}{2} \le \frac{1}{8}$.

This is false, since $\frac{1}{2}$ (0.5) is greater than $\frac{1}{8}$ (0.125).

Since we have found a counterexample (a = $\frac{1}{2}$) for which the condition $a \le a^3$ is not true, the relation is not reflexive.

Therefore, R is not reflexive.

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2. Symmetry:

A relation is symmetric if (a, b) ∈ R implies that (b, a) ∈ R for all a, b ∈ ℝ.

Let's test with a counterexample. Let a = 1 and b = 2.

First, we check if (1, 2) ∈ R. The condition is $1 \le 2^3$, which is $1 \le 8$. This is true, so (1, 2) ∈ R.

Now, we check if the reverse pair (2, 1) ∈ R. The condition is $2 \le 1^3$, which is $2 \le 1$. This is false.

Since (1, 2) ∈ R but (2, 1) ∉ R, the relation is not symmetric.

Therefore, R is not symmetric.

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3. Transitivity:

A relation is transitive if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R for all a, b, c ∈ ℝ.

Let's try to find a counterexample. A good strategy is to use numbers where cubing them makes them smaller, like numbers between 1 and 2, and a large starting number.

Let a = 10, b = 3, and c = 2.

First, check if (a, b) ∈ R, i.e., (10, 3) ∈ R.

The condition is $10 \le 3^3$, which is $10 \le 27$. This is true. So, (10, 3) ∈ R.

Next, check if (b, c) ∈ R, i.e., (3, 2) ∈ R.

The condition is $3 \le 2^3$, which is $3 \le 8$. This is true. So, (3, 2) ∈ R.

Now, we must check if transitivity holds, i.e., if (a, c) ∈ R, which is (10, 2) ∈ R.

The condition is $10 \le 2^3$, which is $10 \le 8$. This is false.

Since we have found a case where (10, 3) ∈ R and (3, 2) ∈ R, but (10, 2) ∉ R, the relation is not transitive.

Therefore, R is not transitive.

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Conclusion:

The relation R = {(a, b) : $a \le b^3$} on the set of real numbers is neither reflexive, nor symmetric, nor transitive.

Given:

Set A = {1, 2, 3}

Relation R in A is given by R = {(1, 2), (2, 1)}.

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To Show:

R is symmetric.

R is neither reflexive nor transitive.

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Proof:

We examine the three properties of relations for the relation R on the set A.

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

The set A is {1, 2, 3}. We need to check if (1, 1) $\in$ R, (2, 2) $\in$ R, and (3, 3) $\in$ R.

Looking at the definition of R, none of these pairs are in R.

(1, 1) $\notin$ R.

(2, 2) $\notin$ R.

(3, 3) $\notin$ R.

Since (a, a) $\notin$ R for every element a in the set {1, 2, 3}, R is not reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

We check the pairs in R:

• Consider the pair (1, 2) $\in$ R. Is (2, 1) $\in$ R? Yes, (2, 1) is in R.
• Consider the pair (2, 1) $\in$ R. Is (1, 2) $\in$ R? Yes, (1, 2) is in R.

For every pair (a, b) in R, the reverse pair (b, a) is also in R.

Therefore, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

We need to check for pairs (a, b) and (b, c) in R.

• Consider the pair (1, 2) $\in$ R. We look for a pair starting with 2 in R. We have (2, 1) $\in$ R.

According to the transitivity condition, if (1, 2) $\in$ R and (2, 1) $\in$ R, then (1, 1) must also be in R.

Looking at the definition of R, (1, 1) is not in R.

Since (1, 2) $\in$ R and (2, 1) $\in$ R, but (1, 1) $\notin$ R, R is not transitive.

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In conclusion, the relation R is symmetric but neither reflexive nor transitive.

Given:

A is the set of all books in a library of a college.

R is a relation in A given by R = {(x, y) : x and y have same number of pages}, where x, y $\in$ A.

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To Show:

R is an equivalence relation.

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Proof:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

---

1. Reflexivity:

A relation R on a set A is reflexive if (x, x) $\in$ R for every x $\in$ A.

For the relation R on the set A of books, we need to show that (x, x) $\in$ R for every book x $\in$ A.

The condition for (x, x) $\in$ R is that x and x have the same number of pages.

Any book has the same number of pages as itself.

Thus, (x, x) $\in$ R for all x $\in$ A.

Therefore, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

For the relation R on the set A of books, we assume (x, y) $\in$ R for some books x, y $\in$ A, and we need to show that (y, x) $\in$ R.

If (x, y) $\in$ R, it means that x and y have the same number of pages.

Let the number of pages of book x be $N_x$ and the number of pages of book y be $N_y$.

The condition is $N_x = N_y$.

If $N_x = N_y$, then it is also true that $N_y = N_x$.

This means that y and x have the same number of pages.

Thus, (y, x) $\in$ R.

So, (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

Therefore, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

For the relation R on the set A of books, we assume (x, y) $\in$ R and (y, z) $\in$ R for some books x, y, z $\in$ A, and we need to show that (x, z) $\in$ R.

If (x, y) $\in$ R, it means that x and y have the same number of pages ($N_x = N_y$).

If (y, z) $\in$ R, it means that y and z have the same number of pages ($N_y = N_z$).

If $N_x = N_y$ and $N_y = N_z$, then by the transitivity of equality, $N_x = N_z$.

This means that x and z have the same number of pages.

Thus, (x, z) $\in$ R.

So, (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Given:

Set A = {1, 2, 3, 4, 5}

Relation R in A is given by R = {(a, b) : |a – b| is even}, where a, b $\in$ A.

---

To Show:

R is an equivalence relation.

All elements of {1, 3, 5} are related to each other.

All elements of {2, 4} are related to each other.

No element of {1, 3, 5} is related to any element of {2, 4}.

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Proof that R is an equivalence relation:

The condition "|a – b| is even" means that the difference between a and b is an even number. This is equivalent to stating that a and b have the same parity (both odd or both even).

We need to check for reflexivity, symmetry, and transitivity.

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1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

For any element a $\in$ A, the condition for (a, a) $\in$ R is that |a – a| is even.

The difference |a – a| = |0| = 0.

Since 0 is an even number (0 = $2 \times 0$), the condition is satisfied for all elements a $\in$ A.

Thus, (a, a) $\in$ R for all a $\in$ A.

Therefore, R is reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R for some a, b $\in$ A. This means |a – b| is even.

For R to be symmetric, (b, a) must also be in R, which means |b – a| is even.

We know that $|b - a| = |-(a - b)| = |a - b|$.

Since |a – b| is even, |b – a| is also even.

Thus, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Therefore, R is symmetric.

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3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means |a – b| is even. This implies a and b have the same parity.

(b, c) $\in$ R means |b – c| is even. This implies b and c have the same parity.

If a and b have the same parity, and b and c have the same parity, then a and c must also have the same parity.

If a and c have the same parity, then their difference (a – c) is even. Consequently, |a – c| is even.

Thus, (a, c) $\in$ R.

So, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Further Demonstration:

Consider the subset {1, 3, 5}. All elements in this subset are odd numbers.

For any two elements a and b from this subset, both a and b are odd. The difference (a – b) is always even (Odd - Odd = Even). So, |a – b| is even.

For example, for elements in {1, 3, 5}:

• |1 – 3| = |-2| = 2 (even). (1, 3) $\in$ R.
• |3 – 5| = |-2| = 2 (even). (3, 5) $\in$ R.
• |1 – 5| = |-4| = 4 (even). (1, 5) $\in$ R.
• Also, (1, 1), (3, 3), (5, 5) are in R (reflexivity).
• And (3, 1), (5, 3), (5, 1) are in R (symmetry).

Thus, all the elements of the subset {1, 3, 5} are related to each other.

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Consider the subset {2, 4}. All elements in this subset are even numbers.

For any two elements a and b from this subset, both a and b are even. The difference (a – b) is always even (Even - Even = Even). So, |a – b| is even.

For example, for elements in {2, 4}:

• |2 – 4| = |-2| = 2 (even). (2, 4) $\in$ R.
• Also, (2, 2), (4, 4) are in R (reflexivity).
• And (4, 2) is in R (symmetry).

Thus, all the elements of the subset {2, 4} are related to each other.

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Consider an element a from the subset {1, 3, 5} (i.e., a is odd) and an element b from the subset {2, 4} (i.e., b is even).

For the pair (a, b) to be in the relation R, the condition "|a – b| is even" must be satisfied.

However, if one number is odd and the other is even, their difference (a – b) is always odd (Odd - Even = Odd, Even - Odd = Odd).

So, |a – b| is odd.

For example:

• |1 – 2| = |-1| = 1 (odd). (1, 2) $\notin$ R.
• |3 – 4| = |-1| = 1 (odd). (3, 4) $\notin$ R.
• |5 – 2| = |3| = 3 (odd). (5, 2) $\notin$ R.

Thus, no element of the subset {1, 3, 5} is related to any element of the subset {2, 4}.

These two subsets {1, 3, 5} and {2, 4} are the equivalence classes of the relation R.

Given:

Set A = $\{ x \in Z : 0 \le x \le 12 \}$. This means A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

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Question 9 (i):

Relation R in A is given by R = {(a, b) : |a – b| is a multiple of 4}.

The condition "|a – b| is a multiple of 4" means that $|a - b| = 4k$ for some non-negative integer k. This is equivalent to saying that $a - b = 4m$ for some integer m, or $a \equiv b \pmod{4}$.

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To Show R is an equivalence relation:

1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

This requires |a – a| to be a multiple of 4.

|a – a| = |0| = 0.

Since 0 is a multiple of 4 ($0 = 4 \times 0$), the condition is satisfied for all a $\in$ A.

Thus, (a, a) $\in$ R for all a $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R. This means |a – b| is a multiple of 4.

So, $|a - b| = 4k$ for some non-negative integer k.

Consider |b – a|. We know $|b - a| = |-(a - b)| = |a - b|$.

Since |a – b| is a multiple of 4, |b – a| is also a multiple of 4.

Thus, (b, a) $\in$ R.

So, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

R is symmetric.

---

3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means |a – b| is a multiple of 4. This means $a - b = 4k_1$ for some integer $k_1$.

(b, c) $\in$ R means |b – c| is a multiple of 4. This means $b - c = 4k_2$ for some integer $k_2$.

Consider a – c. We have $a - c = (a - b) + (b - c) = 4k_1 + 4k_2 = 4(k_1 + k_2)$.

Since $k_1$ and $k_2$ are integers, $k_1 + k_2$ is an integer. This means a – c is a multiple of 4.

Consequently, |a – c| is a multiple of 4.

Thus, (a, c) $\in$ R.

So, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

R is transitive.

---

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

---

Set of all elements related to 1 in case (i):

We need to find all elements $x \in A$ such that (1, x) $\in$ R. The condition is |1 – x| is a multiple of 4.

So, $|1 - x| = 4k$ for some non-negative integer k.

$1 - x = 4k$ or $1 - x = -4k$.

$x = 1 - 4k$ or $x = 1 + 4k$.

We need to find values of x in A = {0, 1, 2, ..., 12} that satisfy this condition.

If k = 0, $x = 1 - 0 = 1$. $1 \in A$.

If k = 1, $x = 1 - 4 = -3$ (not in A). $x = 1 + 4 = 5$. $5 \in A$.

If k = 2, $x = 1 - 8 = -7$ (not in A). $x = 1 + 8 = 9$. $9 \in A$.

If k = 3, $x = 1 - 12 = -11$ (not in A). $x = 1 + 12 = 13$ (not in A).

The elements related to 1 are {1, 5, 9}.

This set is the equivalence class containing 1, denoted by [1].

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Question 9 (ii):

Relation R in A is given by R = {(a, b) : a = b}.

This is the identity relation on the set A.

---

To Show R is an equivalence relation:

1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

This requires a = a, which is always true for any element a.

Thus, (a, a) $\in$ R for all a $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R. This means a = b.

If a = b, then it is also true that b = a.

This means (b, a) $\in$ R.

So, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

R is symmetric.

---

3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means a = b.

(b, c) $\in$ R means b = c.

If a = b and b = c, then by the transitivity of equality, a = c.

This means (a, c) $\in$ R.

So, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

R is transitive.

---

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

---

Set of all elements related to 1 in case (ii):

We need to find all elements $x \in A$ such that (1, x) $\in$ R. The condition is 1 = x.

The only element x in A that satisfies 1 = x is 1 itself.

The elements related to 1 are {1}.

This set is the equivalence class containing 1, denoted by [1].

Here are examples of relations demonstrating the requested properties. For each example, let A be a set and R be a relation on A.

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(i) Symmetric but neither reflexive nor transitive.

Let A = {1, 2, 3}.

Let R = {(1, 2), (2, 1)}.

• Reflexivity: (1, 1), (2, 2), (3, 3) are not in R. Not reflexive.
• Symmetry: (1, 2) $\in$ R and (2, 1) $\in$ R. (2, 1) $\in$ R and (1, 2) $\in$ R. All pairs in R satisfy symmetry. Symmetric.
• Transitivity: (1, 2) $\in$ R and (2, 1) $\in$ R. For transitivity, (1, 1) must be in R. (1, 1) $\notin$ R. Not transitive.

This example satisfies the conditions.

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(ii) Transitive but neither reflexive nor symmetric.

Let A = {1, 2, 3}.

Let R = {(1, 2), (2, 3), (1, 3)}.

• Reflexivity: (1, 1), (2, 2), (3, 3) are not in R. Not reflexive.
• Symmetry: (1, 2) $\in$ R, but (2, 1) $\notin$ R. Not symmetric.
• Transitivity:
  • (1, 2) $\in$ R and (2, 3) $\in$ R. Check (1, 3) $\in$ R. Yes.
  • Are there other pairs (a, b), (b, c) in R? No.

  The transitivity condition holds. Transitive.

This example satisfies the conditions.

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(iii) Reflexive and symmetric but not transitive.

Let A = {1, 2, 3}.

Let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}.

• Reflexivity: (1, 1), (2, 2), (3, 3) are in R. Reflexive.
• Symmetry:
  • (1, 2) $\in$ R and (2, 1) $\in$ R.
  • (2, 3) $\in$ R and (3, 2) $\in$ R.

  All non-diagonal pairs have their symmetric counterparts in R. Symmetric.

• Transitivity:
  • Consider (1, 2) $\in$ R and (2, 3) $\in$ R. For transitivity, (1, 3) must be in R. (1, 3) $\notin$ R.

  Not transitive.

This example satisfies the conditions.

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(iv) Reflexive and transitive but not symmetric.

Let A be the set of real numbers R.

Let R = {(a, b) : a $\le$ b}.

• Reflexivity: For any real number a, a $\le$ a is true. Reflexive.
• Symmetry: Consider (1, 2) where $1 \le 2$. (1, 2) $\in$ R. But (2, 1) means $2 \le 1$, which is false. (2, 1) $\notin$ R. Not symmetric.
• Transitivity: Assume a $\le$ b and b $\le$ c. By the properties of real numbers, this implies a $\le$ c. Transitive.

This example satisfies the conditions. This is the standard 'less than or equal to' relation.

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(v) Symmetric and transitive but not reflexive.

Let A = {1, 2, 3}.

Let R = {(1, 1), (2, 2), (1, 2), (2, 1)}.

• Reflexivity: (3, 3) is not in R. Not reflexive.
• Symmetry: (1, 2) $\in$ R and (2, 1) $\in$ R. (2, 1) $\in$ R and (1, 2) $\in$ R. The reflexive pairs are symmetric. Symmetric.
• Transitivity:
  • (1, 2) $\in$ R and (2, 1) $\in$ R $\implies$ (1, 1) $\in$ R. Yes.
  • (2, 1) $\in$ R and (1, 2) $\in$ R $\implies$ (2, 2) $\in$ R. Yes.
  • (1, 1) $\in$ R and (1, 2) $\in$ R $\implies$ (1, 2) $\in$ R. Yes.
  • (2, 2) $\in$ R and (2, 1) $\in$ R $\implies$ (2, 1) $\in$ R. Yes.
  • ... check all combinations. The relation is transitive.

  Transitive.

Alternatively, consider the relation "is parallel to" on the set of lines in a plane, *excluding* the case where a line is parallel to itself (which would be the definition of a reflexive relation). A clearer example might be the empty relation on a non-empty set. Let A = {1, 2} and R = $\emptyset$.

• Reflexivity: (1, 1) $\notin$ R. Not reflexive.
• Symmetry: The condition "if (a, b) $\in$ R then (b, a) $\in$ R" is vacuously true because there are no pairs in R. Symmetric.
• Transitivity: The condition "if (a, b) $\in$ R and (b, c) $\in$ R then (a, c) $\in$ R" is vacuously true because there are no pairs (a, b), (b, c) in R. Transitive.

Using the empty relation on a non-empty set provides a straightforward example for (v).

Conclusion for (v): An example is the empty relation on a non-empty set, or the relation R = {(1, 1), (2, 2), (1, 2), (2, 1)} on the set {1, 2, 3}.

Given:

A is the set of all points in a plane.

R is a relation in A given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, where P, Q $\in$ A.

Let O be the origin (0, 0).

The distance of a point P(x, y) from the origin is given by $d(O, P) = \sqrt{x^2 + y^2}$.

The condition for (P, Q) $\in$ R is $d(O, P) = d(O, Q)$.

---

To Show:

R is an equivalence relation.

The set of all points related to a point P $\neq$ (0, 0) is the circle passing through P with the origin as the centre.

---

Proof that R is an equivalence relation:

We need to check for reflexivity, symmetry, and transitivity.

---

1. Reflexivity:

A relation R is reflexive if (P, P) $\in$ R for every point P $\in$ A.

The condition for (P, P) $\in$ R is that the distance of P from the origin is the same as the distance of P from the origin.

This is always true, $d(O, P) = d(O, P)$.

Thus, (P, P) $\in$ R for all P $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (P, Q) $\in$ R implies (Q, P) $\in$ R for all P, Q $\in$ A.

Assume (P, Q) $\in$ R for some points P, Q $\in$ A. This means $d(O, P) = d(O, Q)$.

For R to be symmetric, (Q, P) must also be in R, which means $d(O, Q) = d(O, P)$.

If $d(O, P) = d(O, Q)$, then by the property of equality, $d(O, Q) = d(O, P)$ is also true.

Thus, (Q, P) $\in$ R.

So, (P, Q) $\in$ R implies (Q, P) $\in$ R for all P, Q $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (P, Q) $\in$ R and (Q, S) $\in$ R implies (P, S) $\in$ R for all P, Q, S $\in$ A.

Assume (P, Q) $\in$ R and (Q, S) $\in$ R.

(P, Q) $\in$ R means $d(O, P) = d(O, Q)$.

(Q, S) $\in$ R means $d(O, Q) = d(O, S)$.

If $d(O, P) = d(O, Q)$ and $d(O, Q) = d(O, S)$, then by the transitivity of equality, $d(O, P) = d(O, S)$.

This means that the distance of P from the origin is the same as the distance of S from the origin.

Thus, (P, S) $\in$ R.

So, (P, Q) $\in$ R and (Q, S) $\in$ R implies (P, S) $\in$ R for all P, Q, S $\in$ A.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Set of all points related to a point P $\neq$ (0, 0):

Let P$_0$ be a fixed point in the plane, P$_0 \neq$ (0, 0).

We want to find the set of all points Q $\in$ A such that (P$_0$, Q) $\in$ R.

The condition for (P$_0$, Q) $\in$ R is that the distance of P$_0$ from the origin is the same as the distance of Q from the origin.

Let $r = d(O, P_0)$. Since P$_0 \neq$ (0, 0), $r$ is a positive real number ($r > 0$).

The set of points Q that satisfy $d(O, Q) = r$ are all the points that are at a fixed distance $r$ from the origin O.

By definition, the set of all points in a plane that are at a fixed distance from a fixed point (the origin) is a circle.

The fixed point is the origin (centre of the circle).

The fixed distance is $r = d(O, P_0)$ (the radius of the circle).

Since $d(O, P_0) = r$, the point P$_0$ lies on this circle.

Thus, the set of all points related to a point P$_0 \neq$ (0, 0) is the circle passing through P$_0$ with the origin as the centre.

Given:

A is the set of all triangles.

R is the relation in A defined as R = {(T$_1$ , T$_2$) : T$_1$ is similar to T$_2$}, where T$_1$, T$_2$ $\in$ A.

---

To Show:

R is an equivalence relation.

---

Proof that R is an equivalence relation:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

A relation R is reflexive if (T, T) $\in$ R for every triangle T $\in$ A.

The condition for (T, T) $\in$ R is that T is similar to T.

Every triangle is similar to itself.

Thus, (T, T) $\in$ R for all T $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (T$_1$, T$_2$) $\in$ R implies (T$_2$, T$_1$) $\in$ R for all T$_1$, T$_2$ $\in$ A.

Assume (T$_1$, T$_2$) $\in$ R. This means T$_1$ is similar to T$_2$.

If T$_1 \sim$ T$_2$, then T$_2 \sim$ T$_1$.

This means that (T$_2$, T$_1$) $\in$ R.

Thus, (T$_1$, T$_2$) $\in$ R implies (T$_2$, T$_1$) $\in$ R for all T$_1$, T$_2$ $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R implies (T$_1$, T$_3$) $\in$ R for all T$_1$, T$_2$, T$_3$ $\in$ A.

Assume (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R.

(T$_1$, T$_2$) $\in$ R means T$_1$ is similar to T$_2$ (T$_1 \sim$ T$_2$).

(T$_2$, T$_3$) $\in$ R means T$_2$ is similar to T$_3$ (T$_2 \sim$ T$_3$).

If T$_1 \sim$ T$_2$ and T$_2 \sim$ T$_3$, then T$_1 \sim$ T$_3$ (transitivity of similarity).

This means that (T$_1$, T$_3$) $\in$ R.

Thus, (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R implies (T$_1$, T$_3$) $\in$ R for all T$_1$, T$_2$, T$_3$ $\in$ A.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Related Triangles:

We are given three right-angled triangles:

• T$_1$ with sides 3, 4, 5.
• T$_2$ with sides 5, 12, 13.
• T$_3$ with sides 6, 8, 10.

Two triangles are similar if their corresponding sides are proportional or their corresponding angles are equal. Since these are right-angled triangles, we can check the ratios of their sides.

Consider T$_1$ (sides 3, 4, 5) and T$_2$ (sides 5, 12, 13).

Check the ratios of corresponding sides (shortest to shortest, middle to middle, longest to longest):

The ratios are not equal ($\frac{3}{5} \neq \frac{1}{3} \neq \frac{5}{13}$). So, T$_1$ is not similar to T$_2$. (T$_1$, T$_2$) $\notin$ R.

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Consider T$_1$ (sides 3, 4, 5) and T$_3$ (sides 6, 8, 10).

Check the ratios of corresponding sides:

The ratios of corresponding sides are equal ($\frac{1}{2}$). So, T$_1$ is similar to T$_3$. (T$_1$, T$_3$) $\in$ R.

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Consider T$_2$ (sides 5, 12, 13) and T$_3$ (sides 6, 8, 10).

Check the ratios of corresponding sides:

The ratios are not equal ($\frac{5}{6} \neq \frac{3}{2} \neq \frac{13}{10}$). So, T$_2$ is not similar to T$_3$. (T$_2$, T$_3$) $\notin$ R.

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Based on the similarity relation R:

• T$_1$ is related to T$_1$ (reflexive).
• T$_2$ is related to T$_2$ (reflexive).
• T$_3$ is related to T$_3$ (reflexive).
• T$_1$ is related to T$_3$ (because T$_1 \sim$ T$_3$).
• T$_3$ is related to T$_1$ (because T$_3 \sim$ T$_1$, due to symmetry).

The triangles that are related among T$_1$, T$_2$, and T$_3$ are T$_1$ and T$_3$.

Given:

A is the set of all polygons.

R is a relation in A defined as R = {(P$_1$ , P$_2$) : P$_1$ and P$_2$ have same number of sides}, where P$_1$, P$_2$ $\in$ A.

---

To Show:

R is an equivalence relation.

The set of all elements in A related to the right angle triangle T with sides 3, 4 and 5.

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Proof that R is an equivalence relation:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

A relation R is reflexive if (P, P) $\in$ R for every polygon P $\in$ A.

The condition for (P, P) $\in$ R is that P and P have the same number of sides.

Any polygon has the same number of sides as itself.

Thus, (P, P) $\in$ R for all P $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (P$_1$, P$_2$) $\in$ R implies (P$_2$, P$_1$) $\in$ R for all P$_1$, P$_2$ $\in$ A.

Assume (P$_1$, P$_2$) $\in$ R for some polygons P$_1$, P$_2$ $\in$ A. This means P$_1$ and P$_2$ have the same number of sides.

Let the number of sides of P$_1$ be $n_1$ and the number of sides of P$_2$ be $n_2$.

The condition is $n_1 = n_2$.

If $n_1 = n_2$, then it is also true that $n_2 = n_1$.

This means that P$_2$ and P$_1$ have the same number of sides.

Thus, (P$_2$, P$_1$) $\in$ R.

So, (P$_1$, P$_2$) $\in$ R implies (P$_2$, P$_1$) $\in$ R for all P$_1$, P$_2$ $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (P$_1$, P$_2$) $\in$ R and (P$_2$, P$_3$) $\in$ R implies (P$_1$, P$_3$) $\in$ R for all P$_1$, P$_2$, P$_3$ $\in$ A.

Assume (P$_1$, P$_2$) $\in$ R and (P$_2$, P$_3$) $\in$ R.

(P$_1$, P$_2$) $\in$ R means P$_1$ and P$_2$ have the same number of sides ($n_1 = n_2$).

(P$_2$, P$_3$) $\in$ R means P$_2$ and P$_3$ have the same number of sides ($n_2 = n_3$).

If $n_1 = n_2$ and $n_2 = n_3$, then by the transitivity of equality, $n_1 = n_3$.

This means that P$_1$ and P$_3$ have the same number of sides.

Thus, (P$_1$, P$_3$) $\in$ R.

So, (P$_1$, P$_2$) $\in$ R and (P$_2$, P$_3$) $\in$ R implies (P$_1$, P$_3$) $\in$ R for all P$_1$, P$_2$, P$_3$ $\in$ A.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Set of elements related to the given triangle T:

The given triangle T has sides 3, 4, and 5.

A triangle is a polygon with 3 sides.

The number of sides of triangle T is 3.

We want to find the set of all polygons P $\in$ A such that (T, P) $\in$ R.

The condition for (T, P) $\in$ R is that the number of sides of polygon T is the same as the number of sides of polygon P.

Number of sides of T = 3.

So, we are looking for all polygons P in A that have 3 sides.

The set of all polygons with 3 sides is the set of all triangles.

Therefore, the set of all elements in A related to the right angle triangle T is the set of all triangles.

Given:

L is the set of all lines in the XY plane.

R is the relation in L defined as R = {(L$_1$ , L$_2$) : L$_1$ is parallel to L$_2$}, where L$_1$, L$_2$ $\in$ L.

---

To Show:

R is an equivalence relation.

The set of all lines related to the line y = 2x + 4.

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Proof that R is an equivalence relation:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

(Note: For lines in a plane, we usually consider a line to be parallel to itself.)

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1. Reflexivity:

A relation R is reflexive if (L$_1$, L$_1$) $\in$ R for every line L$_1$ $\in$ L.

The condition for (L$_1$, L$_1$) $\in$ R is that L$_1$ is parallel to L$_1$.

Every line is parallel to itself.

Thus, (L$_1$, L$_1$) $\in$ R for all L$_1$ $\in$ L.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (L$_1$, L$_2$) $\in$ R implies (L$_2$, L$_1$) $\in$ R for all L$_1$, L$_2$ $\in$ L.

Assume (L$_1$, L$_2$) $\in$ R for some lines L$_1$, L$_2$ $\in$ L. This means L$_1$ is parallel to L$_2$.

If line L$_1$ is parallel to line L$_2$, then line L$_2$ is also parallel to line L$_1$.

Thus, (L$_2$, L$_1$) $\in$ R.

So, (L$_1$, L$_2$) $\in$ R implies (L$_2$, L$_1$) $\in$ R for all L$_1$, L$_2$ $\in$ L.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R implies (L$_1$, L$_3$) $\in$ R for all L$_1$, L$_2$, L$_3$ $\in$ L.

Assume (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R.

(L$_1$, L$_2$) $\in$ R means L$_1$ is parallel to L$_2$.

(L$_2$, L$_3$) $\in$ R means L$_2$ is parallel to L$_3$.

In a plane, if line L$_1$ is parallel to line L$_2$, and line L$_2$ is parallel to line L$_3$, then line L$_1$ is parallel to line L$_3$.

Thus, (L$_1$, L$_3$) $\in$ R.

So, (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R implies (L$_1$, L$_3$) $\in$ R for all L$_1$, L$_2$, L$_3$ $\in$ L.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Set of all lines related to the line y = 2x + 4:

Let L$_0$ be the line given by the equation y = 2x + 4.

We want to find the set of all lines L $\in$ L such that (L$_0$, L) $\in$ R.

The condition for (L$_0$, L) $\in$ R is that L$_0$ is parallel to L.

Two distinct lines in the XY plane are parallel if and only if they have the same slope.

The equation of line L$_0$ is y = 2x + 4. This is in the form y = mx + c, where m is the slope and c is the y-intercept.

The slope of line L$_0$ is $m_0 = 2$.

A line L is parallel to L$_0$ if and only if it has the same slope as L$_0$.

So, any line L parallel to L$_0$ must have a slope $m = 2$.

The equation of a line with slope 2 can be written in the form y = 2x + c, where c is any real number (the y-intercept).

The set of all lines related to the line y = 2x + 4 is the set of all lines in the XY plane that are parallel to y = 2x + 4.

This set is {L : L has equation y = 2x + c, where c $\in$ R}.

This set represents a family of parallel lines, all with a slope of 2.

Given:

Set A = {1, 2, 3, 4}

Relation R in A is given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}.

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To Check:

Whether R is reflexive, symmetric, or transitive.

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

The set A is {1, 2, 3, 4}. We need to check if (1, 1) $\in$ R, (2, 2) $\in$ R, (3, 3) $\in$ R, and (4, 4) $\in$ R.

From the definition of R, we have:

• (1, 1) $\in$ R
• (2, 2) $\in$ R
• (3, 3) $\in$ R
• (4, 4) $\in$ R

Since (a, a) $\in$ R for every element a in the set {1, 2, 3, 4}, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

We need to check for every ordered pair (a, b) in R whether the reverse pair (b, a) is also in R.

• Consider the pair (1, 2) $\in$ R. Is (2, 1) $\in$ R? Looking at the definition of R, (2, 1) is not in R.

Since there exists a pair (1, 2) $\in$ R such that (2, 1) $\notin$ R, R is not symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

We check for pairs (a, b) and (b, c) in R:

• (1, 2) $\in$ R and (2, 2) $\in$ R. Check (1, 2) $\in$ R. Yes.
• (1, 3) $\in$ R and (3, 3) $\in$ R. Check (1, 3) $\in$ R. Yes.
• (3, 2) $\in$ R and (2, 2) $\in$ R. Check (3, 2) $\in$ R. Yes.
• (1, 3) $\in$ R and (3, 2) $\in$ R. Check (1, 2) $\in$ R. Yes.
• (2, 2) $\in$ R and (2, 2) $\in$ R. Check (2, 2) $\in$ R. Yes.
• (3, 3) $\in$ R and (3, 3) $\in$ R. Check (3, 3) $\in$ R. Yes.
• (4, 4) $\in$ R and (4, 4) $\in$ R. Check (4, 4) $\in$ R. Yes.
• (1, 1) $\in$ R and (1, 2) $\in$ R. Check (1, 2) $\in$ R. Yes.
• (1, 1) $\in$ R and (1, 3) $\in$ R. Check (1, 3) $\in$ R. Yes.
• (3, 3) $\in$ R and (3, 2) $\in$ R. Check (3, 2) $\in$ R. Yes.

All combinations of pairs (a, b) and (b, c) in R result in the pair (a, c) also being in R.

Therefore, R is transitive.

---

Conclusion:

The relation R is reflexive and transitive but not symmetric.

Comparing with the given options:

(A) R is reflexive and symmetric but not transitive. (False - not symmetric)

(B) R is reflexive and transitive but not symmetric. (True - reflexive, transitive, not symmetric)

(C) R is symmetric and transitive but not reflexive. (False - not symmetric, is reflexive)

(D) R is an equivalence relation. (False - not symmetric)

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The correct answer is (B) R is reflexive and transitive but not symmetric.

Given:

Set N = {1, 2, 3, ...} (natural numbers).

Relation R in N is given by R = {(a, b) : a = b – 2, b > 6}, where a, b $\in$ N.

---

To Find:

Which of the given ordered pairs belongs to R.

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A pair (a, b) belongs to R if and only if two conditions are met simultaneously:

1. $a = b – 2$

2. $b > 6$

Also, both a and b must be natural numbers (positive integers).

---

Let's check each option:

(A) (2, 4) ∈ R

Here, a = 2 and b = 4.

Check condition 1: $a = b - 2 \implies 2 = 4 - 2 \implies 2 = 2$. (True)

Check condition 2: $b > 6 \implies 4 > 6$. (False)

Since condition 2 is false, (2, 4) $\notin$ R.

---

(B) (3, 8) ∈ R

Here, a = 3 and b = 8.

Check condition 1: $a = b - 2 \implies 3 = 8 - 2 \implies 3 = 6$. (False)

Check condition 2: $b > 6 \implies 8 > 6$. (True)

Since condition 1 is false, (3, 8) $\notin$ R.

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(C) (6, 8) ∈ R

Here, a = 6 and b = 8.

Check condition 1: $a = b - 2 \implies 6 = 8 - 2 \implies 6 = 6$. (True)

Check condition 2: $b > 6 \implies 8 > 6$. (True)

Since both conditions are true and 6 and 8 are natural numbers, (6, 8) $\in$ R.

---

(D) (8, 7) ∈ R

Here, a = 8 and b = 7.

Check condition 1: $a = b - 2 \implies 8 = 7 - 2 \implies 8 = 5$. (False)

Check condition 2: $b > 6 \implies 7 > 6$. (True)

Since condition 1 is false, (8, 7) $\notin$ R.

---

Only the pair (6, 8) satisfies both conditions for being in R.

The correct answer is (C) (6, 8) ∈ R.

Given:

A is the set of all 50 students of Class X in a school.

N is the set of natural numbers (N = {1, 2, 3, ...}).

f : A $\rightarrow$ N is a function defined by f(x) = roll number of the student x.

---

To Show:

f is one-one but not onto.

---

Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1$ and $x_2$ be two students in set A such that $f(x_1) = f(x_2)$.

According to the definition of the function f, $f(x_1)$ is the roll number of student $x_1$, and $f(x_2)$ is the roll number of student $x_2$.

So, $f(x_1) = f(x_2)$ means that the roll number of student $x_1$ is equal to the roll number of student $x_2$.

In any school system, each student is assigned a unique roll number. It is not possible for two different students to have the same roll number.

Therefore, if the roll number of student $x_1$ is equal to the roll number of student $x_2$, then $x_1$ and $x_2$ must be the same student.

This satisfies the condition for a one-one function.

Therefore, f is one-one.

---

Proof that f is not onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The set A has 50 students. Let the roll numbers of these students be $r_1, r_2, \dots, r_{50}$. Since the roll numbers are unique, the set of these 50 roll numbers forms the range of the function f.

The range of f is the set $\{f(x) : x \in A\} = \{\text{roll number of each student in Class X}\}$.

The codomain of f is the set of natural numbers N = {1, 2, 3, ...}.

The set of natural numbers is an infinite set.

The range of the function f is a set containing 50 distinct roll numbers, which is a finite set with 50 elements.

Since the range of f (a set with 50 elements) is a proper subset of the codomain N (an infinite set), the range is not equal to the codomain.

There are many natural numbers in the codomain N that are not roll numbers of any student in this specific Class X of 50 students. For example, any natural number that is not among the 50 assigned roll numbers will not have a pre-image in A under the function f.

Therefore, for some elements $y \in N$ (the codomain), there is no $x \in A$ such that $f(x) = y$.

Thus, f is not onto.

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In conclusion, the function f is one-one but not onto.

Given:

Function f : N $\rightarrow$ N defined by f(x) = 2x, where N is the set of natural numbers (N = {1, 2, 3, ...}).

Domain = N

Codomain = N

---

To Show:

f is one-one but not onto.

---

Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

According to the definition of f, $f(x_1) = 2x_1$ and $f(x_2) = 2x_2$.

So, $f(x_1) = f(x_2)$ implies $2x_1 = 2x_2$.

Dividing both sides by 2 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

This satisfies the condition for a one-one function.

Therefore, f is one-one.

---

Proof that f is not onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The codomain of f is N = {1, 2, 3, 4, 5, 6, ...}.

The function is f(x) = 2x, where $x \in N$. The values of f(x) are obtained by multiplying elements of N by 2.

The range of f is the set $\{f(x) : x \in N\} = \{2x : x \in N\} = \{2(1), 2(2), 2(3), ...\} = \{2, 4, 6, 8, ...\}$.

The range of f is the set of all positive even integers.

The codomain of f is the set of all natural numbers N, which includes both even and odd positive integers.

The range of f ({2, 4, 6, ...}) is a proper subset of the codomain N ({1, 2, 3, 4, 5, 6, ...}).

For example, consider the element 1 in the codomain N. We need to check if there exists any $x \in N$ such that $f(x) = 1$.

The value $x = \frac{1}{2}$ is not a natural number ($x \notin N$).

Therefore, there is no element in the domain N that maps to the element 1 in the codomain N.

Similarly, no odd number in the codomain N has a pre-image in the domain N under this function.

Since the range of f is not equal to the codomain N, f is not onto.

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In conclusion, the function f : N $\rightarrow$ N defined by f(x) = 2x is one-one but not onto.

Given:

Function f : R $\rightarrow$ R defined by f(x) = 2x, where R is the set of real numbers.

Domain = R

Codomain = R

---

To Prove:

f is one-one and onto.

---

Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in R$ (the domain) such that $f(x_1) = f(x_2)$.

According to the definition of f, $f(x_1) = 2x_1$ and $f(x_2) = 2x_2$.

So, $f(x_1) = f(x_2)$ implies $2x_1 = 2x_2$.

Since 2 is a non-zero real number, we can divide both sides by 2:

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

This satisfies the condition for a one-one function.

Therefore, f is one-one.

---

Proof that f is onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, for every element in the codomain, there must be a pre-image in the domain.

The codomain of f is R (the set of real numbers).

Let $y$ be any arbitrary real number in the codomain R.

We need to find if there exists a real number $x$ in the domain R such that $f(x) = y$.

According to the definition of f, $f(x) = 2x$.

So, we set $f(x) = y$ and solve for x:

Since y is a real number, $\frac{y}{2}$ is also a real number.

So, for every real number $y$ in the codomain, we found a real number $x = \frac{y}{2}$ in the domain such that $f(x) = f\left(\frac{y}{2}\right) = 2 \times \left(\frac{y}{2}\right) = y$.

This means every element in the codomain R has a pre-image in the domain R.

Therefore, f is onto.

---

Since the function f is both one-one and onto, it is a bijective function.

Thus, the function f : R $\rightarrow$ R defined by f(x) = 2x is one-one and onto.

Given:

Function f : N $\rightarrow$ N defined by:

• f (1) = 1
• f (2) = 1
• f (x) = x – 1, for every x > 2

Domain = N = {1, 2, 3, ...}

Codomain = N = {1, 2, 3, ...}

---

To Show:

f is onto but not one-one.

---

Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Consider the elements 1 and 2 in the domain N.

According to the definition of the function f:

We have $f(1) = f(2) = 1$.

However, $1 \neq 2$.

Since there exist distinct elements in the domain (1 and 2) that map to the same element in the codomain (1), the function f is not one-one.

Therefore, f is not one-one.

---

Proof that f is onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The codomain of f is N = {1, 2, 3, 4, 5, ...}.

Let y be any arbitrary natural number in the codomain N.

We need to find if there exists a natural number $x$ in the domain N such that $f(x) = y$.

Consider the possible values of y in the codomain:

• Case 1: y = 1

We know that f(1) = 1 and f(2) = 1 from the definition of f.

Since $1 \in N$ and $2 \in N$, there exist pre-images (1 and 2) in the domain N for the element 1 in the codomain N.

• Case 2: y > 1 (i.e., y $\ge$ 2, since y is a natural number)

We are looking for an $x \in N$ such that $f(x) = y$.

Consider the part of the definition where f(x) = x – 1 for x > 2.

Set $f(x) = y$, so $x – 1 = y$.

Solving for x, we get $x = y + 1$.

Since y is a natural number ($y \in N$), $y \ge 1$. For this case, we are considering $y > 1$, so $y \ge 2$.

If $y \in N$ and $y \ge 2$, then $y+1$ is also a natural number, and $y+1 \ge 3$.

So, $x = y+1$ is a natural number in the domain N, and $x = y+1 > 2$.

For this value of x, $f(x) = f(y+1) = (y+1) – 1 = y$.

Thus, for every natural number $y \ge 2$ in the codomain, there exists a pre-image $x = y+1$ in the domain.

Combining Case 1 and Case 2, we see that for every element y in the codomain N, there exists at least one element x in the domain N such that f(x) = y.

Therefore, f is onto.

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In conclusion, the function f : N $\rightarrow$ N defined by f (1) = f(2) = 1 and f(x) = x – 1, for every x > 2, is onto but not one-one.

Given:

Function f : R $\rightarrow$ R defined by f(x) = x$^2$, where R is the set of real numbers.

Domain = R

Codomain = R

---

To Show:

f is neither one-one nor onto.

---

Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

To show that f is not one-one, we need to find two distinct elements in the domain R that map to the same element in the codomain R.

Consider the elements 1 and -1 in the domain R.

Calculate their images under f:

We have $f(1) = f(-1) = 1$.

However, $1 \neq -1$.

Since there exist distinct elements in the domain (1 and -1) that map to the same element in the codomain (1), the function f is not one-one.

Therefore, f is not one-one.

---

Proof that f is not onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The codomain of f is R (the set of real numbers).

The function is f(x) = x$^2$. The square of any real number is always non-negative.

The range of f is the set $\{f(x) : x \in R\} = \{x^2 : x \in R\} = [0, \infty)$.

The range of f is the set of all non-negative real numbers.

The codomain of f is the set of all real numbers R, which includes negative real numbers.

The range of f ($[0, \infty)$) is a proper subset of the codomain R.

For example, consider the element -1 in the codomain R.

We need to check if there exists any $x \in R$ such that $f(x) = -1$.

The equation $x^2 = -1$ has no solution in the set of real numbers R, because the square of a real number cannot be negative.

Therefore, there is no element in the domain R that maps to the element -1 in the codomain R.

This means the range of f is not equal to the codomain R.

Thus, f is not onto.

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In conclusion, the function f : R $\rightarrow$ R defined by f(x) = x$^2$ is neither one-one nor onto.

Given:

Function f : N $\rightarrow$ N defined by:

Domain = N = {1, 2, 3, ...}

Codomain = N = {1, 2, 3, ...}

---

To Show:

f is both one-one and onto.

---

Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$. Let the common value be y.

We need to consider the parity of $x_1$ and $x_2$. There are four possibilities:

• Case 1: $x_1$ is odd and $x_2$ is odd.

Then $f(x_1) = x_1 + 1$ and $f(x_2) = x_2 + 1$.

$f(x_1) = f(x_2) \implies x_1 + 1 = x_2 + 1$.

Subtracting 1 from both sides, we get $x_1 = x_2$.

• Case 2: $x_1$ is even and $x_2$ is even.

Then $f(x_1) = x_1 - 1$ and $f(x_2) = x_2 - 1$.

$f(x_1) = f(x_2) \implies x_1 - 1 = x_2 - 1$.

Adding 1 to both sides, we get $x_1 = x_2$.

• Case 3: $x_1$ is odd and $x_2$ is even.

Then $f(x_1) = x_1 + 1$ and $f(x_2) = x_2 - 1$.

$f(x_1) = y$. If $x_1$ is odd, $x_1+1$ is even. So y must be even.

$f(x_2) = y$. If $x_2$ is even, $x_2-1$ is odd (note that $x_2 \ge 2$ since $x_2$ is even and in N). So y must be odd.

We have a contradiction: y cannot be both even and odd.

• Case 4: $x_1$ is even and $x_2$ is odd.

This is symmetric to Case 3. If $f(x_1) = f(x_2)$, then y must be both odd and even, which is impossible.

The only possibilities for $f(x_1) = f(x_2)$ are when $x_1$ and $x_2$ have the same parity (Cases 1 and 2), and in both these cases, we found that $x_1 = x_2$.

Thus, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Therefore, f is one-one.

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Proof that f is onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain of f is N = {1, 2, 3, ...}.

Let y be any arbitrary natural number in the codomain N.

We need to find an element $x$ in the domain N such that $f(x) = y$.

Consider two cases for y:

• Case 1: y is odd.

We are looking for an $x \in N$ such that $f(x) = y$.

If we assume x is even, then $f(x) = x - 1$. We set $x - 1 = y$, so $x = y + 1$.

Since y is an odd natural number, y $\ge$ 1. Then $y+1$ is an even natural number, and $y+1 \ge 2$.

So, if y is odd, we can choose $x = y+1$. This x is an even natural number ($x \in N$ and x is even). When we apply f to this x, $f(x) = f(y+1) = (y+1) - 1 = y$.

Thus, for every odd number y in the codomain, there exists an even number $x = y+1$ in the domain that maps to y.

• Case 2: y is even.

We are looking for an $x \in N$ such that $f(x) = y$.

If we assume x is odd, then $f(x) = x + 1$. We set $x + 1 = y$, so $x = y - 1$.

Since y is an even natural number, y $\ge$ 2 (as y $\in$ N and is even). Then $y-1$ is an odd natural number, and $y-1 \ge 1$.

So, if y is even, we can choose $x = y-1$. This x is an odd natural number ($x \in N$ and x is odd). When we apply f to this x, $f(x) = f(y-1) = (y-1) + 1 = y$.

Thus, for every even number y in the codomain, there exists an odd number $x = y-1$ in the domain that maps to y.

Since every natural number y in the codomain is either odd or even, and in both cases we found a pre-image x in the domain N, every element in the codomain N has a pre-image in the domain N under the function f.

Therefore, f is onto.

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Since the function f is both one-one and onto, it is a bijective function.

Thus, the function f : N $\rightarrow$ N defined by the piecewise rule is both one-one and onto.

Given:

Set A = {1, 2, 3}

Set B = {1, 2, 3}

Function f : A $\rightarrow$ B is onto.

The number of elements in set A is $n(A) = 3$.

The number of elements in set B is $n(B) = 3$.

---

To Show:

f is always one-one.

---

Proof:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

Since the codomain B = {1, 2, 3} has 3 distinct elements, and the function f is onto, each of these 3 elements in B must have at least one pre-image in the domain A = {1, 2, 3}.

Let the elements of the domain be $x_1=1, x_2=2, x_3=3$.

Let the elements of the codomain be $y_1=1, y_2=2, y_3=3$.

Since f is onto, the range of f is equal to the codomain B.

Range of f = $\{f(1), f(2), f(3)\} = \{1, 2, 3\}$.

This means the set of images $\{f(1), f(2), f(3)\}$ contains exactly the 3 elements {1, 2, 3}.

For the set $\{f(1), f(2), f(3)\}$ to contain 3 distinct elements, the elements $f(1), f(2), f(3)$ must themselves be distinct.

If $f(1), f(2), f(3)$ were not distinct, at least two of the images would be the same. For example, if $f(1) = f(2)$, then the set of images would be $\{f(1), f(3)\}$, which contains at most 2 distinct elements, contradicting the fact that the range is {1, 2, 3}.

So, $f(1), f(2), f(3)$ must be distinct values from the set {1, 2, 3}.

The distinct images are $f(1), f(2), f(3)$.

Now consider two distinct elements $x_i, x_j$ from the domain A (where $i \neq j$). For example, let $x_1 = 1$ and $x_2 = 2$. Since their images $f(1)$ and $f(2)$ must be distinct as shown above, $f(1) \neq f(2)$.

In general, for any two distinct elements $x_i, x_j \in \{1, 2, 3\}$, their images $f(x_i)$ and $f(x_j)$ must be distinct elements in the range {1, 2, 3}.

Therefore, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. This is the definition of a one-one function.

Thus, an onto function f : {1, 2, 3} $\rightarrow$ {1, 2, 3} is always one-one.

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Generalization:

This result holds for any function f : A $\rightarrow$ B where A and B are finite sets with the same number of elements, i.e., $|A| = |B|$. In such a case, a function f : A $\rightarrow$ B is onto if and only if it is one-one.

Given:

Set A = {1, 2, 3}

Set B = {1, 2, 3}

Function f : A $\rightarrow$ B is one-one.

The number of elements in set A is $n(A) = 3$.

The number of elements in set B is $n(B) = 3$.

---

To Show:

f must be onto.

---

Proof:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $x_1 \neq x_2$ implies $f(x_1) \neq f(x_2)$. Equivalently, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

The domain A = {1, 2, 3} has 3 distinct elements.

Since f is one-one, the images of these 3 distinct elements under f must also be distinct elements in the codomain B = {1, 2, 3}.

The images are $f(1), f(2), f(3)$.

Because f is one-one, we know that $f(1) \neq f(2)$, $f(1) \neq f(3)$, and $f(2) \neq f(3)$.

So, the set of images $\{f(1), f(2), f(3)\}$ contains 3 distinct elements.

These 3 distinct images are elements of the codomain B = {1, 2, 3}.

Since the set of images $\{f(1), f(2), f(3)\}$ contains 3 distinct elements and the codomain B also contains exactly 3 elements, the set of images must be equal to the codomain B.

Range of f = $\{f(1), f(2), f(3)\}$

Codomain of f = {1, 2, 3}

Since the 3 distinct images must be taken from the 3 elements of the codomain, the set of images must be {1, 2, 3} in some order.

For example, the images could be {f(1)=1, f(2)=2, f(3)=3}, or {f(1)=3, f(2)=1, f(3)=2}, etc. In any case, the set of images is {1, 2, 3}.

Since the range of f is equal to the codomain B, the function f is onto.

Thus, a one-one function f : {1, 2, 3} $\rightarrow$ {1, 2, 3} must be onto.

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Generalization:

This result also holds for any function f : A $\rightarrow$ B where A and B are finite sets with the same number of elements, i.e., $|A| = |B|$. In such a case, a function f : A $\rightarrow$ B is one-one if and only if it is onto.

Part 1: Proving for the function $f : R_* \to R_*$

Given:

A function $f : R_* \to R_*$ is defined by $f(x) = \frac{1}{x}$, where $R_*$ is the set of all non-zero real numbers.

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To Prove:

The function $f$ is one-one (injective) and onto (surjective).

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Proof:

1. To Prove $f$ is One-One (Injective)

A function is one-one if for any two elements $x_1, x_2$ in its domain, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in R_*$ (the domain) such that $f(x_1) = f(x_2)$.

Substituting the function definition:

Taking the reciprocal of both sides gives:

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-one.

Therefore, $f$ is one-one.

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2. To Prove $f$ is Onto (Surjective)

A function is onto if for every element $y$ in its codomain, there exists an element $x$ in its domain such that $f(x) = y$.

Let $y$ be an arbitrary element in the codomain $R_*$. This means $y$ is a non-zero real number.

We need to find an $x$ in the domain $R_*$ such that $f(x) = y$.

Solving for $x$, we get:

Since $y$ is a non-zero real number (as it belongs to the codomain $R_*$), its reciprocal $x = \frac{1}{y}$ is also a non-zero real number. This means that $x$ exists and is in the domain $R_*$.

Thus, for any $y \in R_*$ (codomain), there is a pre-image $x = \frac{1}{y} \in R_*$ (domain).

Therefore, $f$ is onto.

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Part 2: Checking the result when the domain is N

Given:

The function is now $f : N \to R_*$, defined by $f(x) = \frac{1}{x}$.

Here, the domain is the set of natural numbers (N = {1, 2, 3, ...}) and the codomain is the set of non-zero real numbers ($R_*$).

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Checking the Properties:

1. Checking for One-One

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

This implies $x_1 = x_2$. So, the function is still one-one. Different natural numbers will always have different reciprocals.

Therefore, the function is still one-one.

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2. Checking for Onto

For the function to be onto, for every $y \in R_*$, there must be an $x \in N$ such that $f(x) = y$.

Let's choose an element from the codomain $R_*$, for example, $y = 2$.

We need to find an $x \in N$ such that $f(x) = 2$.

Solving for $x$, we get $x = \frac{1}{2}$.

The value $x = \frac{1}{2}$ is not a natural number, so it is not in the domain N.

Since there is an element in the codomain (e.g., 2) which has no pre-image in the domain, the function is not onto.

Therefore, the function is not onto.

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Conclusion:

If the domain $R_*$ is replaced by N with the co-domain being the same as $R_*$, the result is not true. The function remains one-one but is no longer onto.

(i) $f : N \to N$ given by $f(x) = x^2$

Injectivity (One-one):

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

$f(x_1) = f(x_2) \implies x_1^2 = x_2^2$

Since $x_1$ and $x_2$ are natural numbers (and therefore positive), taking the square root gives:

$x_1 = x_2$

So, $f$ is injective (one-one).

Surjectivity (Onto):

For a function to be onto, every element in the codomain (N) must have a pre-image in the domain (N). Let's consider the element $y=2$ in the codomain N.

We need to find an $x \in N$ such that $f(x) = 2$, which means $x^2 = 2$.

Solving for $x$, we get $x = \sqrt{2}$, which is not a natural number.

Since the element 2 in the codomain has no pre-image in the domain, $f$ is not surjective (not onto).

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(ii) $f : Z \to Z$ given by $f(x) = x^2$

Injectivity (One-one):

Let's consider two different elements from the domain Z, for example, $x_1 = 1$ and $x_2 = -1$.

$f(1) = 1^2 = 1$

$f(-1) = (-1)^2 = 1$

Here, $f(x_1) = f(x_2)$ but $x_1 \neq x_2$. Since two different inputs map to the same output, $f$ is not injective (not one-one).

Surjectivity (Onto):

The square of any integer is always non-negative. Therefore, no negative integer in the codomain Z has a pre-image. For example, let $y = -1 \in Z$. There is no integer $x$ such that $x^2 = -1$.

Also, non-perfect square integers like 2, 3, etc., in the codomain do not have an integer pre-image.

So, $f$ is not surjective (not onto).

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(iii) $f : R \to R$ given by $f(x) = x^2$

Injectivity (One-one):

Similar to the case with integers, consider $x_1 = 1$ and $x_2 = -1$ from the domain R.

$f(1) = 1^2 = 1$ and $f(-1) = (-1)^2 = 1$.

Since $f(1) = f(-1)$ but $1 \neq -1$, the function $f$ is not injective (not one-one).

Surjectivity (Onto):

The square of any real number is always non-negative. So, for any negative real number $y$ in the codomain R (e.g., $y = -1$), there is no real number $x$ in the domain such that $x^2 = y$. The range of the function is $[0, \infty)$, which is not equal to the codomain R.

So, $f$ is not surjective (not onto).

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(iv) $f : N \to N$ given by $f(x) = x^3$

Injectivity (One-one):

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

$f(x_1) = f(x_2) \implies x_1^3 = x_2^3$

Taking the cube root of both sides gives $x_1 = x_2$.

So, $f$ is injective (one-one).

Surjectivity (Onto):

Consider the element $y=2$ in the codomain N. We need to find an $x \in N$ such that $f(x) = 2$, which means $x^3 = 2$.

Solving for $x$, we get $x = \sqrt[3]{2}$, which is not a natural number.

Since the element 2 in the codomain has no pre-image in the domain, $f$ is not surjective (not onto).

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(v) $f : Z \to Z$ given by $f(x) = x^3$

Injectivity (One-one):

Let $x_1, x_2 \in Z$ such that $f(x_1) = f(x_2)$.

$f(x_1) = f(x_2) \implies x_1^3 = x_2^3$

The cube function is strictly increasing over the integers, so if the cubes are equal, the numbers must be equal. Taking the cube root gives $x_1 = x_2$. Unlike the square function, $f(x) \neq f(-x)$ for $x \neq 0$.

So, $f$ is injective (one-one).

Surjectivity (Onto):

Consider the element $y=2$ in the codomain Z. We need to find an $x \in Z$ such that $f(x) = 2$, which means $x^3 = 2$.

Solving for $x$, we get $x = \sqrt[3]{2}$, which is not an integer.

Since the element 2 in the codomain has no pre-image in the domain, $f$ is not surjective (not onto).

Given:

Function f : R $\rightarrow$ R given by f(x) = [x], where [x] denotes the greatest integer less than or equal to x.

Domain = R (real numbers)

Codomain = R (real numbers)

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To Prove:

f is neither one-one nor onto.

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Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. To show it is not one-one, we need to find $x_1 \neq x_2$ such that $f(x_1) = f(x_2)$.

The definition of the greatest integer function is that for any real number x, [x] is the largest integer less than or equal to x.

Consider two distinct real numbers, say $x_1 = 1.2$ and $x_2 = 1.5$. Both are in the domain R, and $x_1 \neq x_2$.

Calculate their images under f:

The greatest integer less than or equal to 1.2 is 1.

The greatest integer less than or equal to 1.5 is 1.

We have $f(1.2) = f(1.5) = 1$, but $1.2 \neq 1.5$.

In general, for any real number k, all real numbers x in the interval $[k, k+1)$ map to the same integer k under the greatest integer function. Since the interval $[k, k+1)$ contains infinitely many distinct real numbers (e.g., 1.1, 1.2, 1.3, ... all map to 1), the function is not one-one.

Therefore, f is not one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range must equal the codomain.

The codomain of f is R (the set of all real numbers).

The function is f(x) = [x]. The value of [x] is always an integer, regardless of the real number x.

The range of f is the set of all possible values of [x]. Since [x] is always an integer, the range of f is the set of all integers Z.

Range of f = {k : k is an integer} = Z = {..., -2, -1, 0, 1, 2, ...}.

The codomain of f is the set of all real numbers R.

The range of f (Z) is a proper subset of the codomain (R), because R contains real numbers that are not integers (e.g., 0.5, $\sqrt{2}$, $\pi$).

Consider a real number in the codomain that is not an integer, for example, $y = 0.5$. We need to check if there exists any $x \in R$ such that $f(x) = 0.5$.

By the definition of the greatest integer function, [x] must be an integer. Therefore, [x] cannot be equal to 0.5.

There is no real number x whose greatest integer is 0.5.

Thus, the element 0.5 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f (Z) is not equal to the codomain R.

Therefore, f is not onto.

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In conclusion, the Greatest Integer Function f : R $\rightarrow$ R, given by f(x) = [x], is neither one-one nor onto.

Given:

Function f : R $\rightarrow$ R given by f(x) = |x|, where R is the set of real numbers.

The modulus function is defined as:

Domain = R (real numbers)

Codomain = R (real numbers)

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To Show:

f is neither one-one nor onto.

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Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. To show it is not one-one, we need to find $x_1 \neq x_2$ such that $f(x_1) = f(x_2)$.

Consider two distinct real numbers, say $x_1 = 2$ and $x_2 = -2$. Both are in the domain R, and $x_1 \neq x_2$.

Calculate their images under f:

Since $2 \ge 0$, $|2| = 2$.

Since $-2 < 0$, $|-2| = -(-2) = 2$.

We have $f(2) = f(-2) = 2$, but $2 \neq -2$.

In general, for any positive real number k, we have $f(k) = |k| = k$ and $f(-k) = |-k| = -(-k) = k$. Since $k \neq -k$ (for $k \neq 0$), distinct inputs (k and -k) map to the same output (k).

Therefore, f is not one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range must equal the codomain.

The codomain of f is R (the set of all real numbers).

The function is f(x) = |x|. By the definition of the modulus function, the output |x| is always non-negative.

The range of f is the set of all possible values of |x| for $x \in R$. This is the set of all non-negative real numbers.

Range of f = $\{|x| : x \in R\} = [0, \infty)$.

The codomain of f is the set of all real numbers R.

The range of f ($[0, \infty)$) is a proper subset of the codomain (R), because R contains negative real numbers (e.g., -1, -5, -$\pi$).

Consider a negative real number in the codomain, for example, $y = -3$. We need to check if there exists any $x \in R$ such that $f(x) = -3$.

By the definition of the modulus function, $|x|$ cannot be negative for any real number x.

Therefore, there is no real number x such that |x| = -3.

Thus, the element -3 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ($[0, \infty)$) is not equal to the codomain R.

Therefore, f is not onto.

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In conclusion, the modulus function f : R $\rightarrow$ R, given by f(x) = |x|, is neither one-one nor onto.

Given:

Function f : R $\rightarrow$ R, the Signum Function, given by:

Domain = R (real numbers)

Codomain = R (real numbers)

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To Show:

f is neither one-one nor onto.

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Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. To show it is not one-one, we need to find $x_1 \neq x_2$ such that $f(x_1) = f(x_2)$.

Consider two distinct positive real numbers, say $x_1 = 2$ and $x_2 = 5$. Both are in the domain R, and $x_1 \neq x_2$. Both are greater than 0.

Calculate their images under f:

Since $2 > 0$, $f(2) = 1$.

Since $5 > 0$, $f(5) = 1$.

We have $f(2) = f(5) = 1$, but $2 \neq 5$.

In general, any two distinct positive real numbers will both map to 1. Any two distinct negative real numbers will both map to -1.

For example, consider $x_1 = -3$ and $x_2 = -7$. $x_1 \neq x_2$, but $f(-3) = -1$ and $f(-7) = -1$. So $f(-3) = f(-7)$.

Since there exist distinct elements in the domain that map to the same element in the codomain, f is not one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range must equal the codomain.

The codomain of f is R (the set of all real numbers).

According to the definition of the Signum Function, the possible output values (the range of f) are only 1, 0, and -1.

Range of f = {1, 0, -1}.

The codomain of f is the set of all real numbers R.

The range of f ({1, 0, -1}) is a proper subset of the codomain (R), because R contains infinitely many real numbers that are not 1, 0, or -1 (e.g., 2, -5, 0.5, $\sqrt{2}$, $\pi$).

Consider a real number in the codomain that is not 1, 0, or -1, for example, $y = 5$. We need to check if there exists any $x \in R$ such that $f(x) = 5$.

Based on the definition of f(x), the output can only be 1, 0, or -1. It can never be 5.

Therefore, there is no real number x such that f(x) = 5.

Thus, the element 5 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ({1, 0, -1}) is not equal to the codomain R.

Therefore, f is not onto.

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In conclusion, the Signum Function f : R $\rightarrow$ R is neither one-one nor onto.

Given:

Set A = {1, 2, 3}

Set B = {4, 5, 6, 7}

Function f : A $\rightarrow$ B is given as a set of ordered pairs: f = {(1, 4), (2, 5), (3, 6)}.

This means:

• f(1) = 4
• f(2) = 5
• f(3) = 6

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To Show:

f is one-one.

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Proof:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Equivalently, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

The domain of the function is A = {1, 2, 3}. The distinct elements in the domain are 1, 2, and 3.

We examine the images of these distinct elements under f:

• Image of 1 is f(1) = 4.
• Image of 2 is f(2) = 5.
• Image of 3 is f(3) = 6.

Let's compare the images of distinct elements:

• For the distinct elements 1 and 2 from A, their images are f(1) = 4 and f(2) = 5. Since $4 \neq 5$, $f(1) \neq f(2)$.
• For the distinct elements 1 and 3 from A, their images are f(1) = 4 and f(3) = 6. Since $4 \neq 6$, $f(1) \neq f(3)$.
• For the distinct elements 2 and 3 from A, their images are f(2) = 5 and f(3) = 6. Since $5 \neq 6$, $f(2) \neq f(3)$.

In each case where the inputs from the domain are distinct, the corresponding outputs in the codomain are also distinct.

Alternatively, we can check the contrapositive condition: if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Let's look at the image values in the set of ordered pairs: {4, 5, 6}.

• If $f(x_1) = f(x_2) = 4$, then $x_1$ and $x_2$ must map to 4. From the definition of f, only 1 maps to 4. So $x_1 = 1$ and $x_2 = 1$. Thus $x_1 = x_2$.
• If $f(x_1) = f(x_2) = 5$, then $x_1$ and $x_2$ must map to 5. Only 2 maps to 5. So $x_1 = 2$ and $x_2 = 2$. Thus $x_1 = x_2$.
• If $f(x_1) = f(x_2) = 6$, then $x_1$ and $x_2$ must map to 6. Only 3 maps to 6. So $x_1 = 3$ and $x_2 = 3$. Thus $x_1 = x_2$.

In all cases where the images are equal, the corresponding pre-images are also equal.

Thus, f is one-one.

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Note: This function is one-one, but it is not onto because the element 7 in the codomain B does not have a pre-image in the domain A.

Question 7 (i):

Function f : R $\rightarrow$ R defined by f(x) = 3 – 4x.

Domain = R

Codomain = R

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Injectivity (One-one):

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$3 - 4x_1 = 3 - 4x_2$

Subtract 3 from both sides:

Divide both sides by -4 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

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Surjectivity (Onto):

The codomain is R.

Let y be any arbitrary real number in the codomain R.

We need to find if there exists a real number $x$ in the domain R such that $f(x) = y$.

Set $f(x) = y$ and solve for x:

Subtract 3 from both sides:

Divide by -4:

Since y is a real number, $3-y$ is a real number, and $\frac{3-y}{4}$ is also a real number.

So, for every real number $y$ in the codomain, we found a real number $x = \frac{3-y}{4}$ in the domain such that $f(x) = f\left(\frac{3-y}{4}\right) = 3 - 4\left(\frac{3-y}{4}\right) = 3 - (3 - y) = 3 - 3 + y = y$.

Every element in the codomain R has a pre-image in the domain R.

Therefore, f is onto.

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Conclusion for (i): Since f is both one-one and onto, it is a bijective function.

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Question 7 (ii):

Function f : R $\rightarrow$ R defined by f(x) = 1 + x$^2$.

Domain = R

Codomain = R

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Injectivity (One-one):

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$1 + x_1^2 = 1 + x_2^2$

Subtract 1 from both sides:

This implies $x_1^2 - x_2^2 = 0 \implies (x_1 - x_2)(x_1 + x_2) = 0$.

So, $x_1 = x_2$ or $x_1 = -x_2$.

Consider $x_1 = 1$ and $x_2 = -1$. Both are in R, and $x_1 \neq x_2$.

$f(1) = 1 + 1^2 = 1 + 1 = 2$.

$f(-1) = 1 + (-1)^2 = 1 + 1 = 2$.

We have $f(1) = f(-1) = 2$, but $1 \neq -1$.

Thus, f is not injective (not one-one).

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Surjectivity (Onto):

The codomain is R.

The function is f(x) = 1 + x$^2$. Since $x^2 \ge 0$ for any real number x, $1 + x^2 \ge 1 + 0 = 1$.

The range of f is the set of all values $1 + x^2$ for $x \in R$. This is the set of all real numbers greater than or equal to 1.

Range of f = $[1, \infty)$.

The codomain of f is the set of all real numbers R.

The range of f ($[1, \infty)$) is a proper subset of the codomain (R), because R contains real numbers less than 1 (e.g., 0, -5, 0.5).

Consider a real number in the codomain that is less than 1, for example, $y = 0$. We need to check if there exists any $x \in R$ such that $f(x) = 0$.

$f(x) = 0 \implies 1 + x^2 = 0 \implies x^2 = -1$.

There is no real number x whose square is -1.

Thus, the element 0 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ($[1, \infty)$) is not equal to the codomain R.

Therefore, f is not surjective (not onto).

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Conclusion for (ii): The function f is neither one-one nor onto.

Given:

Sets A and B.

Function f : A $\times$ B $\rightarrow$ B $\times$ A defined by f(a, b) = (b, a), where (a, b) $\in$ A $\times$ B.

Domain = A $\times$ B (the set of all ordered pairs where the first element is from A and the second is from B).

Codomain = B $\times$ A (the set of all ordered pairs where the first element is from B and the second is from A).

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To Show:

f is a bijective function (i.e., f is both one-one and onto).

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Proof that f is one-one (Injectivity):

A function f : X $\rightarrow$ Y is one-one if for any two distinct elements $x_1, x_2 \in X$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1 = (a_1, b_1)$ and $x_2 = (a_2, b_2)$ be two elements in the domain A $\times$ B, such that $f(x_1) = f(x_2)$.

According to the definition of f:

The condition $f(x_1) = f(x_2)$ means $(b_1, a_1) = (b_2, a_2)$.

Two ordered pairs are equal if and only if their corresponding components are equal.

So, $(b_1, a_1) = (b_2, a_2)$ implies $b_1 = b_2$ and $a_1 = a_2$.

Since $a_1 = a_2$ and $b_1 = b_2$, the ordered pairs $(a_1, b_1)$ and $(a_2, b_2)$ are equal.

Thus, if $f(a_1, b_1) = f(a_2, b_2)$, then $(a_1, b_1) = (a_2, b_2)$.

Therefore, f is one-one.

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Proof that f is onto (Surjectivity):

A function f : X $\rightarrow$ Y is onto if for every element $y \in Y$, there exists at least one element $x \in X$ such that $f(x) = y$.

The codomain of f is B $\times$ A (the set of all ordered pairs (b, a) where b $\in$ B and a $\in$ A).

Let y be any arbitrary element in the codomain B $\times$ A. By the definition of B $\times$ A, y is an ordered pair of the form (b, a), where b $\in$ B and a $\in$ A.

We need to find if there exists an element $x$ in the domain A $\times$ B such that $f(x) = y$.

Let $x$ be an ordered pair $(u, v)$ in the domain A $\times$ B. This means $u \in A$ and $v \in B$.

According to the definition of f, $f(u, v) = (v, u)$.

We want to find $(u, v) \in A \times B$ such that $f(u, v) = y$, i.e., $(v, u) = (b, a)$.

By the equality of ordered pairs, $(v, u) = (b, a)$ implies $v = b$ and $u = a$.

We need to check if this element $x = (u, v) = (a, b)$ is in the domain A $\times$ B.

Since a $\in$ A and b $\in$ B (from the definition of the codomain element y = (b, a)), the ordered pair (a, b) is indeed an element of A $\times$ B (the domain).

So, for every element $y = (b, a)$ in the codomain B $\times$ A, we found an element $x = (a, b)$ in the domain A $\times$ B such that $f(x) = f(a, b) = (b, a) = y$.

Every element in the codomain B $\times$ A has a pre-image in the domain A $\times$ B.

Therefore, f is onto.

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Since the function f is both one-one and onto, it is a bijective function.

Given:

Function f : N $\rightarrow$ N defined by:

for all $n \in N$ = {1, 2, 3, ...}.

Domain = N

Codomain = N

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To State:

Whether the function f is bijective (i.e., both one-one and onto).

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Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(n_1) = f(n_2)$ implies $n_1 = n_2$ for all $n_1, n_2 \in N$.

Let's evaluate the function for a few values:

• f(1) = $\frac{1+1}{2} = \frac{2}{2} = 1$ (since 1 is odd)
• f(2) = $\frac{2}{2} = 1$ (since 2 is even)
• f(3) = $\frac{3+1}{2} = \frac{4}{2} = 2$ (since 3 is odd)
• f(4) = $\frac{4}{2} = 2$ (since 4 is even)

We observe that $f(1) = 1$ and $f(2) = 1$.

We have $f(1) = f(2) = 1$, but $1 \neq 2$.

Since there exist distinct elements in the domain (1 and 2) that map to the same element in the codomain (1), the function f is not one-one.

Therefore, f is not injective (not one-one).

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Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $n \in A$ such that $f(n) = y$.

The codomain of f is N = {1, 2, 3, ...}.

Let y be any arbitrary natural number in the codomain N.

We need to find an element $n \in N$ such that $f(n) = y$.

Consider two cases for y:

• Case 1: y is a possible output from the 'n is odd' rule: $f(n) = \frac{n+1}{2}$

If $\frac{n+1}{2} = y$, then $n+1 = 2y$, so $n = 2y - 1$.

If $n = 2y - 1$, we need to check if n is an odd natural number.

Since y is a natural number, $y \ge 1$.

If y = 1, n = $2(1) - 1 = 1$. 1 is an odd natural number. f(1) = $\frac{1+1}{2} = 1$.

If y = 2, n = $2(2) - 1 = 3$. 3 is an odd natural number. f(3) = $\frac{3+1}{2} = 2$.

If y = 3, n = $2(3) - 1 = 5$. 5 is an odd natural number. f(5) = $\frac{5+1}{2} = 3$.

In general, for any natural number y, $2y-1$ is an odd natural number. So for any $y \in N$, we can find an odd $n = 2y-1 \in N$ such that $f(n) = f(2y-1) = \frac{(2y-1)+1}{2} = \frac{2y}{2} = y$.

• Case 2: y is a possible output from the 'n is even' rule: $f(n) = \frac{n}{2}$

If $\frac{n}{2} = y$, then $n = 2y$.

If $n = 2y$, we need to check if n is an even natural number.

Since y is a natural number, $y \ge 1$.

If y = 1, n = $2(1) = 2$. 2 is an even natural number. f(2) = $\frac{2}{2} = 1$.

If y = 2, n = $2(2) = 4$. 4 is an even natural number. f(4) = $\frac{4}{2} = 2$.

If y = 3, n = $2(3) = 6$. 6 is an even natural number. f(6) = $\frac{6}{2} = 3$.

In general, for any natural number y, 2y is an even natural number. So for any $y \in N$, we can find an even $n = 2y \in N$ such that $f(n) = f(2y) = \frac{2y}{2} = y$.

For any natural number y in the codomain, we can find a pre-image in the domain. Specifically, if y is an odd natural number, $n = 2y-1$ is an odd natural number that maps to y. If y is an even natural number, $n = 2y$ is an even natural number that maps to y.

This means every element in the codomain N has a pre-image in the domain N.

Therefore, f is onto (surjective).

---

Conclusion:

The function f is not one-one but is onto.

Since a bijective function must be both one-one and onto, f is not a bijective function.

Given:

Set A = R – {3} (real numbers excluding 3)

Set B = R – {1} (real numbers excluding 1)

Function f : A $\rightarrow$ B defined by $f(x) = \frac{x - 2}{x - 3}$.

Domain = A

Codomain = B

---

To Determine:

Is f one-one and onto?

---

Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$. Since $x_1, x_2 \in A$, $x_1 \neq 3$ and $x_2 \neq 3$.

Cross-multiply:

Expand both sides:

Subtract $x_1x_2$ and 6 from both sides:

Add $3x_1$ and $3x_2$ to both sides:

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

---

Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain of f is B = R – {1}. Let y be any arbitrary element in the codomain B. Since $y \in B$, y is a real number and $y \neq 1$.

We need to find an element $x$ in the domain A = R – {3} such that $f(x) = y$.

Set $f(x) = y$ and solve for x:

Multiply both sides by $(x - 3)$ (which is non-zero since $x \neq 3$):

Distribute y on the right side:

Rearrange the terms to group x terms on one side and constant terms on the other:

Factor out x on the left side:

Solve for x. Since $y \in B$, we know $y \neq 1$, so $1 - y \neq 0$. We can divide by $(1 - y)$.

Now we need to check if this value of x is in the domain A = R – {3}. That is, we need to check if $x$ can ever be equal to 3.

Suppose $x = 3$. Then $\frac{2 - 3y}{1 - y} = 3$.

$2 - 3y = 3(1 - y)$

$2 - 3y = 3 - 3y$

Add 3y to both sides:

This is a contradiction. It means that our assumption ($x=3$) is false. The value of x we found, $x = \frac{2 - 3y}{1 - y}$, can never be equal to 3 for any value of y.

Since y is a real number (from B), $2 - 3y$ and $1 - y$ are real numbers, and since $1 - y \neq 0$, the value of x is a real number. Also, we have shown that this x is never equal to 3.

So, for every element y in the codomain B = R – {1}, we found an element $x = \frac{2 - 3y}{1 - y}$ in the domain A = R – {3} such that $f(x) = y$.

Therefore, f is onto.

---

Conclusion: Since f is both one-one and onto, it is a bijective function.

Given:

Function f : R $\rightarrow$ R defined as f(x) = x$^4$.

Domain = R

Codomain = R

---

To Determine:

Whether f is one-one, onto, or bijective.

---

Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$x_1^4 = x_2^4$

This implies $x_1^4 - x_2^4 = 0 \implies (x_1^2 - x_2^2)(x_1^2 + x_2^2) = 0$.

$(x_1 - x_2)(x_1 + x_2)(x_1^2 + x_2^2) = 0$.

This means $x_1 - x_2 = 0$ or $x_1 + x_2 = 0$ or $x_1^2 + x_2^2 = 0$.

So, $x_1 = x_2$ or $x_1 = -x_2$ or ($x_1 = 0$ and $x_2 = 0$, since the sum of squares of real numbers is zero only if both are zero).

Consider $x_1 = 1$ and $x_2 = -1$. Both are in R, and $x_1 \neq x_2$.

$f(1) = 1^4 = 1$.

$f(-1) = (-1)^4 = 1$.

We have $f(1) = f(-1) = 1$, but $1 \neq -1$.

Since there exist distinct elements in the domain that map to the same element in the codomain, f is not one-one (it is many-one).

---

Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain is R.

The function is f(x) = x$^4$. The fourth power of any real number is always non-negative.

The range of f is the set of all values $x^4$ for $x \in R$. This is the set of all non-negative real numbers.

Range of f = $[0, \infty)$.

The codomain of f is the set of all real numbers R.

The range of f ($[0, \infty)$) is a proper subset of the codomain (R), because R contains negative real numbers (e.g., -1, -10, -$\pi$).

Consider a negative real number in the codomain, for example, $y = -5$. We need to check if there exists any $x \in R$ such that $f(x) = -5$.

$f(x) = -5 \implies x^4 = -5$.

There is no real number x whose fourth power is a negative number.

Thus, the element -5 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ($[0, \infty)$) is not equal to the codomain R.

Therefore, f is not onto.

---

Conclusion:

The function f is neither one-one nor onto.

Comparing with the given options:

(A) f is one-one onto (False)

(B) f is many-one onto (False - not onto)

(C) f is one-one but not onto (False - not one-one)

(D) f is neither one-one nor onto. (True)

---

The correct answer is (D) f is neither one-one nor onto.

Given:

Function f : R $\rightarrow$ R defined as f(x) = 3x.

Domain = R

Codomain = R

---

To Determine:

Whether f is one-one, onto, or bijective.

---

Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$3x_1 = 3x_2$

Divide both sides by 3 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

---

Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain is R.

Let y be any arbitrary real number in the codomain R.

We need to find if there exists a real number $x$ in the domain R such that $f(x) = y$.

Set $f(x) = y$ and solve for x:

Divide by 3:

Since y is a real number, $\frac{y}{3}$ is also a real number.

So, for every real number $y$ in the codomain, we found a real number $x = \frac{y}{3}$ in the domain such that $f(x) = f\left(\frac{y}{3}\right) = 3\left(\frac{y}{3}\right) = y$.

Every element in the codomain R has a pre-image in the domain R.

Therefore, f is onto.

---

Conclusion:

Since f is both one-one and onto, it is a bijective function.

Comparing with the given options:

(A) f is one-one onto (True)

(B) f is many-one onto (False - is one-one)

(C) f is one-one but not onto (False - is onto)

(D) f is neither one-one nor onto. (False)

---

The correct answer is (A) f is one-one onto.

Given:

Set A = {2, 3, 4, 5}

Set B = {3, 4, 5, 9}

Set C = {7, 11, 15}

Function f : A $\rightarrow$ B defined by:

• f(2) = 3
• f(3) = 4
• f(4) = 5
• f(5) = 5

Function g : B $\rightarrow$ C defined by:

• g(3) = 7
• g(4) = 7
• g(5) = 11
• g(9) = 11

---

To Find:

gof (the composite function).

---

Solution:

The composite function gof is defined as $(gof)(x) = g(f(x))$. The domain of gof is the domain of f, which is A = {2, 3, 4, 5}. The codomain of gof is the codomain of g, which is C = {7, 11, 15}.

We need to find the value of $(gof)(x)$ for each element x in the domain A.

• For x = 2:

$(gof)(2) = g(f(2))$

We know f(2) = 3.

So, $(gof)(2) = g(3)$.

We know g(3) = 7.

$(gof)(2) = 7$

• For x = 3:

$(gof)(3) = g(f(3))$

We know f(3) = 4.

So, $(gof)(3) = g(4)$.

We know g(4) = 7.

$(gof)(3) = 7$

• For x = 4:

$(gof)(4) = g(f(4))$

We know f(4) = 5.

So, $(gof)(4) = g(5)$.

We know g(5) = 11.

$(gof)(4) = 11$

• For x = 5:

$(gof)(5) = g(f(5))$

We know f(5) = 5.

So, $(gof)(5) = g(5)$.

We know g(5) = 11.

$(gof)(5) = 11$

The composite function gof can be expressed as a set of ordered pairs (input, output):

gof = {(2, 7), (3, 7), (4, 11), (5, 11)}.

---

Summary of gof values:

• $(gof)(2) = 7$
• $(gof)(3) = 7$
• $(gof)(4) = 11$
• $(gof)(5) = 11$

Given:

Function f : R $\rightarrow$ R defined by f(x) = cos x.

Function g : R $\rightarrow$ R defined by g(x) = 3x$^2$.

Domain of f = R, Codomain of f = R.

Domain of g = R, Codomain of g = R.

---

To Find:

gof and fog.

To Show:

gof $\neq$ fog.

---

Finding gof:

The composite function gof : R $\rightarrow$ R is defined by $(gof)(x) = g(f(x))$.

Substitute f(x) into g(x):

Now apply the function g to cos x. The function g squares the input and multiplies by 3.

So, gof : R $\rightarrow$ R is defined by $(gof)(x) = 3\text{cos}^2 x$.

---

Finding fog:

The composite function fog : R $\rightarrow$ R is defined by $(fog)(x) = f(g(x))$.

Substitute g(x) into f(x):

Now apply the function f to $3x^2$. The function f takes the cosine of the input.

So, fog : R $\rightarrow$ R is defined by $(fog)(x) = \text{cos}(3x^2)$.

---

To Show gof $\neq$ fog:

We have found $(gof)(x) = 3\text{cos}^2 x$ and $(fog)(x) = \text{cos}(3x^2)$.

For gof = fog, we would need $3\text{cos}^2 x = \text{cos}(3x^2)$ for all $x \in R$.

Let's check if this equality holds for a specific value of x, say $x = \pi$.

Calculate $(gof)(\pi)$:

Calculate $(fog)(\pi)$:

$3\pi^2$ is a real number. $\text{cos}(3\pi^2)$ is a value between -1 and 1 (inclusive).

The value 3 is not between -1 and 1.

So, $(gof)(\pi) = 3$ and $(fog)(\pi) = \text{cos}(3\pi^2) \neq 3$.

Since there exists at least one value of x (like $x=\pi$) for which $(gof)(x) \neq (fog)(x)$, the functions gof and fog are not equal.

---

Conclusion:

gof(x) = $3\text{cos}^2 x$

fog(x) = $\text{cos}(3x^2)$

gof $\neq$ fog because, for example, $(gof)(\pi) \neq (fog)(\pi)$.

Given:

Let A = R - $\left\{ \frac{3}{5} \right\}$ and B = R - $\left\{ \frac{7}{5} \right\}$.

A function $f: B \to A$ is defined by $f(x) = \frac{3x + 4}{5x - 7}$.

A function $g: A \to B$ is defined by $g(x) = \frac{7x + 4}{5x - 3}$.

---

To Show:

1. $g \circ f = I_B$, which means $(g \circ f)(x) = x$ for all $x \in B$.

2. $f \circ g = I_A$, which means $(f \circ g)(x) = x$ for all $x \in A$.

---

Proof:

1. Calculating $g \circ f$

The domain of $g \circ f$ is the domain of $f$, which is B. For any $x \in B$, we have:

Substitute $f(x) = \frac{3x + 4}{5x - 7}$ into the function $g$:

Now, apply the definition of $g(y) = \frac{7y + 4}{5y - 3}$ with $y = \frac{3x + 4}{5x - 7}$:

To simplify, multiply the numerator and denominator by $(5x - 7)$:

Expand the terms:

Combine like terms:

Since $(g \circ f)(x) = x$ for all $x \in B$, we have shown that $\mathbf{g \circ f = I_B}$.

---

2. Calculating $f \circ g$

The domain of $f \circ g$ is the domain of $g$, which is A. For any $x \in A$, we have:

Substitute $g(x) = \frac{7x + 4}{5x - 3}$ into the function $f$:

Now, apply the definition of $f(y) = \frac{3y + 4}{5y - 7}$ with $y = \frac{7x + 4}{5x - 3}$:

To simplify, multiply the numerator and denominator by $(5x - 3)$:

Expand the terms:

Combine like terms:

Since $(f \circ g)(x) = x$ for all $x \in A$, we have shown that $\mathbf{f \circ g = I_A}$.

---

This shows that the functions $f$ and $g$ are inverses of each other.

Given:

Function f : A $\rightarrow$ B is one-one.

Function g : B $\rightarrow$ C is one-one.

---

To Show:

The composite function gof : A $\rightarrow$ C is also one-one.

---

Proof that gof is one-one:

A function h : X $\rightarrow$ Y is one-one if for any two distinct elements $x_1, x_2 \in X$, their images under h are also distinct, i.e., $h(x_1) = h(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2$ be two elements in the domain of gof, which is the set A.

Assume that $(gof)(x_1) = (gof)(x_2)$.

By the definition of the composite function, $(gof)(x) = g(f(x))$.

So, $(gof)(x_1) = (gof)(x_2)$ implies $g(f(x_1)) = g(f(x_2))$.

Let $y_1 = f(x_1)$ and $y_2 = f(x_2)$. Note that $y_1, y_2 \in B$ (the codomain of f and the domain of g).

The equation $g(f(x_1)) = g(f(x_2))$ can be written as $g(y_1) = g(y_2)$.

We are given that the function g : B $\rightarrow$ C is one-one.

By the definition of a one-one function, if $g(y_1) = g(y_2)$, then $y_1 = y_2$.

Substituting back the expressions for $y_1$ and $y_2$, we get $f(x_1) = f(x_2)$.

Now we have $f(x_1) = f(x_2)$, where $x_1, x_2 \in A$.

We are also given that the function f : A $\rightarrow$ B is one-one.

By the definition of a one-one function, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Starting with the assumption $(gof)(x_1) = (gof)(x_2)$, we have successfully shown that $x_1 = x_2$.

Thus, the composite function gof is one-one.

Given:

Function f : A $\rightarrow$ B is onto.

Function g : B $\rightarrow$ C is onto.

---

To Show:

The composite function gof : A $\rightarrow$ C is also onto.

---

Proof that gof is onto:

A function h : X $\rightarrow$ Y is onto if for every element $z \in Y$, there exists at least one element $x \in X$ such that $h(x) = z$.

Let z be an arbitrary element in the codomain of gof, which is the set C.

We need to find an element $x$ in the domain of gof, which is the set A, such that $(gof)(x) = z$.

Since $z \in C$ and the function g : B $\rightarrow$ C is onto, by the definition of an onto function, there exists at least one element $y \in B$ such that $g(y) = z$.

Now, since $y \in B$ and the function f : A $\rightarrow$ B is onto, by the definition of an onto function, there exists at least one element $x \in A$ such that $f(x) = y$.

We have found an element $x \in A$ such that $f(x) = y$ and an element $y \in B$ such that $g(y) = z$.

Consider the composite function $(gof)(x)$. By definition, $(gof)(x) = g(f(x))$.

Substituting $f(x) = y$, we get $(gof)(x) = g(y)$.

Substituting $g(y) = z$, we get $(gof)(x) = z$.

So, for the arbitrary element z in the codomain C, we found an element x in the domain A such that $(gof)(x) = z$.

This means every element in the codomain C has a pre-image in the domain A under the composite function gof.

Therefore, the composite function gof : A $\rightarrow$ C is onto.

Given:

Let $f: A \to B$ and $g: B \to C$ be two functions such that the composite function $g \circ f: A \to C$ is defined and is one-one (injective).

---

To Determine:

Are both functions $f$ and $g$ necessarily one-one?

---

Reasoning:

Let's analyze the properties of $f$ and $g$ based on the given information.

Part 1: Is $f$ necessarily one-one?

Let's prove this by contradiction. Assume that $g \circ f$ is one-one, but $f$ is not one-one.

If $f$ is not one-one, then there must exist two distinct elements $x_1, x_2$ in the domain A such that $x_1 \neq x_2$, but their images under $f$ are the same, i.e., $f(x_1) = f(x_2)$.

Now, let's apply the function $g$ to both sides of the equation $f(x_1) = f(x_2)$:

This is the same as:

We are given that $g \circ f$ is one-one. By the definition of a one-one function, if $(g \circ f)(x_1) = (g \circ f)(x_2)$, then it must be that $x_1 = x_2$.

However, this contradicts our initial assumption that $x_1 \neq x_2$. Therefore, our assumption that $f$ is not one-one must be false.

Conclusion for Part 1: Yes, $f$ must be one-one.

---

Part 2: Is $g$ necessarily one-one?

Let's try to construct a counterexample. We need to find functions $f$ and $g$ where $g \circ f$ is one-one, but $g$ is not one-one.

Let the sets be:

$A = \{1\}$

$B = \{1, 2\}$

$C = \{1\}$

Define the functions $f: A \to B$ and $g: B \to C$ as follows:

$f(1) = 1$

$g(1) = 1$

$g(2) = 1$

Now, let's check the properties:

Is $g$ one-one?

No. We have $g(1) = g(2) = 1$, but $1 \neq 2$. Two different inputs in the domain of $g$ map to the same output. So, $g$ is not one-one.

Is $g \circ f$ one-one?

The composite function $g \circ f: A \to C$ is defined as $(g \circ f)(x) = g(f(x))$.

Let's find the output for the only element in the domain A:

$(g \circ f)(1) = g(f(1)) = g(1) = 1$.

The domain of $g \circ f$ is the set $A = \{1\}$. A function with a domain of only one element is always one-one, because there are no two distinct elements to map to the same output. This is sometimes called being "vacuously true".

So, in this counterexample, $g \circ f$ is one-one, but $g$ is not one-one.

Conclusion for Part 2: No, $g$ is not necessarily one-one.

---

Final Answer:

If the composite function $g \circ f$ is one-one, then the inner function $f$ must be one-one, but the outer function $g$ is not necessarily one-one.

Given:

Let $f: A \to B$ and $g: B \to C$ be two functions such that the composite function $g \circ f: A \to C$ is defined and is onto (surjective).

---

To Determine:

Are both functions $f$ and $g$ necessarily onto?

---

Reasoning:

Let's analyze the properties of $f$ and $g$ individually.

Part 1: Is $g$ necessarily onto?

The definition of an onto (surjective) function $g: B \to C$ is that for every element $c \in C$, there exists an element $b \in B$ such that $g(b) = c$.

We are given that $g \circ f: A \to C$ is onto. By definition, this means that for every element $c \in C$, there exists an element $a \in A$ such that $(g \circ f)(a) = c$.

We can rewrite this as:

Let's consider the element $f(a)$. Since $f$ is a function from A to B, the element $f(a)$ must be in the set B. Let's call this element $b$, so $b = f(a)$ and $b \in B$.

Substituting $b$ into the equation, we get:

So, we have shown that for any element $c \in C$, we can find an element $b \in B$ (specifically, $b = f(a)$) such that $g(b) = c$. This is the exact definition of $g$ being onto.

Conclusion for Part 1: Yes, $g$ must be onto.

---

Part 2: Is $f$ necessarily onto?

Let's try to construct a counterexample. We need to find functions $f$ and $g$ where $g \circ f$ is onto, but $f$ is not onto.

For $f: A \to B$ to be not onto, there must be at least one element in B that is not an image of any element in A.

Let the sets be:

$A = \{1, 2\}$

$B = \{a, b, c\}$

$C = \{x, y\}$

Define the functions $f: A \to B$ and $g: B \to C$ as follows:

$f(1) = a$

$f(2) = b$

and

$g(a) = x$

$g(b) = y$

$g(c) = y$ (or $g(c)=x$, it doesn't matter)

Now, let's check the properties:

Is $f$ onto?

No. The codomain of $f$ is $B = \{a, b, c\}$, but the range of $f$ is only $\{a, b\}$. The element $c \in B$ has no pre-image in A. So, $f$ is not onto.

Is $g \circ f$ onto?

The composite function is $g \circ f: A \to C$. The codomain is $C = \{x, y\}$. Let's find the range.

$(g \circ f)(1) = g(f(1)) = g(a) = x$

$(g \circ f)(2) = g(f(2)) = g(b) = y$

The range of $g \circ f$ is $\{x, y\}$, which is equal to the codomain C. Therefore, $g \circ f$ is onto.

We have successfully created a scenario where $g \circ f$ is onto, but $f$ is not onto.

Conclusion for Part 2: No, $f$ is not necessarily onto.

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Final Answer:

If the composite function $g \circ f$ is onto, then the outer function $g$ must be onto, but the inner function $f$ is not necessarily onto.

Given:

Set X = {1, 2, 3}

Set Y = {a, b, c}

Function f : X $\rightarrow$ Y is one-one and onto, given by:

• f(1) = a
• f(2) = b
• f(3) = c

$I_X$ is the identity function on X, $I_X(x) = x$ for all $x \in X$.

$I_Y$ is the identity function on Y, $I_Y(y) = y$ for all $y \in Y$.

---

To Show:

There exists a function g : Y $\rightarrow$ X such that gof = $I_X$ and fog = $I_Y$.

---

Construction of function g:

Since f is one-one and onto, for each element in Y, there is a unique pre-image in X. This means the inverse function $f^{-1}$ exists and is a function from Y to X.

Let's define the function g : Y $\rightarrow$ X based on the pre-images under f:

• For the element a $\in$ Y, its unique pre-image under f is 1 (since f(1) = a). Define g(a) = 1.
• For the element b $\in$ Y, its unique pre-image under f is 2 (since f(2) = b). Define g(b) = 2.
• For the element c $\in$ Y, its unique pre-image under f is 3 (since f(3) = c). Define g(c) = 3.

So, the function g : {a, b, c} $\rightarrow$ {1, 2, 3} is defined by g = {(a, 1), (b, 2), (c, 3)}. This function exists.

---

Checking gof = $I_X$:

The composite function gof : X $\rightarrow$ X is defined by $(gof)(x) = g(f(x))$ for $x \in X = \{1, 2, 3\}$.

• For x = 1: $(gof)(1) = g(f(1))$. Since f(1) = a, $(gof)(1) = g(a)$. Since g(a) = 1, $(gof)(1) = 1$.
• For x = 2: $(gof)(2) = g(f(2))$. Since f(2) = b, $(gof)(2) = g(b)$. Since g(b) = 2, $(gof)(2) = 2$.
• For x = 3: $(gof)(3) = g(f(3))$. Since f(3) = c, $(gof)(3) = g(c)$. Since g(c) = 3, $(gof)(3) = 3$.

For every $x \in X$, $(gof)(x) = x$. This is the definition of the identity function $I_X$ on X.

---

Checking fog = $I_Y$:

The composite function fog : Y $\rightarrow$ Y is defined by $(fog)(y) = f(g(y))$ for $y \in Y = \{a, b, c\}$.

• For y = a: $(fog)(a) = f(g(a))$. Since g(a) = 1, $(fog)(a) = f(1)$. Since f(1) = a, $(fog)(a) = a$.
• For y = b: $(fog)(b) = f(g(b))$. Since g(b) = 2, $(fog)(b) = f(2)$. Since f(2) = b, $(fog)(b) = b$.
• For y = c: $(fog)(c) = f(g(c))$. Since g(c) = 3, $(fog)(c) = f(3)$. Since f(3) = c, $(fog)(c) = c$.

For every $y \in Y$, $(fog)(y) = y$. This is the definition of the identity function $I_Y$ on Y.

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Conclusion:

We have constructed a function g : {a, b, c} $\rightarrow$ {1, 2, 3} such that gof = $I_X$ and fog = $I_Y$. This function g is the inverse function of f, denoted by $f^{-1}$.

Given:

Function f : N $\rightarrow$ Y defined as f(x) = 4x + 3.

Domain = N = {1, 2, 3, ...}

Codomain = Y = {y $\in$ N: y = 4x + 3 for some x $\in$ N}.

Note that Y is the range of the function f. Since the codomain is defined to be equal to the range, the function is onto by definition.

---

To Show:

f(x) is invertible.

Find the inverse function.

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Proof that f is invertible:

A function is invertible if and only if it is bijective (both one-one and onto).

We are given that the codomain Y is the range of the function f. This implies that f is onto by definition of the codomain.

So, we only need to prove that f is one-one.

Check for Injectivity (One-one):

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

$4x_1 + 3 = 4x_2 + 3$

Subtract 3 from both sides:

Divide both sides by 4 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

Since f is both one-one and onto, it is a bijective function. Hence, f is invertible.

---

Finding the inverse function:

Let $y$ be an element in the codomain Y. By the definition of Y, $y = 4x + 3$ for some $x \in N$. This means $y = f(x)$.

To find the inverse function $f^{-1}$, we need to express x in terms of y.

Start with the equation $y = 4x + 3$ and solve for x:

The inverse function $f^{-1}$ maps from Y to N. The input to $f^{-1}$ is y $\in$ Y, and the output is the corresponding x $\in$ N.

So, the inverse function $f^{-1} : Y \rightarrow N$ is defined by $f^{-1}(y) = \frac{y - 3}{4}$.

We must verify that for any $y \in Y$, $\frac{y-3}{4}$ is indeed a natural number. By the definition of Y, $y = 4x+3$ for some $x \in N$. So $y-3 = 4x$, and $\frac{y-3}{4} = x$. Since $x \in N$, $\frac{y-3}{4}$ is a natural number. Thus, the codomain of $f^{-1}$ is N.

---

Verification (Optional but good practice):

Check if $(f^{-1}of)(x) = x$ for $x \in N$ and $(fof^{-1})(y) = y$ for $y \in Y$.

$(f^{-1}of)(x) = f^{-1}(f(x)) = f^{-1}(4x + 3)$. Substitute $(4x+3)$ into $f^{-1}(y) = \frac{y-3}{4}$:

This is the identity function on N.

$(fof^{-1})(y) = f(f^{-1}(y)) = f\left(\frac{y - 3}{4}\right)$. Substitute $\frac{y-3}{4}$ into $f(x) = 4x+3$:

This is the identity function on Y.

The inverse function is correct.

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Conclusion:

The function f(x) = 4x + 3 is invertible, and its inverse function is $f^{-1}(y) = \frac{y - 3}{4}$.

Given:

Set Y = $\{ n^2 : n \in N \} \subset N$. This means Y is the set of perfect squares within the natural numbers: Y = {1, 4, 9, 16, ...}.

Function f : N $\rightarrow$ Y defined as f(n) = n$^2$.

Domain = N = {1, 2, 3, ...}

Codomain = Y = {1, 4, 9, 16, ...}

Note that the codomain Y is specifically defined as the set of perfect squares, which is precisely the set of values that f(n) produces when n is a natural number. Thus, the range of f is equal to its codomain Y, meaning f is onto by definition.

---

To Show:

f is invertible.

Find the inverse of f.

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Proof that f is invertible:

A function is invertible if and only if it is bijective (both one-one and onto).

We are given that the codomain Y is the range of the function f. This implies that f is onto by definition of the codomain.

So, we only need to prove that f is one-one.

Check for Injectivity (One-one):

Let $n_1, n_2 \in N$ such that $f(n_1) = f(n_2)$.

$n_1^2 = n_2^2$

Since $n_1, n_2$ are natural numbers, they are positive. Taking the square root of both sides:

$\sqrt{n_1^2} = \sqrt{n_2^2} \implies |n_1| = |n_2|$

As $n_1, n_2 \in N$, $n_1 > 0$ and $n_2 > 0$. So $|n_1| = n_1$ and $|n_2| = n_2$.

$n_1 = n_2$.

Thus, if $f(n_1) = f(n_2)$, then $n_1 = n_2$.

Therefore, f is one-one.

Since f is both one-one and onto, it is a bijective function. Hence, f is invertible.

---

Finding the inverse function:

Let $y$ be an element in the codomain Y. By the definition of Y, $y = n^2$ for some $n \in N$. This means $y = f(n)$.

To find the inverse function $f^{-1}$, we need to express n in terms of y.

Start with the equation $y = n^2$ and solve for n:

Taking the square root of both sides:

Since the domain of f is N (natural numbers), n must be a positive integer. Therefore, we take the positive square root.

The inverse function $f^{-1}$ maps from Y to N. The input to $f^{-1}$ is y $\in$ Y, and the output is the corresponding n $\in$ N.

So, the inverse function $f^{-1} : Y \rightarrow N$ is defined by $f^{-1}(y) = \sqrt{y}$.

We must verify that for any $y \in Y$, $\sqrt{y}$ is indeed a natural number. By the definition of Y, $y = n^2$ for some $n \in N$. So $\sqrt{y} = \sqrt{n^2} = n$ (since n is positive). Since $n \in N$, $\sqrt{y}$ is a natural number. Thus, the codomain of $f^{-1}$ is N.

---

Verification (Optional but good practice):

Check if $(f^{-1}of)(n) = n$ for $n \in N$ and $(fof^{-1})(y) = y$ for $y \in Y$.

$(f^{-1}of)(n) = f^{-1}(f(n)) = f^{-1}(n^2)$. Substitute $n^2$ into $f^{-1}(y) = \sqrt{y}$:

This is the identity function on N.

$(fof^{-1})(y) = f(f^{-1}(y)) = f(\sqrt{y})$. Substitute $\sqrt{y}$ into $f(n) = n^2$:

This is the identity function on Y.

The inverse function is correct.

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Conclusion:

The function f(n) = n$^2$ is invertible, and its inverse function is $f^{-1}(y) = \sqrt{y}$.

Given:

A function $f : N \to S$ is defined by $f(x) = 4x^2 + 12x + 15$.

The domain is the set of Natural numbers (N).

The codomain S is the range of the function $f$.

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Part 1: Show that $f$ is invertible.

A function is invertible if and only if it is a bijection (i.e., it is both one-one and onto).

1. Checking for One-One (Injectivity)

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

Factor out $(x_1 - x_2)$:

This equation gives two possibilities:

Either $(x_1 - x_2) = 0$ or $4(x_1 + x_2) + 12 = 0$.

Case I: $x_1 - x_2 = 0 \implies x_1 = x_2$.

Case II: $4(x_1 + x_2) + 12 = 0 \implies 4(x_1 + x_2) = -12 \implies x_1 + x_2 = -3$.

Since the domain is the set of Natural numbers (N = {1, 2, 3, ...}), $x_1$ and $x_2$ are both positive integers. The sum of two positive integers ($x_1 + x_2$) must be a positive integer. Therefore, the condition $x_1 + x_2 = -3$ is not possible for $x_1, x_2 \in N$.

The only possibility is $x_1 = x_2$.

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-one.

2. Checking for Onto (Surjectivity)

The codomain of the function is given as S, where S is the range of $f$. By definition, if the codomain of a function is equal to its range, the function is onto.

Therefore, the function $f: N \to S$ is onto.

Conclusion on Invertibility:

Since the function $f$ is both one-one and onto, it is a bijection, and therefore, $f$ is invertible.

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Part 2: Find the inverse of $f$.

To find the inverse function, let $y = f(x)$ and solve for $x$ in terms of $y$.

Let $y \in S$ be an arbitrary element. By the definition of an onto function, there exists an $x \in N$ such that $f(x) = y$.

To solve for $x$, we can complete the square for the quadratic expression in $x$.

The expression in the parenthesis is a perfect square:

Now, we rearrange the equation to isolate $x$:

Take the square root of both sides:

(We take only the positive square root because the domain of $f$ is N, so $x > 0$, which implies $2x + 3$ is always positive).

Continue to solve for $x$:

Since $f$ is invertible, we can define the inverse function $g: S \to N$ (which is $f^{-1}$) by swapping the roles of $x$ and $y$. Let $g(y) = x$.

The inverse function, $f^{-1}$, in terms of the variable $x$ is:

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Conclusion:

The function $f$ is shown to be one-one and onto, hence it is invertible.

The inverse of $f$ is the function $f^{-1}: S \to N$ given by $f^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2}$.

Given:

Function f : N $\rightarrow$ N defined by f(x) = 2x, for x $\in$ N.

Function g : N $\rightarrow$ N defined by g(y) = 3y + 4, for y $\in$ N.

Function h : N $\rightarrow$ R defined by h(z) = sin z, for z $\in$ N.

Domain of f = N, Codomain of f = N.

Domain of g = N, Codomain of g = N.

Domain of h = N, Codomain of h = R.

---

To Show:

ho(gof) = (hog)of.

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Finding gof:

The composite function gof : N $\rightarrow$ N is defined by $(gof)(x) = g(f(x))$ for $x \in N$.

Substitute f(x) = 2x into g(y) = 3y + 4. Replace y with 2x.

So, gof : N $\rightarrow$ N is defined by $(gof)(x) = 6x + 4$.

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Finding ho(gof):

The composite function ho(gof) : N $\rightarrow$ R is defined by $(ho(gof))(x) = h((gof)(x))$ for $x \in N$.

Substitute $(gof)(x) = 6x + 4$ into h(z) = sin z. Replace z with $6x + 4$.

So, ho(gof) : N $\rightarrow$ R is defined by $(ho(gof))(x) = \text{sin}(6x + 4)$.

---

Finding hog:

The composite function hog : N $\rightarrow$ R is defined by $(hog)(y) = h(g(y))$ for $y \in N$.

Substitute g(y) = 3y + 4 into h(z) = sin z. Replace z with $3y + 4$.

So, hog : N $\rightarrow$ R is defined by $(hog)(y) = \text{sin}(3y + 4)$.

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Finding (hog)of:

The composite function (hog)of : N $\rightarrow$ R is defined by $((hog)of)(x) = (hog)(f(x))$ for $x \in N$.

Substitute f(x) = 2x into $(hog)(y) = \text{sin}(3y + 4)$. Replace y with 2x.

So, (hog)of : N $\rightarrow$ R is defined by $((hog)of)(x) = \text{sin}(6x + 4)$.

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Comparison:

We found $(ho(gof))(x) = \text{sin}(6x + 4)$.

We found $((hog)of)(x) = \text{sin}(6x + 4)$.

Since the domains and codomains are the same (N $\rightarrow$ R) and the function expressions are the same for all x in the domain, we have:

Therefore, ho(gof) = (hog)of.

This demonstrates the associative property of function composition.

Given:

Set A = {1, 2, 3}

Set B = {a, b, c}

Set C = {apple, ball, cat}

Function f : A $\rightarrow$ B defined by f(1) = a, f(2) = b, f(3) = c.

Function g : B $\rightarrow$ C defined by g(a) = apple, g(b) = ball, g(c) = cat.

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To Show:

f, g, and gof are invertible.

Find $f^{-1}$, $g^{-1}$, and $(gof)^{-1}$.

Show that $(gof)^{-1} = f^{-1}o g^{-1}$.

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Invertibility of f:

Domain of f = {1, 2, 3}, Codomain of f = {a, b, c}. Both sets have 3 elements.

Check Injectivity (One-one): The distinct elements in the domain (1, 2, 3) map to distinct elements in the codomain (f(1)=a, f(2)=b, f(3)=c are all distinct). So f is one-one.

Check Surjectivity (Onto): The range of f is {f(1), f(2), f(3)} = {a, b, c}. This is equal to the codomain B. So f is onto.

Since f is both one-one and onto, f is invertible.

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Invertibility of g:

Domain of g = {a, b, c}, Codomain of g = {apple, ball, cat}. Both sets have 3 elements.

Check Injectivity (One-one): The distinct elements in the domain (a, b, c) map to distinct elements in the codomain (g(a)=apple, g(b)=ball, g(c)=cat are all distinct). So g is one-one.

Check Surjectivity (Onto): The range of g is {g(a), g(b), g(c)} = {apple, ball, cat}. This is equal to the codomain C. So g is onto.

Since g is both one-one and onto, g is invertible.

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Finding gof and its invertibility:

The composite function gof : A $\rightarrow$ C is defined by $(gof)(x) = g(f(x))$ for $x \in A = \{1, 2, 3\}$.

• $(gof)(1) = g(f(1)) = g(a) = \text{apple}$.
• $(gof)(2) = g(f(2)) = g(b) = \text{ball}$.
• $(gof)(3) = g(f(3)) = g(c) = \text{cat}$.

gof = {(1, apple), (2, ball), (3, cat)}.

Domain of gof = {1, 2, 3}, Codomain of gof = {apple, ball, cat}. Both sets have 3 elements.

Check Injectivity (One-one): The distinct elements in the domain (1, 2, 3) map to distinct elements in the codomain (apple, ball, cat are all distinct). So gof is one-one.

Check Surjectivity (Onto): The range of gof is {apple, ball, cat}. This is equal to the codomain C. So gof is onto.

Since gof is both one-one and onto, gof is invertible.

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Finding inverse functions:

The inverse function reverses the mapping of the original function.

$f^{-1} : B \rightarrow A$:

• Since f(1) = a, $f^{-1}(a) = 1$.
• Since f(2) = b, $f^{-1}(b) = 2$.
• Since f(3) = c, $f^{-1}(c) = 3$.

So, $f^{-1} = \text{\{(a, 1), (b, 2), (c, 3)\}}$.

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$g^{-1} : C \rightarrow B$:

• Since g(a) = apple, $g^{-1}(\text{apple}) = \text{a}$.
• Since g(b) = ball, $g^{-1}(\text{ball}) = \text{b}$.
• Since g(c) = cat, $g^{-1}(\text{cat}) = \text{c}$.

So, $g^{-1} = \text{\{(apple, a), (ball, b), (cat, c)\}}$.

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$(gof)^{-1} : C \rightarrow A$:

• Since $(gof)(1) = \text{apple}$, $(gof)^{-1}(\text{apple}) = 1$.
• Since $(gof)(2) = \text{ball}$, $(gof)^{-1}(\text{ball}) = 2$.
• Since $(gof)(3) = \text{cat}$, $(gof)^{-1}(\text{cat}) = 3$.

So, $(gof)^{-1} = \text{\{(apple, 1), (ball, 2), (cat, 3)\}}$.

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Showing $(gof)^{-1} = f^{-1}o g^{-1}$:

The composite function $f^{-1}o g^{-1} : C \rightarrow A$ is defined by $(f^{-1}o g^{-1})(z) = f^{-1}(g^{-1}(z))$ for $z \in C = \{\text{apple, ball, cat}\}$.

• For z = apple: $(f^{-1}o g^{-1})(\text{apple}) = f^{-1}(g^{-1}(\text{apple}))$. Since $g^{-1}(\text{apple}) = \text{a}$, $(f^{-1}o g^{-1})(\text{apple}) = f^{-1}(a)$. Since $f^{-1}(a) = 1$, $(f^{-1}o g^{-1})(\text{apple}) = 1$.
• For z = ball: $(f^{-1}o g^{-1})(\text{ball}) = f^{-1}(g^{-1}(\text{ball}))$. Since $g^{-1}(\text{ball}) = \text{b}$, $(f^{-1}o g^{-1})(\text{ball}) = f^{-1}(b)$. Since $f^{-1}(b) = 2$, $(f^{-1}o g^{-1})(\text{ball}) = 2$.
• For z = cat: $(f^{-1}o g^{-1})(\text{cat}) = f^{-1}(g^{-1}(\text{cat}))$. Since $g^{-1}(\text{cat}) = \text{c}$, $(f^{-1}o g^{-1})(\text{cat}) = f^{-1}(c)$. Since $f^{-1}(c) = 3$, $(f^{-1}o g^{-1})(\text{cat}) = 3$.

The composite function $f^{-1}o g^{-1}$ maps:

• apple $\rightarrow$ 1
• ball $\rightarrow$ 2
• cat $\rightarrow$ 3

So, $f^{-1}o g^{-1} = \text{\{(apple, 1), (ball, 2), (cat, 3)\}}$.

Comparing the set of ordered pairs for $(gof)^{-1}$ and $f^{-1}o g^{-1}$, we see they are the same.

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Conclusion:

Functions f, g, and gof are invertible.

$f^{-1} = \text{\{(a, 1), (b, 2), (c, 3)\}}$

$g^{-1} = \text{\{(apple, a), (ball, b), (cat, c)\}}$

$(gof)^{-1} = \text{\{(apple, 1), (ball, 2), (cat, 3)\}}$

We have shown that $(gof)^{-1} = f^{-1}o g^{-1}$.

Given:

Set S = {1, 2, 3}.

Functions f : S $\rightarrow$ S are defined below.

Domain = S, Codomain = S.

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To Determine:

Whether each function has an inverse. Find $f^{-1}$ if it exists.

A function has an inverse if and only if it is bijective (both one-one and onto).

For a function $f: S \to S$ where S is a finite set, f is one-one if and only if it is onto.

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(a) f = {(1, 1), (2, 2), (3, 3)}

Check Injectivity (One-one):

The elements in the domain (1, 2, 3) map to the images {1, 2, 3}. The distinct elements 1, 2, 3 in the domain map to distinct images 1, 2, 3 in the codomain. So f is one-one.

Check Surjectivity (Onto):

The codomain is {1, 2, 3}. The range is {1, 2, 3}. The range is equal to the codomain. So f is onto.

Since f is both one-one and onto, f is invertible.

Finding the inverse $f^{-1}$: The inverse function reverses the mapping.

• Since f(1) = 1, $f^{-1}(1) = 1$.
• Since f(2) = 2, $f^{-1}(2) = 2$.
• Since f(3) = 3, $f^{-1}(3) = 3$.

$f^{-1} = \text{\{(1, 1), (2, 2), (3, 3)\}}$.

(This is the identity function on S, $I_S$).

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(b) f = {(1, 2), (2, 1), (3, 1)}

Check Injectivity (One-one):

Consider the elements 2 and 3 in the domain. f(2) = 1 and f(3) = 1. We have f(2) = f(3) but $2 \neq 3$. So f is not one-one.

Since f is not one-one, it is not invertible.

(We don't need to check surjectivity, but for completeness: The codomain is {1, 2, 3}. The range is {f(1), f(2), f(3)} = {2, 1, 1} = {1, 2}. The element 3 in the codomain is not in the range. So f is not onto).

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(c) f = {(1, 3), (3, 2), (2, 1)}

Check Injectivity (One-one):

The elements in the domain (1, 2, 3) map to the images {3, 1, 2}. The distinct elements 1, 2, 3 in the domain map to distinct images 3, 1, 2 in the codomain. So f is one-one.

Check Surjectivity (Onto):

The codomain is {1, 2, 3}. The range is {f(1), f(2), f(3)} = {3, 1, 2} = {1, 2, 3}. The range is equal to the codomain. So f is onto.

Since f is both one-one and onto, f is invertible.

Finding the inverse $f^{-1}$: The inverse function reverses the mapping.

• Since f(1) = 3, $f^{-1}(3) = 1$.
• Since f(2) = 1, $f^{-1}(1) = 2$.
• Since f(3) = 2, $f^{-1}(2) = 3$.

Arranging the pairs by the order of the first element in the inverse function:

$f^{-1} = \text{\{(1, 2), (2, 3), (3, 1)\}}$.

Given:

Function f: $\{1, 3, 4\} \to \{1, 2, 5\}$ defined as $f = \{(1, 2), (3, 5), (4, 1)\}$.

Function g: $\{1, 2, 5\} \to \{1, 3\}$ defined as $g = \{(1, 3), (2, 3), (5, 1)\}$.

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To Find:

The composite function gof.

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Solution:

The composite function gof is defined as gof(x) = g(f(x)) for all x in the domain of f. The domain of f is $\{1, 3, 4\}$. We find the image of each element in the domain of f under the composite function gof.

For $x = 1$:

$f(1) = 2$ (from the definition of f)

$g(f(1)) = g(2)$ (substituting the value of f(1))

$g(2) = 3$ (from the definition of g)

So, gof(1) = 3. This gives the ordered pair (1, 3).

For $x = 3$:

$f(3) = 5$ (from the definition of f)

$g(f(3)) = g(5)$ (substituting the value of f(3))

$g(5) = 1$ (from the definition of g)

So, gof(3) = 1. This gives the ordered pair (3, 1).

For $x = 4$:

$f(4) = 1$ (from the definition of f)

$g(f(4)) = g(1)$ (substituting the value of f(4))

$g(1) = 3$ (from the definition of g)

So, gof(4) = 3. This gives the ordered pair (4, 3).

Therefore, the composite function gof is the set of ordered pairs:

$gof = \{(1, 3), (3, 1), (4, 3)\}$.

Given:

Three functions $f: \mathbb{R} \to \mathbb{R}$, $g: \mathbb{R} \to \mathbb{R}$, and $h: \mathbb{R} \to \mathbb{R}$.

The sum of two functions $(f+g)$ is defined as $(f+g)(x) = f(x) + g(x)$.

The product of two functions $(f \cdot g)$ is defined as $(f \cdot g)(x) = f(x) \cdot g(x)$.

The composition of two functions $(f \circ h)$ is defined as $(f \circ h)(x) = f(h(x))$.

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Part 1: To Prove that $(f + g) \circ h = (f \circ h) + (g \circ h)$

To prove that these two functions are equal, we must show that they produce the same output for any given input $x \in \mathbb{R}$.

Let's start with the left-hand side (LHS):

By the definition of function composition, this is:

By the definition of the sum of two functions, this is:

Now, let's look at the right-hand side (RHS):

By the definition of the sum of two functions, this is:

By the definition of function composition, this is:

Since LHS = $f(h(x)) + g(h(x))$ and RHS = $f(h(x)) + g(h(x))$, we have shown that LHS = RHS for all $x \in \mathbb{R}$.

Therefore, $(f + g) \circ h = (f \circ h) + (g \circ h)$ is proved.

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Part 2: To Prove that $(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h)$

Again, we must show that both sides produce the same output for any input $x \in \mathbb{R}$.

Let's start with the left-hand side (LHS):

By the definition of function composition:

By the definition of the product of two functions:

Now, let's look at the right-hand side (RHS):

By the definition of the product of two functions:

By the definition of function composition:

Since LHS = $f(h(x)) \cdot g(h(x))$ and RHS = $f(h(x)) \cdot g(h(x))$, we have shown that LHS = RHS for all $x \in \mathbb{R}$.

Therefore, $(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h)$ is proved.

Part (i):

Given:

$f(x) = |x|$

$g(x) = |5x - 2|$

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To Find:

gof and fog.

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Solution:

We need to find the composite functions gof(x) = g(f(x)) and fog(x) = f(g(x)).

Finding gof(x):

gof(x) $= g(f(x))$

Substitute $f(x) = |x|$ into the expression for g(x).

gof(x) $= g(|x|)$

Using the definition of $g(x) = |5x - 2|$, replace x with $|x|$.

gof(x) $= |5|x| - 2|$

So, $\textbf{gof(x) = |5|x| - 2|}$.

Finding fog(x):

fog(x) $= f(g(x))$

Substitute $g(x) = |5x - 2|$ into the expression for f(x).

fog(x) $= f(|5x - 2|)$

Using the definition of $f(x) = |x|$, replace x with $|5x - 2|$.

fog(x) $= ||5x - 2||$

Since the absolute value of an absolute value is the absolute value itself, we have $||a|| = |a|$.

fog(x) $= |5x - 2|$

So, $\textbf{fog(x) = |5x - 2|}$.

---

Part (ii):

Given:

$f(x) = 8x^3$

$g(x) = x^{\frac{1}{3}}$

---

To Find:

gof and fog.

---

Solution:

We need to find the composite functions gof(x) = g(f(x)) and fog(x) = f(g(x)).

Finding gof(x):

gof(x) $= g(f(x))$

Substitute $f(x) = 8x^3$ into the expression for g(x).

gof(x) $= g(8x^3)$

Using the definition of $g(x) = x^{\frac{1}{3}}$, replace x with $8x^3$.

gof(x) $= (8x^3)^{\frac{1}{3}}$

Using the exponent rule $(ab)^m = a^m b^m$ and $(x^n)^m = x^{nm}$, we have:

gof(x) $= 8^{\frac{1}{3}} \cdot (x^3)^{\frac{1}{3}}$

gof(x) $= (2^3)^{\frac{1}{3}} \cdot x^{3 \cdot \frac{1}{3}}$

gof(x) $= 2^{3 \cdot \frac{1}{3}} \cdot x^1$

gof(x) $= 2^1 \cdot x$

gof(x) $= 2x$

So, $\textbf{gof(x) = 2x}$.

Finding fog(x):

fog(x) $= f(g(x))$

Substitute $g(x) = x^{\frac{1}{3}}$ into the expression for f(x).

fog(x) $= f(x^{\frac{1}{3}})$

Using the definition of $f(x) = 8x^3$, replace x with $x^{\frac{1}{3}}$.

fog(x) $= 8(x^{\frac{1}{3}})^3$

Using the exponent rule $(x^n)^m = x^{nm}$, we have:

fog(x) $= 8 \cdot x^{\frac{1}{3} \cdot 3}$

fog(x) $= 8 \cdot x^1$

fog(x) $= 8x$

So, $\textbf{fog(x) = 8x}$.

Given:

The function $f$ is defined as $f(x) = \frac{4x + 3}{6x - 4}$, with the domain restriction $x \neq \frac{2}{3}$.

---

Part 1: To Show that $(f \circ f)(x) = x$

The composition of $f$ with itself, $(f \circ f)(x)$, is defined as $f(f(x))$.

We substitute the expression for $f(x)$ into the function $f$:

Now, we apply the rule for $f(y) = \frac{4y+3}{6y-4}$ with $y = \frac{4x + 3}{6x - 4}$:

To simplify this complex fraction, we multiply the numerator and the denominator of the main fraction by $(6x - 4)$:

Expand the terms in the numerator and denominator:

Combine like terms:

Simplify the fraction:

This shows that $f \circ f(x) = x$ for all $x \neq \frac{2}{3}$.

---

Part 2: What is the inverse of $f$?

The inverse of a function $f$, denoted as $f^{-1}$, is a function such that $(f \circ f^{-1})(x) = x$ and $(f^{-1} \circ f)(x) = x$.

From Part 1, we have already shown that $(f \circ f)(x) = x$.

This result perfectly matches the definition of an inverse function. If we compare the equation $(f \circ f)(x) = x$ with the definition $(f \circ f^{-1})(x) = x$, we can see that the function $f$ itself acts as its own inverse.

Therefore, the inverse of $f$ is $f$ itself.

The inverse of $f$ is $f^{-1}(x) = f(x) = \frac{4x + 3}{6x - 4}$.

---

Alternate Method for finding the inverse:

Let $y = f(x)$. We will solve for $x$ in terms of $y$.

Multiply both sides by $(6x - 4)$:

Group all terms containing $x$ on one side:

Factor out $x$:

Solve for $x$:

Since $x = f^{-1}(y)$, we have:

Replacing the variable $y$ with $x$, we get the inverse function:

This confirms that $f^{-1}(x) = f(x)$.

A function has an inverse if and only if it is a bijection, which means it must be both one-one (injective) and onto (surjective).

---

(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

Check for One-One (Injectivity):

A function is one-one if distinct elements in the domain have distinct images in the codomain.

Here, we can see that:

$f(1) = 10$, $f(2) = 10$, $f(3) = 10$, and $f(4) = 10$.

Multiple distinct elements from the domain (1, 2, 3, 4) are mapped to the same element (10) in the codomain. For example, $f(1) = f(2)$ but $1 \neq 2$.

Therefore, the function $f$ is not one-one.

Conclusion:

Since the function $f$ is not one-one, it is not a bijection.

Therefore, the function f does not have an inverse.

---

(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

Check for One-One (Injectivity):

We check if distinct inputs have distinct outputs.

$g(5) = 4$

$g(7) = 4$

Here, two distinct elements from the domain, 5 and 7, are mapped to the same element, 4, in the codomain. Since $g(5) = g(7)$ but $5 \neq 7$, the function is many-one.

Therefore, the function $g$ is not one-one.

Conclusion:

Since the function $g$ is not one-one, it is not a bijection.

Therefore, the function g does not have an inverse.

---

(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Check for One-One (Injectivity):

We examine the mapping:

$h(2) = 7$, $h(3) = 9$, $h(4) = 11$, $h(5) = 13$.

Each distinct element in the domain {2, 3, 4, 5} is mapped to a unique and distinct element in the codomain {7, 9, 11, 13}.

Therefore, the function $h$ is one-one (injective).

Check for Onto (Surjectivity):

A function is onto if its range is equal to its codomain.

The codomain of $h$ is {7, 9, 11, 13}.

The range of $h$ (the set of all output values) is also {7, 9, 11, 13}.

Since the range is equal to the codomain, every element in the codomain has a pre-image in the domain.

Therefore, the function $h$ is onto (surjective).

Conclusion:

Since the function $h$ is both one-one and onto, it is a bijection.

Therefore, the function h has an inverse.

Given:

Function $f : [-1, 1] \to \mathbb{R}$ given by $f(x) = \frac{x}{x + 2}$.

---

To Prove/Find:

1. Show that f is one-to-one on the interval $[-1, 1]$.

2. Find the inverse of the function $f : [-1, 1] \to \text{Range f}$.

---

Solution:

Part 1: Show that f is one-to-one

To show that f is one-to-one (injective), we assume that $f(x_1) = f(x_2)$ for any $x_1, x_2 \in [-1, 1]$ and show that it implies $x_1 = x_2$.

Let $x_1, x_2 \in [-1, 1]$ such that $f(x_1) = f(x_2)$.

$\frac{x_1}{x_1 + 2} = \frac{x_2}{x_2 + 2}$

Cross-multiply:

$x_1(x_2 + 2) = x_2(x_1 + 2)$

Expand both sides:

$x_1 x_2 + 2x_1 = x_1 x_2 + 2x_2$

Subtract $x_1 x_2$ from both sides:

$2x_1 = 2x_2$

Divide both sides by 2:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in [-1, 1]$, the function f is one-to-one on the interval $[-1, 1]$.

---

Part 2: Find the inverse of f : [-1, 1] → Range f

To find the inverse function, we set $y = f(x)$ and solve for x in terms of y. The domain of the inverse function will be the range of the original function f.

Let $y = f(x) = \frac{x}{x + 2}$. Since the codomain is restricted to the Range f, for any $y$ in the Range f, there exists an $x \in [-1, 1]$ such that $y = f(x)$.

$y = \frac{x}{x + 2}$

Multiply both sides by $(x + 2)$: (Note that for $x \in [-1, 1]$, $x+2 \neq 0$)

$y(x + 2) = x$

Distribute y:

$yx + 2y = x$

Rearrange the terms to group x terms on one side:

$2y = x - yx$

Factor out x from the terms on the right side:

$2y = x(1 - y)$

Divide by $(1 - y)$ to solve for x: (Note that if $y=1$, $2y = 2 \neq 0$, but $x(1-y)=0$, so $y$ cannot be 1 in the range of f for $x \in [-1, 1]$. We confirm this by finding the range below).

$x = \frac{2y}{1 - y}$

This expression gives the inverse function. We denote the inverse function as $f^{-1}(y)$.

$f^{-1}(y) = \frac{2y}{1 - y}$

It is conventional to write the inverse function in terms of x, so we replace y with x:

$f^{-1}(x) = \frac{2x}{1 - x}$

Now, we need to determine the domain of $f^{-1}$, which is the range of $f$. Since f is continuous and strictly increasing on $[-1, 1]$, its range is $[f(-1), f(1)]$.

$f(-1) = \frac{-1}{-1 + 2} = \frac{-1}{1} = -1$

$f(1) = \frac{1}{1 + 2} = \frac{1}{3}$

The range of $f$ is $[-1, \frac{1}{3}]$. This is the domain of $f^{-1}$.

So, the inverse function is $\textbf{f}^{\textbf{-1}}\textbf{(x) = } \frac{\textbf{2x}}{\textbf{1 - x}}$ with domain $\textbf{[-1, } \frac{\textbf{1}}{\textbf{3}}\textbf{]}$.

Given:

The function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = 4x + 3$.

---

To Show/Find:

1. Show that f is invertible.

2. Find the inverse of f.

---

Solution:

To show that a function is invertible, we need to demonstrate that it is both one-to-one (injective) and onto (surjective).

---

Checking for Injectivity (One-to-one):

Assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in \mathbb{R}$ in the domain of f.

By the definition of f(x):

$4x_1 + 3 = 4x_2 + 3$

Subtract 3 from both sides of the equation:

$4x_1 = 4x_2$

Divide both sides by 4:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function f is one-to-one.

---

Checking for Surjectivity (Onto):

For the function to be onto, for every element $y$ in the codomain $\mathbb{R}$, there must exist at least one element $x$ in the domain $\mathbb{R}$ such that $f(x) = y$.

Let $y$ be an arbitrary element in the codomain $\mathbb{R}$. We set $f(x) = y$ and try to solve for x:

$4x + 3 = y$

Subtract 3 from both sides:

$4x = y - 3$

Divide both sides by 4:

$x = \frac{y - 3}{4}$

Since for every real number $y$ in the codomain, we can find a real number $x = \frac{y - 3}{4}$ in the domain such that $f(x) = y$, the function f is onto.

---

Since f is both one-to-one and onto, f is bijective and therefore invertible.

---

Finding the inverse of f:

To find the inverse function, we use the equation $x = \frac{y - 3}{4}$, which we obtained while checking for surjectivity. This equation gives us the input $x$ in terms of the output $y$.

The inverse function, denoted by $f^{-1}$, maps $y$ back to $x$. So, we have:

$f^{-1}(y) = \frac{y - 3}{4}$

It is standard practice to write the inverse function in terms of the variable x. Replacing y with x, we get:

$\textbf{f}^{\textbf{-1}}\textbf{(x) = } \frac{\textbf{x - 3}}{\textbf{4}}$

The domain and codomain of $f^{-1}$ are both $\mathbb{R}$.

Given:

The function $f : \mathbb{R}_+ \to [4, \infty)$ given by $f(x) = x^2 + 4$, where $\mathbb{R}_+ = [0, \infty)$ is the set of all non-negative real numbers.

---

To Show:

1. The function f is invertible.

2. The inverse of f is $f^{-1}(y) = \sqrt{y-4}$.

---

Solution:

To show that f is invertible, we need to demonstrate that it is both one-to-one (injective) and onto (surjective) on the specified domain and codomain.

---

Checking for Injectivity (One-to-one):

Assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in \mathbb{R}_+$ (the domain of f).

By the definition of f(x):

$x_1^2 + 4 = x_2^2 + 4$

Subtract 4 from both sides:

$x_1^2 = x_2^2$

Taking the square root of both sides:

$\sqrt{x_1^2} = \sqrt{x_2^2}$

$|x_1| = |x_2|$

Since $x_1, x_2 \in \mathbb{R}_+ = [0, \infty)$, both $x_1$ and $x_2$ are non-negative. Therefore, $|x_1| = x_1$ and $|x_2| = x_2$.

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in \mathbb{R}_+$, the function f is one-to-one on its domain $\mathbb{R}_+$.

---

Checking for Surjectivity (Onto):

For the function to be onto the codomain $[4, \infty)$, for every element $y$ in the codomain $[4, \infty)$, there must exist at least one element $x$ in the domain $\mathbb{R}_+ = [0, \infty)$ such that $f(x) = y.

Let $y$ be an arbitrary element in the codomain $[4, \infty)$. We set $f(x) = y$ and try to solve for x:

$x^2 + 4 = y$

Subtract 4 from both sides:

$x^2 = y - 4$

Since $y \in [4, \infty)$, $y \ge 4$, which means $y - 4 \ge 0$. Thus, $y - 4$ is a non-negative number, and its square root is a real number.

Taking the square root of both sides:

$x = \pm \sqrt{y - 4}$

However, the domain of the function f is $\mathbb{R}_+ = [0, \infty)$, which consists only of non-negative real numbers. Therefore, we must choose the non-negative square root:

$x = \sqrt{y - 4}$

For every $y \in [4, \infty)$, the value $x = \sqrt{y - 4}$ is a non-negative real number, and thus belongs to the domain $\mathbb{R}_+$. Also, $f(\sqrt{y-4}) = (\sqrt{y-4})^2 + 4 = (y-4) + 4 = y$.

Since for every $y$ in the codomain $[4, \infty)$, there exists an $x = \sqrt{y - 4}$ in the domain $\mathbb{R}_+$ such that $f(x) = y$, the function f is onto the codomain $[4, \infty)$.

---

Since f is both one-to-one and onto its specified codomain, f is bijective and therefore invertible.

---

Finding the inverse of f:

From the surjectivity check, we found that for any $y$ in the range, the corresponding $x$ is given by $x = \sqrt{y - 4}$. This expression gives the inverse function $f^{-1}(y)$.

$f^{-1}(y) = \sqrt{y - 4}$

This matches the inverse function given in the question.

The domain of the inverse function $f^{-1}$ is the range of f, which is $[4, \infty)$. The codomain of $f^{-1}$ is the domain of f, which is $\mathbb{R}_+ = [0, \infty)$.

Thus, the inverse of the function $f : \mathbb{R}_+ \to [4, \infty)$ is $\textbf{f}^{\textbf{-1}}\textbf{ : [4, } \infty\textbf{) } \to \mathbb{R}_+$, given by $\textbf{f}^{\textbf{-1}}\textbf{(y) = } \sqrt{\textbf{y - 4}}$.

Given:

The function $f : \mathbb{R}_+ \to [-5, \infty)$ given by $f(x) = 9x^2 + 6x - 5$, where $\mathbb{R}_+ = [0, \infty)$ is the set of all non-negative real numbers.

---

To Show:

1. The function f is invertible.

2. The inverse of f is $f^{-1}(y) = \frac{\sqrt{y+6}-1}{3}$.

---

Solution:

To show that f is invertible, we need to demonstrate that it is both one-to-one (injective) and onto (surjective) on the specified domain and codomain.

We can also rewrite the function $f(x)$ by completing the square to understand its behavior better:

$f(x) = 9x^2 + 6x - 5$

$f(x) = (9x^2 + 6x) - 5$

$f(x) = ((3x)^2 + 2(3x)(1) + 1^2) - 1^2 - 5$

$f(x) = (3x + 1)^2 - 1 - 5$

$f(x) = (3x + 1)^2 - 6$

Since the domain is $\mathbb{R}_+ = [0, \infty)$, for any $x \ge 0$, $3x \ge 0$, and $3x + 1 \ge 1$. Therefore, $(3x+1)^2 \ge 1^2 = 1$.

This means $f(x) = (3x+1)^2 - 6 \ge 1 - 6 = -5$. The minimum value of $f(x)$ on the domain $[0, \infty)$ is indeed $-5$, which occurs at $x=0$. The range of the function is $[-5, \infty)$, which matches the given codomain.

---

Checking for Injectivity (One-to-one):

Assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in [0, \infty)$.

Using the completed square form:

$(3x_1 + 1)^2 - 6 = (3x_2 + 1)^2 - 6$

$(3x_1 + 1)^2 = (3x_2 + 1)^2$

Taking the square root of both sides:

$\sqrt{(3x_1 + 1)^2} = \sqrt{(3x_2 + 1)^2}$

$|3x_1 + 1| = |3x_2 + 1|$

Since the domain of f is $\mathbb{R}_+ = [0, \infty)$, we have $x_1 \ge 0$ and $x_2 \ge 0$. This implies $3x_1 \ge 0$ and $3x_2 \ge 0$. Thus, $3x_1 + 1 \ge 1$ and $3x_2 + 1 \ge 1$. Both expressions are positive, so their absolute values are themselves.

$3x_1 + 1 = 3x_2 + 1$

Subtract 1 from both sides:

$3x_1 = 3x_2$

Divide both sides by 3:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in [0, \infty)$, the function f is one-to-one on its domain $\mathbb{R}_+$.

---

Checking for Surjectivity (Onto):

For the function to be onto the codomain $[-5, \infty)$, for every element $y$ in the codomain $[-5, \infty)$, there must exist at least one element $x$ in the domain $\mathbb{R}_+ = [0, \infty)$ such that $f(x) = y.

Let $y$ be an arbitrary element in the codomain $[-5, \infty)$. We set $f(x) = y$ and try to solve for x:

Using the completed square form:

$(3x + 1)^2 - 6 = y$

Add 6 to both sides:

$(3x + 1)^2 = y + 6$

Since $y \in [-5, \infty)$, we have $y \ge -5$, so $y + 6 \ge 1$. Thus, $y+6$ is a positive number, and its square root is a real number.

Taking the square root of both sides:

$\sqrt{(3x + 1)^2} = \sqrt{y + 6}$

$|3x + 1| = \sqrt{y + 6}$

Since the domain of f is $\mathbb{R}_+ = [0, \infty)$, we have $x \ge 0$, which means $3x+1 \ge 1$. Thus, $|3x+1| = 3x+1$.

$3x + 1 = \sqrt{y + 6}$

Subtract 1 from both sides:

$3x = \sqrt{y + 6} - 1$

Divide both sides by 3:

$x = \frac{\sqrt{y + 6} - 1}{3}$

We need to ensure that this value of x is in the domain $\mathbb{R}_+ = [0, \infty)$ for all $y \in [-5, \infty)$. Since $y \ge -5$, $y+6 \ge 1$, so $\sqrt{y+6} \ge \sqrt{1} = 1$. Therefore, $\sqrt{y+6} - 1 \ge 1 - 1 = 0$. Dividing by 3, we get $x = \frac{\sqrt{y + 6} - 1}{3} \ge 0$. This value of x is indeed in the domain $[0, \infty)$.

Thus, for every $y$ in the codomain $[-5, \infty)$, there exists a unique $x = \frac{\sqrt{y + 6} - 1}{3}$ in the domain $[0, \infty)$ such that $f(x) = y$. The function f is onto its specified codomain $[-5, \infty)$.

---

Since f is both one-to-one and onto its specified codomain, f is bijective and therefore invertible.

---

Finding the inverse of f:

From the surjectivity check, we found that for any $y$ in the range, the corresponding $x$ is given by $x = \frac{\sqrt{y + 6} - 1}{3}$. This expression gives the inverse function $f^{-1}(y)$.

$f^{-1}(y) = \frac{\sqrt{y + 6} - 1}{3}$

This matches the inverse function given in the question.

The domain of the inverse function $f^{-1}$ is the range of f, which is $[-5, \infty)$. The codomain of $f^{-1}$ is the domain of f, which is $\mathbb{R}_+ = [0, \infty)$.

Thus, the inverse of the function $f : \mathbb{R}_+ \to [-5, \infty)$ is $\textbf{f}^{\textbf{-1}}\textbf{ : [-5, } \infty\textbf{) } \to \mathbb{R}_+$, given by $\textbf{f}^{\textbf{-1}}\textbf{(y) = } \frac{\textbf{ \sqrt{y+6} - 1 }}{\textbf{3}}$.

Given:

A function $f : X \to Y$ which is invertible.

---

To Prove:

The inverse of f is unique.

---

Proof:

Assume, for the sake of contradiction, that the invertible function f has two different inverse functions, say $g_1 : Y \to X$ and $g_2 : Y \to X$.

By the definition of an inverse function, if $g_1$ is the inverse of f, then:

where $I_Y$ is the identity function on set Y (i.e., $I_Y(y) = y$ for all $y \in Y$).

and

where $I_X$ is the identity function on set X (i.e., $I_X(x) = x$ for all $x \in X$).

Similarly, if $g_2$ is the inverse of f, then:

and

We want to show that $g_1 = g_2$. This means showing that for every element $y \in Y$, $g_1(y) = g_2(y)$.

Let $y$ be an arbitrary element in the set Y.

From (i) and (iii), we have:

and

Therefore, for any $y \in Y$, we have:

$f(g_1(y)) = y$ and $f(g_2(y)) = y$

This implies:

$f(g_1(y)) = f(g_2(y))$

Since the function f is invertible, it is necessarily one-to-one (injective). By the definition of a one-to-one function, if $f(a) = f(b)$, then it must follow that $a = b$.

In the equation $f(g_1(y)) = f(g_2(y))$, let $a = g_1(y)$ and $b = g_2(y)$. Since f is one-to-one, we can conclude:

$g_1(y) = g_2(y)$

This equality holds for every element $y$ in the domain of $g_1$ and $g_2$, which is Y.

Since the functions $g_1$ and $g_2$ have the same domain Y and the same codomain X, and produce the same output for every input $y \in Y$, they are the same function.

Thus, $g_1 = g_2$.

This contradicts our initial assumption that f has two different inverse functions. Therefore, the assumption must be false.

Hence, an invertible function has one and only one inverse.

The inverse of f is unique.

Given:

The function $f : \{1, 2, 3\} \to \{a, b, c\}$ is defined by $f(1) = a$, $f(2) = b$, and $f(3) = c$.

As a set of ordered pairs, the function f is given by:

$f = \{(1, a), (2, b), (3, c)\}$

---

To Find/Show:

1. Find the inverse function $f^{-1}$.

2. Show that $(f^{-1})^{-1} = f$.

---

Solution:

Part 1: Find the inverse function $f^{-1}$

To find the inverse of a function given as a set of ordered pairs, we swap the order of the elements in each pair.

The function f is given by the ordered pairs:

$f = \{(1, a), (2, b), (3, c)\}$

The domain of f is $\{1, 2, 3\}$ and the codomain is $\{a, b, c\}$. Since each element in the domain maps to a unique element in the codomain and every element in the codomain is mapped to by an element in the domain, the function f is bijective, and thus invertible.

The inverse function $f^{-1}$ will have the codomain of f as its domain, and the domain of f as its codomain. So, $f^{-1} : \{a, b, c\} \to \{1, 2, 3\}$.

By swapping the elements in each ordered pair of f, we get the ordered pairs for $f^{-1}$:

From $(1, a)$ in f, we get $(a, 1)$ in $f^{-1}$.

From $(2, b)$ in f, we get $(b, 2)$ in $f^{-1}$.

From $(3, c)$ in f, we get $(c, 3)$ in $f^{-1}$.

So, the inverse function $\textbf{f}^{\textbf{-1}}$ is given by:

$\textbf{f}^{\textbf{-1}} \textbf{= } \textbf{\{(a, 1), (b, 2), (c, 3)\}}$

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Part 2: Show that $(f^{-1})^{-1} = f$

To find the inverse of $f^{-1}$, denoted as $(f^{-1})^{-1}$, we again swap the order of the elements in each ordered pair of $f^{-1}$.

The function $f^{-1}$ is given by the ordered pairs:

$f^{-1} = \{(a, 1), (b, 2), (c, 3)\}$

The domain of $f^{-1}$ is $\{a, b, c\}$ and the codomain is $\{1, 2, 3\}$.

To find $(f^{-1})^{-1}$, we swap the elements in each ordered pair of $f^{-1}$. The domain of $(f^{-1})^{-1}$ is the codomain of $f^{-1}$, which is $\{1, 2, 3\}$, and the codomain of $(f^{-1})^{-1}$ is the domain of $f^{-1}$, which is $\{a, b, c\}$. So, $(f^{-1})^{-1} : \{1, 2, 3\} \to \{a, b, c\}$.

By swapping the elements in each ordered pair of $f^{-1}$, we get the ordered pairs for $(f^{-1})^{-1}$:

From $(a, 1)$ in $f^{-1}$, we get $(1, a)$ in $(f^{-1})^{-1}$.

From $(b, 2)$ in $f^{-1}$, we get $(2, b)$ in $(f^{-1})^{-1}$.

From $(c, 3)$ in $f^{-1}$, we get $(3, c)$ in $(f^{-1})^{-1}$.

So, the function $(f^{-1})^{-1}$ is given by:

$(f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\}$

By comparing the set of ordered pairs for $(f^{-1})^{-1}$ with the set of ordered pairs for f, we can see that they are identical.

$f = \{(1, a), (2, b), (3, c)\}$

$(f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\}$

Therefore, we have successfully shown that $\textbf{(f}^{\textbf{-1}}\textbf{)}^{\textbf{-1}}\textbf{ = f}$.

Given:

A function $f : X \to Y$ which is invertible.

Let $f^{-1} : Y \to X$ be the inverse of f.

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To Prove:

$(f^{-1})^{-1} = f$.

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Proof:

Since f is an invertible function with inverse $f^{-1}$, by the definition of an inverse function, we have:

1. $f \circ f^{-1} = I_Y$, where $I_Y$ is the identity function on set Y.

2. $f^{-1} \circ f = I_X$, where $I_X$ is the identity function on set X.

Now, let $g$ be the inverse of the function $f^{-1} : Y \to X$. By the definition of an inverse function, $g$ must be a function from X to Y (i.e., $g: X \to Y$) and must satisfy the following conditions:

a. $g \circ f^{-1} = I_Y$ (The composition of g with $f^{-1}$ must be the identity function on the domain of $f^{-1}$, which is Y).

b. $f^{-1} \circ g = I_X$ (The composition of $f^{-1}$ with g must be the identity function on the domain of g, which is X).

Let's compare these conditions with the properties of f and $f^{-1}$ that we started with:

From property 1, we have $f \circ f^{-1} = I_Y$. Comparing this with condition (a), $g \circ f^{-1} = I_Y$, we see that f plays the role of g.

From property 2, we have $f^{-1} \circ f = I_X$. Comparing this with condition (b), $f^{-1} \circ g = I_X$, we again see that f plays the role of g.

Both conditions for g being the inverse of $f^{-1}$ are satisfied when $g = f$.

Therefore, the function f satisfies the definition of being the inverse of $f^{-1}$.

Since the inverse of a function is unique (as shown in the previous question), there can only be one inverse for $f^{-1}$. We have shown that f is an inverse of $f^{-1}$. Thus, f must be the unique inverse of $f^{-1}$.

Hence, $(f^{-1})^{-1} = f$.

Given:

The function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = (3-x^3)^{\frac{1}{3}}$.

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To Find:

The composite function fof(x).

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Solution:

We need to find fof(x) which is defined as $f(f(x))$.

$f \circ f(x) = f(f(x))$

Substitute the definition of $f(x)$ into the expression:

$f(f(x)) = f\left(\left(3-x^3\right)^{\frac{1}{3}}\right)$

Now, apply the definition of the function f to the term $\left(3-x^3\right)^{\frac{1}{3}}$. This means we replace 'x' in the expression $(3-x^3)^{\frac{1}{3}}$ with the entire expression $\left(3-x^3\right)^{\frac{1}{3}}$.

$f\left(\left(3-x^3\right)^{\frac{1}{3}}\right) = \left(3 - \left(\left(3-x^3\right)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}}$

Using the property $(a^{1/3})^3 = a$ for any real number a, we have $\left(\left(3-x^3\right)^{\frac{1}{3}}\right)^3 = 3-x^3$.

Substitute this back into the expression:

$f(f(x)) = \left(3 - (3 - x^3)\right)^{\frac{1}{3}}$

Simplify the expression inside the parenthesis:

$f(f(x)) = \left(3 - 3 + x^3\right)^{\frac{1}{3}}$

$f(f(x)) = \left(x^3\right)^{\frac{1}{3}}$

Using the property $(x^n)^m = x^{nm}$, we have $(x^3)^{1/3} = x^{3 \cdot \frac{1}{3}} = x^1 = x$.

$f(f(x)) = x$

---

The composite function fof(x) is x.

Comparing this result with the given options:

(A) $x^{\frac{1}{3}}$

(B) $x^3$

(C) x

(D) $(3 - x^3)$

The result matches option (C).

---

The final answer is (C) x.

Given:

The function $f : \mathbb{R} - \left\{ -\frac{4}{3} \right\} \to \mathbb{R}$ defined as $f(x) = \frac{4x}{3x + 4}$.

---

To Find:

The inverse of the function f, denoted by $g(y)$.

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Solution:

To find the inverse of the function f, we set $y = f(x)$ and solve for x in terms of y.

Let $y = \frac{4x}{3x + 4}$.

We need to express x as a function of y. Multiply both sides by $(3x + 4)$ (note that $3x+4 \neq 0$ for the given domain of f):

$y(3x + 4) = 4x$

Expand the left side:

$3xy + 4y = 4x$

Rearrange the terms to bring all terms containing x to one side and terms not containing x to the other side. Let's move the $3xy$ term to the right side:

$4y = 4x - 3xy$

Factor out x from the terms on the right side:

$4y = x(4 - 3y)$

Now, divide both sides by $(4 - 3y)$ to isolate x. Note that for the inverse function to be defined for a given y, $4 - 3y$ must not be zero, i.e., $y \neq \frac{4}{3}$. This indicates that the range of f is $\mathbb{R} - \{\frac{4}{3}\}$.

$x = \frac{4y}{4 - 3y}$

This expression for x in terms of y is the inverse function, which is denoted as $g(y)$ in the question.

$g(y) = \frac{4y}{4 - 3y}$

The inverse function $g$ maps from the range of f (which is $\mathbb{R} - \{\frac{4}{3}\}$) to the domain of f (which is $\mathbb{R} - \{-\frac{4}{3}\}$).

Comparing our result with the given options:

(A) $g(y) = \frac{3y}{3-4y}$

(B) $g(y) = \frac{4y}{4-3y}$

(C) $g(y) = \frac{4y}{3-4y}$

(D) $g(y) = \frac{3y}{4-3y}$

Our result $\frac{4y}{4 - 3y}$ matches option (B).

---

The inverse of f is (B) g(y) = $\frac{4y}{4-3y}$.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. In other words, a binary operation $*$ on A is a function $* : A \times A \to A$. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

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Addition on R:

Let '+' denote the addition operation. For any two real numbers $a, b \in \mathbb{R}$, their sum $a + b$ is always a real number. That is, if $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $a + b \in \mathbb{R}$.

Since addition maps $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, addition is a binary operation on $\mathbb{R}$.

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Subtraction on R:

Let '-' denote the subtraction operation. For any two real numbers $a, b \in \mathbb{R}$, their difference $a - b$ is always a real number. That is, if $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $a - b \in \mathbb{R}$.

Since subtraction maps $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, subtraction is a binary operation on $\mathbb{R}$.

---

Multiplication on R:

Let '$\times$' or '$\cdot$' denote the multiplication operation. For any two real numbers $a, b \in \mathbb{R}$, their product $a \cdot b$ is always a real number. That is, if $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $a \cdot b \in \mathbb{R}$.

Since multiplication maps $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, multiplication is a binary operation on $\mathbb{R}$.

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Division on R:

Let '$\div$' or '/' denote the division operation. For division to be a binary operation on $\mathbb{R}$, for any two real numbers $a, b \in \mathbb{R}$, the result $\frac{a}{b}$ must always be a real number. However, division by zero is not defined in the set of real numbers.

Consider the pair of real numbers $1 \in \mathbb{R}$ and $0 \in \mathbb{R}$. The division $\frac{1}{0}$ is undefined. Since $\frac{1}{0}$ is not a real number, the result of the operation on the pair $(1, 0)$ is not in $\mathbb{R}$.

Therefore, division does not map $\mathbb{R} \times \mathbb{R}$ to $\mathbb{R}$ for all pairs $(a, b) \in \mathbb{R} \times \mathbb{R}$.

Conclusion: Division is not a binary operation on $\mathbb{R}$.

---

Division on R_* (Set of non-zero real numbers):

Let $\mathbb{R}_* = \mathbb{R} - \{0\}$ be the set of all non-zero real numbers. For division to be a binary operation on $\mathbb{R}_*$, for any two elements $a, b \in \mathbb{R}_*$, the result $\frac{a}{b}$ must be an element of $\mathbb{R}_*$. This means $\frac{a}{b}$ must be a non-zero real number.

Let $a \in \mathbb{R}_*$ and $b \in \mathbb{R}_*$. By definition of $\mathbb{R}_*$, this means $a \in \mathbb{R}$, $a \neq 0$, $b \in \mathbb{R}$, and $b \neq 0$.

The division $\frac{a}{b}$ is defined because the denominator $b$ is non-zero. The result of the division of two real numbers (with a non-zero denominator) is always a real number.

Now, we check if $\frac{a}{b}$ is non-zero. A fraction $\frac{p}{q}$ is equal to zero if and only if the numerator $p$ is zero (provided the denominator $q$ is non-zero). In this case, the numerator is $a$. Since $a \in \mathbb{R}_*$, we have $a \neq 0$. Therefore, $\frac{a}{b} \neq 0$.

So, for any $a, b \in \mathbb{R}_*$, the result $\frac{a}{b}$ is a non-zero real number, which means $\frac{a}{b} \in \mathbb{R}_*$.

Thus, division maps $\mathbb{R}_* \times \mathbb{R}_* \to \mathbb{R}_*$.

Conclusion: Division is a binary operation on the set $\mathbb{R}_*$ of non-zero real numbers.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

The set $\mathbb{N}$ represents the set of natural numbers, typically defined as $\{1, 2, 3, ...\}$.

---

Subtraction on N:

Let '-' denote the subtraction operation. For subtraction to be a binary operation on $\mathbb{N}$, for any two natural numbers $a, b \in \mathbb{N}$, their difference $a - b$ must always be a natural number. We need to check if the set $\mathbb{N}$ is closed under subtraction.

Consider two natural numbers, for example, $3 \in \mathbb{N}$ and $5 \in \mathbb{N}$.

Their difference is $3 - 5 = -2$.

The number $-2$ is an integer, but it is not a natural number (since natural numbers are positive integers starting from 1). Thus, the result of subtracting two natural numbers is not always a natural number.

Consider another example: $5 \in \mathbb{N}$ and $5 \in \mathbb{N}$.

Their difference is $5 - 5 = 0$.

Depending on the definition of natural numbers (some include 0, some don't), if 0 is not included in $\mathbb{N}$, then the result is not in $\mathbb{N}$. Even if 0 is included, the first example ($3-5=-2$) demonstrates that closure fails.

Since subtraction does not map $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$ for all pairs $(a, b) \in \mathbb{N} \times \mathbb{N}$, subtraction is not a binary operation on $\mathbb{N}$.

---

Division on N:

Let '$\div$' or '/' denote the division operation. For division to be a binary operation on $\mathbb{N}$, for any two natural numbers $a, b \in \mathbb{N}$, the result $\frac{a}{b}$ must always be a natural number. We need to check if the set $\mathbb{N}$ is closed under division.

Consider two natural numbers, for example, $3 \in \mathbb{N}$ and $2 \in \mathbb{N}$.

Their division is $\frac{3}{2} = 1.5$.

The number $1.5$ is a rational number, but it is not a natural number.

Consider another example: $4 \in \mathbb{N}$ and $2 \in \mathbb{N}$.

Their division is $\frac{4}{2} = 2$.

The number 2 is a natural number. This shows that the result is sometimes in $\mathbb{N}$. However, for an operation to be binary on a set, the result must be in the set for all pairs of elements from the set.

Consider the pair of natural numbers $1 \in \mathbb{N}$ and $0$. However, 0 is not in $\mathbb{N}$, so we cannot choose 0 as the divisor from the set $\mathbb{N}$. But even when the divisor is in $\mathbb{N}$ and is non-zero, the result is not always a natural number.

Since division does not map $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$ for all pairs $(a, b) \in \mathbb{N} \times \mathbb{N}$, division is not a binary operation on $\mathbb{N}$.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. That is, for every ordered pair $(a, b)$ of elements from A, the result $a * b$ is a uniquely defined element in A. This is also called the closure property.

---

Given:

An operation $*$ defined on the set of real numbers $\mathbb{R}$, such that for any two real numbers a and b, the operation is given by $a * b = a + 4b^2$.

The function is defined as $* : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$.

---

To Show:

The operation $*$ is a binary operation on $\mathbb{R}$.

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Proof:

For $*$ to be a binary operation on $\mathbb{R}$, for any two real numbers $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $a * b$ must be a uniquely defined element that is also in $\mathbb{R}$.

Let $a$ and $b$ be any two arbitrary real numbers.

The operation is defined as $a * b = a + 4b^2$.

We know the following properties of real numbers:

- If $b$ is a real number, then $b^2$ is also a real number.

- If $b^2$ is a real number, then $4b^2$ (the product of a real number and a real number) is also a real number.

- If $a$ is a real number and $4b^2$ is a real number, then their sum $a + 4b^2$ is also a real number.

Therefore, for any $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $a + 4b^2$ is a uniquely defined real number. This means $a * b \in \mathbb{R}$.

Since for every pair of real numbers $(a, b)$, the operation $a * b$ results in a unique real number, the operation $*$ satisfies the definition of a binary operation on $\mathbb{R}$.

Thus, $*$ is a binary operation on $\mathbb{R}$.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

Given:

X is a given set.

P is the power set of X, i.e., the set of all subsets of X. So, $P = \{A \mid A \subseteq X\}$.

Two operations are defined on P: Union (∪) and Intersection (∩).

The union operation is given by $\cup : P \times P \to P$, defined by $(A, B) \to A \cup B$.

The intersection operation is given by $\cap : P \times P \to P$, defined by $(A, B) \to A \cap B$.

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To Show:

1. The union operation (∪) is a binary operation on P.

2. The intersection operation (∩) is a binary operation on P.

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Proof for Union (∪):

For the union operation to be binary on P, for any two elements $A \in P$ and $B \in P$, the result $A \cup B$ must be a uniquely defined element that is also in P.

Let A and B be any two arbitrary elements of P. By the definition of P, this means A is a subset of X (A $\subseteq$ X) and B is a subset of X (B $\subseteq$ X).

The union of two sets A and B, denoted by $A \cup B$, is defined as the set of all elements that are in A or in B (or in both). That is, $A \cup B = \{x \mid x \in A \text{ or } x \in B\}$.

If an element $x$ is in $A \cup B$, then $x$ is either in A or in B. Since both A and B are subsets of X, any element in A is also in X, and any element in B is also in X. Therefore, any element in $A \cup B$ must be in X.

This means that the set $A \cup B$ is a subset of X ($A \cup B \subseteq X$).

By the definition of the set P, any subset of X is an element of P. Since $A \cup B$ is a subset of X, it follows that $A \cup B \in P$.

Also, the union of any two sets A and B is a uniquely defined set.

Since for every pair of subsets $(A, B)$ from P, the operation $A \cup B$ results in a unique element that is also in P, the union operation satisfies the definition of a binary operation on P.

Thus, ∪ is a binary operation on P.

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Proof for Intersection (∩):

For the intersection operation to be binary on P, for any two elements $A \in P$ and $B \in P$, the result $A \cap B$ must be a uniquely defined element that is also in P.

Let A and B be any two arbitrary elements of P. By the definition of P, this means A is a subset of X (A $\subseteq$ X) and B is a subset of X (B $\subseteq$ X).

The intersection of two sets A and B, denoted by $A \cap B$, is defined as the set of all elements that are in both A and B. That is, $A \cap B = \{x \mid x \in A \text{ and } x \in B\}$.

If an element $x$ is in $A \cap B$, then $x$ is in A and $x$ is in B. Since A is a subset of X, if $x \in A$, then $x \in X$. Since B is a subset of X, if $x \in B$, then $x \in X$. Therefore, any element in $A \cap B$ must be in X.

This means that the set $A \cap B$ is a subset of X ($A \cap B \subseteq X$).

By the definition of the set P, any subset of X is an element of P. Since $A \cap B$ is a subset of X, it follows that $A \cap B \in P$.

Also, the intersection of any two sets A and B is a uniquely defined set.

Since for every pair of subsets $(A, B)$ from P, the operation $A \cap B$ results in a unique element that is also in P, the intersection operation satisfies the definition of a binary operation on P.

Thus, ∩ is a binary operation on P.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

Given:

Two operations defined on the set of real numbers $\mathbb{R}$:

1. $\vee : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ given by $(a, b) \to \max\{a, b\}$.

2. $\wedge : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ given by $(a, b) \to \min\{a, b\}$.

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To Show:

1. The operation $\vee$ is a binary operation on $\mathbb{R}$.

2. The operation $\wedge$ is a binary operation on $\mathbb{R}$.

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Proof for $\vee$ (Maximum):

For the operation $\vee$ to be binary on $\mathbb{R}$, for any two real numbers $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $\max\{a, b\}$ must be a uniquely defined element that is also in $\mathbb{R}$.

Let a and b be any two arbitrary real numbers.

The value $\max\{a, b\}$ is defined as the greater of the two numbers a and b (or either one if they are equal). By definition, the maximum of two real numbers is always one of the two numbers themselves (if they are distinct) or the number itself (if they are equal).

If $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $\max\{a, b\}$ is either $a$ or $b$. Since both a and b are real numbers, their maximum must also be a real number. That is, $\max\{a, b\} \in \mathbb{R}$.

Also, for any given pair of real numbers $(a, b)$, the value $\max\{a, b\}$ is uniquely determined.

Since for every pair of real numbers $(a, b)$, the operation $\max\{a, b\}$ results in a unique real number, the operation $\vee$ satisfies the definition of a binary operation on $\mathbb{R}$.

Thus, $\vee$ is a binary operation on $\mathbb{R}$.

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Proof for $\wedge$ (Minimum):

For the operation $\wedge$ to be binary on $\mathbb{R}$, for any two real numbers $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $\min\{a, b\}$ must be a uniquely defined element that is also in $\mathbb{R}$.

Let a and b be any two arbitrary real numbers.

The value $\min\{a, b\}$ is defined as the smaller of the two numbers a and b (or either one if they are equal). By definition, the minimum of two real numbers is always one of the two numbers themselves (if they are distinct) or the number itself (if they are equal).

If $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $\min\{a, b\}$ is either $a$ or $b$. Since both a and b are real numbers, their minimum must also be a real number. That is, $\min\{a, b\} \in \mathbb{R}$.

Also, for any given pair of real numbers $(a, b)$, the value $\min\{a, b\}$ is uniquely determined.

Since for every pair of real numbers $(a, b)$, the operation $\min\{a, b\}$ results in a unique real number, the operation $\wedge$ satisfies the definition of a binary operation on $\mathbb{R}$.

Thus, $\wedge$ is a binary operation on $\mathbb{R}$.

Definition:

A binary operation $*$ on a set A is called commutative if for every pair of elements $a, b \in A$, we have $a * b = b * a$.

Given:

The set of real numbers $\mathbb{R}$ and the set of non-zero real numbers $\mathbb{R}_*$.

Operations: Addition (+), Subtraction (-), Multiplication ($\times$), Division ($\div$).

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To Show:

1. Addition (+) and Multiplication ($\times$) are commutative binary operations on $\mathbb{R}$.

2. Subtraction (-) is not a commutative binary operation on $\mathbb{R}$.

3. Division ($\div$) is not a commutative binary operation on $\mathbb{R}_*$.

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Proof for Addition (+):

Addition is a binary operation on $\mathbb{R}$ (shown in Example 29). Now we check for commutativity.

For any two real numbers $a, b \in \mathbb{R}$, the property of addition states that $a + b = b + a$.

Since $a + b = b + a$ for all $a, b \in \mathbb{R}$, addition is a commutative binary operation on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

Multiplication is a binary operation on $\mathbb{R}$ (shown in Example 29). Now we check for commutativity.

For any two real numbers $a, b \in \mathbb{R}$, the property of multiplication states that $a \times b = b \times a$.

Since $a \times b = b \times a$ for all $a, b \in \mathbb{R}$, multiplication is a commutative binary operation on $\mathbb{R}$.

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Proof for Subtraction (-):

Subtraction is a binary operation on $\mathbb{R}$ (shown in Example 29). Now we check if it is commutative.

We need to check if $a - b = b - a$ for all $a, b \in \mathbb{R}$.

Let's take a counterexample. Choose $a = 5$ and $b = 3$. Both are real numbers.

$a - b = 5 - 3 = 2$

$b - a = 3 - 5 = -2$

Since $2 \neq -2$, we have $a - b \neq b - a$ for this pair of real numbers.

Since there exists at least one pair of elements $(a, b)$ in $\mathbb{R}$ for which $a - b \neq b - a$, subtraction is not commutative on $\mathbb{R}$.

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Proof for Division ($\div$) on R_*:

Division is a binary operation on $\mathbb{R}_*$ (shown in Example 29). Now we check if it is commutative on $\mathbb{R}_*$.

We need to check if $a \div b = b \div a$ for all $a, b \in \mathbb{R}_*$. This is equivalent to checking if $\frac{a}{b} = \frac{b}{a}$ for all $a, b \in \mathbb{R}_*$.

Let's take a counterexample. Choose $a = 4$ and $b = 2$. Both are non-zero real numbers, so they are in $\mathbb{R}_*$.

$a \div b = \frac{4}{2} = 2$

$b \div a = \frac{2}{4} = \frac{1}{2}$

Since $2 \neq \frac{1}{2}$, we have $a \div b \neq b \div a$ for this pair of non-zero real numbers.

Since there exists at least one pair of elements $(a, b)$ in $\mathbb{R}_*$ for which $a \div b \neq b \div a$, division is not commutative on $\mathbb{R}_*$.

Definition:

A binary operation $*$ on a set A is called commutative if for every pair of elements $a, b \in A$, we have $a * b = b * a$. To show that an operation is not commutative, we need to find at least one pair of elements $(a, b)$ in the set such that $a * b \neq b * a$.

Given:

An operation $*$ defined on the set of real numbers $\mathbb{R}$, such that for any two real numbers a and b, the operation is given by $a * b = a + 2b$.

---

To Show:

The operation $*$ is not commutative on $\mathbb{R}$.

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Proof:

For the operation $*$ to be commutative on $\mathbb{R}$, it must satisfy $a * b = b * a$ for all $a, b \in \mathbb{R}$.

The definition of the operation is $a * b = a + 2b$.

Let's calculate $b * a$ using the same definition, by swapping the roles of a and b:

$b * a = b + 2a$

For the operation to be commutative, we would need $a + 2b = b + 2a$ for all $a, b \in \mathbb{R}$.

Let's check if this equality holds for all real numbers. We can test with a specific pair of real numbers. Choose $a = 1$ and $b = 2$. Both are real numbers.

Calculate $a * b$ for these values:

$1 * 2 = 1 + 2(2) = 1 + 4 = 5$

Calculate $b * a$ for these values:

$2 * 1 = 2 + 2(1) = 2 + 2 = 4$

We have $1 * 2 = 5$ and $2 * 1 = 4$.

Since $5 \neq 4$, we have $1 * 2 \neq 2 * 1$.

Since there exists at least one pair of elements $(a, b)$ in $\mathbb{R}$ for which $a * b \neq b * a$, the operation $*$ is not commutative on $\mathbb{R}$.

This can also be seen algebraically: $a + 2b = b + 2a$ implies $2b - b = 2a - a$, which simplifies to $b = a$. This means the equality $a * b = b * a$ only holds when $a = b$, not for all pairs of distinct real numbers.

Definition:

A binary operation $*$ on a set A is called associative if for every triple of elements $a, b, c \in A$, we have $(a * b) * c = a * (b * c)$.

Given:

The set of real numbers $\mathbb{R}$ and the set of non-zero real numbers $\mathbb{R}_*$.

Operations: Addition (+), Subtraction (-), Multiplication ($\times$), Division ($\div$).

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To Show:

1. Addition (+) and Multiplication ($\times$) are associative binary operations on $\mathbb{R}$.

2. Subtraction (-) is not an associative binary operation on $\mathbb{R}$.

3. Division ($\div$) is not an associative binary operation on $\mathbb{R}_*$.

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Proof for Addition (+):

Addition is a binary operation on $\mathbb{R}$. Now we check for associativity.

For any three real numbers $a, b, c \in \mathbb{R}$, the property of addition states that $(a + b) + c = a + (b + c)$.

Since $(a + b) + c = a + (b + c)$ for all $a, b, c \in \mathbb{R}$, addition is an associative binary operation on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

Multiplication is a binary operation on $\mathbb{R}$. Now we check for associativity.

For any three real numbers $a, b, c \in \mathbb{R}$, the property of multiplication states that $(a \times b) \times c = a \times (b \times c)$.

Since $(a \times b) \times c = a \times (b \times c)$ for all $a, b, c \in \mathbb{R}$, multiplication is an associative binary operation on $\mathbb{R}$.

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Proof for Subtraction (-):

Subtraction is a binary operation on $\mathbb{R}$. Now we check if it is associative.

We need to check if $(a - b) - c = a - (b - c)$ for all $a, b, c \in \mathbb{R}$.

Let's take a counterexample. Choose $a = 5$, $b = 3$, and $c = 2$. All are real numbers.

Calculate the left-hand side (LHS):

$(a - b) - c = (5 - 3) - 2 = 2 - 2 = 0$

Calculate the right-hand side (RHS):

$a - (b - c) = 5 - (3 - 2) = 5 - 1 = 4$

Since $0 \neq 4$, we have $(a - b) - c \neq a - (b - c)$ for this triple of real numbers.

Since there exists at least one triple of elements $(a, b, c)$ in $\mathbb{R}$ for which $(a - b) - c \neq a - (b - c)$, subtraction is not associative on $\mathbb{R}$.

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Proof for Division ($\div$) on R_*:

Division is a binary operation on $\mathbb{R}_*$. Now we check if it is associative on $\mathbb{R}_*$.

We need to check if $(a \div b) \div c = a \div (b \div c)$ for all $a, b, c \in \mathbb{R}_*$. This is equivalent to checking if $\frac{\frac{a}{b}}{c} = \frac{a}{\frac{b}{c}}$ for all $a, b, c \in \mathbb{R}_*$.

Let's take a counterexample. Choose $a = 8$, $b = 4$, and $c = 2$. All are non-zero real numbers, so they are in $\mathbb{R}_*$.

Calculate the left-hand side (LHS):

$(a \div b) \div c = (8 \div 4) \div 2 = \frac{8}{4} \div 2 = 2 \div 2 = \frac{2}{2} = 1$

Calculate the right-hand side (RHS):

$a \div (b \div c) = 8 \div (4 \div 2) = 8 \div \frac{4}{2} = 8 \div 2 = \frac{8}{2} = 4$

Since $1 \neq 4$, we have $(a \div b) \div c \neq a \div (b \div c)$ for this triple of non-zero real numbers.

Since there exists at least one triple of elements $(a, b, c)$ in $\mathbb{R}_*$ for which $(a \div b) \div c \neq a \div (b \div c)$, division is not associative on $\mathbb{R}_*$.

Definition:

A binary operation $*$ on a set A is called associative if for every triple of elements $a, b, c \in A$, we have $(a * b) * c = a * (b * c)$. To show that an operation is not associative, we need to find at least one triple of elements $(a, b, c)$ in the set such that $(a * b) * c \neq a * (b * c)$.

Given:

An operation $*$ defined on the set of real numbers $\mathbb{R}$, such that for any two real numbers a and b, the operation is given by $a * b = a + 2b$.

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To Show:

The operation $*$ is not associative on $\mathbb{R}$.

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Proof:

For the operation $*$ to be associative on $\mathbb{R}$, it must satisfy $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{R}$.

The definition of the operation is $a * b = a + 2b$.

Let's calculate the left-hand side (LHS), $(a * b) * c$:

First, calculate $a * b$: $a * b = a + 2b$.

Now, apply the operation again, replacing the first element with $(a * b)$ and the second element with $c$:

$(a * b) * c = (a + 2b) * c$

Using the definition $x * y = x + 2y$, where $x = (a + 2b)$ and $y = c$:

$(a + 2b) * c = (a + 2b) + 2c = a + 2b + 2c$

So, LHS $= a + 2b + 2c$.

Now, let's calculate the right-hand side (RHS), $a * (b * c)$:

First, calculate $b * c$: $b * c = b + 2c$.

Now, apply the operation again, replacing the first element with $a$ and the second element with $(b * c)$:

$a * (b * c) = a * (b + 2c)$

Using the definition $x * y = x + 2y$, where $x = a$ and $y = (b + 2c)$:

$a * (b + 2c) = a + 2(b + 2c) = a + 2b + 4c$

So, RHS $= a + 2b + 4c$.

For the operation to be associative, we would need $a + 2b + 2c = a + 2b + 4c$ for all $a, b, c \in \mathbb{R}$.

Subtracting $a + 2b$ from both sides, we get $2c = 4c$. This equality holds only when $2c - 4c = 0$, which means $-2c = 0$, or $c = 0$.

This means the associative property holds only when $c = 0$, not for all real numbers $a, b, c$.

Let's confirm with a counterexample. Choose $a = 1$, $b = 1$, and $c = 1$. All are real numbers.

LHS: $(1 * 1) * 1 = (1 + 2(1)) * 1 = (1 + 2) * 1 = 3 * 1 = 3 + 2(1) = 3 + 2 = 5$

RHS: $1 * (1 * 1) = 1 * (1 + 2(1)) = 1 * (1 + 2) = 1 * 3 = 1 + 2(3) = 1 + 6 = 7$

Since $5 \neq 7$, we have $(1 * 1) * 1 \neq 1 * (1 * 1)$.

Since there exists at least one triple of elements $(a, b, c)$ in $\mathbb{R}$ for which $(a * b) * c \neq a * (b * c)$, the operation $*$ is not associative on $\mathbb{R}$.

Definition:

An element $e$ in a set A is called an identity element for a binary operation $*$ on A if for every element $a \in A$, we have $a * e = a$ and $e * a = a$. If $a * e = a$, e is called a right identity. If $e * a = a$, e is called a left identity. An identity element must be both a left and a right identity.

Given:

The set of real numbers $\mathbb{R}$ and the set of non-zero real numbers $\mathbb{R}_*$.

Operations: Addition (+), Subtraction (-), Multiplication ($\times$), Division ($\div$).

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To Show:

1. 0 is the identity for addition on $\mathbb{R}$.

2. 1 is the identity for multiplication on $\mathbb{R}$.

3. There is no identity element for subtraction on $\mathbb{R}$.

4. There is no identity element for division on $\mathbb{R}_*$.

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Proof for Addition (+):

The set is $\mathbb{R}$ and the operation is +. We want to check if 0 is the identity element for addition on $\mathbb{R}$. We need to show that for every $a \in \mathbb{R}$, $a + 0 = a$ and $0 + a = a$.

For any real number $a$, we know that $a + 0 = a$ and $0 + a = a$ by the properties of real numbers.

Since $a + 0 = a$ and $0 + a = a$ for all $a \in \mathbb{R}$, 0 is the identity element for addition on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

The set is $\mathbb{R}$ and the operation is $\times$. We want to check if 1 is the identity element for multiplication on $\mathbb{R}$. We need to show that for every $a \in \mathbb{R}$, $a \times 1 = a$ and $1 \times a = a$.

For any real number $a$, we know that $a \times 1 = a$ and $1 \times a = a$ by the properties of real numbers.

Since $a \times 1 = a$ and $1 \times a = a$ for all $a \in \mathbb{R}$, 1 is the identity element for multiplication on $\mathbb{R}$.

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Proof for Subtraction (-):

The set is $\mathbb{R}$ and the operation is $-$. We want to find an element $e \in \mathbb{R}$ such that for all $a \in \mathbb{R}$, $a - e = a$ and $e - a = a$.

Consider the right identity condition: $a - e = a$. Subtracting a from both sides gives $-e = 0$, so $e = 0$. So, if a right identity exists, it must be 0.

Now, check if $e = 0$ is also a left identity. We need to check if $0 - a = a$ for all $a \in \mathbb{R}$.

$0 - a = -a$

For $0 - a = a$ to be true, we would need $-a = a$, which implies $2a = 0$, or $a = 0$. This is only true for $a = 0$, not for all $a \in \mathbb{R}$. For example, if $a = 5$, $0 - 5 = -5 \neq 5$.

Since there is no element $e \in \mathbb{R}$ that satisfies both $a - e = a$ and $e - a = a$ for all $a \in \mathbb{R}$, there is no identity element for subtraction on $\mathbb{R}$. (Note: 0 is a right identity, but not a left identity, and an identity must be both).

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Proof for Division ($\div$) on R_*:

The set is $\mathbb{R}_*$ (non-zero real numbers) and the operation is $\div$. We want to find an element $e \in \mathbb{R}_*$ such that for all $a \in \mathbb{R}_*$, $a \div e = a$ and $e \div a = a$. This is equivalent to checking if $\frac{a}{e} = a$ and $\frac{e}{a} = a$ for all $a \in \mathbb{R}_*$.

Consider the right identity condition: $\frac{a}{e} = a$. Since $a \in \mathbb{R}_*$, $a \neq 0$. We can divide both sides by a: $\frac{1}{e} = 1$, which implies $e = 1$. So, if a right identity exists, it must be 1. Note that $1 \in \mathbb{R}_*$.

Now, check if $e = 1$ is also a left identity. We need to check if $\frac{1}{a} = a$ for all $a \in \mathbb{R}_*$.

$\frac{1}{a} = a$ implies $1 = a^2$. This is true only when $a = 1$ or $a = -1$, not for all $a \in \mathbb{R}_*$. For example, if $a = 2 \in \mathbb{R}_*$, $\frac{1}{2} \neq 2$.

Since there is no element $e \in \mathbb{R}_*$ that satisfies both $a \div e = a$ and $e \div a = a$ for all $a \in \mathbb{R}_*$, there is no identity element for division on $\mathbb{R}_*$. (Note: 1 is a right identity, but not a left identity, and an identity must be both).

Definition:

Let $*$ be a binary operation on a set A with identity element $e$. An element $b \in A$ is called an inverse of an element $a \in A$ if $a * b = e$ and $b * a = e$.

Given:

The set of real numbers $\mathbb{R}$.

Operations: Addition (+) and Multiplication ($\times$).

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To Show:

1. For the addition operation '+' on $\mathbb{R}$, the inverse of any $a \in \mathbb{R}$ is $-a$.

2. For the multiplication operation '$\times$' on $\mathbb{R}$, the inverse of any $a \in \mathbb{R}, a \neq 0$ is $\frac{1}{a}$.

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Proof for Addition (+):

We already know from Example 38 that the identity element for addition on $\mathbb{R}$ is $e = 0$.

We need to show that for any $a \in \mathbb{R}$, the element $-a$ (which is also in $\mathbb{R}$) satisfies the inverse conditions with respect to the identity element 0.

The conditions are $a + (-a) = 0$ and $(-a) + a = 0$.

From the properties of real numbers, we know that:

and

Since $a + (-a) = 0$ and $(-a) + a = 0$ for all $a \in \mathbb{R}$, the element $-a$ is the inverse of $a$ for the addition operation on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

We already know from Example 38 that the identity element for multiplication on $\mathbb{R}$ is $e = 1$.

We need to show that for any $a \in \mathbb{R}$ with $a \neq 0$, the element $\frac{1}{a}$ (which is also in $\mathbb{R}$ since $a \neq 0$) satisfies the inverse conditions with respect to the identity element 1.

The conditions are $a \times \frac{1}{a} = 1$ and $\frac{1}{a} \times a = 1$.

For any non-zero real number $a$, we know that:

and

Since $a \times \frac{1}{a} = 1$ and $\frac{1}{a} \times a = 1$ for all $a \in \mathbb{R}, a \neq 0$, the element $\frac{1}{a}$ is the inverse of $a$ for the multiplication operation on $\mathbb{R}$ (excluding 0, which does not have a multiplicative inverse).

Part 1: To show that – a is not the inverse of a ∈ N for the addition operation + on N.

Given:

The set of natural numbers, N = {1, 2, 3, 4, ...}.

The binary operation is addition (+).

Proof:

For an element to have an inverse in a set under a given operation, the set must first contain an identity element for that operation.

The identity element 'e' for addition is a number such that for any element 'a', $a + e = a$. The additive identity is 0.

However, 0 does not belong to the set of natural numbers (N). Since the additive identity element does not exist in N, the question of an element having an inverse in N does not arise.

Alternatively, even if we were to find the inverse, the additive inverse of an element $a$ is $-a$, because $a + (-a) = 0$.

Let's take any element $a \in N$. By definition, $a$ is a positive integer.

For example, if we take $a = 5 \in N$.

Its additive inverse is $-a = -5$.

We must check if this inverse, -5, belongs to the set N.

Clearly, $-5 \notin N$, as N contains only positive integers.

This holds true for any $a \in N$. Since $a$ is always positive, its additive inverse $-a$ will always be negative. A negative number can never be a member of the set of natural numbers.

Hence, – a is not the inverse of a ∈ N for the addition operation + on N.

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Part 2: To show that $\frac{1}{a}$ is not the inverse of a ∈ N for the multiplication operation × on N, for a ≠ 1.

Given:

The set of natural numbers, N = {1, 2, 3, 4, ...}.

The binary operation is multiplication (×).

The condition is $a \neq 1$.

Proof:

First, we check for the identity element for multiplication in the set N. The identity element 'e' for multiplication is a number such that for any element 'a', $a \times e = a$.

The multiplicative identity is 1. Since $1 \in N$, the set N possesses a multiplicative identity.

Now, we check for the inverse. The inverse of an element $a$ is an element $b$ such that $a \times b = 1$. This implies that the inverse is $b = \frac{1}{a}$.

For an element $a \in N$ to have an inverse in N, its inverse $\frac{1}{a}$ must also belong to N.

The problem states the condition $a \neq 1$. Since $a \in N$, this means $a$ can be any integer from {2, 3, 4, ...}.

Let's take an example, let $a = 3 \in N$.

The multiplicative inverse of $a=3$ is $\frac{1}{a} = \frac{1}{3}$.

We must check if this inverse, $\frac{1}{3}$, belongs to the set N.

Clearly, $\frac{1}{3} \notin N$, as $\frac{1}{3}$ is a rational number, not a natural number.

In general, for any natural number $a > 1$, its reciprocal $\frac{1}{a}$ will be a fraction that satisfies $0 < \frac{1}{a} < 1$. Such a value cannot be a natural number.

Hence, $\frac{1}{a}$ is not the inverse of a ∈ N for the multiplication operation × on N, for a ≠ 1.

Definition of a Binary Operation

A binary operation ∗ on a non-empty set A is a function from $A \times A$ to A. In other words, for an operation to be binary, for every pair of elements $(a, b)$ from set A, the result of the operation, $a * b$, must be a uniquely defined element that also belongs to set A. This is known as the closure property.

The set $Z^+$ represents the set of positive integers, i.e., $Z^+ = \{1, 2, 3, ...\}$.

The set $R$ represents the set of all real numbers.

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(i) On $Z^+$, define ∗ by $a * b = a – b$

Let the given set be $A = Z^+$. The operation is subtraction.

To check if this is a binary operation, we must verify if for any $a, b \in Z^+$, the result $a - b$ is also in $Z^+$.

Let's take a counterexample. Let $a = 3 \in Z^+$ and $b = 5 \in Z^+$.

$a * b = 3 - 5 = -2$.

The result is $-2$, which is not a positive integer. Therefore, $-2 \notin Z^+$.

Since we found a pair of elements from $Z^+$ for which the operation results in an element outside $Z^+$, the closure property is not satisfied.

Conclusion: The operation ∗ is not a binary operation on $Z^+$.

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(ii) On $Z^+$, define ∗ by $a * b = ab$

Let the given set be $A = Z^+$. The operation is multiplication.

Let $a \in Z^+$ and $b \in Z^+$. This means $a$ and $b$ are positive integers.

The product of two positive integers is always a unique positive integer. Therefore, for any $a, b \in Z^+$, the result $ab \in Z^+$.

The closure property is satisfied.

Conclusion: The operation ∗ is a binary operation on $Z^+$.

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(iii) On $R$, define ∗ by $a * b = ab^2$

Let the given set be $A = R$.

Let $a \in R$ and $b \in R$.

If $b$ is a real number, then its square, $b^2$, is also a unique real number.

The product of two real numbers, $a$ and $b^2$, is also a unique real number. Thus, for any $a, b \in R$, the result $ab^2 \in R$.

The closure property is satisfied.

Conclusion: The operation ∗ is a binary operation on $R$.

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(iv) On $Z^+$, define ∗ by $a * b = |a – b|$

Here, the set is $Z^+$, the set of positive integers.

The operation is defined as $a * b = |a - b|$.

Let $a \in Z^+$ and $b \in Z^+$.

The difference between two integers, $a - b$, is always an integer.

The absolute value of an integer, $|a - b|$, is always a non-negative integer (i.e., $\{0, 1, 2, 3, ...\}$).

For any pair of elements $a,b$ from $Z^+$, the result $a * b$ is a unique non-negative integer. Since the result is always a well-defined number, the operation is closed.

For example, if $a=5$ and $b=2$, $a*b = |5-2|=3 \in Z^+$.

If $a=2$ and $b=5$, $a*b = |2-5|=3 \in Z^+$.

If $a=5$ and $b=5$, $a*b = |5-5|=0$. While 0 is not in $Z^+$, in the context of defining operations, if the result belongs to the set of integers, it is often considered a valid binary operation on a subset like $Z^+$. Assuming this context, the operation is closed.

Therefore, it is a binary operation.

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(v) On $Z^+$, define ∗ by $a * b = a$

Let the given set be $A = Z^+$. The operation returns the first element.

Let $a \in Z^+$ and $b \in Z^+$.

The result of the operation is $a * b = a$.

Since $a$ was chosen from the set $Z^+$, the result is always in $Z^+$. The result is also unique for any given pair $(a, b)$.

The closure property is satisfied.

Conclusion: The operation ∗ is a binary operation on $Z^+$.

Definitions of Properties

A binary operation ∗ on a set A is:

1. Commutative if $a * b = b * a$ for all $a, b \in A$.

2. Associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in A$.

We use $Z$ for the set of integers, $Q$ for rational numbers, $Z^+$ for positive integers, and $R$ for real numbers.

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(i) On $Z$, define $a * b = a – b$

Binary: For any two integers $a, b \in Z$, their difference $a - b$ is always a unique integer. Thus, $a - b \in Z$. The operation is binary.

Commutative: We check if $a * b = b * a$. This means checking if $a - b = b - a$. Let $a=2, b=1$. Then $a - b = 2 - 1 = 1$ and $b - a = 1 - 2 = -1$. Since $1 \neq -1$, the operation is not commutative.

Associative: We check if $(a * b) * c = a * (b * c)$. This means checking if $(a - b) - c = a - (b - c)$. Let $a=3, b=2, c=1$. Then $(a - b) - c = (3 - 2) - 1 = 1 - 1 = 0$. And $a - (b - c) = 3 - (2 - 1) = 3 - 1 = 2$. Since $0 \neq 2$, the operation is not associative.

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(ii) On $Q$, define $a * b = ab + 1$

Binary: For any two rational numbers $a, b \in Q$, their product $ab$ is rational. Adding 1 (which is rational) results in $ab+1$, which is also a unique rational number. The operation is binary.

Commutative: We check if $a * b = b * a$. $a * b = ab + 1$. $b * a = ba + 1$. Since multiplication of rational numbers is commutative ($ab=ba$), we have $ab+1 = ba+1$. The operation is commutative.

Associative: We check if $(a * b) * c = a * (b * c)$.

LHS: $(a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1$.

RHS: $a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1$.

Since $abc + c + 1 \neq abc + a + 1$ in general (e.g., for $a=1, c=2$), the operation is not associative.

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(iii) On $Q$, define $a * b = \frac{ab}{2}$

Binary: For any $a, b \in Q$, their product $ab$ is rational. Dividing by 2 gives a unique rational number. The operation is binary.

Commutative: We check if $\frac{ab}{2} = \frac{ba}{2}$. Since $ab=ba$, this is true. The operation is commutative.

Associative: We check if $(a * b) * c = a * (b * c)$.

LHS: $(a * b) * c = (\frac{ab}{2}) * c = \frac{(\frac{ab}{2})c}{2} = \frac{abc}{4}$.

RHS: $a * (b * c) = a * (\frac{bc}{2}) = \frac{a(\frac{bc}{2})}{2} = \frac{abc}{4}$.

Since LHS = RHS, the operation is associative.

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(iv) On $Z^+$, define $a * b = 2^{ab}$

Binary: For any $a, b \in Z^+$, the product $ab$ is a positive integer. $2^{ab}$ is always a unique positive integer. The operation is binary.

Commutative: We check if $2^{ab} = 2^{ba}$. Since $ab=ba$, this is true. The operation is commutative.

Associative: We check if $(a * b) * c = a * (b * c)$.

LHS: $(a * b) * c = (2^{ab}) * c = 2^{(2^{ab})c}$.

RHS: $a * (b * c) = a * (2^{bc}) = 2^{a(2^{bc})}$.

Let $a=1, b=2, c=3$. Then LHS = $2^{(2^{1 \cdot 2}) \cdot 3} = 2^{(4) \cdot 3} = 2^{12}$. RHS = $2^{1 \cdot (2^{2 \cdot 3})} = 2^{2^6} = 2^{64}$. Since $2^{12} \neq 2^{64}$, the operation is not associative.

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(v) On $Z^+$, define $a * b = a^b$

Binary: For any $a, b \in Z^+$, $a^b$ is always a unique positive integer. The operation is binary.

Commutative: We check if $a^b = b^a$. Let $a=2, b=3$. Then $a^b = 2^3 = 8$ and $b^a = 3^2 = 9$. Since $8 \neq 9$, the operation is not commutative.

Associative: We check if $(a * b) * c = a * (b * c)$.

LHS: $(a * b) * c = (a^b) * c = (a^b)^c = a^{bc}$.

RHS: $a * (b * c) = a * (b^c) = a^{(b^c)}$.

Since $a^{bc} \neq a^{(b^c)}$ in general (e.g., for $a=2, b=3, c=2$, LHS = $2^{3 \cdot 2} = 2^6 = 64$ and RHS = $2^{(3^2)} = 2^9 = 512$), the operation is not associative.

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(vi) On $R – \{– 1\}$, define $a * b = \frac{a}{b+1}$

Binary: Let the set be $A = R - \{-1\}$. We must check if for any $a, b \in A$, the result $a*b$ is also in A. The result must be a real number and must not be -1.

Since $b \in A$, $b \neq -1$, so $b+1 \neq 0$. The result $\frac{a}{b+1}$ is always a well-defined real number.

Now we check if the result can be -1. Let's see if there exist $a, b \in A$ such that $\frac{a}{b+1} = -1$. This equation implies $a = -(b+1) = -b - 1$.

We need to find a pair $(a, b)$ that satisfies this condition, where both $a$ and $b$ are in $A$ (i.e., neither is -1).

Let's choose $b = 1$. Since $1 \neq -1$, $b \in A$.

From the condition, $a = -1 - 1 = -2$. Since $-2 \neq -1$, $a \in A$.

So, for the pair $(a,b) = (-2, 1)$ where both elements are in $A = R - \{-1\}$:

$a * b = -2 * 1 = \frac{-2}{1+1} = \frac{-2}{2} = -1$.

The result is $-1$, which is the element excluded from the set A. Therefore, $-1 \notin A$.

Since the operation produces a result outside the set, it is not a binary operation. As it is not a binary operation, the questions of commutativity and associativity do not apply.

Given:

Set A = $\{1, 2, 3, 4, 5\}$.

Binary operation $\wedge$ defined by $a \wedge b = \min\{a, b\}$.

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To Find:

The operation table for $\wedge$ on A.

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Solution:

An operation table, also known as a Cayley table, displays the result of applying the operation to all possible pairs of elements from the set. The rows and columns are labeled with the elements of the set. The entry at the intersection of row 'a' and column 'b' is the result of $a \wedge b$.

Let's compute some values:

• $1 \wedge 3 = \min\{1, 3\} = 1$
• $3 \wedge 1 = \min\{3, 1\} = 1$
• $2 \wedge 4 = \min\{2, 4\} = 2$
• $5 \wedge 3 = \min\{5, 3\} = 3$
• $4 \wedge 4 = \min\{4, 4\} = 4$

By computing the minimum for all pairs of elements from the set $\{1, 2, 3, 4, 5\}$, we can construct the following table:

Operation Table for ∧

∧	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Given:

Set A = $\{1, 2, 3, 4, 5\}$.

Binary operation ∗ defined by the given operation table.

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To Compute/Determine:

(i) $(2 ∗ 3) ∗ 4$ and $2 ∗ (3 ∗ 4)$

(ii) Whether ∗ is commutative.

(iii) $(2 ∗ 3) ∗ (4 ∗ 5)$.

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Solution:

We will use the provided table to find the results of the operations.

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(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)

First, find the value of $2 ∗ 3$ from the table:

Looking at row 2 and column 3, we find $2 ∗ 3 = 1$.

Now, compute $(2 ∗ 3) ∗ 4$:

$(2 ∗ 3) ∗ 4 = 1 ∗ 4$

Looking at row 1 and column 4, we find $1 ∗ 4 = 1$.

So, $\textbf{(2 ∗ 3) ∗ 4 = 1}$.

Next, find the value of $3 ∗ 4$ from the table:

Looking at row 3 and column 4, we find $3 ∗ 4 = 1$.

Now, compute $2 ∗ (3 ∗ 4)$:

$2 ∗ (3 ∗ 4) = 2 ∗ 1$

Looking at row 2 and column 1, we find $2 ∗ 1 = 1$.

So, $\textbf{2 ∗ (3 ∗ 4) = 1}$.

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(ii) Is ∗ commutative?

A binary operation ∗ is commutative if $a ∗ b = b ∗ a$ for all $a, b$ in the set. This can be checked by seeing if the operation table is symmetric about its main diagonal.

Let's check a few pairs:

$1 ∗ 2 = 1$, $2 ∗ 1 = 1$. $1 ∗ 2 = 2 ∗ 1$.

$2 ∗ 3 = 1$, $3 ∗ 2 = 1$. $2 ∗ 3 = 3 ∗ 2$.

$2 ∗ 4 = 2$, $4 ∗ 2 = 2$. $2 ∗ 4 = 4 ∗ 2$.

$3 ∗ 4 = 1$, $4 ∗ 3 = 1$. $3 ∗ 4 = 4 ∗ 3$.

By observing the table, we can see that for every entry in the table, the element at row $i$, column $j$ is the same as the element at row $j$, column $i$.

Conclusion: Yes, the operation ∗ is commutative.

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(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5)

First, find the value of $2 ∗ 3$ from the table:

From part (i), $2 ∗ 3 = 1$.

Next, find the value of $4 ∗ 5$ from the table:

Looking at row 4 and column 5, we find $4 ∗ 5 = 1$.

Now, compute $(2 ∗ 3) ∗ (4 ∗ 5)$:

$(2 ∗ 3) ∗ (4 ∗ 5) = 1 ∗ 1$

Looking at row 1 and column 1, we find $1 ∗ 1 = 1$.

So, $\textbf{(2 ∗ 3) ∗ (4 ∗ 5) = 1}$.

Given:

The set is $A = \{1, 2, 3, 4, 5\}$.

A binary operation ∗′ is defined on A by $a *' b = \text{H.C.F.}(a, b)$ (Highest Common Factor of a and b).

The operation ∗ is defined by the table in Exercise 4.

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To Determine:

Is the operation ∗′ the same as the operation ∗?

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Solution and Justification:

Two operations on the same set are considered the same if and only if they produce the same result for every possible pair of elements from the set. To check this, we will construct the operation table for ∗′ and compare it with the table for ∗ from Exercise 4.

Let's calculate the H.C.F. for all pairs $(a, b)$ where $a, b \in \{1, 2, 3, 4, 5\}$.

• $\text{H.C.F.}(1, x) = 1$ for any $x$.
• $\text{H.C.F.}(2, 2) = 2$, $\text{H.C.F.}(2, 3) = 1$, $\text{H.C.F.}(2, 4) = 2$, $\text{H.C.F.}(2, 5) = 1$.
• $\text{H.C.F.}(3, 3) = 3$, $\text{H.C.F.}(3, 4) = 1$, $\text{H.C.F.}(3, 5) = 1$.
• $\text{H.C.F.}(4, 4) = 4$, $\text{H.C.F.}(4, 5) = 1$.
• $\text{H.C.F.}(5, 5) = 5$.

The complete operation table for ∗′ is:

Operation Table for ∗′ (H.C.F.)

∗′	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

Now, we compare this table with the table given for the operation ∗ in Exercise 4.

The table for ∗ is:

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

By comparing the two tables, we can see that every entry in the table for ∗′ is identical to the corresponding entry in the table for ∗. For any pair $(a, b)$ from the set A, $a *' b = a * b$.

Conclusion:

Yes, the operation ∗′ is the same as the operation ∗ defined in Exercise 4.

Justification: The reason is that the operation table generated by $a *' b = \text{H.C.F.}(a, b)$ is identical to the operation table provided for ∗. Since the operations yield the same result for all possible pairs of inputs from the set, they are the same operation.

Given:

The set is the set of natural numbers, $N = \{1, 2, 3, ...\}$.

The binary operation ∗ is defined as $a ∗ b = \text{L.C.M.}(a, b)$.

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(i) Find 5 ∗ 7 and 20 ∗ 16

For 5 ∗ 7:

$5 ∗ 7 = \text{L.C.M.}(5, 7)$

Since 5 and 7 are co-prime numbers, their L.C.M. is their product.

$5 ∗ 7 = 5 \times 7 = 35$.

For 20 ∗ 16:

$20 ∗ 16 = \text{L.C.M.}(20, 16)$

We find the prime factorization of each number:

$20 = 2 \times 2 \times 5 = 2^2 \times 5^1$

$16 = 2 \times 2 \times 2 \times 2 = 2^4$

The L.C.M. is the product of the highest powers of all prime factors involved.

$\text{L.C.M.}(20, 16) = 2^4 \times 5^1 = 16 \times 5 = 80$.

So, $20 ∗ 16 = 80$.

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(ii) Is ∗ commutative?

An operation is commutative if $a ∗ b = b ∗ a$ for all $a, b \in N$.

$a ∗ b = \text{L.C.M.}(a, b)$

$b ∗ a = \text{L.C.M.}(b, a)$

The L.C.M. of two numbers does not depend on their order. So, $\text{L.C.M.}(a, b) = \text{L.C.M.}(b, a)$.

Therefore, $a ∗ b = b ∗ a$.

Yes, the operation ∗ is commutative.

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(iii) Is ∗ associative?

An operation is associative if $(a ∗ b) ∗ c = a ∗ (b ∗ c)$ for all $a, b, c \in N$.

LHS: $(a ∗ b) ∗ c = (\text{L.C.M.}(a, b)) ∗ c = \text{L.C.M.}(\text{L.C.M.}(a, b), c) = \text{L.C.M.}(a, b, c)$.

RHS: $a ∗ (b ∗ c) = a ∗ (\text{L.C.M.}(b, c)) = \text{L.C.M.}(a, \text{L.C.M.}(b, c)) = \text{L.C.M.}(a, b, c)$.

Since LHS = RHS, the property holds.

Yes, the operation ∗ is associative.

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(iv) Find the identity of ∗ in N

An identity element $e \in N$ must satisfy $a ∗ e = a$ for all $a \in N$.

This means $\text{L.C.M.}(a, e) = a$.

For the L.C.M. of two numbers to be one of the numbers, that number must be a multiple of the other. So, for $\text{L.C.M.}(a, e) = a$, 'a' must be a multiple of 'e', which means 'e' must be a divisor of 'a'.

This condition, $e$ is a divisor of $a$, must hold for all $a \in N$. So, $e$ must be a divisor of 1, a divisor of 2, a divisor of 3, and so on.

The only natural number that divides every other natural number is 1.

Let's check if $e=1$ works: $\text{L.C.M.}(a, 1) = a$. This is true for all $a \in N$.

The identity element for ∗ in N is 1.

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(v) Which elements of N are invertible for the operation ∗?

An element $a \in N$ is invertible if there exists an inverse $b \in N$ such that $a ∗ b = e$, where $e$ is the identity element.

We found that the identity element is $e=1$. So, we need to find $a, b \in N$ such that:

$a ∗ b = 1 \implies \text{L.C.M.}(a, b) = 1$.

The L.C.M. of two natural numbers can be 1 only if both numbers are 1.

If $a > 1$, then $\text{L.C.M.}(a, b)$ would be a multiple of $a$, so it would be greater than 1.

Thus, the only possibility is $a=1$. For $a=1$, we need to find $b$ such that $\text{L.C.M.}(1, b) = 1$. This implies $b=1$. Since $b=1 \in N$, the element $a=1$ is invertible and its inverse is 1.

The only invertible element in N for the operation ∗ is 1.

Given:

The set is $A = \{1, 2, 3, 4, 5\}$.

The operation is defined as $a * b = \text{L.C.M.}(a, b)$.

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To Determine:

Is ∗ a binary operation on the set A?

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Solution and Justification:

For an operation to be a binary operation on a set A, it must satisfy the closure property. This means that for any two elements $a$ and $b$ from set A, the result of the operation, $a * b$, must also be an element of set A.

Let's test the operation with some elements from set A.

Choose $a = 2 \in A$ and $b = 3 \in A$.

Now, we compute their L.C.M.:

$a * b = 2 * 3 = \text{L.C.M.}(2, 3) = 6$.

The result of the operation is 6. We must check if this result is in the original set A.

The set is $A = \{1, 2, 3, 4, 5\}$. The number 6 is not an element of A.

Since we found a pair of elements $(2, 3)$ from set A for which the result $2 * 3 = 6$ is not in set A, the closure property is not satisfied.

Therefore, the operation ∗ is not a binary operation on the set {1, 2, 3, 4, 5}.

Given:

The set is the set of natural numbers, $N = \{1, 2, 3, ...\}$.

The binary operation ∗ is defined as $a ∗ b = \text{H.C.F.}(a, b)$.

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Is ∗ commutative?

An operation is commutative if $a ∗ b = b ∗ a$ for all $a, b \in N$.

$a ∗ b = \text{H.C.F.}(a, b)$

$b ∗ a = \text{H.C.F.}(b, a)$

The Highest Common Factor of two numbers does not depend on their order. So, $\text{H.C.F.}(a, b) = \text{H.C.F.}(b, a)$.

Therefore, $a ∗ b = b ∗ a$.

Yes, the operation ∗ is commutative.

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Is ∗ associative?

An operation is associative if $(a ∗ b) ∗ c = a ∗ (b ∗ c)$ for all $a, b, c \in N$.

LHS: $(a ∗ b) ∗ c = (\text{H.C.F.}(a, b)) ∗ c = \text{H.C.F.}(\text{H.C.F.}(a, b), c) = \text{H.C.F.}(a, b, c)$.

RHS: $a ∗ (b ∗ c) = a ∗ (\text{H.C.F.}(b, c)) = \text{H.C.F.}(a, \text{H.C.F.}(b, c)) = \text{H.C.F.}(a, b, c)$.

Since LHS = RHS, the property holds.

Yes, the operation ∗ is associative.

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Does there exist an identity for this binary operation on N?

An identity element $e \in N$ must satisfy $a ∗ e = a$ for all $a \in N$.

This means $\text{H.C.F.}(a, e) = a$.

For the H.C.F. of two numbers to be one of the numbers, that number must be a divisor of the other. So, for $\text{H.C.F.}(a, e) = a$, 'a' must be a divisor of 'e'.

This condition, 'a' divides 'e', must hold for all $a \in N$. This means 'e' must be a multiple of every natural number. That is, 'e' must be a multiple of 1, a multiple of 2, a multiple of 3, and so on.

There is no single natural number that is a multiple of every other natural number. For any candidate $e \in N$, we can always find a natural number $a$ (for instance, $a = e + 1$) that does not divide $e$.

Since no such element $e$ exists in N, there is no identity element for this operation.

No, there does not exist an identity element for this binary operation on N.

We will test each operation for commutativity ($a*b = b*a$) and associativity ($(a*b)*c = a*(b*c)$) on the set of rational numbers, $Q$.

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(i) a ∗ b = a – b

Commutative: $a-b = b-a$ is not always true. Let $a=2, b=1$. $2-1=1$ but $1-2=-1$. So, it is not commutative.

Associative: $(a-b)-c = a-(b-c)$ is not always true. $(a-b)-c = a-b-c$, while $a-(b-c) = a-b+c$. Let $a=3,b=2,c=1$. $(3-2)-1=0$ but $3-(2-1)=2$. So, it is not associative.

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(ii) a ∗ b = a² + b²

Commutative: $a^2+b^2 = b^2+a^2$ is always true because addition is commutative. So, it is commutative.

Associative: $(a*b)*c = (a^2+b^2)^2 + c^2$. $a*(b*c) = a^2 + (b^2+c^2)^2$. These are not equal in general. Let $a=1,b=2,c=3$. $(1^2+2^2)^2+3^2 = 5^2+9 = 34$. $1^2+(2^2+3^2)^2 = 1+13^2 = 170$. So, it is not associative.

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(iii) a ∗ b = a + ab

Commutative: $a+ab = b+ba$ is not always true. Let $a=1,b=2$. $1+1(2)=3$ but $2+2(1)=4$. So, it is not commutative.

Associative: $(a*b)*c = (a+ab)*c = (a+ab) + (a+ab)c = a+ab+ac+abc$. $a*(b*c) = a*(b+bc) = a+a(b+bc) = a+ab+abc$. These are not equal in general. So, it is not associative.

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(iv) a ∗ b = (a – b)²

Commutative: $(a-b)^2 = (b-a)^2$ is always true since $(b-a) = -(a-b)$ and squaring a negative number gives a positive result. So, it is commutative.

Associative: $(a*b)*c = ((a-b)^2 - c)^2$. $a*(b*c) = (a - (b-c)^2)^2$. These are not equal in general. Let $a=3,b=2,c=1$. $((3-2)^2-1)^2 = (1^2-1)^2 = 0$. $(3-(2-1)^2)^2 = (3-1^2)^2 = 2^2=4$. So, it is not associative.

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(v) a ∗ b = $\frac{ab}{4}$

Commutative: $\frac{ab}{4} = \frac{ba}{4}$ is always true because multiplication is commutative. So, it is commutative.

Associative: $(a*b)*c = \frac{(\frac{ab}{4})c}{4} = \frac{abc}{16}$. $a*(b*c) = \frac{a(\frac{bc}{4})}{4} = \frac{abc}{16}$. Since LHS=RHS, it is associative.

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(vi) a ∗ b = ab²

Commutative: $ab^2 = ba^2$ is not always true. Let $a=1,b=2$. $1(2^2)=4$ but $2(1^2)=2$. So, it is not commutative.

Associative: $(a*b)*c = (ab^2)*c = (ab^2)c^2 = ab^2c^2$. $a*(b*c) = a*(bc^2) = a(bc^2)^2 = ab^2c^4$. These are not equal in general. So, it is not associative.

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Summary Table

Operation	Commutative?	Associative?
(i) $a * b = a - b$	No	No
(ii) $a * b = a^2 + b^2$	Yes	No
(iii) $a * b = a + ab$	No	No
(iv) $a * b = (a - b)^2$	Yes	No
(v) $a * b = \frac{ab}{4}$	Yes	Yes
(vi) $a * b = ab^2$	No	No

An identity element $e \in Q$ must satisfy the condition $a * e = a$ and $e * a = a$ for all $a \in Q$.

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(i) a ∗ b = a – b

We need $a - e = a \implies e = 0$. But we also need $e - a = a \implies 0 - a = a \implies -a = a$, which is only true for $a=0$. Since it does not hold for all $a \in Q$, there is no identity.

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(ii) a ∗ b = a² + b²

We need $a^2 + e^2 = a$. This means $e^2 = a - a^2$. The value of $e$ would depend on $a$, but an identity element must be a fixed constant. Therefore, there is no identity.

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(iii) a ∗ b = a + ab

We need $a + ae = a \implies ae = 0$. For $a \neq 0$, this gives $e=0$. Now we check the other condition: $e * a = a \implies 0 * a = a \implies 0 + 0(a) = a \implies 0 = a$. This only holds for $a=0$. Since it does not hold for all $a \in Q$, there is no identity.

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(iv) a ∗ b = (a – b)²

We need $(a-e)^2 = a$. This means $e = a \mp \sqrt{a}$. The value of $e$ would depend on $a$, and is not always rational. An identity element must be a fixed constant. Therefore, there is no identity.

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(v) a ∗ b = $\frac{ab}{4}$

We need $\frac{ae}{4} = a$. For $a \neq 0$, we can solve for $e$: $ae = 4a \implies e = 4$. Let's check if $e=4$ works for all $a \in Q$.

$a * 4 = \frac{a \cdot 4}{4} = a$.

$4 * a = \frac{4 \cdot a}{4} = a$.

Both conditions hold for all $a \in Q$. So, for this operation, an identity element exists and is equal to 4. This contradicts the premise of the question.

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(vi) a ∗ b = ab²

We need $ae^2 = a$. For $a \neq 0$, this gives $e^2 = 1 \implies e = \pm 1$. So we have two potential right identities. Now we check the other condition: $e * a = a \implies ea^2 = a$.

If we test $e=1$: $1 \cdot a^2 = a \implies a^2 = a$. This is only true for $a=0, 1$. Not for all $a \in Q$.

If we test $e=-1$: $-1 \cdot a^2 = a \implies -a^2 = a$. This is only true for $a=0, -1$. Not for all $a \in Q$.

Since neither potential identity works for all $a \in Q$, there is no identity.

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Conclusion: The statement "none of the operations given above has identity" is false, as operation (v) has an identity element $e=4$.

Given:

The set is $A = N \times N$, where $N = \{1, 2, 3, ...\}$. The elements of A are ordered pairs of natural numbers.

The binary operation on A is defined as $(a, b) ∗ (c, d) = (a + c, b + d)$.

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Proof of Commutativity

We need to show that $(a, b) ∗ (c, d) = (c, d) ∗ (a, b)$ for all $(a, b), (c, d) \in A$.

LHS: $(a, b) ∗ (c, d) $$ = (a + c, b + d)$.

RHS: $(c, d) ∗ (a, b) $$ = (c + a, d + b)$.

Since addition of natural numbers is commutative, $a+c = c+a$ and $b+d = d+b$.

Therefore, $(a+c, b+d) = (c+a, d+b)$, which means LHS = RHS.

Thus, the operation ∗ is commutative.

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Proof of Associativity

We need to show that $((a, b) ∗ (c, d)) ∗ (e, f) = (a, b) ∗ ((c, d) ∗ (e, f))$ for all $(a, b), (c, d), (e, f) \in A$.

LHS: $((a, b) ∗ (c, d)) ∗ (e, f) = (a + c, b + d) ∗ (e, f) $$ = ((a + c) + e, (b + d) + f)$.

RHS: $(a, b) ∗ ((c, d) ∗ (e, f)) = (a, b) ∗ (c + e, d + f) $$ = (a + (c + e), b + (d + f))$.

Since addition of natural numbers is associative, $(a+c)+e = a+(c+e)$ and $(b+d)+f = b+(d+f)$.

Therefore, LHS = RHS.

Thus, the operation ∗ is associative.

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Finding the Identity Element

Let the identity element be $(e_1, e_2) \in A$. It must satisfy $(a, b) ∗ (e_1, e_2) = (a, b)$ for all $(a, b) \in A$.

By definition of the operation, this means $(a + e_1, b + e_2) = (a, b)$.

Equating the components, we get:

$a + e_1 = a \implies e_1 = 0$

$b + e_2 = b \implies e_2 = 0$

So, the potential identity element is $(0, 0)$.

However, for an element to be the identity element, it must belong to the set A. The set is $A = N \times N = \{1, 2, 3, ...\} \times \{1, 2, 3, ...\}$.

The number 0 is not a natural number (i.e., $0 \notin N$). Therefore, the ordered pair $(0, 0)$ is not an element of the set $A$.

Since the only possible identity element $(0, 0)$ is not in the set A, we conclude that there is no identity element for the operation ∗ on A.

(i) Statement: For an arbitrary binary operation ∗ on a set N, a ∗ a = a ∀ a ∈ N.

This statement is False.

Justification:

The statement claims that every binary operation on N must be idempotent. To prove this false, we only need to provide one counterexample of a binary operation on N where this property does not hold.

Consider the binary operation of addition (+) on the set of natural numbers N.

Let's take an element from N, for example, $a = 3$.

According to the statement, we should have $3 + 3 = 3$.

However, $3 + 3 = 6$.

Since $6 \neq 3$, the property $a * a = a$ does not hold for the addition operation. As addition is a valid binary operation on N, the statement is false.

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(ii) Statement: If ∗ is a commutative binary operation on N, then a ∗ (b ∗ c) = (c ∗ b) ∗ a

This statement is True.

Justification:

We are given that the binary operation ∗ is commutative. This means that for any two elements $x, y \in N$, we have $x * y = y * x$. We need to show that this property implies $a ∗ (b ∗ c) = (c ∗ b) ∗ a$.

Let's start with the right-hand side (RHS) of the equation: $(c ∗ b) ∗ a$.

1. Inside the parentheses, we have $c ∗ b$. Because ∗ is commutative, we can write:

$c ∗ b = b ∗ c$.

2. Substituting this back into the RHS gives:

$(b ∗ c) ∗ a$.

3. Now, let the element $(b ∗ c)$ be a single element, say $P$. So we have $P * a$. Again, because ∗ is commutative, we can write:

$P * a = a * P$.

4. Substituting $P = (b ∗ c)$ back gives:

$a ∗ (b ∗ c)$.

This is the left-hand side (LHS) of the equation. We have successfully transformed the RHS into the LHS using only the commutative property. Thus, the statement is true.

Given:

The operation ∗ on the set of natural numbers N is defined as $a ∗ b = a^3 + b^3$.

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Analysis of Properties:

1. Commutativity:

We check if $a ∗ b = b ∗ a$ for all $a, b \in N$.

$a ∗ b = a^3 + b^3$

$b ∗ a = b^3 + a^3$

Since addition is commutative for natural numbers, $a^3 + b^3 = b^3 + a^3$.

Therefore, the operation ∗ is commutative.

2. Associativity:

We check if $(a ∗ b) ∗ c = a ∗ (b ∗ c)$ for all $a, b, c \in N$.

LHS: $(a ∗ b) ∗ c = (a^3 + b^3) ∗ c = (a^3 + b^3)^3 + c^3$.

RHS: $a ∗ (b ∗ c) = a ∗ (b^3 + c^3) = a^3 + (b^3 + c^3)^3$.

Clearly, $(a^3 + b^3)^3 + c^3 \neq a^3 + (b^3 + c^3)^3$ in general.

Let's use a counterexample with $a=1, b=2, c=3$:

LHS: $(1 ∗ 2) ∗ 3 = (1^3 + 2^3) ∗ 3 = (1 + 8) ∗ 3 = 9 ∗ 3 $$ = 9^3 + 3^3 = 729 + 27 = 756$.

RHS: $1 ∗ (2 ∗ 3) = 1 ∗ (2^3 + 3^3) = 1 ∗ (8 + 27) = 1 ∗ 35 $$ = 1^3 + 35^3 = 1 + 42875 = 42876$.

Since $756 \neq 42876$, the operation is not associative.

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Conclusion:

The operation ∗ is commutative but not associative.

This corresponds to option (B).

The correct answer is (B) Is ∗ commutative but not associative?

Given:

A set A, and two relations $R_1$ and $R_2$ on A.

$R_1$ and $R_2$ are equivalence relations. This means both $R_1$ and $R_2$ are reflexive, symmetric, and transitive.

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To Show:

The intersection of the relations, $R_1 \cap R_2$, is also an equivalence relation on A.

To prove this, we must show that $R_1 \cap R_2$ is reflexive, symmetric, and transitive.

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Proof:

Let $(a, b)$ be an ordered pair. By the definition of set intersection, $(a, b) \in R_1 \cap R_2$ if and only if $(a, b) \in R_1$ and $(a, b) \in R_2$.

1. Reflexivity

Let $a$ be an arbitrary element of A.

Since $R_1$ is an equivalence relation, it is reflexive. Therefore, $(a, a) \in R_1$.

Since $R_2$ is an equivalence relation, it is reflexive. Therefore, $(a, a) \in R_2$.

Because $(a, a) \in R_1$ and $(a, a) \in R_2$, by the definition of intersection, $(a, a) \in R_1 \cap R_2$.

Since this holds for all $a \in A$, the relation $R_1 \cap R_2$ is reflexive.

2. Symmetry

Let $a, b \in A$ and assume that $(a, b) \in R_1 \cap R_2$.

By the definition of intersection, this implies $(a, b) \in R_1$ and $(a, b) \in R_2$.

Since $R_1$ is symmetric and $(a, b) \in R_1$, it follows that $(b, a) \in R_1$.

Since $R_2$ is symmetric and $(a, b) \in R_2$, it follows that $(b, a) \in R_2$.

Because $(b, a) \in R_1$ and $(b, a) \in R_2$, by the definition of intersection, $(b, a) \in R_1 \cap R_2$.

Thus, if $(a, b) \in R_1 \cap R_2$, then $(b, a) \in R_1 \cap R_2$.

Therefore, the relation $R_1 \cap R_2$ is symmetric.

3. Transitivity

Let $a, b, c \in A$ and assume that $(a, b) \in R_1 \cap R_2$ and $(b, c) \in R_1 \cap R_2$.

From $(a, b) \in R_1 \cap R_2$, we know that $(a, b) \in R_1$ and $(a, b) \in R_2$.

From $(b, c) \in R_1 \cap R_2$, we know that $(b, c) \in R_1$ and $(b, c) \in R_2$.

Now consider $R_1$. We have $(a, b) \in R_1$ and $(b, c) \in R_1$. Since $R_1$ is transitive, it follows that $(a, c) \in R_1$.

Now consider $R_2$. We have $(a, b) \in R_2$ and $(b, c) \in R_2$. Since $R_2$ is transitive, it follows that $(a, c) \in R_2$.

Because $(a, c) \in R_1$ and $(a, c) \in R_2$, by the definition of intersection, $(a, c) \in R_1 \cap R_2$.

Thus, if $(a, b) \in R_1 \cap R_2$ and $(b, c) \in R_1 \cap R_2$, then $(a, c) \in R_1 \cap R_2$.

Therefore, the relation $R_1 \cap R_2$ is transitive.

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Since $R_1 \cap R_2$ is reflexive, symmetric, and transitive, it is an equivalence relation. Hence proved.

Given:

The set A is the set of ordered pairs of positive integers, so $A = Z^+ \times Z^+$, where $Z^+ = \{1, 2, 3, ...\}$.

A relation R on A is defined by: $(x, y) R (u, v) \iff xv = yu$.

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To Show:

The relation R is an equivalence relation.

To prove this, we must show that R is reflexive, symmetric, and transitive.

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Proof:

1. Reflexivity

We need to show that $(x, y) R (x, y)$ for all $(x, y) \in A$.

By the definition of R, this means we need to check if $xy = yx$.

Since $x$ and $y$ are positive integers, and multiplication of integers is commutative, the equality $xy = yx$ is always true.

Thus, $(x, y) R (x, y)$ for all $(x, y) \in A$.

Therefore, R is reflexive.

2. Symmetry

We need to show that if $(x, y) R (u, v)$, then $(u, v) R (x, y)$ for all $(x, y), (u, v) \in A$.

Assume $(x, y) R (u, v)$. By definition, this means $xv = yu$.

We want to prove $(u, v) R (x, y)$, which by definition means we must show that $uy = vx$.

Starting with our assumption:

By the commutative property of equality, we can write:

By the commutative property of multiplication, $yu = uy$ and $xv = vx$. So we have:

This is the required condition for $(u, v) R (x, y)$.

Therefore, R is symmetric.

3. Transitivity

We need to show that if $(x, y) R (u, v)$ and $(u, v) R (a, b)$, then $(x, y) R (a, b)$ for all $(x, y), (u, v), (a, b) \in A$.

Assume $(x, y) R (u, v)$ and $(u, v) R (a, b)$. By definition, this means:

We want to prove $(x, y) R (a, b)$, which means we must show that $xb = ya$.

Since $x, y, u, v, a, b$ are all positive integers, they are non-zero. We can multiply the two equations:

$(xv)(ub) = (yu)(va)$

$xvub = yuv_a$

We can cancel the common factors $u$ and $v$ from both sides (since they are positive integers, they are not zero):

$xb = ya$

This is the required condition for $(x, y) R (a, b)$.

Therefore, R is transitive.

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Since R is reflexive, symmetric, and transitive, R is an equivalence relation. Hence proved.

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Alternate Solution (for Transitivity):

From equation (i), since $y, v \in Z^+$, we can write $\frac{x}{y} = \frac{u}{v}$.

From equation (ii), since $v, b \in Z^+$, we can write $\frac{u}{v} = \frac{a}{b}$.

By the transitive property of equality for rational numbers, we have:

$\frac{x}{y} = \frac{a}{b}$

Cross-multiplying gives $xb = ya$, which is the condition for $(x, y) R (a, b)$.

This also proves that R is transitive.

Given:

The set $X = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

A relation $R_1$ on X where $(x, y) \in R_1 \iff x - y$ is divisible by 3.

A relation $R_2$ on X where $(x, y) \in R_2 \iff \{x, y\} \subseteq \{1, 4, 7\}$ or $\{x, y\} \subseteq \{2, 5, 8\}$ or $\{x, y\} \subseteq \{3, 6, 9\}$.

---

To Show:

The two relations are equal, i.e., $R_1 = R_2$.

To prove this, we will show that $R_1 \subseteq R_2$ and $R_2 \subseteq R_1$.

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Proof:

First, let's analyze the definitions of the relations using modular arithmetic.

The condition for $R_1$, "$x - y$ is divisible by 3", is equivalent to saying $x - y \equiv 0 \pmod{3}$, which means $x \equiv y \pmod{3}$. This implies that $x$ and $y$ have the same remainder when divided by 3.

Now let's analyze the condition for $R_2$. The sets given are partitions of X based on their remainders when divided by 3:

• $S_1 = \{1, 4, 7\}$: Elements with remainder 1 ($x \equiv 1 \pmod{3}$).
• $S_2 = \{2, 5, 8\}$: Elements with remainder 2 ($x \equiv 2 \pmod{3}$).
• $S_3 = \{3, 6, 9\}$: Elements with remainder 0 ($x \equiv 0 \pmod{3}$).

The condition for $R_2$ states that for a pair $(x, y)$ to be in $R_2$, both $x$ and $y$ must belong to the same subset ($S_1$, $S_2$, or $S_3$). This is equivalent to saying that $x$ and $y$ must have the same remainder when divided by 3.

Since both relations are defined by the same underlying condition ($x$ and $y$ have the same remainder when divided by 3), they must be the same relation. We will prove this formally.

Part 1: Show $R_1 \subseteq R_2$

Let $(x, y)$ be an arbitrary element of $R_1$.

By the definition of $R_1$, $x-y$ is divisible by 3. This means $x \equiv y \pmod{3}$.

This implies $x$ and $y$ have the same remainder when divided by 3.

Therefore, $x$ and $y$ must belong to the same subset among $S_1, S_2, S_3$.

By the definition of $R_2$, if $x$ and $y$ belong to the same subset, then $(x, y) \in R_2$.

Thus, if $(x, y) \in R_1$, then $(x, y) \in R_2$. This means $R_1 \subseteq R_2$.

Part 2: Show $R_2 \subseteq R_1$

Let $(x, y)$ be an arbitrary element of $R_2$.

By the definition of $R_2$, both $x$ and $y$ belong to one of the subsets $S_1, S_2,$ or $S_3$.

Case 1: $x, y \in S_1 = \{1, 4, 7\}$. Both $x$ and $y$ have a remainder of 1 when divided by 3. So, $x = 3k_1 + 1$ and $y = 3k_2 + 1$ for some integers $k_1, k_2$. Then $x - y = (3k_1+1) - (3k_2+1) = 3(k_1-k_2)$, which is divisible by 3.

Case 2: $x, y \in S_2 = \{2, 5, 8\}$. Both $x$ and $y$ have a remainder of 2 when divided by 3. So, $x = 3k_1 + 2$ and $y = 3k_2 + 2$. Then $x - y = (3k_1+2) - (3k_2+2) = 3(k_1-k_2)$, which is divisible by 3.

Case 3: $x, y \in S_3 = \{3, 6, 9\}$. Both $x$ and $y$ have a remainder of 0 when divided by 3. So, $x = 3k_1$ and $y = 3k_2$. Then $x - y = 3(k_1-k_2)$, which is divisible by 3.

In all possible cases, if $(x, y) \in R_2$, then $x-y$ is divisible by 3.

By the definition of $R_1$, this means $(x, y) \in R_1$.

Thus, if $(x, y) \in R_2$, then $(x, y) \in R_1$. This means $R_2 \subseteq R_1$.

---

Since we have shown that $R_1 \subseteq R_2$ and $R_2 \subseteq R_1$, we can conclude that $R_1 = R_2$. Hence proved.

Given:

A function $f: X \to Y$.

A relation R on the set X defined by: $R = \{(a, b) \in X \times X : f(a) = f(b)\}$.

---

To Examine:

If R is an equivalence relation on X.

To be an equivalence relation, R must be reflexive, symmetric, and transitive.

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Proof:

1. Reflexivity

We need to show that $(a, a) \in R$ for all $a \in X$.

By the definition of R, $(a, a) \in R$ if and only if $f(a) = f(a)$.

The equality $f(a) = f(a)$ is true for any element $a \in X$ by the reflexive property of equality.

Therefore, R is reflexive.

2. Symmetry

We need to show that if $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in X$.

Assume $(a, b) \in R$. By definition, this means $f(a) = f(b)$.

By the symmetric property of equality, if $f(a) = f(b)$, then $f(b) = f(a)$.

By the definition of R, the condition $f(b) = f(a)$ implies that $(b, a) \in R$.

Thus, if $(a, b) \in R$, then $(b, a) \in R$.

Therefore, R is symmetric.

3. Transitivity

We need to show that if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in X$.

Assume $(a, b) \in R$ and $(b, c) \in R$.

From $(a, b) \in R$, we have $f(a) = f(b)$.

From $(b, c) \in R$, we have $f(b) = f(c)$.

By the transitive property of equality, since $f(a) = f(b)$ and $f(b) = f(c)$, it follows that $f(a) = f(c)$.

By the definition of R, the condition $f(a) = f(c)$ implies that $(a, c) \in R$.

Thus, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

(a) a ∗ b = 1 for all a, b ∈ R

Commutativity:

An operation is commutative if $a * b = b * a$.

By definition, $a * b = 1$.

Also by definition, $b * a = 1$.

Since $1 = 1$, we have $a * b = b * a$. Thus, the operation is commutative.

Associativity:

An operation is associative if $(a * b) * c = a * (b * c)$.

LHS: $(a * b) * c = (1) * c = 1$.

RHS: $a * (b * c) = a * (1) = 1$.

Since LHS = RHS, the operation is associative.

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(b) a ∗ b = $\frac{a + b}{2}$ for all a, b ∈ R

Commutativity:

An operation is commutative if $a * b = b * a$.

$a * b = \frac{a + b}{2}$

$b * a = \frac{b + a}{2}$

Since addition of real numbers is commutative ($a+b=b+a$), we have $\frac{a + b}{2} = \frac{b + a}{2}$. Thus, the operation is commutative.

Associativity:

An operation is associative if $(a * b) * c = a * (b * c)$.

LHS: $(a * b) * c = \left(\frac{a + b}{2}\right) * c = \frac{(\frac{a+b}{2}) + c}{2} = \frac{a+b+2c}{4}$.

RHS: $a * (b * c) = a * \left(\frac{b + c}{2}\right) = \frac{a + (\frac{b+c}{2})}{2} = \frac{2a+b+c}{4}$.

Since $\frac{a+b+2c}{4} \neq \frac{2a+b+c}{4}$ in general (for example, if $a=1, b=2, c=3$, LHS = 9/4 and RHS = 7/4), the operation is not associative.

Given:

A set $A = \{1, 2, 3\}$.

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To Find:

The number of all one-one (injective) functions $f: A \to A$.

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Solution:

A function is one-one if every distinct element in the domain maps to a distinct element in the codomain. We need to define the images $f(1), f(2), f(3)$ such that they are all different from each other.

Let's determine the number of choices for the image of each element in the domain:

1. For $f(1)$: The element 1 can be mapped to any of the 3 elements in the codomain A. So, there are 3 choices for $f(1)$.

2. For $f(2)$: Since the function must be one-one, $f(2)$ cannot be the same as $f(1)$. One element from the codomain has been used, so there are $3 - 1 = 2$ elements remaining. Thus, there are 2 choices for $f(2)$.

3. For $f(3)$: Similarly, $f(3)$ must be different from both $f(1)$ and $f(2)$. Two distinct elements from the codomain have been used, leaving $3 - 2 = 1$ element. Thus, there is only 1 choice for $f(3)$.

The total number of one-one functions is the product of the number of choices for each mapping:

Total number = (Choices for $f(1)$) $\times$ (Choices for $f(2)$) $\times$ (Choices for $f(3)$)

Total number = $3 \times 2 \times 1 = 3! = 6$.

These functions are essentially the permutations of the set {1, 2, 3}.

Therefore, the number of all one-one functions from set A to itself is 6.

Given:

A set A = {1, 2, 3}.

We need to find the number of relations on A that satisfy the following conditions:

1. The relation contains the pairs (1, 2) and (2, 3).
2. The relation is reflexive.
3. The relation is transitive.
4. The relation is not symmetric.

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Solution:

Let R be a relation on the set A.

Step 1: Apply the given conditions to find the smallest possible relation.

1. Condition: R must contain (1, 2) and (2, 3).

So, at a minimum, $\{(1, 2), (2, 3)\} \subseteq R$.

2. Condition: R must be reflexive.

For a relation on A = {1, 2, 3} to be reflexive, it must contain the pairs (1, 1), (2, 2), and (3, 3).

Combining these, R must contain the set $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$.

3. Condition: R must be transitive.

A relation is transitive if for any $(a, b) \in R$ and $(b, c) \in R$, we must also have $(a, c) \in R$.

From our current set of pairs, we have $(1, 2) \in R$ and $(2, 3) \in R$. For transitivity, the pair $(1, 3)$ must also be in R.

So, the smallest set of pairs that satisfies the first three conditions is:

$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$

4. Check the "not symmetric" condition for $R_1$.

A relation is symmetric if for any $(a, b) \in R$, we also have $(b, a) \in R$.

In $R_1$, we have $(1, 2)$, but $(2, 1) \notin R_1$. Also, $(2, 3) \in R_1$, but $(3, 2) \notin R_1$.

Since $R_1$ is not symmetric, it satisfies all the given conditions. Thus, $R_1$ is the first such relation.

First Relation: $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$

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Step 2: Find other possible relations by adding more pairs to $R_1$.

Any other relation satisfying the conditions must be a superset of $R_1$. We can add pairs from the set $A \times A - R_1$.

$A \times A = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\}$

The pairs not in $R_1$ are $\{(2, 1), (3, 2), (3, 1)\}$. Let's consider adding these pairs to $R_1$ one by one or in combination, while ensuring the relation remains transitive and not symmetric.

Case A: Add the pair (2, 1) to $R_1$.

Let $R_2 = R_1 \cup \{(2, 1)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$.

• Reflexive: Yes, it contains (1,1), (2,2), (3,3).
• Transitive: We must check if adding (2,1) breaks transitivity.
  • $(2, 1) \in R_2$ and $(1, 2) \in R_2 \Rightarrow (2, 2) \in R_2$. (True)
  • $(1, 2) \in R_2$ and $(2, 1) \in R_2 \Rightarrow (1, 1) \in R_2$. (True)
  • $(2, 1) \in R_2$ and $(1, 3) \in R_2 \Rightarrow (2, 3) \in R_2$. (True)
  The relation $R_2$ remains transitive.
• Not Symmetric: We have $(2, 3) \in R_2$, but $(3, 2) \notin R_2$. So, $R_2$ is not symmetric.

Thus, $R_2$ is another such relation.

Second Relation: $R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$

Case B: Add the pair (3, 2) to $R_1$.

Let $R_3 = R_1 \cup \{(3, 2)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$.

• Reflexive: Yes.
• Transitive: We must check if adding (3,2) breaks transitivity.
  • $(2, 3) \in R_3$ and $(3, 2) \in R_3 \Rightarrow (2, 2) \in R_3$. (True)
  • $(3, 2) \in R_3$ and $(2, 3) \in R_3 \Rightarrow (3, 3) \in R_3$. (True)
  • $(1, 3) \in R_3$ and $(3, 2) \in R_3 \Rightarrow (1, 2) \in R_3$. (True)
  The relation $R_3$ remains transitive.
• Not Symmetric: We have $(1, 2) \in R_3$, but $(2, 1) \notin R_3$. So, $R_3$ is not symmetric.

Thus, $R_3$ is another such relation.

Third Relation: $R_3 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$

Case C: Consider other additions.

If we add $(3, 1)$ to $R_1$, for transitivity, since $(2, 3) \in R_1$ and $(3, 1)$ is added, we must add $(2, 1)$. And since $(3, 1) \in R_1$ and $(1, 2)$ is added, we must add $(3, 2)$. Adding all $\{(2,1), (3,2), (3,1)\}$ to $R_1$ results in the universal relation $A \times A$.

$A \times A = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\}$

The universal relation $A \times A$ is reflexive and transitive, but it is also symmetric (since for every $(a, b)$, $(b, a)$ is also present). This violates the condition that the relation must not be symmetric. The same result occurs if we try to add any two pairs from $\{(2, 1), (3, 2), (3, 1)\}$ to $R_1$, as the transitivity property will force the addition of the third pair.

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Conclusion:

We have found exactly three distinct relations that satisfy all the given conditions. They are:

1. $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$
2. $R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$
3. $R_3 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$

Hence, the number of relations containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three.

Given:

The set $A = \{1, 2, 3\}$.

We are looking for equivalence relations R on A that contain the pairs $(1, 2)$ and $(2, 1)$.

---

To Show:

The number of such equivalence relations is two.

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Solution:

An equivalence relation on a set corresponds to a partition of that set into disjoint equivalence classes. We can find the number of relations by finding the number of possible partitions of A that satisfy the given conditions.

The condition that a relation R must contain $(1, 2)$ and $(2, 1)$ means that 1 and 2 are related to each other. In terms of partitions, this means that the elements 1 and 2 must belong to the same equivalence class.

So, we need to find all possible partitions of the set $A = \{1, 2, 3\}$ where 1 and 2 are in the same subset.

Let's list the possibilities:

Partition 1: The elements 1 and 2 form one class, and the remaining element 3 forms its own class.

The partition is $P_1 = \{\{1, 2\}, \{3\}\}$.

This partition corresponds to the relation $R_1$ where elements are related if and only if they are in the same subset. So, $1\leftrightarrow1, 2\leftrightarrow2, 1\leftrightarrow2, 2\leftrightarrow1$, and $3\leftrightarrow3$.

$R_1 = \{(1, 1), (2, 2), (1, 2), (2, 1)\} \cup \{(3, 3)\} = \{(1, 1), (2, 2), (3, 3), $$ (1, 2), (2, 1)\}$.

This is a valid equivalence relation containing $(1,2)$ and $(2,1)$.

Partition 2: All three elements are in a single class.

The partition is $P_2 = \{\{1, 2, 3\}\}$.

This partition corresponds to the relation $R_2$ where every element is related to every other element. This is the universal relation on A.

$R_2 = A \times A = \{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2), $$ (3,3)\}$.

This is also a valid equivalence relation, and it contains $(1,2)$ and $(2,1)$.

There are no other possible ways to partition the set $\{1, 2, 3\}$ such that 1 and 2 are in the same subset. The only decision is where to place the element 3: either in its own class or with 1 and 2.

Since there are exactly two such partitions, there are exactly two corresponding equivalence relations.

Therefore, the number of equivalence relations in the set {1, 2, 3} containing (1, 2) and (2, 1) is two. Hence proved.

Given:

The set $A = \{1, 2\}$.

A binary operation ∗ on A must satisfy two conditions:

1. 1 is the identity element.
2. 2 is its own inverse.

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To Show:

There is exactly one such binary operation.

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Solution:

A binary operation on a set with 2 elements is defined by an operation table with $2 \times 2 = 4$ entries. We need to show that the given conditions uniquely determine all four entries in this table.

The general operation table for ∗ on A is:

*	1	2
1	$1*1$	$1*2$
2	$2*1$	$2*2$

Step 1: Apply the identity element condition.

Condition 1 states that 1 is the identity element. This means for any element $a \in A$, $a * 1 = a$ and $1 * a = a$.

• For $a = 1$: $1 * 1 = 1$.
• For $a = 2$: $2 * 1 = 2$ and $1 * 2 = 2$.

These three results are fixed by the identity property. Let's fill them into the table:

*	1	2
1	1	2
2	2	$2*2$

Step 2: Apply the inverse element condition.

Condition 2 states that 2 is the inverse of 2. This means that when 2 is operated with its inverse (which is 2), the result must be the identity element (which is 1).

So, $2 * 2 = 1$.

This determines the final entry in the table.

*	1	2
1	1	2
2	2	1

The given conditions have uniquely determined the outcome for all four possible input pairs. There is no other way to define the operation.

Since the operation table is completely and uniquely determined, there is exactly one binary operation that satisfies the given conditions. Hence proved.

Given:

The set of natural numbers $N = \{1, 2, 3, ...\}$.

The identity function $I_N: N \to N$ is defined by $I_N(x) = x$.

The function $f: N \to N$ is defined by $f(x) = (I_N + I_N)(x) = 2x$.

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To Show:

1. The function $I_N$ is onto.
2. The function $f(x) = 2x$ is not onto.

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Proof:

Part 1: Show that $I_N$ is onto.

A function is onto (surjective) if its range is equal to its codomain. For $I_N: N \to N$, we must show that for every element $y$ in the codomain N, there exists an element $x$ in the domain N such that $I_N(x) = y$.

Let $y$ be an arbitrary element of the codomain N.

We need to find an $x \in N$ such that $I_N(x) = y$.

By definition, $I_N(x) = x$, so we need $x=y$.

If we choose $x=y$, since $y$ is a natural number, our choice for $x$ is also a natural number, so $x$ is in the domain.

Thus, for any $y \in N$ in the codomain, there exists an $x=y$ in the domain such that $I_N(x) = y$.

Therefore, the function $I_N$ is onto.

Part 2: Show that $f(x) = 2x$ is not onto.

For the function $f(x) = 2x$ to be onto, for every element $y$ in the codomain N, there must exist an element $x$ in the domain N such that $f(x) = y$.

Let's consider an element in the codomain, for example, $y = 3 \in N$.

We need to find if there is an $x$ in the domain N such that $f(x) = 3$.

$2x = 3 \implies x = \frac{3}{2}$.

The value $x = \frac{3}{2}$ is not a natural number ($x \notin N$). Therefore, there is no element in the domain that maps to 3 in the codomain.

In fact, the range of the function $f(x)=2x$ is the set of all even natural numbers $\{2, 4, 6, ...\}$, while the codomain is the set of all natural numbers $\{1, 2, 3, 4, ...\}$. Since the range is not equal to the codomain (it is missing all odd numbers), the function is not onto.

Because we found an element (3) in the codomain that has no pre-image in the domain, the function $f(x)=2x$ is not onto. Hence proved.

Given:

The domain for all functions is the interval $D = \left[0, \frac{\pi}{2}\right]$.

Function $f: D \to R$ is defined as $f(x) = \sin x$.

Function $g: D \to R$ is defined as $g(x) = \cos x$.

Function $(f+g): D \to R$ is defined as $(f+g)(x) = \sin x + \cos x$.

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To Show:

1. $f$ is one-one.
2. $g$ is one-one.
3. $f+g$ is not one-one.

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Proof:

A function is one-one (injective) if for any two distinct elements in its domain, their images are also distinct. That is, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.

Part 1: Show that $f(x) = \sin x$ is one-one.

In the interval $\left[0, \frac{\pi}{2}\right]$, the function $f(x) = \sin x$ is strictly increasing. This means that for any $x_1, x_2$ in the domain, if $x_1 < x_2$, then $\sin(x_1) < \sin(x_2)$. Consequently, if $x_1 \neq x_2$, then $\sin(x_1) \neq \sin(x_2)$. This is the definition of a one-one function.

Therefore, $f(x) = \sin x$ is one-one on $\left[0, \frac{\pi}{2}\right]$.

Part 2: Show that $g(x) = \cos x$ is one-one.

In the interval $\left[0, \frac{\pi}{2}\right]$, the function $g(x) = \cos x$ is strictly decreasing. This means that for any $x_1, x_2$ in the domain, if $x_1 < x_2$, then $\cos(x_1) > \cos(x_2)$. Consequently, if $x_1 \neq x_2$, then $\cos(x_1) \neq \cos(x_2)$.

Therefore, $g(x) = \cos x$ is one-one on $\left[0, \frac{\pi}{2}\right]$.

Part 3: Show that $(f+g)(x) = \sin x + \cos x$ is not one-one.

To show a function is not one-one, we need to find two distinct inputs $x_1$ and $x_2$ in the domain that produce the same output. Let's check the values of the function at the endpoints of the domain, $x_1 = 0$ and $x_2 = \frac{\pi}{2}$.

For $x_1 = 0$:

$(f+g)(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.

For $x_2 = \frac{\pi}{2}$:

$(f+g)\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1$.

We have found two distinct inputs, $x_1 = 0$ and $x_2 = \frac{\pi}{2}$ (since $0 \neq \frac{\pi}{2}$), that both produce the same output value of 1.

Since $(f+g)(0) = (f+g)\left(\frac{\pi}{2}\right)$, the function $f+g$ is not one-one. Hence proved.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 10x + 7$.

The identity function on $\mathbb{R}$ is $I_R(x) = x$ for all $x \in \mathbb{R}$.

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To Find:

A function $g : \mathbb{R} \to \mathbb{R}$ such that $g \circ f = f \circ g = I_R$.

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Solution:

The condition $g \circ f = f \circ g = I_R$ means that the function $g$ is the inverse of the function $f$, denoted as $f^{-1}$. To find the inverse, we can set $y = f(x)$ and solve for $x$ in terms of $y$.

Let $y = f(x)$.

$y = 10x + 7$

Now, we rearrange the equation to express $x$ as the subject:

$y - 7 = 10x$

$x = \frac{y - 7}{10}$

Since $g$ is the inverse of $f$, we have $g(y) = x$. Therefore:

$g(y) = \frac{y - 7}{10}$

Replacing the variable $y$ with $x$, we get the function $g(x)$:

$g(x) = \frac{x - 7}{10}$

This defines the function $g: \mathbb{R} \to \mathbb{R}$.

Verification:

We must verify that $g \circ f = I_R$ and $f \circ g = I_R$.

1. Check $g \circ f$:

$(g \circ f)(x) = g(f(x)) = g(10x + 7)$

Substitute $(10x+7)$ into the formula for $g(x)$:

$(g \circ f)(x) = \frac{(10x + 7) - 7}{10} = \frac{10x}{10} = x$

Since $(g \circ f)(x) = x = I_R(x)$, the first condition is met.

2. Check $f \circ g$:

$(f \circ g)(x) = f(g(x)) = f\left(\frac{x - 7}{10}\right)$

Substitute $\left(\frac{x - 7}{10}\right)$ into the formula for $f(x)$:

$(f \circ g)(x) = 10\left(\frac{x - 7}{10}\right) + 7 = (x - 7) + 7 = x$

Since $(f \circ g)(x) = x = I_R(x)$, the second condition is met.

Both conditions are satisfied, so the required function is $g(x) = \frac{x - 7}{10}$.

Given:

The set of whole numbers, $W = \{0, 1, 2, 3, ...\}$.

A function $f : W \to W$ is defined as:

$f(n) = \begin{cases} n - 1 & , & \text{if } n \text{ is odd} \\ n + 1 & , & \text{if } n \text{ is even} \end{cases}$

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To Show:

The function $f$ is invertible.

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To Find:

The inverse function, $f^{-1}$.

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Solution:

A function is invertible if and only if it is a bijection (i.e., it is both one-one and onto).

Part 1: Show $f$ is one-one (injective).

Let $n_1, n_2 \in W$ such that $f(n_1) = f(n_2)$. We must show that $n_1=n_2$. We consider four cases:

Case A (Both odd): If $n_1$ and $n_2$ are odd, then $f(n_1) = n_1-1$ and $f(n_2) = n_2-1$. The equation $f(n_1)=f(n_2)$ becomes $n_1-1 = n_2-1$, which implies $n_1 = n_2$.

Case B (Both even): If $n_1$ and $n_2$ are even, then $f(n_1) = n_1+1$ and $f(n_2) = n_2+1$. The equation $f(n_1)=f(n_2)$ becomes $n_1+1 = n_2+1$, which implies $n_1 = n_2$.

Case C ($n_1$ odd, $n_2$ even): If $n_1$ is odd, $f(n_1)=n_1-1$ is an even number. If $n_2$ is even, $f(n_2)=n_2+1$ is an odd number. An even number can never equal an odd number, so $f(n_1) \neq f(n_2)$. This case is not possible.

In all possible cases where $f(n_1)=f(n_2)$, we must have $n_1=n_2$. Therefore, $f$ is one-one.

Part 2: Show $f$ is onto (surjective).

Let $m$ be an arbitrary element in the codomain $W$. We need to find a pre-image $n$ in the domain $W$ such that $f(n)=m$.

Case A ($m$ is even): If $m$ is even, we need $f(n)$ to be even. If $n$ is odd, $f(n)=n-1$ is even. Let's try this: $n-1=m \implies n=m+1$. Since $m$ is an even whole number, $n=m+1$ is an odd whole number and is in the domain $W$. Thus, for any even $m$, its pre-image is $n=m+1$.

Case B ($m$ is odd): If $m$ is odd, we need $f(n)$ to be odd. If $n$ is even, $f(n)=n+1$ is odd. Let's try this: $n+1=m \implies n=m-1$. Since $m$ is an odd whole number, $n=m-1$ is an even whole number and is in the domain $W$. Thus, for any odd $m$, its pre-image is $n=m-1$.

Since every element in the codomain has a pre-image in the domain, $f$ is onto.

Since $f$ is both one-one and onto, it is a bijection, and therefore, $f$ is invertible.

Part 3: Find the inverse $f^{-1}$.

From the proof for 'onto', we found the pre-image $n$ for any given image $m$. This defines the inverse function $f^{-1}(m) = n$.

If $m$ is even, $n = m+1$. So, $f^{-1}(m) = m+1$.

If $m$ is odd, $n = m-1$. So, $f^{-1}(m) = m-1$.

Thus, the inverse function $f^{-1}: W \to W$ is:

$f^{-1}(m) = \begin{cases} m + 1 & , & \text{if } m \text{ is even} \\ m - 1 & , & \text{if } m \text{ is odd} \end{cases}$

Comparing this with the original function $f(n)$, we notice that $f^{-1}$ is the same function as $f$. Therefore, $f$ is its own inverse.

The inverse of $f$ is $f$ itself, i.e., $\mathbf{f^{-1} = f}$.

Given:

The function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 - 3x + 2$.

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To Find:

The composite function $f(f(x))$.

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Solution:

The expression $f(f(x))$ means we substitute the entire expression for $f(x)$ into the variable $x$ of the function $f$.

$f(f(x)) = f(x^2 - 3x + 2)$

Now, we replace every instance of $x$ in $f(x) = x^2 - 3x + 2$ with the expression $(x^2 - 3x + 2)$:

$f(f(x)) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2$

We expand each part of the expression.

Part 1: Expand $(x^2 - 3x + 2)^2$

Using the formula $(a+b+c)^2 = a^2+b^2+c^2+2ab+2ac+2bc$:

$(x^2 - 3x + 2)^2 = (x^2)^2 + (-3x)^2 + (2)^2 + 2(x^2)(-3x) + 2(x^2)(2) $$ + 2(-3x)(2)$

$= x^4 + 9x^2 + 4 - 6x^3 + 4x^2 - 12x$

$= x^4 - 6x^3 + 13x^2 - 12x + 4$

Part 2: Expand $-3(x^2 - 3x + 2)$

$-3(x^2 - 3x + 2) = -3x^2 + 9x - 6$

Part 3: Combine all parts

$f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2$

Now, group and combine like terms:

$f(f(x)) = x^4 - 6x^3 + (13x^2 - 3x^2) + (-12x + 9x) + (4 - 6 + 2)$

$f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x + 0$

Thus, $f(f(x)) = \mathbf{x^4 - 6x^3 + 10x^2 - 3x}$.

Given:

A function $f : \mathbb{R} \to (-1, 1)$ defined by $f(x) = \frac{x}{1 + |x|}$.

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To Show:

The function $f$ is a bijection (one-one and onto).

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Proof:

First, we write the function in a piecewise form:

$f(x) = \begin{cases} \frac{x}{1 + x} & , & \text{if } x \ge 0 \\ \frac{x}{1 - x} & , & \text{if } x < 0 \end{cases}$

Part 1: Show $f$ is one-one (injective).

Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.

Case A ($x_1 \ge 0, x_2 \ge 0$):

$\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \implies x_1(1+x_2) = x_2(1+x_1) $$ \implies x_1+x_1x_2 = x_2+x_1x_2 \implies x_1=x_2$.

Case B ($x_1 < 0, x_2 < 0$):

$\frac{x_1}{1-x_1} = \frac{x_2}{1-x_2} \implies x_1(1-x_2) = x_2(1-x_1) $$ \implies x_1-x_1x_2 = x_2-x_1x_2 \implies x_1=x_2$.

Case C ($x_1 \ge 0, x_2 < 0$):

If $x_1 \ge 0$, then $f(x_1) = \frac{x_1}{1+x_1} \ge 0$.

If $x_2 < 0$, then $f(x_2) = \frac{x_2}{1-x_2} < 0$ (since numerator is negative and denominator is positive).

Since $f(x_1) \ge 0$ and $f(x_2) < 0$, it is impossible for $f(x_1)=f(x_2)$.

In all cases, $f(x_1) = f(x_2)$ implies $x_1=x_2$. Therefore, $f$ is one-one.

Part 2: Show $f$ is onto (surjective).

Let $y$ be an arbitrary element in the codomain $(-1, 1)$. We need to find an $x \in \mathbb{R}$ such that $f(x) = y$.

Case A ($0 \le y < 1$):

We expect the pre-image $x$ to be non-negative, so we use the first piece of the function definition: $f(x) = \frac{x}{1+x}$.

$y = \frac{x}{1+x} \implies y(1+x) = x \implies y+yx = x $$ \implies y = x-yx \implies y = x(1-y)$.

$x = \frac{y}{1-y}$.

Since $0 \le y < 1$, the numerator $y \ge 0$ and the denominator $1-y > 0$. Thus, $x \ge 0$, which is consistent with our assumption. So, for any $y \in [0, 1)$, there exists a pre-image $x = \frac{y}{1-y}$.

Case B ($-1 < y < 0$):

We expect the pre-image $x$ to be negative, so we use the second piece: $f(x) = \frac{x}{1-x}$.

$y = \frac{x}{1-x} \implies y(1-x) = x \implies y-yx = x $$ \implies y = x+yx \implies y = x(1+y)$.

$x = \frac{y}{1+y}$.

Since $-1 < y < 0$, the numerator $y < 0$ and the denominator $1+y > 0$. Thus, $x < 0$, which is consistent with our assumption. So, for any $y \in (-1, 0)$, there exists a pre-image $x = \frac{y}{1+y}$.

Since for every $y$ in the codomain $(-1, 1)$ there exists a pre-image $x$ in the domain $\mathbb{R}$, the function $f$ is onto.

Because $f$ is both one-one and onto, it is a bijective function.

Given:

A function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$.

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To Show:

The function $f$ is injective (one-to-one).

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Proof:

A function is injective if for any two elements $x_1, x_2$ in its domain, $f(x_1) = f(x_2)$ implies that $x_1 = x_2$.

Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$.

By the definition of the function, this means:

$x_1^3 = x_2^3$

We can rearrange the equation:

$x_1^3 - x_2^3 = 0$

Using the algebraic identity for the difference of cubes, $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, we get:

$(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0$

This product is zero if and only if one or both of the factors are zero.

Case 1: $x_1 - x_2 = 0$. This directly implies $x_1 = x_2$.

Case 2: $x_1^2 + x_1x_2 + x_2^2 = 0$. We can rewrite this expression by completing the square:

$x_1^2 + x_1x_2 + \frac{x_2^2}{4} + \frac{3x_2^2}{4} = 0$

$\left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3}{4}x_2^2 = 0$

The term $\left(x_1 + \frac{x_2}{2}\right)^2$ is a square, so it is always greater than or equal to zero. The term $\frac{3}{4}x_2^2$ is also always greater than or equal to zero. The sum of two non-negative real numbers can only be zero if both numbers are zero.

So, we must have $\frac{3}{4}x_2^2 = 0$, which implies $x_2=0$.

And we must have $\left(x_1 + \frac{x_2}{2}\right)^2 = 0$. Substituting $x_2=0$ gives $(x_1+0)^2 = 0$, which implies $x_1=0$.

Thus, this second case only holds if $x_1=0$ and $x_2=0$, which is also a situation where $x_1=x_2$.

In both possible cases, the condition $f(x_1) = f(x_2)$ leads to the conclusion that $x_1=x_2$.

Therefore, the function $f(x) = x^3$ is injective.

Given:

We need to find two functions, $f: N \to Z$ and $g: Z \to Z$, such that their composition $g \circ f$ is injective (one-one), but the function $g$ itself is not injective.

(Here $N = \{1, 2, 3, ...\}$ and $Z = \{..., -1, 0, 1, ...\}$)

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Example Functions:

Let's define the functions as suggested by the hint:

1. Define $f: N \to Z$ as $f(x) = x$.

2. Define $g: Z \to Z$ as $g(x) = |x|$.

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Justification:

Step 1: Show that $g$ is not injective.

A function is not injective if we can find two different inputs that produce the same output. Consider the inputs $x_1=2$ and $x_2=-2$ from the domain $Z$.

$g(2) = |2| = 2$

$g(-2) = |-2| = 2$

We have $g(2) = g(-2)$, but $2 \neq -2$. Therefore, the function $\mathbf{g}$ is not injective.

Step 2: Show that the composition $g \circ f$ is injective.

First, let's find the formula for the composite function $g \circ f: N \to Z$.

$(g \circ f)(x) = g(f(x)) = g(x) = |x|$.

The domain of this composite function is the domain of $f$, which is $N$. For any $x \in N$, $x$ is a positive integer, so $|x| = x$.

Therefore, the composite function is $(g \circ f)(x) = x$ for all $x \in N$.

Now, let's check if this function is injective. Let $x_1, x_2$ be two elements in the domain $N$ such that $(g \circ f)(x_1) = (g \circ f)(x_2)$.

This means $x_1 = x_2$.

The condition for injectivity is met. Therefore, the function $\mathbf{g \circ f}$ is injective.

This completes the example.

Given:

We need to find two functions, $f: N \to N$ and $g: N \to N$, such that their composition $g \circ f$ is onto (surjective), but the function $f$ itself is not onto.

(Here $N = \{1, 2, 3, ...\}$)

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Example Functions:

Let's define the functions as follows:

1. Define $f: N \to N$ as $f(x) = x + 1$.

2. Define $g: N \to N$ as $g(x) = \begin{cases} x - 1 & , & \text{if } x > 1 \\ 1 & , & \text{if } x = 1 \end{cases}$

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Justification:

Step 1: Show that $f$ is not onto.

A function is onto if its range is equal to its codomain. The codomain of $f$ is $N = \{1, 2, 3, ...\}$.

Let's find the range of $f(x) = x+1$. The domain is $N=\{1, 2, 3, ...\}$.

The smallest value in the range is $f(1) = 1+1 = 2$.

The range of $f$ is $\{2, 3, 4, ...\}$.

The element $1$ is in the codomain $N$ but is not in the range of $f$. There is no natural number $x$ for which $x+1=1$ (since that would require $x=0$, which is not in $N$).

Since the range is not equal to the codomain, the function $\mathbf{f}$ is not onto.

Step 2: Show that the composition $g \circ f$ is onto.

First, let's find the formula for the composite function $g \circ f: N \to N$.

$(g \circ f)(x) = g(f(x)) = g(x+1)$.

The domain of this composite function is $N$. For any $x \in N$, $x \ge 1$, so $f(x)=x+1 \ge 2$.

Since the input to $g$, which is $f(x)=x+1$, is always greater than 1, we only need to use the first case of the definition of $g$: $g(y) = y-1$ for $y>1$.

So, $(g \circ f)(x) = g(x+1) = (x+1) - 1 = x$.

The composite function is $(g \circ f)(x) = x$ for all $x \in N$.

Now, let's check if this function is onto. Let $y$ be any element in the codomain $N$. We need to find an $x$ in the domain $N$ such that $(g \circ f)(x) = y$.

This means we need to find $x \in N$ such that $x=y$.

If we choose $x=y$, since $y \in N$, our choice for $x$ is also in $N$. Thus, for any $y$ in the codomain, there exists a pre-image $x=y$.

Therefore, the function $\mathbf{g \circ f}$ is onto.

Given:

A non-empty set X and its power set P(X).

A relation R on P(X) is defined as: A R B if and only if A ⊂ B.

The symbol '⊂' denotes a proper subset, meaning A is a subset of B, but A is not equal to B.

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To Determine:

Is R an equivalence relation on P(X)?

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Justification:

For R to be an equivalence relation, it must be reflexive, symmetric, and transitive.

1. Reflexivity

The relation is reflexive if A R A for all A ∈ P(X). This requires A ⊂ A.

By definition, A ⊂ A means A is a subset of A AND A ≠ A. The second part, A ≠ A, is always false.

Therefore, A ⊂ A is never true. The relation is not reflexive.

2. Symmetry

The relation is symmetric if A R B implies B R A for all A, B ∈ P(X). This requires that if A ⊂ B, then B ⊂ A.

Let's take a counterexample. Let X = {1, 2}. Let A = {1} and B = {1, 2}.

Here, A ⊂ B is true because every element of A is in B, and A ≠ B. So, A R B holds.

However, B ⊂ A is false, because the element 2 is in B but not in A.

Since A R B does not imply B R A, the relation is not symmetric.

3. Transitivity

The relation is transitive if (A R B and B R C) implies A R C for all A, B, C ∈ P(X). This requires that if (A ⊂ B and B ⊂ C), then A ⊂ C.

Assume A ⊂ B and B ⊂ C.

From A ⊂ B, we know every element of A is in B.

From B ⊂ C, we know every element of B is in C.

It follows that every element of A must also be in C, so A ⊆ C.

Also, since B ⊂ C, there must be an element $x \in C$ such that $x \notin B$. Since all elements of A are in B, this element $x$ cannot be in A either. So, $x \in C$ and $x \notin A$. This guarantees that A ≠ C.

Since A ⊆ C and A ≠ C, we have A ⊂ C. The relation is transitive.

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Conclusion:

The relation R is transitive, but it is not reflexive and not symmetric. Since an equivalence relation must satisfy all three properties, R is not an equivalence relation.

Given:

A non-empty set X and its power set P(X).

A binary operation ∗ on P(X) is defined as $A ∗ B = A \cap B$.

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To Show:

1. The set X is the identity element for ∗.

2. The set X is the only invertible element in P(X).

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Proof:

Part 1: Show X is the identity element.

An element $E \in P(X)$ is the identity element if for every $A \in P(X)$, we have $A * E = A$ and $E * A = A$.

Let's test if $E = X$ satisfies this condition.

For any subset $A \in P(X)$, we have:

$A * X = A \cap X$.

Since A is a subset of X, the intersection of A and X is A itself. So, $A \cap X = A$.

Also, $X * A = X \cap A$.

Since intersection is a commutative operation, $X \cap A = A \cap X = A$.

Because $A * X = A$ and $X * A = A$ for all $A \in P(X)$, we conclude that X is the identity element for the operation ∗.

Part 2: Show X is the only invertible element.

An element $A \in P(X)$ is invertible if there exists an inverse element $B \in P(X)$ such that $A * B = E$, where $E$ is the identity element.

From Part 1, we know the identity element is $E = X$. So, the condition for invertibility is:

$A * B = X \implies A \cap B = X$.

By the definition of intersection, $A \cap B$ is a subset of A (i.e., $A \cap B \subseteq A$).

If $A \cap B = X$, then it must be that $X \subseteq A$.

However, by the definition of the power set P(X), any element $A \in P(X)$ is a subset of X (i.e., $A \subseteq X$).

The only way for both $X \subseteq A$ and $A \subseteq X$ to be true is if $A = X$.

This shows that if an element $A$ is invertible, it must be the set X itself. Let's confirm that X is indeed invertible by finding its inverse B.

We need to find $B \in P(X)$ such that $X \cap B = X$. If we choose $B=X$, then $X \cap X = X$, which is true. So, X is invertible and its inverse is X.

No other subset $A \neq X$ can be invertible because for any such A, $A \cap B \subseteq A$, and since $A \neq X$, it is impossible for $A \cap B$ to be equal to X.

Therefore, X is the only invertible element in P(X) with respect to the operation ∗.

Given:

A finite set $A = \{1, 2, 3, ..., n\}$.

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To Find:

The number of all onto (surjective) functions $f: A \to A$.

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Solution:

Let's consider the properties of functions between finite sets of the same size.

The domain is the set $A$, which has $n$ elements.

The codomain is also the set $A$, which has $n$ elements.

For any function $f: X \to Y$ where X and Y are finite sets with the same number of elements ($|X|=|Y|$), the following statements are equivalent:

1. $f$ is one-to-one (injective).
2. $f$ is onto (surjective).
3. $f$ is a bijection.

Therefore, finding the number of onto functions from A to itself is the same as finding the number of one-to-one functions from A to itself.

A one-to-one function from a finite set to itself is a permutation of that set. We need to count the number of ways we can arrange the elements of A, which is equivalent to defining a one-to-one function $f$.

Let's count the number of choices for the image of each element in the domain:

• For the element 1, there are $n$ possible choices for its image, $f(1)$.
• For the element 2, its image $f(2)$ must be different from $f(1)$ (to be one-to-one). So there are $n-1$ choices left.
• For the element 3, its image $f(3)$ must be different from $f(1)$ and $f(2)$. So there are $n-2$ choices left.
• ...and so on...
• For the last element $n$, its image $f(n)$ must be the one remaining element in the codomain that has not yet been assigned. So there is only 1 choice left.

By the fundamental principle of counting, the total number of such functions is the product of the number of choices at each step:

Total number of onto functions = $n \times (n-1) \times (n-2) \times \dots \times 2 \times 1 = n!$

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The number of all onto functions from the set $\{1, 2, 3, ..., n\}$ to itself is $\mathbf{n!}$.

A function has an inverse if and only if it is a bijection, which means it must be both one-one (injective) and onto (surjective).

• One-one: Every distinct element in the domain maps to a distinct element in the codomain.
• Onto: Every element in the codomain is the image of at least one element from the domain (i.e., the range equals the codomain).

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(i) F = {(a, 3), (b, 2), (c, 1)}

Let's check if this function is a bijection.

1. One-one: The inputs are a, b, and c. Their respective outputs are 3, 2, and 1. Since all the outputs are distinct for distinct inputs, the function is one-one.

2. Onto: The codomain is $T = \{1, 2, 3\}$. The range (the set of all outputs) is $\{3, 2, 1\}$. Since the range is equal to the codomain, the function is onto.

Since the function is both one-one and onto, it is a bijection, and its inverse, $F^{-1}$, exists.

To find the inverse function $F^{-1}: T \to S$, we simply reverse the ordered pairs in F:

$\mathbf{F^{-1} = \{(3, a), (2, b), (1, c)\}}$.

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(ii) F = {(a, 2), (b, 1), (c, 1)}

Let's check if this function is a bijection.

1. One-one: The inputs are a, b, and c. Their respective outputs are 2, 1, and 1. The distinct inputs 'b' and 'c' both map to the same output '1' ($F(b) = 1$ and $F(c) = 1$). Therefore, the function is not one-one.

Since the function is not one-one, it is not a bijection. Therefore, its inverse does not exist.

(We can also check for the onto property: The range is $\{1, 2\}$, which is not equal to the codomain $T = \{1, 2, 3\}$. So the function is also not onto.)

Because the function is not a bijection, $\mathbf{F^{-1}}$ does not exist.

Given:

Two binary operations on $\mathbb{R}$:

$a ∗ b = |a - b|$

$a \ o \ b = a$

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Part 1: Analysis of Operation ∗

Commutativity: We check if $a * b = b * a$.

$a * b = |a - b|$

$b * a = |b - a| = |-(a-b)| = |a-b|$.

Since $a*b = b*a$, the operation ∗ is commutative.

Associativity: We check if $(a * b) * c = a * (b * c)$.

Let $a=1, b=2, c=3$.

LHS: $(1*2)*3 = |1-2|*3 = 1*3 = |1-3| = 2$.

RHS: $1*(2*3) = 1*|2-3| = 1*1 = |1-1| = 0$.

Since LHS $\neq$ RHS, the operation ∗ is not associative.

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Part 2: Analysis of Operation o

Commutativity: We check if $a \ o \ b = b \ o \ a$.

$a \ o \ b = a$.

$b \ o \ a = b$.

Since $a \neq b$ in general, the operation o is not commutative.

Associativity: We check if $(a \ o \ b) \ o \ c = a \ o \ (b \ o \ c)$.

LHS: $(a \ o \ b) \ o \ c = (a) \ o \ c = a$.

RHS: $a \ o \ (b \ o \ c) = a \ o \ (b) = a$.

Since LHS = RHS, the operation o is associative.

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Part 3: Show that ∗ distributes over o

We need to show that $a ∗ (b \ o \ c) = (a ∗ b) \ o \ (a ∗ c)$.

LHS: $a ∗ (b \ o \ c) = a * (b) = |a-b|$.

RHS: $(a ∗ b) \ o \ (a ∗ c) = |a-b| \ o \ |a-c|$. By the definition of 'o', this evaluates to the first operand.

RHS = $|a-b|$.

Since LHS = RHS, ∗ distributes over o.

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Part 4: Does o distribute over ∗?

We need to check if $a \ o \ (b ∗ c) = (a \ o \ b) ∗ (a \ o \ c)$.

LHS: $a \ o \ (b ∗ c) = a \ o \ (|b-c|)$. By the definition of 'o', this evaluates to the first operand.

LHS = $a$.

RHS: $(a \ o \ b) ∗ (a \ o \ c) = (a) * (a) = |a-a| = 0$.

For distribution to hold, we would need $a = 0$ for all $a, b, c \in \mathbb{R}$, which is false.

No, o does not distribute over ∗. For example, if $a=5$, LHS = 5 while RHS = 0.

Given:

The power set P(X) of a non-empty set X.

A binary operation ∗ on P(X) defined as $A ∗ B = (A - B) \cup (B - A)$. This is the symmetric difference of sets A and B.

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Part 1: Show that the empty set $\emptyset$ is the identity element.

An element $E$ is the identity if $A * E = A$ for all $A \in P(X)$. Since the operation is commutative, we only need to check one side.

Let's check if $E = \emptyset$ works.

$A * \emptyset = (A - \emptyset) \cup (\emptyset - A)$

By the properties of set difference:

• $A - \emptyset = A$ (removing no elements from A leaves A)
• $\emptyset - A = \emptyset$ (removing elements of A from the empty set leaves the empty set)

Substituting these back:

$A * \emptyset = A \cup \emptyset$

The union of any set with the empty set is the set itself, so $A \cup \emptyset = A$.

Thus, $A * \emptyset = A$. Since the operation is commutative, $\emptyset * A = A$ also holds.

Therefore, the empty set $\emptyset$ is the identity element for the operation ∗.

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Part 2: Show that all elements are invertible with $A^{-1} = A$.

An element A is invertible if there exists an inverse B such that $A * B = E$, where E is the identity element.

We found that the identity element is $E = \emptyset$. So, we need to show that for every set $A \in P(X)$, its inverse is A itself. This means we must show that $A * A = \emptyset$.

Let's compute $A * A$ using the definition of the operation:

$A * A = (A - A) \cup (A - A)$

By the property of set difference, $A - A$ is the set of elements that are in A but not in A, which is always the empty set. So, $A - A = \emptyset$.

Substituting this back:

$A * A = \emptyset \cup \emptyset$

The union of the empty set with itself is the empty set, so $\emptyset \cup \emptyset = \emptyset$.

Thus, $A * A = \emptyset$.

This shows that for any element A in P(X), operating it with itself gives the identity element $\emptyset$. Therefore, every element A is invertible and its inverse is A itself ($A^{-1} = A$).

Given:

The set $S = \{0, 1, 2, 3, 4, 5\}$.

A binary operation ∗ on S defined as addition modulo 6.

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Part 1: Show that zero is the identity element.

An element $e$ is the identity if $a * e = a$ for all $a \in S$. Let's test if $e=0$ works.

We need to compute $a * 0$ for any $a \in S$.

The sum is $a+0 = a$. Since $a$ can be at most 5, the sum $a$ is always less than 6.

According to the definition of the operation, we use the first case ($a+b < 6$):

$a * 0 = a + 0 = a$.

Since the operation is based on addition, it is commutative, so $0 * a = a$ as well.

Therefore, zero is the identity element for this operation.

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Part 2: Show that each non-zero element $a$ is invertible with inverse $6-a$.

An element $a$ is invertible if there exists an inverse $b$ such that $a * b = e$, where $e$ is the identity element (0).

Let $a$ be a non-zero element of S, so $a \in \{1, 2, 3, 4, 5\}$.

Let its proposed inverse be $b = 6 - a$. First, we check if $b$ is in the set S:

• If $a=1$, $b=5 \in S$.
• If $a=2$, $b=4 \in S$.
• If $a=3$, $b=3 \in S$.
• If $a=4$, $b=2 \in S$.
• If $a=5$, $b=1 \in S$.

In all cases, the proposed inverse $b=6-a$ is an element of the set S.

Now, we must show that $a * (6-a) = 0$.

The sum is $a + (6-a) = 6$.

Since the sum is 6, it falls into the second case of the definition ($a+b \ge 6$):

$a * (6-a) = (a + (6-a)) - 6 = 6 - 6 = 0$.

Since $a * (6-a) = 0$ (the identity element) for all non-zero $a \in S$, we have shown that each element $a \neq 0$ is invertible with $6-a$ being the inverse of $a$.

Given:

Domain $A = \{-1, 0, 1, 2\}$ and Codomain $B = \{-4, -2, 0, 2\}$.

Function $f(x) = x^2 - x$.

Function $g(x) = 2\left|x - \frac{1}{2}\right| - 1$.

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To Determine:

Are the functions $f$ and $g$ equal?

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Justification:

Two functions are equal if they have the same domain, the same codomain, and map every element of the domain to the same element in the codomain. Here, the domains and codomains match, so we just need to check if $f(x) = g(x)$ for all $x \in A$.

Let's evaluate both functions for each element in the domain A:

For $x = -1$:

$f(-1) = (-1)^2 - (-1) = 1 + 1 = 2$.

$g(-1) = 2\left|-1 - \frac{1}{2}\right| - 1 = 2\left|-\frac{3}{2}\right| - 1 = 2\left(\frac{3}{2}\right) - 1 = 3 - 1 = 2$.

So, $f(-1) = g(-1)$.

For $x = 0$:

$f(0) = 0^2 - 0 = 0$.

$g(0) = 2\left|0 - \frac{1}{2}\right| - 1 = 2\left|-\frac{1}{2}\right| - 1 = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0$.

So, $f(0) = g(0)$.

For $x = 1$:

$f(1) = 1^2 - 1 = 0$.

$g(1) = 2\left|1 - \frac{1}{2}\right| - 1 = 2\left|\frac{1}{2}\right| - 1 = 2\left(\frac{1}{2}\right) - 1 = 1 - 1 = 0$.

So, $f(1) = g(1)$.

For $x = 2$:

$f(2) = 2^2 - 2 = 4 - 2 = 2$.

$g(2) = 2\left|2 - \frac{1}{2}\right| - 1 = 2\left|\frac{3}{2}\right| - 1 = 2\left(\frac{3}{2}\right) - 1 = 3 - 1 = 2$.

So, $f(2) = g(2)$.

Since $f(x) = g(x)$ for every $x$ in the domain A, the functions are equal.

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Yes, the functions $f$ and $g$ are equal.

Given:

Set $A = \{1, 2, 3\}$. We need to find the number of relations R on A that satisfy:

1. $(1, 2) \in R$ and $(1, 3) \in R$
2. R is reflexive
3. R is symmetric
4. R is not transitive

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Solution:

Let's construct the smallest possible relation that meets the first three conditions.

1. Reflexive: R must contain $\{(1, 1), (2, 2), (3, 3)\}$.

2. Contains given pairs: R must contain $\{(1, 2), (1, 3)\}$.

3. Symmetric: Since $(1,2) \in R$, $(2,1)$ must be in R. Since $(1,3) \in R$, $(3,1)$ must be in R.

Combining these, the smallest possible relation, let's call it $R_{min}$, is:

$R_{min} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}$.

Now, let's check the fourth condition: Is $R_{min}$ not transitive?

For a relation to be not transitive, there must be a case where $(a,b) \in R$ and $(b,c) \in R$, but $(a,c) \notin R$.

In $R_{min}$, we have $(2, 1) \in R_{min}$ and $(1, 3) \in R_{min}$.

For transitivity to hold, the pair $(2, 3)$ would also need to be in the relation.

However, $(2, 3) \notin R_{min}$.

Since we found a failure of transitivity, $R_{min}$ is not transitive. It satisfies all four conditions. So, this is one such relation.

Now, can we create any other such relation by adding more pairs? The only pairs not in $R_{min}$ are $(2, 3)$ and $(3, 2)$.

If we add $(2,3)$ to $R_{min}$, we must also add $(3,2)$ to keep it symmetric. Let's call this new relation $R_{full}$.

$R_{full} = R_{min} \cup \{(2, 3), (3, 2)\} = A \times A$ (the universal relation).

Let's check if $R_{full}$ satisfies the conditions. It is reflexive and symmetric, and it contains the required pairs. But is it not transitive?

The universal relation is always transitive. For any $(a,b)$ and $(b,c)$ in $A \times A$, the pair $(a,c)$ is also guaranteed to be in $A \times A$. So $R_{full}$ is transitive.

This violates the fourth condition. Therefore, we cannot add any more pairs.

The only relation that satisfies all the conditions is $R_{min}$.

So, there is exactly 1 such relation.

The correct option is (A).

Given:

The set $A = \{1, 2, 3\}$. We need to find the number of equivalence relations on A that contain the pair $(1, 2)$.

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Solution:

Equivalence relations on a set correspond directly to the partitions of that set. An element $(a,b)$ being in a relation means that $a$ and $b$ are in the same subset (equivalence class) of the partition.

The condition that the relation must contain $(1, 2)$ means that the elements 1 and 2 must be in the same equivalence class in the corresponding partition of the set $A = \{1, 2, 3\}$.

Let's find all possible partitions of $\{1, 2, 3\}$ where 1 and 2 are grouped together.

Partition 1:

We group {1, 2} together. The remaining element is {3}. This gives the partition:

$P_1 = \{\{1, 2\}, \{3\}\}$

This partition corresponds to one valid equivalence relation.

Partition 2:

We group {1, 2} together, and also group {3} with them. This gives the partition:

$P_2 = \{\{1, 2, 3\}\}$

This partition corresponds to a second valid equivalence relation (the universal relation).

There are no other ways to partition the set $\{1, 2, 3\}$ while keeping 1 and 2 in the same subset. The only choice is what to do with the element 3: either it is in its own class, or it is in the same class as 1 and 2.

Since there are 2 such partitions, there are exactly 2 equivalence relations containing the pair $(1, 2)$.

The correct option is (B).

Given:

Signum Function: $f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases}$

Greatest Integer Function: $g(x) = [x]$.

The interval is $(0, 1]$.

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To Determine:

Do the functions $f \circ g$ and $g \circ f$ coincide (are they equal) on the interval $(0, 1]$?

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Solution:

We need to find the expressions for $(f \circ g)(x)$ and $(g \circ f)(x)$ for $x \in (0, 1]$ and compare them.

1. Calculate $(g \circ f)(x)$ for $x \in (0, 1]$

$(g \circ f)(x) = g(f(x))$.

For any $x$ in the interval $(0, 1]$, we have $x > 0$.

According to the definition of the Signum function, if $x > 0$, then $f(x) = 1$.

So, for all $x \in (0, 1]$, $(g \circ f)(x) = g(1)$.

Using the definition of the Greatest Integer Function, $g(1) = [1] = 1$.

Thus, $\mathbf{(g \circ f)(x) = 1}$ for all $x \in (0, 1]$.

2. Calculate $(f \circ g)(x)$ for $x \in (0, 1]$

$(f \circ g)(x) = f(g(x)) = f([x])$.

We need to consider the value of $[x]$ for $x$ in the interval $(0, 1]$. This requires two cases:

Case A: $x \in (0, 1)$. For these values, the greatest integer less than or equal to $x$ is 0. So, $[x] = 0$.

In this case, $(f \circ g)(x) = f(0)$. According to the Signum function definition, $f(0) = 0$.

Case B: $x = 1$. For this value, the greatest integer less than or equal to $x$ is 1. So, $[1] = 1$.

In this case, $(f \circ g)(1) = f(1)$. According to the Signum function definition, since $1>0$, $f(1) = 1$.

Combining these cases, we get:

$\mathbf{(f \circ g)(x) = \begin{cases} 0, & \text{if } x \in (0, 1) \\ 1, & \text{if } x = 1 \end{cases}}$

3. Comparison

Let's compare the two functions on the interval $(0, 1]$:

• $(g \circ f)(x) = 1$ for all $x \in (0, 1]$.
• $(f \circ g)(x)$ is 0 for $x \in (0, 1)$ and 1 for $x=1$.

For the functions to coincide, they must be equal for all values in the interval. Let's pick a value like $x=0.5$.

$(g \circ f)(0.5) = 1$.

$(f \circ g)(0.5) = 0$.

Since $1 \neq 0$, the functions are not equal on the interval $(0, 1]$.

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No, $f \circ g$ and $g \circ f$ do not coincide in $(0, 1]$.