Chapter 1 Relations And Functions

Given:

Set A = {all students of a boys school}

Relation R = {(a, b) : a is sister of b}, where $a, b \in A$

Relation R’ = {(a, b) : the difference between heights of a and b is less than 3 meters}, where $a, b \in A$

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To Show:

R is an empty relation.

R’ is a universal relation.

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Solution:

We are given that A is the set of all students of a boys school.

This means that set A contains only male students.

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Consider the relation R = {(a, b) : a is sister of b}.

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

If $a \in A$, then 'a' is a student of a boys school, which implies 'a' is male.

A male person cannot be a sister.

Therefore, for any element 'a' in set A, 'a' cannot be a sister of any student 'b' in set A.

Thus, there is no pair $(a, b)$ from set A such that 'a' is the sister of 'b'.

Hence, the relation R contains no elements.

A relation that contains no elements is called an empty relation.

So, R is an empty relation on A.

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Consider the relation R’ = {(a, b) : the difference between heights of a and b is less than 3 meters}.

For an ordered pair $(a, b)$ to be in R’, the absolute difference between the height of 'a' and the height of 'b' must be less than 3 meters.

Let $h_a$ be the height of student 'a' and $h_b$ be the height of student 'b'.

The condition is $|h_a - h_b| < 3$ meters.

In a boys school, the height of any student is typically in the range of about 0.5 meters (for very young students) to around 2 meters (for very tall students).

The maximum possible difference between the heights of any two students in a school would be the difference between the tallest and the shortest student.

This difference is highly unlikely to be as large as 3 meters.

For example, if the tallest student is 2 meters and the shortest is 0.5 meters, the difference is $2 - 0.5 = 1.5$ meters, which is less than 3 meters.

It is a practical certainty that for any two students 'a' and 'b' in a school, the difference in their heights will always be less than 3 meters.

Thus, for every possible pair $(a, b)$ from set A, the condition $|h_a - h_b| < 3$ meters will be satisfied.

Therefore, the relation R’ contains all possible ordered pairs from set A $\times$ A.

A relation that contains all possible ordered pairs from $A \times A$ is called a universal relation.

So, R’ is a universal relation on A.

Given:

Set T = {all triangles in a plane}

Relation R = {(T₁ , T₂ ) : T₁ is congruent to T₂ }, where T₁, T₂ $\in$ T

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To Show:

R is an equivalence relation on T.

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Solution:

To show that R is an equivalence relation, we need to verify that R satisfies the following three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

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

In this case, for R to be reflexive, we must have $(T_1, T_1) \in R$ for every $T_1 \in T$.

$(T_1, T_1) \in R$ means that $T_1$ is congruent to $T_1$.

Every triangle is congruent to itself.

Thus, $T_1$ is congruent to $T_1$ for all $T_1 \in T$.

Therefore, $(T_1, T_1) \in R$ for all $T_1 \in T$.

Hence, R is reflexive.

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2. Symmetry:

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

In this case, for R to be symmetric, we must show that if $(T_1, T_2) \in R$, then $(T_2, T_1) \in R$ for all $T_1, T_2 \in T$.

Suppose $(T_1, T_2) \in R$. By the definition of R, this means that $T_1$ is congruent to $T_2$.

If $T_1$ is congruent to $T_2$, it implies that $T_2$ is also congruent to $T_1$.

Thus, if $T_1$ is congruent to $T_2$, then $T_2$ is congruent to $T_1$.

This means that if $(T_1, T_2) \in R$, then $(T_2, T_1) \in R$.

Hence, R is symmetric.

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3. Transitivity:

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

In this case, for R to be transitive, we must show that if $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$, then $(T_1, T_3) \in R$ for all $T_1, T_2, T_3 \in T$.

Suppose $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$.

$(T_1, T_2) \in R$ means that $T_1$ is congruent to $T_2$.

$(T_2, T_3) \in R$ means that $T_2$ is congruent to $T_3$.

If $T_1$ is congruent to $T_2$ and $T_2$ is congruent to $T_3$, then it follows that $T_1$ is congruent to $T_3$.

This means that if $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$, then $(T_1, T_3) \in R$.

Hence, 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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Conclusion:

The relation R = {(T₁ , T₂ ) : T₁ is congruent to T₂ } on the set T of all triangles in a plane is an equivalence relation.

Given:

Set L = {all lines in a plane}

Relation R = {(L₁ , L₂ ) : L₁ is perpendicular to L₂ }, where L₁, L₂ $\in$ L

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To Show:

R is symmetric.

R is neither reflexive nor transitive.

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Solution:

We analyze the relation R for reflexivity, symmetry, and transitivity.

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1. Reflexivity:

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

For R to be reflexive, we must have $(L_1, L_1) \in R$ for every $L_1 \in L$.

$(L_1, L_1) \in R$ means that $L_1$ is perpendicular to $L_1$.

However, a line cannot be perpendicular to itself.

For example, consider a line L$_1$. The angle between L$_1$ and itself is $0^\circ$, not $90^\circ$.

Therefore, $(L_1, L_1) \notin R$ for any line $L_1 \in L$.

Hence, R is not reflexive.

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2. Symmetry:

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

For R to be symmetric, we must show that if $(L_1, L_2) \in R$, then $(L_2, L_1) \in R$ for all $L_1, L_2 \in L$.

Suppose $(L_1, L_2) \in R$. By the definition of R, this means that $L_1$ is perpendicular to $L_2$.

If line $L_1$ is perpendicular to line $L_2$, it is geometrically true that line $L_2$ is also perpendicular to line $L_1$.

Thus, if $L_1$ is perpendicular to $L_2$, then $L_2$ is perpendicular to $L_1$.

This means that if $(L_1, L_2) \in R$, then $(L_2, L_1) \in R$.

Hence, R is symmetric.

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3. Transitivity:

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

For R to be transitive, we must show that if $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$, then $(L_1, L_3) \in R$ for all $L_1, L_2, L_3 \in L$.

Suppose $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$.

$(L_1, L_2) \in R$ means that $L_1$ is perpendicular to $L_2$.

$(L_2, L_3) \in R$ means that $L_2$ is perpendicular to $L_3$.

If $L_1$ is perpendicular to $L_2$, and $L_2$ is perpendicular to $L_3$, then $L_1$ and $L_3$ must be parallel to each other (assuming $L_1 \ne L_3$).

Parallel lines are not perpendicular to each other (unless they are the same line, which contradicts the condition for reflexivity).

For example, let $L_1$ be the x-axis and $L_2$ be the y-axis. $L_1 \perp L_2$. Let $L_3$ be the line $y=5$. $L_2 \perp L_3$ (y-axis is perpendicular to a horizontal line). However, $L_1$ (x-axis) and $L_3$ (horizontal line) are parallel, not perpendicular. So, $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$, but $(L_1, L_3) \notin R$.

Therefore, the relation R is not transitive.

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Conclusion:

The relation R is symmetric because if $L_1$ is perpendicular to $L_2$, then $L_2$ is perpendicular to $L_1$.

The relation R is not reflexive because a line is not perpendicular to itself.

The relation R is not transitive because if $L_1$ is perpendicular to $L_2$ and $L_2$ is perpendicular to $L_3$, then $L_1$ is parallel to $L_3$, not perpendicular to $L_3$ (unless $L_1=L_3$).

Hence, R is symmetric but neither reflexive nor transitive.

Given:

Set A = {1, 2, 3}

Relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} on set A.

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To Show:

R is reflexive.

R is neither symmetric nor transitive.

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Solution:

We examine the properties of reflexivity, symmetry, and transitivity for the given relation R on the set A = {1, 2, 3}.

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1. Reflexivity:

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

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

From the definition of R, we have:

(1, 1) $\in$ R

(2, 2) $\in$ R

(3, 3) $\in$ R

Since $(a, a) \in R$ for all $a \in A = \{1, 2, 3\}$, the relation R is reflexive.

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2. Symmetry:

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

We need to check if for every pair $(a, b)$ in R, the reversed pair $(b, a)$ is also in R.

Consider the pair (1, 2) which is in R.

According to the definition of symmetry, the pair (2, 1) must also be in R for R to be symmetric.

However, the pair (2, 1) is not in the given relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

Therefore, the condition for symmetry is not satisfied.

Hence, R is not symmetric.

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3. Transitivity:

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

We need to check if for every case where $(a, b) \in R$ and $(b, c) \in R$, it follows that $(a, c) \in R$.

Consider the pairs (1, 2) and (2, 3) which are both in R.

Here, $a=1$, $b=2$, and $c=3$. We have $(1, 2) \in R$ and $(2, 3) \in R$.

According to the definition of transitivity, the pair $(a, c) = (1, 3)$ must also be in R for R to be transitive.

However, the pair (1, 3) is not in the given relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

Therefore, the condition for transitivity is not satisfied.

Hence, R is not transitive.

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Conclusion:

We have shown that the relation R on the set {1, 2, 3} is reflexive because (1, 1), (2, 2), and (3, 3) are in R.

We have shown that R is not symmetric because (1, 2) is in R but (2, 1) is not in R.

We have shown that R is not transitive because (1, 2) is in R and (2, 3) is in R, but (1, 3) is not in R.

Thus, R is reflexive but neither symmetric nor transitive.

Given:

Set Z = {all integers}

Relation R = {(a, b) : 2 divides a – b}, where $a, b \in Z$

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To Show:

R is an equivalence relation on Z.

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Solution:

To show that R is an equivalence relation, we must verify that R satisfies the following three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

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

According to the definition of R, $(a, a) \in R$ if 2 divides the difference $a - a$.

Consider the difference $a - a$.

$a - a = 0$

We know that 2 divides 0, since $0 = 2 \times 0$.

Thus, 2 divides $a - a$ for all $a \in Z$.

Therefore, $(a, a) \in R$ for all $a \in Z$.

Hence, R is reflexive.

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2. Symmetry:

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

Suppose $(a, b) \in R$. By the definition of R, this means that 2 divides $a - b$.

If 2 divides $a - b$, we can write $a - b = 2k$ for some integer $k$.

We need to check if $(b, a) \in R$, which means we need to check if 2 divides $b - a$.

Consider the difference $b - a$.

$b - a = -(a - b)$

Substitute the expression for $a - b$ from our assumption:

$b - a = -(2k) = 2(-k)$

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

So, the difference $b - a$ can be expressed as 2 times an integer.

This means that 2 divides $b - a$.

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

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

Hence, R is symmetric.

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3. Transitivity:

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

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

$(a, b) \in R$ means that 2 divides $a - b$. So, $a - b = 2k_1$ for some integer $k_1$.

$(b, c) \in R$ means that 2 divides $b - c$. So, $b - c = 2k_2$ for some integer $k_2$.

We need to check if $(a, c) \in R$, which means we need to check if 2 divides $a - c$.

Consider the difference $a - c$. We can write $a - c$ as the sum of $(a - b)$ and $(b - c)$.

$a - c = (a - b) + (b - c)$

Substitute the expressions for $(a - b)$ and $(b - c)$ from our assumptions:

$a - c = 2k_1 + 2k_2$

Factor out the common factor 2:

$a - c = 2(k_1 + k_2)$

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

So, the difference $a - c$ can be expressed as 2 times an integer.

This means that 2 divides $a - c$.

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

Thus, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Hence, R is transitive.

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Conclusion:

Since the relation R on the set of integers Z is reflexive, symmetric, and transitive, it is an equivalence relation.

Given:

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

Relation R = {(a, b) : both a and b are either odd or even}, where $a, b \in A$.

This condition "both a and b are either odd or even" is equivalent to saying that $a$ and $b$ have the same parity, which means their difference $a-b$ is an even number. Thus, R can also be defined as R = {(a, b) : $a-b$ is even}.

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To Show:

R is an equivalence relation on A.

All elements in {1, 3, 5, 7} are related to each other.

All elements in {2, 4, 6} are related to each other.

No element from {1, 3, 5, 7} is related to any element from {2, 4, 6}.

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Solution:

To show that R is an equivalence relation, we must verify that R satisfies the properties of reflexivity, symmetry, and transitivity.

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1. Reflexivity:

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

For any element $a \in A$, we consider the pair $(a, a)$. The condition for $(a, a) \in R$ is that both $a$ and $a$ are either odd or even. This is always true, as $a$ has the same parity as itself.

Alternatively, we check if $a-a$ is even. $a-a = 0$, and 0 is an even number (since $0 = 2 \times 0$).

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

Hence, R is reflexive.

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2. Symmetry:

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

Suppose $(a, b) \in R$. This means that $a$ and $b$ have the same parity (either both odd or both even).

If $a$ and $b$ have the same parity, then it is also true that $b$ and $a$ have the same parity.

Alternatively, suppose $(a, b) \in R$. This means $a-b$ is even. We can write $a-b = 2k$ for some integer $k$.

We need to check if $(b, a) \in R$, which means we check if $b-a$ is even.

$b-a = -(a-b) = -(2k) = 2(-k)$. Since $-k$ is an integer, $b-a$ is even.

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

Hence, R is symmetric.

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3. Transitivity:

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

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

$(a, b) \in R$ means $a$ and $b$ have the same parity.

$(b, c) \in R$ means $b$ and $c$ have the same parity.

If $a$ and $b$ have the same parity, and $b$ and $c$ have the same parity, it implies that $a$ and $c$ must also have the same parity.

(Case 1: a, b are odd, b, c are odd $\implies$ a, c are odd. Same parity.)

(Case 2: a, b are even, b, c are even $\implies$ a, c are even. Same parity.)

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

$(a, b) \in R \implies a-b$ is even, so $a-b = 2k_1$ for some integer $k_1$.

$(b, c) \in R \implies b-c$ is even, so $b-c = 2k_2$ for some integer $k_2$.

Consider the difference $a-c$. $a-c = (a-b) + (b-c) = 2k_1 + 2k_2 = 2(k_1 + k_2)$.

Since $k_1 + k_2$ is an integer, $a-c$ is even.

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

Therefore, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Hence, R is transitive.

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

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Now, consider the subset {1, 3, 5, 7}. These are all the odd numbers in set A.

Let $a, b$ be any two elements from {1, 3, 5, 7}. Both $a$ and $b$ are odd.

According to the definition of R, $(a, b) \in R$ if both $a$ and $b$ are odd or both are even. Since both are odd, the condition is satisfied.

Thus, any two elements in {1, 3, 5, 7} are related to each other under R. This confirms that all elements of the subset {1, 3, 5, 7} are related to each other.

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Next, consider the subset {2, 4, 6}. These are all the even numbers in set A.

Let $a, b$ be any two elements from {2, 4, 6}. Both $a$ and $b$ are even.

According to the definition of R, $(a, b) \in R$ if both $a$ and $b$ are odd or both are even. Since both are even, the condition is satisfied.

Thus, any two elements in {2, 4, 6} are related to each other under R. This confirms that all elements of the subset {2, 4, 6} are related to each other.

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Finally, consider an element $a$ from {1, 3, 5, 7} and an element $b$ from {2, 4, 6}.

Element $a$ is an odd number, and element $b$ is an even number.

According to the definition of R, $(a, b) \in R$ if both $a$ and $b$ are odd or both are even.

In this case, one is odd and the other is even, so they do not have the same parity.

Thus, $(a, b) \notin R$ for any $a \in \{1, 3, 5, 7\}$ and any $b \in \{2, 4, 6\}$.

This confirms that no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6}.

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Conclusion:

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

The elements in {1, 3, 5, 7} are all odd and thus related to each other.

The elements in {2, 4, 6} are all even and thus related to each other.

An odd number from {1, 3, 5, 7} cannot be related to an even number from {2, 4, 6} because they have different parity.

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

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Given set A = {1, 2, 3, ..., 14}.

Relation R = {(x, y) : $3x - y = 0$} = {(x, y) : $y = 3x$}, where $x, y \in A$.

Let's list the elements in R based on the condition $y=3x$ and $x, y \in A$:

• If $x=1$, $y = 3 \times 1 = 3$. So, (1, 3) $\in$ R.
• If $x=2$, $y = 3 \times 2 = 6$. So, (2, 6) $\in$ R.
• If $x=3$, $y = 3 \times 3 = 9$. So, (3, 9) $\in$ R.
• If $x=4$, $y = 3 \times 4 = 12$. So, (4, 12) $\in$ R.
• If $x=5$, $y = 3 \times 5 = 15$. But $15 \notin A$.

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

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Check for Reflexivity:

For R to be reflexive, $(x, x)$ must be in R for every $x \in A$. This means $3x - x = 0$, which implies $2x = 0$, or $x=0$.

However, $0 \notin A$. For any element $x \in A$, $x \neq 0$, so $2x \neq 0$.

For example, for $1 \in A$, we check if $(1, 1) \in R$. The condition is $3(1) - 1 = 0$, which is $3 - 1 = 0$, or $2 = 0$. This is false.

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

Hence, R is not reflexive.

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Check for Symmetry:

For R to be symmetric, if $(x, y) \in R$, then $(y, x)$ must be in R for all $x, y \in A$.

Suppose $(x, y) \in R$. This means $3x - y = 0$, or $y = 3x$.

For $(y, x)$ to be in R, we must have $3y - x = 0$, or $x = 3y$.

Let's take an element from R, for instance, (1, 3) $\in$ R.

We check if $(3, 1) \in R$. The condition for $(3, 1) \in R$ is $3(3) - 1 = 0$, which is $9 - 1 = 0$, or $8 = 0$. This is false.

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

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Check for Transitivity:

For R to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z)$ must be in R for all $x, y, z \in A$.

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

$(x, y) \in R \implies 3x - y = 0 \implies y = 3x$.

$(y, z) \in R \implies 3y - z = 0 \implies z = 3y$.

Substitute $y = 3x$ into the second equation: $z = 3(3x) = 9x$.

For $(x, z)$ to be in R, we must have $3x - z = 0$, or $z = 3x$.

We need to check if $9x = 3x$ always holds when $(x,y) \in R$ and $(y,z) \in R$. This implies $6x=0$, so $x=0$, which is not in A.

Let's take specific elements from R.

We have (1, 3) $\in$ R and (3, 9) $\in$ R.

Here, $x=1$, $y=3$, $z=9$. Both (1, 3) and (3, 9) are in R.

According to the definition of transitivity, $(1, 9)$ must be in R.

We check if $(1, 9) \in R$. The condition is $3(1) - 9 = 0$, which is $3 - 9 = 0$, or $-6 = 0$. This is false.

Since (1, 3) $\in$ R and (3, 9) $\in$ R, but (1, 9) $\notin$ R, the relation 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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(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}

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Given set N = {1, 2, 3, ...}.

Relation R = {(x, y) : $y = x + 5$ and $x < 4$}, where $x, y \in N$.

Since $x \in N$ and $x < 4$, the possible values for $x$ are {1, 2, 3}.

Let's list the elements in R:

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

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

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Check for Reflexivity:

For R to be reflexive, $(x, x)$ must be in R for every $x \in N$. This means $x = x+5$ and $x < 4$.

$x = x+5$ implies $0 = 5$, which is false for all $x$.

Thus, $(x, x) \notin R$ for any $x \in N$.

Hence, R is not reflexive.

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Check for Symmetry:

For R to be symmetric, if $(x, y) \in R$, then $(y, x)$ must be in R for all $x, y \in N$.

Let's take an element from R, for instance, (1, 6) $\in$ R.

We check if $(6, 1) \in R$. The condition for $(6, 1) \in R$ is $1 = 6 + 5$ and $6 < 4$. Both conditions $1 = 11$ and $6 < 4$ are false.

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

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Check for Transitivity:

For R to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z)$ must be in R for all $x, y, z \in N$.

We examine the elements in R: {(1, 6), (2, 7), (3, 8)}.

We look for pairs $(x, y)$ and $(y, z)$ where the second element of the first pair is the same as the first element of the second pair.

The second elements in R are 6, 7, 8.

The first elements allowed in R must be less than 4 (i.e., 1, 2, or 3).

Since the second elements (6, 7, 8) are not among the allowed first elements (1, 2, 3), it is impossible to find two pairs $(x, y) \in R$ and $(y, z) \in R$ such that the 'y' values match.

The premise "if $(x, y) \in R$ and $(y, z) \in R$" is never true for the relation R.

In logic, an implication "if P then Q" is considered true if the premise P is false.

Therefore, the condition for transitivity is vacuously satisfied.

Hence, R is transitive.

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

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

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Given set A = {1, 2, 3, 4, 5, 6}.

Relation R = {(x, y) : y is divisible by x}, where $x, y \in A$. This means $x | y$.

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Check for Reflexivity:

For R to be reflexive, $(x, x)$ must be in R for every $x \in A$. This means $x$ must be divisible by $x$.

For any integer $x \ne 0$, $x$ is divisible by $x$ (since $x = 1 \times x$).

Since all elements in A are non-zero, $x | x$ for all $x \in A$.

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

Hence, R is reflexive.

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Check for Symmetry:

For R to be symmetric, if $(x, y) \in R$, then $(y, x)$ must be in R for all $x, y \in A$.

Suppose $(x, y) \in R$. This means $y$ is divisible by $x$. We can write $y = kx$ for some integer $k$.

For $(y, x)$ to be in R, we must have $x$ divisible by $y$. We can write $x = my$ for some integer $m$.

Let's take an example. Consider $x=2, y=4$. $(2, 4) \in R$ because 4 is divisible by 2 ($4 = 2 \times 2$).

We check if $(4, 2) \in R$. The condition for $(4, 2) \in R$ is that 2 is divisible by 4. This is false.

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

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Check for Transitivity:

For R to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z)$ must be in R for all $x, y, z \in A$.

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

$(x, y) \in R \implies y$ is divisible by $x$. So, $y = k_1 x$ for some integer $k_1$.

$(y, z) \in R \implies z$ is divisible by $y$. So, $z = k_2 y$ for some integer $k_2$.

Substitute the first equation into the second: $z = k_2 (k_1 x) = (k_1 k_2) x$.

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

The equation $z = (k_1 k_2) x$ means that $z$ is divisible by $x$.

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

Therefore, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.

Hence, R is transitive.

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

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

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Given set Z = {..., -2, -1, 0, 1, 2, ...}.

Relation R = {(x, y) : $x - y$ is an integer}, where $x, y \in Z$.

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Check for Reflexivity:

For R to be reflexive, $(x, x)$ must be in R for every $x \in Z$. This means $x - x$ must be an integer.

$x - x = 0$. We know that 0 is an integer.

Thus, $x - x$ is an integer for all $x \in Z$.

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

Hence, R is reflexive.

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Check for Symmetry:

For R to be symmetric, if $(x, y) \in R$, then $(y, x)$ must be in R for all $x, y \in Z$.

Suppose $(x, y) \in R$. This means $x - y$ is an integer.

We need to check if $(y, x) \in R$, which means $y - x$ must be an integer.

Consider $y - x$. $y - x = -(x - y)$.

Since $x - y$ is an integer, its negative, $-(x - y)$, is also an integer.

Thus, $y - x$ is an integer.

Therefore, if $(x, y) \in R$, then $(y, x) \in R$.

Hence, R is symmetric.

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Check for Transitivity:

For R to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z)$ must be in R for all $x, y, z \in Z$.

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

$(x, y) \in R \implies x - y$ is an integer.

$(y, z) \in R \implies y - z$ is an integer.

We need to check if $(x, z) \in R$, which means $x - z$ must be an integer.

Consider $x - z$. We can write $x - z = (x - y) + (y - z)$.

Since $(x - y)$ is an integer and $(y - z)$ is an integer, their sum $(x - y) + (y - z)$ is also an integer.

Thus, $x - z$ is an integer.

Therefore, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.

Hence, 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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(v) Relation R in the set A of human beings in a town at a particular time

Let A be the set of all 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}

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Check for Reflexivity:

For any person $x \in A$, is $(x, x) \in R$? This means $x$ works at the same place as $x$. This is true.

Hence, R is reflexive.

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Check for Symmetry:

If $(x, y) \in R$, is $(y, x) \in R$? If $x$ and $y$ work at the same place, then $y$ and $x$ also work at the same place. This is true.

Hence, R is symmetric.

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Check for Transitivity:

If $(x, y) \in R$ and $(y, z) \in R$, is $(x, z) \in R$? If $x$ and $y$ work at the same place, and $y$ and $z$ work at the same place, then $x$ and $z$ must work at the same place. This is true.

Hence, R is transitive.

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Conclusion for (va): The relation R is reflexive, symmetric, and transitive.

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

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Check for Reflexivity:

For any person $x \in A$, is $(x, x) \in R$? This means $x$ lives in the same locality as $x$. This is true.

Hence, R is reflexive.

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Check for Symmetry:

If $(x, y) \in R$, is $(y, x) \in R$? If $x$ and $y$ live in the same locality, then $y$ and $x$ also live in the same locality. This is true.

Hence, R is symmetric.

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Check for Transitivity:

If $(x, y) \in R$ and $(y, z) \in R$, is $(x, z) \in R$? If $x$ and $y$ live in the same locality, and $y$ and $z$ live in the same locality, then $x$ and $z$ must live in the same locality. This is true.

Hence, R is transitive.

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Conclusion for (vb): The relation R is reflexive, symmetric, and transitive.

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

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Let $h(p)$ denote the height of person $p$. The relation is $(x, y) \in R$ if $h(x) = h(y) + 7$ cm.

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Check for Reflexivity:

For any person $x \in A$, is $(x, x) \in R$? This means $x$ is exactly 7 cm taller than $x$. $h(x) = h(x) + 7$ cm, which implies $0 = 7$ cm. This is false.

Hence, R is not reflexive.

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Check for Symmetry:

If $(x, y) \in R$, is $(y, x) \in R$? Suppose $(x, y) \in R$. This means $h(x) = h(y) + 7$ cm.

For $(y, x) \in R$, we must have $h(y) = h(x) + 7$ cm.

Substitute the first equation into the second: $h(y) = (h(y) + 7 \text{ cm}) + 7 \text{ cm} = h(y) + 14 \text{ cm}$. This implies $0 = 14$ cm, which is false.

Alternatively, if $x$ is exactly 7 cm taller than $y$, then $y$ is exactly 7 cm shorter than $x$, not 7 cm taller than $x$.

Hence, R is not symmetric.

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Check for Transitivity:

If $(x, y) \in R$ and $(y, z) \in R$, is $(x, z) \in R$? Suppose $(x, y) \in R$ and $(y, z) \in R$.

$(x, y) \in R \implies h(x) = h(y) + 7$ cm.

$(y, z) \in R \implies h(y) = h(z) + 7$ cm.

Substitute the second equation into the first: $h(x) = (h(z) + 7 \text{ cm}) + 7 \text{ cm} = h(z) + 14 \text{ cm}$.

For $(x, z) \in R$, we require $h(x) = h(z) + 7$ cm. However, we found $h(x) = h(z) + 14$ cm.

Since $14 \neq 7$, $(x, z) \notin R$ if $(x, y) \in R$ and $(y, z) \in R$.

Hence, R is not transitive.

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

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

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Check for Reflexivity:

For any person $x \in A$, is $(x, x) \in R$? This means $x$ is the wife of $x$. This is false.

Hence, R is not reflexive.

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Check for Symmetry:

If $(x, y) \in R$, is $(y, x) \in R$? Suppose $(x, y) \in R$. This means $x$ is the wife of $y$. This implies $x$ is female and $y$ is male.

For $(y, x) \in R$, $y$ must be the wife of $x$. This implies $y$ is female and $x$ is male.

This contradicts the assumption that $y$ is male and $x$ is female. So, $(y, x) \notin R$ if $(x, y) \in R$.

Hence, R is not symmetric.

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Check for Transitivity:

If $(x, y) \in R$ and $(y, z) \in R$, is $(x, z) \in R$? Suppose $(x, y) \in R$ and $(y, z) \in R$.

$(x, y) \in R \implies x$ is wife of $y$. This means $y$ is the husband of $x$ (assuming a traditional sense where wife implies husband). So, $y$ is male.

$(y, z) \in R \implies y$ is wife of $z$. This means $y$ is female.

A person cannot be both male and female simultaneously. Thus, the premise "$(x, y) \in R$ and $(y, z) \in R$" is never satisfied.

Therefore, the condition for transitivity is vacuously satisfied.

Hence, R is transitive.

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

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

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Check for Reflexivity:

For any person $x \in A$, is $(x, x) \in R$? This means $x$ is the father of $x$. This is false.

Hence, R is not reflexive.

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Check for Symmetry:

If $(x, y) \in R$, is $(y, x) \in R$? Suppose $(x, y) \in R$. This means $x$ is the father of $y$. Then $y$ is the child of $x$. $y$ cannot be the father of $x$.

Hence, R is not symmetric.

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Check for Transitivity:

If $(x, y) \in R$ and $(y, z) \in R$, is $(x, z) \in R$? Suppose $(x, y) \in R$ and $(y, z) \in R$.

$(x, y) \in R \implies x$ is the father of $y$.

$(y, z) \in R \implies y$ 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 $z$.

Thus, $(x, z) \notin R$ if $(x, y) \in R$ and $(y, z) \in R$.

Hence, R is not transitive.

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

Given:

Set $\mathbb{R}$ = {all real numbers}

Relation R = {(a, b) : a $\leq$ b$^2$}, where $a, b \in \mathbb{R}$.

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To Show:

R is neither reflexive nor symmetric nor transitive.

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Solution:

To show that R is neither reflexive, symmetric, nor transitive, we need to demonstrate that at least one condition for each property is not met by providing counterexamples.

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1. Reflexivity:

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

For R to be reflexive on $\mathbb{R}$, we must have $(a, a) \in R$ for every $a \in \mathbb{R}$. This means $a \leq a^2$ must be true for all real numbers $a$.

Consider a real number $a$ between 0 and 1, for example, $a = 0.5$.

We check if $(0.5, 0.5) \in R$. This requires $0.5 \leq (0.5)^2$.

Calculating the square: $(0.5)^2 = 0.25$.

The condition becomes $0.5 \leq 0.25$. This is false.

Since we found a real number $0.5$ for which $(0.5, 0.5) \notin R$, the relation R is not reflexive.

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2. Symmetry:

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

For R to be symmetric on $\mathbb{R}$, we must show that if $a \leq b^2$, then $b \leq a^2$ for all $a, b \in \mathbb{R}$.

Let's take $a = 1$ and $b = 2$.

Check if $(1, 2) \in R$: Is $1 \leq 2^2$? $1 \leq 4$. Yes, this is true. So, $(1, 2) \in R$.

Now, check if $(2, 1) \in R$. This requires $2 \leq 1^2$.

Calculating the square: $1^2 = 1$.

The condition becomes $2 \leq 1$. This is false.

Since we found a pair $(1, 2) \in R$ such that $(2, 1) \notin R$, the relation R is not symmetric.

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3. Transitivity:

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

For R to be transitive on $\mathbb{R}$, we must show that if $a \leq b^2$ and $b \leq c^2$, then $a \leq c^2$ for all $a, b, c \in \mathbb{R}$.

Let's try to find a counterexample. We need $(a, b) \in R$ and $(b, c) \in R$, but $(a, c) \notin R$. This means $a \leq b^2$, $b \leq c^2$, but $a > c^2$.

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

Check if $(a, b) = (3, -2) \in R$: Is $3 \leq (-2)^2$? $3 \leq 4$. Yes, this is true. So, $(3, -2) \in R$.

Check if $(b, c) = (-2, 1.5) \in R$: Is $-2 \leq (1.5)^2$? $-2 \leq 2.25$. Yes, this is true. So, $(-2, 1.5) \in R$.

Now, check if $(a, c) = (3, 1.5) \in R$. This requires $3 \leq (1.5)^2$.

Calculating the square: $(1.5)^2 = 2.25$.

The condition becomes $3 \leq 2.25$. This is false.

Since we found $a=3, b=-2, c=1.5$ such that $(3, -2) \in R$ and $(-2, 1.5) \in R$, but $(3, 1.5) \notin R$, the relation R is not transitive.

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Conclusion:

The relation R = {(a, b) : a $\leq$ b$^2$} on the set of real numbers is not reflexive (e.g., $0.5 \not\leq (0.5)^2$), not symmetric (e.g., $1 \leq 2^2$ but $2 \not\leq 1^2$), and not transitive (e.g., $3 \leq (-2)^2$ and $-2 \leq (1.5)^2$ but $3 \not\leq (1.5)^2$).

Therefore, R is neither reflexive nor symmetric nor transitive.

Given:

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

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

Let's list the elements in R based on the condition $b = a + 1$ and $a, b \in A$:

• 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 $a=6$, $b = 6+1 = 7$. But $7 \notin A$.

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

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Check for Reflexivity:

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

For R to be reflexive, we must have $(a, a) \in R$ for all $a \in \{1, 2, 3, 4, 5, 6\}$. This means $a = a + 1$.

$a = a + 1$ implies $0 = 1$, which is false for all values of $a$.

For example, for $1 \in A$, we check if $(1, 1) \in R$. The condition is $1 = 1 + 1$, or $1 = 2$. This is false.

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

Hence, R is not reflexive.

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Check for Symmetry:

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

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

For $(b, a)$ to be in R, we must have $a = b + 1$.

Let's take an element from R, for instance, (1, 2) $\in$ R.

We check if $(2, 1) \in R$. 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, the relation R is not symmetric.

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Check for Transitivity:

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

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

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

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

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

For $(a, c)$ to be in R, we must have $c = a + 1$.

We need to check if $a + 2 = a + 1$ always holds when $(a, b) \in R$ and $(b, c) \in R$. This implies $2=1$, which is false.

Let's take specific elements from R.

We have (1, 2) $\in$ R and (2, 3) $\in$ R.

Here, $a=1$, $b=2$, $c=3$. Both (1, 2) and (2, 3) are in R.

According to the definition of transitivity, $(1, 3)$ must be in R.

We check if $(1, 3) \in R$. The condition 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, the relation R is not transitive.

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Conclusion:

The relation R in the set {1, 2, 3, 4, 5, 6} defined by R = {(a, b) : b = a + 1} is neither reflexive, nor symmetric, nor transitive.

Given:

Set $\mathbb{R}$ = {all real numbers}

Relation R = {(a, b) : a $\leq$ b}, where $a, b \in \mathbb{R}$.

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To Show:

R is reflexive.

R is transitive.

R is not symmetric.

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Solution:

We examine the properties of reflexivity, symmetry, and transitivity for the given relation R on the set of real numbers $\mathbb{R}$.

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1. Reflexivity:

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

For R to be reflexive on $\mathbb{R}$, we must have $(a, a) \in R$ for every $a \in \mathbb{R}$. According to the definition of R, this means $a \leq a$ must be true for all real numbers $a$.

The statement $a \leq a$ means $a$ is less than or equal to $a$. This is always true for any real number $a$ because $a$ is equal to $a$.

Thus, $a \leq a$ for all $a \in \mathbb{R}$.

Therefore, $(a, a) \in R$ for all $a \in \mathbb{R}$.

Hence, R is reflexive.

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2. Symmetry:

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

For R to be symmetric on $\mathbb{R}$, we must show that if $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in \mathbb{R}$. This means if $a \leq b$, then $b \leq a$.

Let's test with specific real numbers. Consider $a = 1$ and $b = 2$.

We check if $(1, 2) \in R$: Is $1 \leq 2$? Yes, this is true. So, $(1, 2) \in R$.

Now, check if $(2, 1) \in R$: Is $2 \leq 1$? This is false.

Since we found a pair $(1, 2) \in R$ such that $(2, 1) \notin R$, the relation R is not symmetric.

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3. Transitivity:

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

For R to be transitive on $\mathbb{R}$, we must show that if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in \mathbb{R}$.

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

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

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

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

From the properties of real numbers, if $a \leq b$ and $b \leq c$, it logically follows that $a \leq c$.

Thus, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Hence, R is transitive.

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Conclusion:

The relation R = {(a, b) : a $\leq$ b} on the set of real numbers $\mathbb{R}$ is reflexive because $a \leq a$ is always true, and transitive because if $a \leq b$ and $b \leq c$, then $a \leq c$. However, it is not symmetric because $a \leq b$ does not imply $b \leq a$ (e.g., $1 \leq 2$ but $2 \not\leq 1$).

Therefore, R is reflexive and transitive but not symmetric.

Given:

Set $\mathbb{R}$ = {all real numbers}

Relation R = {(a, b) : a $\leq$ b$^3$}, where $a, b \in \mathbb{R}$.

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To Check:

Whether R is reflexive, symmetric, or transitive.

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Solution:

We examine the properties of reflexivity, symmetry, and transitivity for the given relation R on the set of real numbers $\mathbb{R}$.

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1. Reflexivity:

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

For R to be reflexive on $\mathbb{R}$, we must have $(a, a) \in R$ for every $a \in \mathbb{R}$. This means $a \leq a^3$ must be true for all real numbers $a$.

Consider a real number $a = 0.5$.

We check if $(0.5, 0.5) \in R$. This requires $0.5 \leq (0.5)^3$.

$(0.5)^3 = 0.5 \times 0.5 \times 0.5 = 0.25 \times 0.5 = 0.125$.

The condition is $0.5 \leq 0.125$, which is false.

Consider another real number $a = -2$.

We check if $(-2, -2) \in R$. This requires $-2 \leq (-2)^3$.

$(-2)^3 = (-2) \times (-2) \times (-2) = 4 \times (-2) = -8$.

The condition is $-2 \leq -8$, which is false.

Since we found real numbers ($0.5$ and $-2$) for which $a \not\leq a^3$, the relation R is not reflexive.

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2. Symmetry:

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

For R to be symmetric on $\mathbb{R}$, we must show that if $a \leq b^3$, then $b \leq a^3$ for all $a, b \in \mathbb{R}$.

Let's try to find a counterexample. We need $a, b$ such that $a \leq b^3$ is true, but $b \leq a^3$ is false.

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

Check if $(1, 2) \in R$: Is $1 \leq 2^3$? $1 \leq 8$. Yes, this is true.

Now, check if $(2, 1) \in R$. This requires $2 \leq 1^3$.

$1^3 = 1 \times 1 \times 1 = 1$.

The condition is $2 \leq 1$, which is false.

Since we found a pair $(1, 2) \in R$ such that $(2, 1) \notin R$, the relation R is not symmetric.

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3. Transitivity:

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

For R to be transitive on $\mathbb{R}$, we must show that if $a \leq b^3$ and $b \leq c^3$, then $a \leq c^3$ for all $a, b, c \in \mathbb{R}$.

Let's try to find a counterexample. We need $a, b, c$ such that $(a, b) \in R$ and $(b, c) \in R$, but $(a, c) \notin R$. This means $a \leq b^3$, $b \leq c^3$, but $a > c^3$.

Consider $a = 3$, $b = 1.5$, and $c = 1.2$.

Check if $(a, b) = (3, 1.5) \in R$: Is $3 \leq (1.5)^3$? $1.5^3 = 3.375$. Is $3 \leq 3.375$? Yes, this is true. So, $(3, 1.5) \in R$.

Check if $(b, c) = (1.5, 1.2) \in R$: Is $1.5 \leq (1.2)^3$? $1.2^3 = 1.728$. Is $1.5 \leq 1.728$? Yes, this is true. So, $(1.5, 1.2) \in R$.

Now, check if $(a, c) = (3, 1.2) \in R$. This requires $3 \leq (1.2)^3$.

$(1.2)^3 = 1.728$.

The condition is $3 \leq 1.728$, which is false.

Since we found $a=3, b=1.5, c=1.2$ such that $(3, 1.5) \in R$ and $(1.5, 1.2) \in R$, but $(3, 1.2) \notin R$, the relation R is not transitive.

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Conclusion:

The relation R = {(a, b) : a $\leq$ b$^3$} on the set of real numbers $\mathbb{R}$ is not reflexive (e.g., $0.5 \not\leq (0.5)^3$), not symmetric (e.g., $1 \leq 2^3$ but $2 \not\leq 1^3$), and not transitive (e.g., $3 \leq (1.5)^3$ and $1.5 \leq (1.2)^3$ but $3 \not\leq (1.2)^3$).

Therefore, R is neither reflexive nor symmetric nor transitive.

Given:

Set A = {1, 2, 3}

Relation R = {(1, 2), (2, 1)} on set A.

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To Show:

R is symmetric.

R is neither reflexive nor transitive.

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Solution:

We examine the properties of reflexivity, symmetry, and transitivity for the given relation R on the set A = {1, 2, 3}.

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1. Reflexivity:

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

The set A is {1, 2, 3}. For R to be reflexive, the pairs (1, 1), (2, 2), and (3, 3) must all be in R.

Looking at the given relation R = {(1, 2), (2, 1)}, we see that none of the pairs (1, 1), (2, 2), or (3, 3) are present in R.

For instance, $(1, 1) \notin R$.

Thus, the condition $(a, a) \in R$ is not satisfied for all $a \in A$.

Hence, R is not reflexive.

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2. Symmetry:

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

We need to check if for every pair $(a, b)$ in R, the reversed pair $(b, a)$ is also in R.

Consider the pairs in R:

• Take the pair (1, 2) $\in$ R. The reversed pair is (2, 1). Is (2, 1) $\in$ R? Yes, it is.
• Take the pair (2, 1) $\in$ R. The reversed pair is (1, 2). Is (1, 2) $\in$ R? Yes, it is.

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

Therefore, the relation R is symmetric.

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3. Transitivity:

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

We need to check if for every case where $(a, b) \in R$ and $(b, c) \in R$, it follows that $(a, c) \in R$.

Consider the pairs in R: {(1, 2), (2, 1)}.

We look for paths of length 2:

• Take $(a, b) = (1, 2) \in R$ and $(b, c) = (2, 1) \in R$. Here $a=1, b=2, c=1$. Both conditions $(1, 2) \in R$ and $(2, 1) \in R$ are true. According to transitivity, $(a, c) = (1, 1)$ must be in R. However, $(1, 1) \notin R$.
• Take $(a, b) = (2, 1) \in R$ and $(b, c) = (1, 2) \in R$. Here $a=2, b=1, c=2$. Both conditions $(2, 1) \in R$ and $(1, 2) \in R$ are true. According to transitivity, $(a, c) = (2, 2)$ must be in R. However, $(2, 2) \notin R$.

Since we found cases where $(a, b) \in R$ and $(b, c) \in R$, but $(a, c) \notin R$, the relation R is not transitive.

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Conclusion:

The relation R on the set {1, 2, 3} is not reflexive because (1, 1), (2, 2), (3, 3) are not in R. The relation R is symmetric because for every pair $(a, b)$ in R, $(b, a)$ is also in R. The relation R is not transitive because $(1, 2) \in R$ and $(2, 1) \in R$, but $(1, 1) \notin R$ (and similarly $(2, 1) \in R$ and $(1, 2) \in R$, but $(2, 2) \notin R$).

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

Given:

Set A = {all books in a library of a college}

Relation 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 on A.

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Solution:

To show that R is an equivalence relation, we must verify that R satisfies the following three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

Let $N(x)$ denote the number of pages in book $x$. The relation R can be written as R = {(x, y) : $N(x) = N(y)$}.

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1. Reflexivity:

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

For R to be reflexive, we must have $(x, x) \in R$ for every book $x$ in the library.

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

This is always true; any book has the same number of pages as itself.

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

Hence, R is reflexive.

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2. Symmetry:

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

Suppose $(x, y) \in R$. By the definition of R, this means that book $x$ and book $y$ have the same number of pages, i.e., $N(x) = N(y)$.

For $(y, x)$ to be in R, we must have that book $y$ and book $x$ have the same number of pages, i.e., $N(y) = N(x)$.

If $N(x) = N(y)$, then it is also true that $N(y) = N(x)$.

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

Hence, R is symmetric.

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3. Transitivity:

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

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

$(x, y) \in R$ means that book $x$ and book $y$ have the same number of pages, i.e., $N(x) = N(y)$.

$(y, z) \in R$ means that book $y$ and book $z$ have the same number of pages, i.e., $N(y) = N(z)$.

We need to check if $(x, z) \in R$, which means that book $x$ and book $z$ have the same number of pages, i.e., $N(x) = N(z)$.

If $N(x) = N(y)$ and $N(y) = N(z)$, by the transitive property of equality, it follows that $N(x) = N(z)$.

Thus, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.

Hence, R is transitive.

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Conclusion:

Since the relation R on the set A of all books in the library is reflexive, symmetric, and transitive, it is an equivalence relation.

Given:

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

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

Note that $|a - b|$ is even if and only if $a - b$ is even. This happens if and only if $a$ and $b$ have the same parity (both odd or both even).

So, the relation R is equivalent to R = {(a, b) : a and b have the same parity}.

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To Show:

R is an equivalence relation on A.

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

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

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

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Solution:

To show that R is an equivalence relation, we must verify that R satisfies the properties of reflexivity, symmetry, and transitivity.

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1. Reflexivity:

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

For any element $a \in A$, we consider the pair $(a, a)$. The condition for $(a, a) \in R$ is that $|a - a|$ is even.

Consider the difference $|a - a| = |0| = 0$. We know that 0 is an even number (since $0 = 2 \times 0$).

Thus, $|a - a|$ is even for all $a \in A$.

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

Hence, R is reflexive.

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2. Symmetry:

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

Suppose $(a, b) \in R$. This means that $|a - b|$ is even.

We need to check if $(b, a) \in R$, which means we need to check if $|b - a|$ is even.

Consider the difference $|b - a|$. We know that $|b - a| = |-(a - b)| = |-1 \times (a - b)| = |-1| \times |a - b| = 1 \times |a - b| = |a - b|$.

Since $|a - b|$ is even (by assumption that $(a, b) \in R$), $|b - a|$ is also even.

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

Hence, R is symmetric.

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3. Transitivity:

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

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

$(a, b) \in R$ means $|a - b|$ is even. This implies $a - b$ is even. So, $a - b = 2k_1$ for some integer $k_1$.

$(b, c) \in R$ means $|b - c|$ is even. This implies $b - c$ is even. So, $b - c = 2k_2$ for some integer $k_2$.

We need to check if $(a, c) \in R$, which means $|a - c|$ must be even (or $a-c$ must be even).

Consider the difference $a - c$. We can write $a - c = (a - b) + (b - c)$.

Substitute the expressions for $(a - b)$ and $(b - c)$ from our assumptions:

$a - c = 2k_1 + 2k_2$

Factor out the common factor 2:

$a - c = 2(k_1 + k_2)$

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

So, the difference $a - c$ can be expressed as 2 times an integer, which means $a - c$ is even.

If $a-c$ is even, then $|a-c|$ is also even.

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

Therefore, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Hence, R is transitive.

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

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Now, consider the subset {1, 3, 5}. These are all the odd numbers in set A.

Let $a, b$ be any two elements from {1, 3, 5}. Both $a$ and $b$ are odd numbers.

The difference between two odd numbers is always even (e.g., $5 - 3 = 2$, $3 - 1 = 2$, $|1 - 5| = 4$).

Since $|a - b|$ is even for any $a, b \in \{1, 3, 5\}$, $(a, b) \in R$ for all $a, b \in \{1, 3, 5\}$.

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

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Next, consider the subset {2, 4}. These are all the even numbers in set A.

Let $a, b$ be any two elements from {2, 4}. Both $a$ and $b$ are even numbers.

The difference between two even numbers is always even (e.g., $4 - 2 = 2$, $|2 - 4| = 2$).

Since $|a - b|$ is even for any $a, b \in \{2, 4\}$, $(a, b) \in R$ for all $a, b \in \{2, 4\}$.

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

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Finally, consider an element $a$ from {1, 3, 5} and an element $b$ from {2, 4}.

Element $a$ is an odd number, and element $b$ is an even number.

The difference between an odd number and an even number is always odd (e.g., $3 - 2 = 1$, $4 - 1 = 3$, $|1 - 4| = 3$).

Since $|a - b|$ is odd for any $a \in \{1, 3, 5\}$ and any $b \in \{2, 4\}$, $|a - b|$ is not even.

Thus, $(a, b) \notin R$ for any $a \in \{1, 3, 5\}$ and any $b \in \{2, 4\}$.

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

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Conclusion:

The relation R = {(a, b) : |a – b| is even} on the set {1, 2, 3, 4, 5} is an equivalence relation.

The set {1, 3, 5} forms an equivalence class where all elements are related (have the same parity - odd).

The set {2, 4} forms another equivalence class where all elements are related (have the same parity - even).

Elements from different equivalence classes are not related.

Given:

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

Relation (i) R$_1$ = {(a, b) : $|a – b|$ is a multiple of 4}, where $a, b \in A$.

Relation (ii) R$_2$ = {(a, b) : a = b}, where $a, b \in A$.

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To Show:

R$_1$ is an equivalence relation.

R$_2$ is an equivalence relation.

Find the set of all elements related to 1 for R$_1$ and R$_2$.

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Solution for R$_1$:

Relation R$_1$ = {(a, b) : $|a – b|$ is a multiple of 4}. This means $|a - b| = 4k$ for some non-negative integer $k$. Equivalently, $a - b$ is a multiple of 4, which means $a \equiv b \pmod{4}$.

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Check for Reflexivity (R$_1$):

For any element $a \in A$, we check if $(a, a) \in R_1$. This means $|a - a|$ must be a multiple of 4.

$|a - a| = |0| = 0$. We know that 0 is a multiple of 4 (since $0 = 4 \times 0$).

Thus, $|a - a|$ is a multiple of 4 for all $a \in A$.

Therefore, $(a, a) \in R_1$ for all $a \in A$.

Hence, R$_1$ is reflexive.

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Check for Symmetry (R$_1$):

Suppose $(a, b) \in R_1$. This means $|a - b|$ is a multiple of 4. So, $|a - b| = 4k$ for some non-negative integer $k$. This implies $a - b = 4k$ or $a - b = -4k$. In either case, $a - b$ is a multiple of 4.

We need to check if $(b, a) \in R_1$, which means $|b - a|$ must be a multiple of 4.

$|b - a| = |-(a - b)| = |a - b|$.

Since $|a - b|$ is a multiple of 4 (by assumption), $|b - a|$ is also a multiple of 4.

Thus, if $(a, b) \in R_1$, then $(b, a) \in R_1$.

Hence, R$_1$ is symmetric.

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Check for Transitivity (R$_1$):

Suppose $(a, b) \in R_1$ and $(b, c) \in R_1$.

$(a, b) \in R_1$ means $|a - b|$ is a multiple of 4. This implies $a - b$ is a multiple of 4. So, $a - b = 4k_1$ for some integer $k_1$.

$(b, c) \in R_1$ means $|b - c|$ is a multiple of 4. This implies $b - c$ is a multiple of 4. So, $b - c = 4k_2$ for some integer $k_2$.

We need to check if $(a, c) \in R_1$, which means $|a - c|$ must be a multiple of 4 (or $a-c$ must be a multiple of 4).

Consider the difference $a - c$. We can write $a - c = (a - b) + (b - c)$.

Substitute the expressions for $(a - b)$ and $(b - c)$:

$a - c = 4k_1 + 4k_2 = 4(k_1 + k_2)$.

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

So, $a - c$ is a multiple of 4. This implies $|a - c|$ is a multiple of 4.

Thus, $(a, c) \in R_1$.

Therefore, if $(a, b) \in R_1$ and $(b, c) \in R_1$, then $(a, c) \in R_1$.

Hence, R$_1$ is transitive.

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Since R$_1$ is reflexive, symmetric, and transitive, R$_1$ is an equivalence relation.

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Find the set of all elements related to 1 in R$_1$:

An element $x \in A$ is related to 1 if $(x, 1) \in R_1$. This means $|x - 1|$ is a multiple of 4.

So, $|x - 1|$ can be 0, 4, 8, 12, 16, ...

This means $x - 1$ can be 0, 4, -4, 8, -8, 12, -12, ...

So, $x$ can be $1 + 0$, $1 + 4$, $1 - 4$, $1 + 8$, $1 - 8$, $1 + 12$, $1 - 12$, ...

$x$ can be 1, 5, -3, 9, -7, 13, -11, ...

We need to find the elements $x$ in set A = {0, 1, 2, ..., 12} that are in this list.

• If $x = 1$, $|1 - 1| = 0$, which is a multiple of 4. $1 \in A$. So, 1 is related to 1.
• If $x = 5$, $|5 - 1| = 4$, which is a multiple of 4. $5 \in A$. So, 5 is related to 1.
• If $x = 9$, $|9 - 1| = 8$, which is a multiple of 4. $9 \in A$. So, 9 is related to 1.
• If $x = 13$, $|13 - 1| = 12$, which is a multiple of 4. $13 \notin A$.
• If $x = -3$, $|-3 - 1| = |-4| = 4$, which is a multiple of 4. $-3 \notin A$.
• If $x = -7$, $|-7 - 1| = |-8| = 8$, which is a multiple of 4. $-7 \notin A$.
• If $x = -11$, $|-11 - 1| = |-12| = 12$, which is a multiple of 4. $-11 \notin A$.

The elements in A related to 1 are those whose difference with 1 is a multiple of 4. These are elements $x \in A$ such that $x \equiv 1 \pmod{4}$.

The elements in A are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

Checking each element:

• $0 - 1 = -1$ (not mult of 4)
• $1 - 1 = 0$ (mult of 4)
• $2 - 1 = 1$ (not mult of 4)
• $3 - 1 = 2$ (not mult of 4)
• $4 - 1 = 3$ (not mult of 4)
• $5 - 1 = 4$ (mult of 4)
• $6 - 1 = 5$ (not mult of 4)
• $7 - 1 = 6$ (not mult of 4)
• $8 - 1 = 7$ (not mult of 4)
• $9 - 1 = 8$ (mult of 4)
• $10 - 1 = 9$ (not mult of 4)
• $11 - 1 = 10$ (not mult of 4)
• $12 - 1 = 11$ (not mult of 4)

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

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Solution for R$_2$:

Relation R$_2$ = {(a, b) : a = b}, where $a, b \in A$.

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Check for Reflexivity (R$_2$):

For any element $a \in A$, we check if $(a, a) \in R_2$. This means $a = a$. This is true for all $a \in A$.

Hence, R$_2$ is reflexive.

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Check for Symmetry (R$_2$):

Suppose $(a, b) \in R_2$. This means $a = b$.

We need to check if $(b, a) \in R_2$, which means $b = a$.

If $a = b$, then $b = a$. This is true.

Thus, if $(a, b) \in R_2$, then $(b, a) \in R_2$.

Hence, R$_2$ is symmetric.

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Check for Transitivity (R$_2$):

Suppose $(a, b) \in R_2$ and $(b, c) \in R_2$.

$(a, b) \in R_2$ means $a = b$.

$(b, c) \in R_2$ means $b = c$.

If $a = b$ and $b = c$, then by the transitive property of equality, $a = c$.

Thus, $(a, c) \in R_2$.

Therefore, if $(a, b) \in R_2$ and $(b, c) \in R_2$, then $(a, c) \in R_2$.

Hence, R$_2$ is transitive.

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Since R$_2$ is reflexive, symmetric, and transitive, R$_2$ is an equivalence relation.

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Find the set of all elements related to 1 in R$_2$:

An element $x \in A$ is related to 1 if $(x, 1) \in R_2$. This means $x = 1$.

The only element $x$ in set A = {0, 1, 2, ..., 12} that satisfies $x = 1$ is 1 itself.

The set of all elements related to 1 in R$_2$ is {1}.

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Conclusion:

Both relation R$_1$ (|$a - b$| is a multiple of 4) and relation R$_2$ ($a = b$) defined on the set A = {0, 1, ..., 12} are equivalence relations.

The set of elements related to 1 under R$_1$ is {1, 5, 9}.

The set of elements related to 1 under R$_2$ is {1}.

Given:

A set and a relation defined on it.

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To Find:

Examples of relations with specific combinations of reflexivity, symmetry, and transitivity.

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Solution:

We will provide an example for each required type of relation. For simplicity, we will use a small finite set, such as A = {1, 2, 3}.

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(i) Symmetric but neither reflexive nor transitive.

Consider the set A = {1, 2, 3}.

Let the relation be R = {(1, 2), (2, 1)}.

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Check for Reflexivity:

For R to be reflexive, (1, 1), (2, 2), and (3, 3) must be in R. None of these are in R.

So, R is not reflexive.

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Check for Symmetry:

The pairs in R are (1, 2) and (2, 1).

If (1, 2) $\in$ R, is (2, 1) $\in$ R? Yes.

If (2, 1) $\in$ R, is (1, 2) $\in$ R? Yes.

So, R is symmetric.

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Check for Transitivity:

Consider the path (1, 2) $\in$ R and (2, 1) $\in$ R. For transitivity, (1, 1) must be in R. But (1, 1) $\notin$ R.

Consider the path (2, 1) $\in$ R and (1, 2) $\in$ R. For transitivity, (2, 2) must be in R. But (2, 2) $\notin$ R.

So, R is not transitive.

Example for (i): R = {(1, 2), (2, 1)} on the set {1, 2, 3}.

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(ii) Transitive but neither reflexive nor symmetric.

Consider the set A = {1, 2, 3}.

Let the relation be R = {(1, 2)}.

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Check for Reflexivity:

For R to be reflexive, (1, 1), (2, 2), and (3, 3) must be in R. None of these are in R.

So, R is not reflexive.

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Check for Symmetry:

Consider the pair (1, 2) $\in$ R. For symmetry, (2, 1) must be in R. But (2, 1) $\notin$ R.

So, R is not symmetric.

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Check for Transitivity:

We need to check if for every case where $(a, b) \in R$ and $(b, c) \in R$, it follows that $(a, c) \in R$.

The only pair in R is (1, 2).

If we start with (1, 2) as $(a, b)$, we need a pair $(2, c)$ in R. There is no such pair in R whose first element is 2.

Thus, the premise "if $(a, b) \in R$ and $(b, c) \in R$" is never true.

The condition for transitivity is vacuously satisfied.

So, R is transitive.

Example for (ii): R = {(1, 2)} on the set {1, 2, 3}.

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(iii) Reflexive and symmetric but not transitive.

Consider the set A = {1, 2, 3}.

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

(This relation represents being "related" if the numbers are close, specifically within a distance of 1, considering numbers as points on a line, plus the reflexive pairs.)

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Check for Reflexivity:

The set is {1, 2, 3}. The pairs (1, 1), (2, 2), and (3, 3) are all in R.

So, R is reflexive.

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Check for Symmetry:

We check the non-reflexive pairs:

(1, 2) $\in$ R, is (2, 1) $\in$ R? Yes.

(2, 3) $\in$ R, is (3, 2) $\in$ R? Yes.

So, R is symmetric.

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Check for Transitivity:

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

For transitivity, $(1, 3)$ must be in R. But (1, 3) $\notin$ R.

So, R is not transitive.

Example for (iii): R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} on the set {1, 2, 3}.

Another example on a larger set: R = {(a, b) : |a - b| $\leq$ 1} on the set of integers Z. Reflexive: $|a-a|=0 \leq 1$. Symmetric: $|a-b| = |b-a|$, so if $|a-b| \leq 1$, then $|b-a| \leq 1$. Not Transitive: (1, 2) $\in$ R ($|1-2|=1 \leq 1$), (2, 3) $\in$ R ($|2-3|=1 \leq 1$), but (1, 3) $\notin$ R ($|1-3|=2 \not\leq 1$).

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(iv) Reflexive and transitive but not symmetric.

Consider the set A = {1, 2, 3}.

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

(This relation is based on the order $\leq$ on the set {1, 2, 3}, but restricted to certain pairs. This specific relation is the standard $\leq$ relation on {1, 2, 3}).

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Check for Reflexivity:

The set is {1, 2, 3}. The pairs (1, 1), (2, 2), and (3, 3) are all in R.

So, R is reflexive.

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Check for Symmetry:

Consider the pair (1, 2) $\in$ R. For symmetry, (2, 1) must be in R. But (2, 1) $\notin$ R.

So, R is not symmetric.

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Check for Transitivity:

We check for paths of length 2:

• (1, 2) $\in$ R and (2, 3) $\in$ R. Check (1, 3) $\in$ R. Yes, it is.
• Any other path $(a, b) \in R$ and $(b, c) \in R$? For example, (1, 1) and (1, 2) $\implies$ (1, 2) $\in$ R. (1, 2) and (2, 2) $\implies$ (1, 2) $\in$ R. (1, 2) and (2, 3) $\implies$ (1, 3) $\in$ R. (1, 3) and (3, 3) $\implies$ (1, 3) $\in$ R. And so on.

For every possible path $(a, b) \in R$ and $(b, c) \in R$, the pair $(a, c)$ is also in R.

So, R is transitive.

Example for (iv): R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} (which is the relation $\leq$) on the set {1, 2, 3}.

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(v) Symmetric and transitive but not reflexive.

Consider the set A = {1, 2, 3}.

Let the relation be R = {(1, 1), (2, 2)}.

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Check for Reflexivity:

For R to be reflexive on {1, 2, 3}, (1, 1), (2, 2), and (3, 3) must be in R. The pair (3, 3) is not in R.

So, R is not reflexive.

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Check for Symmetry:

The pairs in R are (1, 1) and (2, 2).

If (1, 1) $\in$ R, is (1, 1) $\in$ R? Yes.

If (2, 2) $\in$ R, is (2, 2) $\in$ R? Yes.

So, R is symmetric.

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Check for Transitivity:

We need to check if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$.

Consider (1, 1) $\in$ R and (1, 1) $\in$ R. Does this imply (1, 1) $\in$ R? Yes.

Consider (2, 2) $\in$ R and (2, 2) $\in$ R. Does this imply (2, 2) $\in$ R? Yes.

Are there any other pairs $(a, b) \in R$ and $(b, c) \in R$ where $b$ is the second element of the first pair and the first element of the second pair?

The second elements in R are 1 and 2. The first elements in R are 1 and 2.

We already checked $b=1$ and $b=2$ with $a=b=c$. Any other combination? No, because there are no pairs like (1, x) with $x \ne 1$ or (2, y) with $y \ne 2$ in R.

Thus, the premise "if $(a, b) \in R$ and $(b, c) \in R$" is only true when $a=b$ and $b=c$, which implies $a=c$, and the relation is satisfied $(a, a) \in R$. The transitivity holds.

So, R is transitive.

Example for (v): R = {(1, 1), (2, 2)} on the set {1, 2, 3}.

Another example: The empty relation on a non-empty set is symmetric and transitive but not reflexive.

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Conclusion:

We have provided examples for relations satisfying the requested properties.

(i) Symmetric but neither reflexive nor transitive: R = {(1, 2), (2, 1)} on {1, 2, 3}.

(ii) Transitive but neither reflexive nor symmetric: R = {(1, 2)} on {1, 2, 3}.

(iii) Reflexive and symmetric but not transitive: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} on {1, 2, 3}.

(iv) Reflexive and transitive but not symmetric: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on {1, 2, 3}.

(v) Symmetric and transitive but not reflexive: R = {(1, 1), (2, 2)} on {1, 2, 3}.

Given:

Set A = {all points in a plane}

Origin O = (0, 0)

Relation R = {(P, Q) : distance of P from O is same as distance of Q from O}, where P, Q $\in$ A.

Let $d(X, Y)$ denote the distance between points X and Y.

The relation is R = {(P, Q) : $d(P, O) = d(Q, O)$}.

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To Show:

R is an equivalence relation on A.

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

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Solution:

To show that R is an equivalence relation, we must verify the properties of reflexivity, symmetry, and transitivity.

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1. Reflexivity:

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

For R to be reflexive, we must have $(P, P) \in R$ for every point $P$ in the plane. This means the distance of P from the origin must be the same as the distance of P from the origin.

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

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

Hence, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if whenever $(P, Q) \in R$, then $(Q, P) \in R$ for all $P, Q \in A$.

Suppose $(P, Q) \in R$. By the definition of R, this means the distance of P from the origin is the same as the distance of Q from the origin, i.e., $d(P, O) = d(Q, O)$.

For $(Q, P)$ to be in R, we must have the distance of Q from the origin is the same as the distance of P from the origin, i.e., $d(Q, O) = d(P, O)$.

If $d(P, O) = d(Q, O)$, then it is also true that $d(Q, O) = d(P, O)$.

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

Hence, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if whenever $(P, Q) \in R$ and $(Q, S) \in R$, then $(P, S) \in R$ for all $P, Q, S \in A$.

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

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

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

We need to check if $(P, S) \in R$, which means $d(P, O) = d(S, O)$.

If $d(P, O) = d(Q, O)$ and $d(Q, O) = d(S, O)$, by the transitive property of equality, it follows that $d(P, O) = d(S, O)$.

Thus, if $(P, Q) \in R$ and $(Q, S) \in R$, then $(P, S) \in R$.

Hence, R is transitive.

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

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Now, let's find the set of all points related to a point P $\neq$ (0, 0).

Let S be the set of all points related to P $\in$ A, where P $\neq$ (0, 0).

By the definition of the relation R, a point Q $\in$ A is related to P if $(Q, P) \in R$.

$(Q, P) \in R$ means that the distance of Q from the origin is the same as the distance of P from the origin, i.e., $d(Q, O) = d(P, O)$.

Let the distance of point P from the origin be $r$. Since P $\neq$ (0, 0), $r > 0$.

So, the set of points Q related to P is the set of all points Q such that $d(Q, O) = r$.

The set of all points in a plane whose distance from a fixed point (the origin O) is a constant positive value $r$ is the definition of a circle.

This circle has its centre at the origin O and its radius is $r$, which is the distance of point P from the origin.

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

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

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Conclusion:

The relation R defined by having the same distance from the origin is an equivalence relation because it is reflexive, symmetric, and transitive.

The equivalence class of a point P $\neq$ (0, 0) under this relation is the set of all points Q such that $d(Q, O) = d(P, O)$. This set describes a circle centered at the origin passing through P.

Given:

Set A = {all triangles}

Relation R = {(T₁ , T₂ ) : T₁ is similar to T₂ }, where T₁, T₂ $\in$ A.

Three right angle triangles:

T$_1$ with sides 3, 4, 5.

T$_2$ with sides 5, 12, 13.

T$_3$ with sides 6, 8, 10.

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To Show:

R is an equivalence relation on A.

Identify which of T$_1$, T$_2$, T$_3$ are related under R.

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Solution:

To show that R is an equivalence relation, we must verify that R satisfies the properties of reflexivity, symmetry, and transitivity.

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1. Reflexivity:

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

For R to be reflexive, we must have $(T, T) \in R$ for every triangle $T$. This means that triangle T must be similar to triangle T.

Every triangle is similar to itself (with a scale factor of 1).

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

Hence, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if whenever $(T_1, T_2) \in R$, then $(T_2, T_1) \in R$ for all $T_1, T_2 \in A$.

Suppose $(T_1, T_2) \in R$. By the definition of R, this means that triangle T$_1$ is similar to triangle T$_2$.

If triangle T$_1$ is similar to triangle T$_2$, it implies that triangle T$_2$ is also similar to triangle T$_1$. The similarity relationship is mutual.

Thus, if $(T_1, T_2) \in R$, then $(T_2, T_1) \in R$.

Hence, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if whenever $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$, then $(T_1, T_3) \in R$ for all $T_1, T_2, T_3 \in A$.

Suppose $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$.

$(T_1, T_2) \in R$ means that T$_1$ is similar to T$_2$.

$(T_2, T_3) \in R$ means that T$_2$ is similar to T$_3$.

If T$_1$ is similar to T$_2$, there exists a scale factor $k_1$ such that the side lengths of T$_1$ are $k_1$ times the corresponding side lengths of T$_2$ (or vice versa). More precisely, the ratios of corresponding sides are equal, and corresponding angles are equal.

If T$_2$ is similar to T$_3$, there exists a scale factor $k_2$ such that the side lengths of T$_2$ are $k_2$ times the corresponding side lengths of T$_3$ (or vice versa), and corresponding angles are equal.

If T$_1$ is similar to T$_2$ and T$_2$ is similar to T$_3$, then T$_1$ is also similar to T$_3$. (The ratios of corresponding sides will be multiplied, e.g., if side $a_1$ of T$_1$ corresponds to $a_2$ of T$_2$ and $a_2$ of T$_2$ corresponds to $a_3$ of T$_3$, then $a_1/a_2 = k_1$ and $a_2/a_3 = k_2$. Then $a_1/a_3 = (a_1/a_2) \times (a_2/a_3) = k_1 k_2$. The ratio is constant for all corresponding sides, and angles are preserved through similarity).

Thus, if $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$, then $(T_1, T_3) \in R$.

Hence, R is transitive.

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

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Now, consider the three right angle 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 the ratios of their corresponding sides are equal.

Check if T$_1$ is similar to T$_2$:

Compare the ratios of corresponding sides (smallest to smallest, medium to medium, largest to largest):

$\frac{3}{5}$, $\frac{4}{12} = \frac{1}{3}$, $\frac{5}{13}$.

The ratios are not equal ($\frac{3}{5} \neq \frac{1}{3} \neq \frac{5}{13}$).

So, T$_1$ is not related to T$_2$.

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Check if T$_1$ is similar to T$_3$:

Compare the ratios of corresponding sides:

$\frac{3}{6} = \frac{1}{2}$

$\frac{4}{8} = \frac{1}{2}$

$\frac{5}{10} = \frac{1}{2}$

The ratios of corresponding sides are equal ($\frac{1}{2}$).

So, T$_1$ is related to T$_3$.

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Check if T$_2$ is similar to T$_3$:

Compare the ratios of corresponding sides:

$\frac{5}{6}$, $\frac{12}{8} = \frac{3}{2}$, $\frac{13}{10}$.

The ratios are not equal ($\frac{5}{6} \neq \frac{3}{2} \neq \frac{13}{10}$).

So, T$_2$ is not related to T$_3$.

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Since the relation is symmetric, if T$_1$ is related to T$_3$, then T$_3$ is related to T$_1$.

The pairs of triangles that are related are (T$_1$, T$_3$) and (T$_3$, T$_1$).

Also, since the relation is reflexive, each triangle is related to itself: (T$_1$, T$_1$), (T$_2$, T$_2$), (T$_3$, T$_3$) are in R.

When the question asks "Which triangles among T$_1$, T$_2$ and T$_3$ are related?", it typically refers to relationships between distinct triangles. In that sense, only T$_1$ and T$_3$ are related to each other (excluding the reflexive case).

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Conclusion:

The relation of similarity between triangles is an equivalence relation.

Among the given triangles T$_1$, T$_2$, and T$_3$, triangle T$_1$ and triangle T$_3$ are related because they are similar. Triangle T$_2$ is not related to either T$_1$ or T$_3$.

Given:

Set A = {all polygons}

Relation R = {(P₁ , P₂ ) : P₁ and P₂ have same number of sides}, where P₁, P₂ $\in$ A.

A right angle triangle T with sides 3, 4, and 5.

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To Show:

R is an equivalence relation on A.

Find the set of all elements in A related to the triangle T.

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Solution:

To show that R is an equivalence relation, we must verify the properties of reflexivity, symmetry, and transitivity.

Let $N(P)$ denote the number of sides of a polygon P.

The relation is R = {(P$_1$, P$_2$) : $N(P_1) = N(P_2)$}.

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1. Reflexivity:

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

For R to be reflexive, we must have $(P, P) \in R$ for every polygon $P$. This means that polygon P and polygon P have the same number of sides.

This is always true: $N(P) = N(P)$.

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

Hence, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if whenever $(P_1, P_2) \in R$, then $(P_2, P_1) \in R$ for all P$_1$, P$_2$ $\in$ A.

Suppose $(P_1, P_2) \in R$. By the definition of R, this means that polygon P$_1$ and polygon P$_2$ have the same number of sides, i.e., $N(P_1) = N(P_2)$.

For $(P_2, P_1)$ to be in R, we must have that polygon P$_2$ and polygon P$_1$ have the same number of sides, i.e., $N(P_2) = N(P_1)$.

If $N(P_1) = N(P_2)$, then it is also true that $N(P_2) = N(P_1)$.

Thus, if $(P_1, P_2) \in R$, then $(P_2, P_1) \in R$.

Hence, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if whenever $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$, then $(P_1, P_3) \in R$ for all P$_1$, P$_2$, P$_3$ $\in$ A.

Suppose $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$.

$(P_1, P_2) \in R$ means that P$_1$ and P$_2$ have the same number of sides, i.e., $N(P_1) = N(P_2)$.

$(P_2, P_3) \in R$ means that P$_2$ and P$_3$ have the same number of sides, i.e., $N(P_2) = N(P_3)$.

If $N(P_1) = N(P_2)$ and $N(P_2) = N(P_3)$, by the transitive property of equality, it follows that $N(P_1) = N(P_3)$.

Thus, if $(P_1, P_2) \in R$ and $(P_2, P_3) \in R$, then $(P_1, P_3) \in R$.

Hence, R is transitive.

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

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

A triangle is a polygon with 3 sides.

So, the given triangle T has 3 sides. $N(T) = 3$.

An element P' in A is related to the triangle T if $(P', T) \in R$.

$(P', T) \in R$ means that polygon P' and triangle T have the same number of sides, i.e., $N(P') = N(T)$.

Since $N(T) = 3$, the condition is $N(P') = 3$.

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.

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Conclusion:

The relation R defined by having the same number of sides is an equivalence relation on the set of all polygons.

The set of all elements related to the right angle triangle T is the set of all polygons that have the same number of sides as T. Since T is a triangle (a 3-sided polygon), the set of related elements is the set of all 3-sided polygons, which is the set of all triangles.

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

Given:

Set L = {all lines in XY plane}

Relation R = {(L₁ , L₂ ) : L₁ is parallel to L₂ }, where L₁, L₂ $\in$ L.

A specific line $L: y = 2x + 4$.

Note: We consider a line to be parallel to itself. Two distinct lines are parallel if they have the same slope.

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To Show:

R is an equivalence relation on L.

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

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Solution:

To show that R is an equivalence relation, we must verify the properties of reflexivity, symmetry, and transitivity.

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1. Reflexivity:

A relation R on a set L is reflexive if $(L_1, L_1) \in R$ for every element $L_1 \in L$.

For R to be reflexive, we must have $(L_1, L_1) \in R$ for every line $L_1$. This means that line L$_1$ must be parallel to itself.

By convention in mathematics (and for the definition of parallel lines having the same slope), a line is considered parallel to itself.

Thus, $(L_1, L_1) \in R$ for all $L_1 \in L$.

Hence, R is reflexive.

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2. Symmetry:

A relation R on a set L is symmetric if whenever $(L_1, L_2) \in R$, then $(L_2, L_1) \in R$ for all $L_1, L_2 \in L$.

Suppose $(L_1, L_2) \in R$. By the definition of R, this means that line L$_1$ is parallel to line L$_2$.

If line L$_1$ is parallel to line L$_2$, it is also true that line L$_2$ is parallel to line L$_1$.

Thus, if $(L_1, L_2) \in R$, then $(L_2, L_1) \in R$.

Hence, R is symmetric.

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3. Transitivity:

A relation R on a set L is transitive if whenever $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$, then $(L_1, L_3) \in R$ for all $L_1, L_2, L_3 \in L$.

Suppose $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$.

$(L_1, L_2) \in R$ means that L$_1$ is parallel to L$_2$.

$(L_2, L_3) \in R$ means that L$_2$ is parallel to L$_3$.

In geometry, 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$_3$.

This can also be seen in terms of slopes. If L$_1$ has slope $m_1$, L$_2$ has slope $m_2$, and L$_3$ has slope $m_3$. Parallel lines have the same slope. So, $m_1 = m_2$ and $m_2 = m_3$. By transitivity of equality, $m_1 = m_3$, which means L$_1$ is parallel to L$_3$. (Vertical lines also follow this property).

Thus, if $(L_1, L_2) \in R$ and $(L_2, L_3) \in R$, then $(L_1, L_3) \in R$.

Hence, R is transitive.

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

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Now, we need to find the set of all lines related to the line $L: y = 2x + 4$.

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

The slope of line L is $m = 2$.

A line L' in L is related to line L if $(L', L) \in R$.

$(L', L) \in R$ means that line L' is parallel to line L.

Two distinct lines are parallel if they have the same slope. A line is parallel to itself.

So, a line L' is related to line L if L' has the same slope as L.

The slope of L is 2. So, any line L' related to L must have a slope of 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 have a slope of 2.

This set includes the line $y = 2x + 4$ itself (when $c'=4$).

The set of all lines related to the line $y = 2x + 4$ is $\{L' \in L : L' \text{ has slope } 2\}$.

This set can be described by the equation $y = 2x + c$, where $c$ is any real number.

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Conclusion:

The relation of being parallel between lines is an equivalence relation.

The set of all lines related to the line $y = 2x + 4$ is the set of all lines parallel to it, which are lines with the equation $y = 2x + c$, where $c$ is any real number.

Given:

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

Relation R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)} on set A.

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To Choose:

Which of the given options correctly describes the relation R.

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Solution:

We examine the properties of reflexivity, symmetry, and transitivity for the given relation R on the set A = {1, 2, 3, 4}.

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1. Reflexivity:

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

The set A is {1, 2, 3, 4}. For R to be reflexive, the pairs (1, 1), (2, 2), (3, 3), and (4, 4) must all be in R.

Looking at the given relation R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}, we see that:

(1, 1) $\in$ R

(2, 2) $\in$ R

(3, 3) $\in$ R

(4, 4) $\in$ R

Since $(a, a) \in R$ for all $a \in A = \{1, 2, 3, 4\}$, the relation R is reflexive.

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2. Symmetry:

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

We need to check if for every pair $(a, b)$ in R, the reversed pair $(b, a)$ is also in R.

Consider the pair (1, 2) $\in$ R. For symmetry, (2, 1) must be in R. But (2, 1) $\notin$ R.

Consider the pair (1, 3) $\in$ R. For symmetry, (3, 1) must be in R. But (3, 1) $\notin$ R.

Since we found pairs (1, 2) and (1, 3) in R whose reversed pairs (2, 1) and (3, 1) are not in R, the relation R is not symmetric.

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3. Transitivity:

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

We check for paths of length 2 in R:

• (1, 2) $\in$ R. Are there pairs starting with 2? Yes, (2, 2) $\in$ R. Consider $(a, b) = (1, 2)$ and $(b, c) = (2, 2)$. Here $a=1, b=2, c=2$. We need to check if $(a, c) = (1, 2) \in R$. Yes, it is.
• (1, 3) $\in$ R. Are there pairs starting with 3? Yes, (3, 3) $\in$ R and (3, 2) $\in$ R.
  • Consider $(a, b) = (1, 3)$ and $(b, c) = (3, 3)$. Here $a=1, b=3, c=3$. We need to check if $(a, c) = (1, 3) \in R$. Yes, it is.
  • Consider $(a, b) = (1, 3)$ and $(b, c) = (3, 2)$. Here $a=1, b=3, c=2$. We need to check if $(a, c) = (1, 2) \in R$. Yes, it is.
• (3, 2) $\in$ R. Are there pairs starting with 2? Yes, (2, 2) $\in$ R.
  • Consider $(a, b) = (3, 2)$ and $(b, c) = (2, 2)$. Here $a=3, b=2, c=2$. We need to check if $(a, c) = (3, 2) \in R$. Yes, it is.
• We also need to check paths involving reflexive pairs like (1, 1), (2, 2), (3, 3), (4, 4).
  • (1, 1) $\in$ R and (1, 2) $\in$ R $\implies$ (1, 2) $\in$ R (True)
  • (1, 1) $\in$ R and (1, 3) $\in$ R $\implies$ (1, 3) $\in$ R (True)
  • (2, 2) $\in$ R and any pair (2, x) $\in$ R $\implies$ (2, x) $\in$ R. The only such pair is (2, 2). (2, 2) and (2, 2) $\implies$ (2, 2) $\in$ R (True).
  • Any pair (x, 2) $\in$ R and (2, 2) $\in$ R $\implies$ (x, 2) $\in$ R. Pairs ending in 2 are (1, 2), (2, 2), (3, 2). (1, 2) and (2, 2) $\implies$ (1, 2) $\in$ R (True). (2, 2) and (2, 2) $\implies$ (2, 2) $\in$ R (True). (3, 2) and (2, 2) $\implies$ (3, 2) $\in$ R (True).
  • (3, 3) $\in$ R and (3, 2) $\in$ R $\implies$ (3, 2) $\in$ R (True).
  • Any pair (x, 3) $\in$ R and (3, 3) $\in$ R $\implies$ (x, 3) $\in$ R. Pairs ending in 3 are (1, 3), (3, 3). (1, 3) and (3, 3) $\implies$ (1, 3) $\in$ R (True). (3, 3) and (3, 3) $\implies$ (3, 3) $\in$ R (True).
  • For (4, 4) there are no other pairs starting or ending with 4. (4, 4) and (4, 4) $\implies$ (4, 4) $\in$ R (True).

In all cases where $(a, b) \in R$ and $(b, c) \in R$, we found that $(a, c) \in R$.

So, R is transitive.

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Summary of properties:

R is reflexive.

R is not symmetric.

R is transitive.

---

Now let's look at the options:

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

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

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

(D) R is an equivalence relation. (False, R is not symmetric, so it's not an equivalence relation)

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Conclusion:

Based on our analysis, the relation R is reflexive and transitive but not symmetric.

The correct option is (B).

Given:

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

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

---

To Choose:

The correct ordered pair that belongs to the relation R.

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Solution:

We are given the relation R defined by two conditions that an ordered pair $(a, b)$ must satisfy to be in R:

1. $a = b - 2$

2. $b > 6$

Both $a$ and $b$ must also be natural numbers ($a \in N$ and $b \in N$).

Let's check each given option:

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(A) (2, 4) $\in$ 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 the second condition is false, the pair (2, 4) is not in R.

---

(B) (3, 8) $\in$ R

Here, $a = 3$ and $b = 8$.

Check if $a, b \in N$: $3 \in N$, $8 \in N$ (True)

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

Since the first condition is false, the pair (3, 8) is not in R.

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(C) (6, 8) $\in$ R

Here, $a = 6$ and $b = 8$.

Check if $a, b \in N$: $6 \in N$, $8 \in N$ (True)

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, the pair (6, 8) is in R.

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(D) (8, 7) $\in$ R

Here, $a = 8$ and $b = 7$.

Check if $a, b \in N$: $8 \in N$, $7 \in N$ (True)

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

Since the first condition is false, the pair (8, 7) is not in R.

---

Only option (C) satisfies both conditions required for an ordered pair to be in the relation R.

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Answer:

The correct answer is (C) (6, 8) $\in$ R.

Given:

Set A = {all 50 students of Class X in a school}

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

Function f : A $\to$ N defined by f(x) = roll number of the student x.

---

To Show:

f is one-one (injective).

f is not onto (surjective).

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Solution:

---

Check if f is One-one:

A function f : A $\to$ B is one-one if for any two distinct elements $x_1, x_2 \in A$, their images under f are distinct, i.e., if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Equivalently, f is one-one if whenever $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$, it must imply that $x_1 = x_2$.

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

By 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, the condition $f(x_1) = f(x_2)$ 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. Therefore, if two students have the same roll number, they must be the same student.

Thus, $x_1 = x_2$.

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$, the function f is one-one.

---

Check if f is Onto:

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

The domain A is the set of 50 students. The codomain is N, the set of all natural numbers {1, 2, 3, ...}.

The range of the function f is the set of all roll numbers assigned to the 50 students in Class X.

Since there are exactly 50 students in set A, and each student has a unique roll number (as f is one-one), the set of roll numbers of these 50 students contains exactly 50 distinct numbers.

The range of f is a set containing 50 elements, which are the 50 unique roll numbers.

The codomain N is the set of all natural numbers {1, 2, 3, ...}, which is an infinite set.

Since the range of f is a finite set with 50 elements, and the codomain N is an infinite set, the range of f cannot be equal to the codomain N.

For example, if the roll numbers are assigned from 1 to 50, then the number 51 is in the codomain N, but there is no student in A whose roll number is 51, so 51 is not in the range of f. Thus, there exists an element in the codomain (N) which is not the image of any element in the domain (A).

Therefore, the function f is not onto.

---

Conclusion:

The function f : A $\to$ N, defined by f(x) = roll number of student x, where A is the set of 50 students of Class X, is one-one because each student has a unique roll number.

The function f is not onto because the set of 50 roll numbers (the range) is a finite set and cannot cover the entire set of natural numbers (the codomain), which is infinite.

Hence, f is one-one but not onto.

Given:

The function $f : N \to N$, where N is the set of natural numbers (i.e., $N = \{1, 2, 3, ...\}$).

The function is defined by $f(x) = 2x$ for all $x \in N$.

---

To Show:

The function f is one-one (injective).

The function f is not onto (surjective).

---

Proof:

---

(i) Check if f is One-one:

A function $f: A \to B$ is defined as one-one if for any two elements $x_1, x_2$ in the domain A, the equality of their images $f(x_1) = f(x_2)$ implies the equality of the elements themselves, i.e., $x_1 = x_2$.

Let $x_1, x_2$ be any two elements in the domain N such that $f(x_1) = f(x_2)$.

By the definition of the function $f(x) = 2x$, we have:

$f(x_1) = 2x_1$

$f(x_2) = 2x_2$

Since $f(x_1) = f(x_2)$, we have the equation:

Since $x_1$ and $x_2$ are natural numbers, they are integers. We can divide both sides of the equation by 2.

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in N$, the function f is indeed one-one.

---

(ii) Check if f is Onto:

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

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

We need to check if for every natural number $y$ in the codomain N, there exists a natural number $x$ in the domain N such that $f(x) = y$.

The equation $f(x) = y$ is equivalent to $2x = y$.

Solving for $x$, we get $x = y/2$.

For the function to be onto, for every $y \in N$, $y/2$ must also be an element of N.

Let's consider an element in the codomain N, for example, $y = 1$.

We look for an $x \in N$ such that $f(x) = 1$, i.e., $2x = 1$.

Solving for $x$, we get $x = 1/2$.

However, $1/2$ is not a natural number ($1/2 \notin N$). This means that the element 1 in the codomain does not have a corresponding pre-image in the domain N under the function f.

We can observe that for any odd natural number $y$ (such as 1, 3, 5, ...), the value of $y/2$ will not be an integer, and thus will not be a natural number.

The range of the function f is the set of all possible values of $f(x)$ for $x \in N$.

Range(f) = $\{f(1), f(2), f(3), ...\} = \{2(1), 2(2), 2(3), ...\} = \{2, 4, 6, ...\}$.

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

Since the set of even natural numbers is a proper subset of the set of all natural numbers, the range of f is not equal to its codomain ($Range(f) \neq N$).

Therefore, the function f is not onto.

---

Conclusion:

We have shown that for any $x_1, x_2 \in N$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Hence, f is one-one.

We have also shown that there exist elements in the codomain N (e.g., odd numbers) that do not have a pre-image in the domain N. Hence, f is not onto.

Thus, the function $f : N \to N$, given by $f(x) = 2x$, is one-one but not onto.

Given:

The function $f : \mathbb{R} \to \mathbb{R}$.

The domain is the set of all real numbers, $\mathbb{R}$.

The codomain is the set of all real numbers, $\mathbb{R}$.

The function is defined by $f(x) = 2x$ for all $x \in \mathbb{R}$.

---

To Prove:

The function f is one-one (injective).

The function f is onto (surjective).

---

Proof:

---

(i) To prove that f is One-one:

A function $f: A \to B$ is one-one if for any two elements $x_1, x_2$ in the domain A, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2$ be any two real numbers in the domain $\mathbb{R}$ such that $f(x_1) = f(x_2)$.

By the definition of the function f, $f(x_1) = 2x_1$ and $f(x_2) = 2x_2$.

Setting the function values equal, we have:

Since 2 is a non-zero real number, we can divide both sides of the equation by 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-one.

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(ii) To prove that f is Onto:

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

The domain of f is $\mathbb{R}$. The codomain of f is $\mathbb{R}$.

Let $y$ be an arbitrary real number in the codomain $\mathbb{R}$.

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

Using the definition of the function $f(x)$, we set $f(x) = y$:

Now, we solve for $x$ in terms of $y$. Dividing both sides by 2 (which is non-zero), we get:

Since $y$ is a real number, and the division of a real number by a non-zero real number (2) results in a real number, $x = y/2$ is also a real number. Thus, $x \in \mathbb{R}$ (the domain of f).

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

This shows that every element in the codomain $\mathbb{R}$ has a pre-image in the domain $\mathbb{R}$.

Therefore, the function f is onto.

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Conclusion:

Since the function f has been proven to be both one-one and onto, we conclude that the function $f : \mathbb{R} \to \mathbb{R}$, given by $f(x) = 2x$, is one-one and onto.

Given:

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

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

$f(1) = 1$

$f(2) = 1$

$f(x) = x - 1$ for $x > 2$

---

To Show:

f is onto (surjective).

f is not one-one (injective).

---

Solution:

---

Check if f is One-one:

A function f : A $\to$ B is one-one if for any two distinct elements $x_1, x_2 \in A$, their images under f are distinct, i.e., if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Equivalently, f is one-one if whenever $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$, it must imply that $x_1 = x_2$.

Consider the elements 1 and 2 from the domain N.

$f(1) = 1$ (given)

$f(2) = 1$ (given)

Here, we have two distinct elements in the domain, $1 \neq 2$, but their images are the same, $f(1) = f(2) = 1$.

Since there exist distinct elements in the domain that have the same image in the codomain, the function f is not one-one.

---

Check if f is Onto:

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

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

We need to check if for every $y \in N$, there exists an $x \in N$ such that $f(x) = y$.

Let $y$ be an arbitrary natural number in the codomain N.

We consider two cases for $y$:

Case 1: $y = 1$

We need to find an $x \in N$ such that $f(x) = 1$. From the definition, we know that $f(1) = 1$ and $f(2) = 1$. So, for $y=1$, there exist elements $x=1$ and $x=2$ in the domain such that $f(x)=1$. Since we found at least one element (in fact, two), $y=1$ is in the range.

Case 2: $y > 1$ (i.e., $y = 2, 3, 4, ...$)

We need to find an $x \in N$ such that $f(x) = y$. We use the definition $f(x) = x - 1$ for $x > 2$.

Set $f(x) = y$: $x - 1 = y$.

Solving for $x$: $x = y + 1$.

Since $y$ is a natural number and $y > 1$, $y+1$ will be a natural number greater than 2 (e.g., if $y=2$, $x=3$; if $y=3$, $x=4$; etc.). So $x = y+1 \in N$ and $x > 2$.

For this value of $x = y+1$, we have $f(x) = f(y+1) = (y+1) - 1 = y$.

So, for any $y \in N$ with $y > 1$, there exists an $x = y+1 \in N$ (with $x > 2$) such that $f(x) = y$.

Combining both cases: For $y=1$, $x=1$ or $x=2$ exists in the domain. For any $y > 1$, $x = y+1$ exists in the domain.

Therefore, for every $y$ in the codomain N, there exists an $x$ in the domain N such that $f(x) = y$.

Hence, the function f is onto.

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Conclusion:

The function f : N $\to$ N is not one-one because $f(1) = f(2) = 1$ but $1 \neq 2$.

The function f is onto because for every natural number $y$ in the codomain, we can find a natural number $x$ in the domain such that $f(x) = y$. Specifically, if $y=1$, $x=1$ or $x=2$ works; if $y>1$, $x=y+1$ works.

Thus, f is onto but not one-one.

Given:

Set $\mathbb{R}$ = {all real numbers}

Function f : $\mathbb{R}$ $\to$ $\mathbb{R}$ defined by f(x) = x$^2$.

---

To Show:

f is neither one-one (injective) nor onto (surjective).

---

Solution:

---

Check if f is One-one:

A function f : A $\to$ B is one-one if for any two distinct elements $x_1, x_2 \in A$, their images under f are distinct, i.e., if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Equivalently, f is one-one if whenever $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$, it must imply that $x_1 = x_2$.

Let's try to find two distinct real numbers that have the same image under f.

Consider $x_1 = 1$ and $x_2 = -1$ from the domain $\mathbb{R}$. Note that $x_1 \neq x_2$.

Calculate their images:

$f(x_1) = f(1) = 1^2 = 1$

$f(x_2) = f(-1) = (-1)^2 = 1$

Here, we have two distinct elements in the domain, $1 \neq -1$, but their images are the same, $f(1) = f(-1) = 1$.

Since there exist distinct elements in the domain that have the same image in the codomain, the function f is not one-one.

---

Check if f is Onto:

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

The domain of f is $\mathbb{R}$. The codomain of f is $\mathbb{R}$.

We need to check if for every $y \in \mathbb{R}$ (codomain), there exists an $x \in \mathbb{R}$ (domain) such that $f(x) = y$.

Set $f(x) = y$: $x^2 = y$.

For $y$ to be in the range of f, there must be a real number $x$ whose square is $y$.

Consider a negative number in the codomain $\mathbb{R}$, for example, $y = -1$.

We need to find if there exists an $x \in \mathbb{R}$ such that $f(x) = -1$, which means $x^2 = -1$.

However, the square of any real number is always non-negative ($x^2 \geq 0$). There is no real number $x$ such that $x^2 = -1$.

So, for the element $y=-1$ in the codomain $\mathbb{R}$, there is no element $x$ in the domain $\mathbb{R}$ such that $f(x) = -1$.

Similarly, for any negative number $y$ in the codomain $\mathbb{R}$, there is no real number $x$ such that $x^2 = y$.

The range of f is the set of squares of real numbers, which is $\{y \in \mathbb{R} : y \geq 0\} = [0, \infty)$.

The codomain is $\mathbb{R} = (-\infty, \infty)$.

Since the range of f is the set of non-negative real numbers and the codomain is the set of all real numbers, the range is a proper subset of the codomain.

Range(f) $\neq$ $\mathbb{R}$.

Therefore, the function f is not onto.

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Conclusion:

The function f : $\mathbb{R}$ $\to$ $\mathbb{R}$, defined by f(x) = x$^2$, is not one-one because distinct inputs like 1 and -1 produce the same output (1).

The function f is not onto because negative numbers in the codomain do not have corresponding real numbers in the domain that map to them.

Thus, f is neither one-one nor onto.

Given:

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

Function f : N $\to$ N defined by the piecewise rule given above.

---

To Show:

f is both one-one (injective) and onto (surjective).

---

Solution:

---

Check if f is One-one:

A function f : A $\to$ B is one-one if for any two distinct elements $x_1, x_2 \in A$, their images under f are distinct, i.e., if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Equivalently, f is one-one if whenever $f(x_1) = f(x_2)$ for $x_1, x_2 \in A$, it must imply that $x_1 = x_2$.

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$. We need to show that $x_1 = x_2$.

We consider the possible parities of $x_1$ and $x_2$:

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

$f(x_1) = x_1 + 1$ (since $x_1$ is odd)

$f(x_2) = x_2 + 1$ (since $x_2$ is odd)

If $f(x_1) = f(x_2)$, then $x_1 + 1 = x_2 + 1$. Subtracting 1 from both sides gives $x_1 = x_2$. In this case, the function is one-one.

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

$f(x_1) = x_1 - 1$ (since $x_1$ is even)

$f(x_2) = x_2 - 1$ (since $x_2$ is even)

If $f(x_1) = f(x_2)$, then $x_1 - 1 = x_2 - 1$. Adding 1 to both sides gives $x_1 = x_2$. In this case, the function is one-one.

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

$f(x_1) = x_1 + 1$ (since $x_1$ is odd)

$f(x_2) = x_2 - 1$ (since $x_2$ is even)

If $f(x_1) = f(x_2)$, then $x_1 + 1 = x_2 - 1$.

This implies $x_2 - x_1 = 2$.

If $x_1$ is odd, $x_1 + 1$ is even. So $f(x_1)$ is even.

If $x_2$ is even, $x_2 - 1$ is odd. So $f(x_2)$ is odd.

If $f(x_1) = f(x_2)$, it means an even number is equal to an odd number. This is only possible if they are equal to 0, but the codomain is N = {1, 2, 3, ...}, so the values must be natural numbers. Thus, an even natural number cannot be equal to an odd natural number.

Therefore, the equality $f(x_1) = f(x_2)$ is impossible in this case (where $x_1$ is odd and $x_2$ is even). This means there are no distinct $x_1, x_2$ of different parity such that $f(x_1) = f(x_2)$.

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

This case is symmetric to Case 3, and the equality $f(x_1) = f(x_2)$ is also impossible.

In all possible cases where $f(x_1) = f(x_2)$, it leads to $x_1 = x_2$.

Therefore, the function f is one-one.

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Check if f is Onto:

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

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

We need to check if for every $y \in N$ (codomain), there exists an $x \in N$ (domain) such that $f(x) = y$.

Let $y$ be an arbitrary natural number in the codomain N.

We consider two cases for $y$ based on its parity:

Case 1: $y$ is odd.

We need to find an $x \in N$ such that $f(x) = y$. We can try to see if $f(x)$ could be $y$ when $x$ is even. If $x$ is even, $f(x) = x - 1$. Set $x - 1 = y$. Solving for $x$, we get $x = y + 1$.

Since $y$ is an odd natural number, $y+1$ is an even natural number. Since $y \ge 1$, $y+1 \ge 2$. So $x = y+1$ is a natural number and is even.

For this value of $x$, $f(x) = f(y+1) = (y+1) - 1 = y$ (since $y+1$ is even and $\ge 2$).

So, for every odd $y$ in the codomain, there exists an even $x = y+1$ in the domain such that $f(x) = y$.

Case 2: $y$ is even.

We need to find an $x \in N$ such that $f(x) = y$. We can try to see if $f(x)$ could be $y$ when $x$ is odd. If $x$ is odd, $f(x) = x + 1$. Set $x + 1 = y$. Solving for $x$, we get $x = y - 1$.

Since $y$ is an even natural number, $y \ge 2$. If $y = 2$, $x = 2 - 1 = 1$, which is an odd natural number.

If $y$ is an even natural number and $y \ge 2$, then $y-1$ is an odd natural number and $y-1 \ge 1$. So $x = y-1$ is a natural number and is odd.

For this value of $x = y-1$, $f(x) = f(y-1) = (y-1) + 1 = y$ (since $y-1$ is odd and $\ge 1$).

So, for every even $y$ in the codomain, there exists an odd $x = y-1$ in the domain such that $f(x) = y$.

In both cases (y is odd or y is even), we found an element $x$ in the domain N such that $f(x) = y$.

Therefore, for every $y$ in the codomain N, there exists an $x$ in the domain N such that $f(x) = y$.

Hence, the function f is onto.

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Conclusion:

The function f : N $\to$ N, defined by $f(x) = x+1$ if x is odd and $f(x) = x-1$ if x is even, is both one-one and onto.

It is one-one because distinct inputs map to distinct outputs. It is onto because every natural number in the codomain has a pre-image in the domain (odd numbers come from even numbers, and even numbers come from odd numbers).

Given:

Domain set $A = \{1, 2, 3\}$.

Codomain set $B = \{1, 2, 3\}$.

A function $f : A \to B$ which is onto (surjective).

---

To Show:

The function f is always one-one (injective).

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Proof:

We are given that the function $f : \{1, 2, 3\} \to \{1, 2, 3\}$ is onto.

Let the domain be $A = \{1, 2, 3\}$ and the codomain be $B = \{1, 2, 3\}$.

The number of elements in the domain is $|A| = 3$.

The number of elements in the codomain is $|B| = 3$.

We are given that f is an onto function.

By the definition of an onto function, every element in the codomain B must have at least one pre-image in the domain A.

This means that for each element $y \in \{1, 2, 3\}$ (in the codomain), there is at least one $x \in \{1, 2, 3\}$ (in the domain) such that $f(x) = y$.

Let's consider the elements in the codomain: 1, 2, and 3.

Since f is onto:

• There is at least one element in the domain that maps to 1.
• There is at least one element in the domain that maps to 2.
• There is at least one element in the domain that maps to 3.

The range of the function f is the set of all images of the elements in the domain. Since f is onto, the range is equal to the codomain.

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

This means there are exactly 3 distinct values in the range of f.

Now, let's consider the definition of a one-one function.

A function f is one-one if for any two distinct elements in the domain, their images are also distinct.

In other words, if $x_1, x_2 \in \{1, 2, 3\}$ and $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.

The domain has only 3 elements: 1, 2, and 3.

When we apply the function f to these 3 distinct elements, we get their images: $f(1), f(2), f(3)$.

The set of these images is the range: $\{f(1), f(2), f(3)\}$.

We already established that the range is $\{1, 2, 3\}$, which contains exactly 3 distinct elements.

Since the 3 elements in the domain {1, 2, 3} must map to 3 distinct elements in the codomain {1, 2, 3} (because the range is exactly {1, 2, 3}), it is not possible for any two distinct elements in the domain to map to the same element in the codomain.

For example, if $f(1) = f(2)$, then the set of images $\{f(1), f(2), f(3)\}$ would have at most 2 distinct elements ($f(1)$ and $f(3)$), which contradicts the fact that the range is $\{1, 2, 3\}$.

Therefore, it must be true that if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.

Hence, the function f is one-one.

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Conclusion:

For a function between two finite sets of the same size, being onto implies being one-one.

In this case, since the function $f : \{1, 2, 3\} \to \{1, 2, 3\}$ is onto, every element in the codomain has a pre-image. As the domain and codomain have the same number of elements (3), this means each element in the domain must map to a unique element in the codomain.

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

Given:

Set A = {1, 2, 3}

Set B = {1, 2, 3}

A function f : A $\to$ B that is one-one (injective).

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To Show:

f is always onto (surjective).

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Proof:

We are given that the function f : {1, 2, 3} $\to$ {1, 2, 3} is one-one.

This means that for any two distinct elements $x_1, x_2 \in \{1, 2, 3\}$, their images under f are distinct, i.e., if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.

The domain A has 3 elements: 1, 2, and 3.

Since f is one-one, the images of these 3 distinct elements must also be distinct.

Let the images be $f(1)$, $f(2)$, and $f(3)$.

Because f is one-one, we have:

• $f(1) \neq f(2)$ (since $1 \neq 2$)
• $f(1) \neq f(3)$ (since $1 \neq 3$)
• $f(2) \neq f(3)$ (since $2 \neq 3$)

So, the three values $f(1)$, $f(2)$, and $f(3)$ are all distinct elements in the codomain B = {1, 2, 3}.

The set of these distinct image values forms the range of the function:

Since $f(1), f(2), f(3)$ are 3 distinct elements, and the codomain B contains exactly 3 elements {1, 2, 3}, the set $\{f(1), f(2), f(3)\}$ must be exactly the set {1, 2, 3}. The order might be different, but the elements are the same.

The definition of an onto function is that the range of the function is equal to its codomain.

In this case, the codomain is {1, 2, 3}.

From (ii), we have Range(f) = {1, 2, 3}.

So, the range of f is equal to the codomain.

Therefore, the function f is onto.

This proof relies on the fact that the domain and codomain are finite sets of the same size. If a function from a finite set to another finite set of the same size is one-one, it must map each element in the domain to a unique element in the codomain. Since the number of elements in the domain equals the number of elements in the codomain, all elements in the codomain must be images of elements from the domain, which is the definition of onto.

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Conclusion:

For a function f from a finite set A to a finite set B where the number of elements in A is equal to the number of elements in B (i.e., $|A| = |B|$), if f is one-one, it means that distinct elements in A map to distinct elements in B. Since $|A| = |B|$, this means that all elements in B must be covered by the images of elements from A, which is the definition of onto.

Specifically for f : {1, 2, 3} $\to$ {1, 2, 3}, if it is one-one, the 3 distinct elements in the domain must map to 3 distinct elements in the codomain. Since the codomain only has 3 elements, the range must be the entire codomain, making the function onto.

Given:

Function $f : \mathbf{R}_* \to \mathbf{R}_*$ defined by $f(x) = \frac{1}{x}$.

$\mathbf{R}_*$ is the set of all non-zero real numbers.

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To Show (Part 1):

f is one-one (injective).

f is onto (surjective).

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Solution (Part 1):

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Check if f is One-one:

A function $f : A \to B$ is one-one if for any two elements $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in \mathbf{R}_*$ such that $f(x_1) = f(x_2)$.

By the definition of the function f, $f(x_1) = \frac{1}{x_1}$ and $f(x_2) = \frac{1}{x_2}$.

So, the condition $f(x_1) = f(x_2)$ means $\frac{1}{x_1} = \frac{1}{x_2}$.

Since $x_1, x_2 \in \mathbf{R}_*$, they are non-zero, so we can take reciprocals or cross-multiply.

$1 \times x_2 = 1 \times x_1$

$x_2 = x_1$

Thus, $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in \mathbf{R}_*$.

Therefore, the function f is one-one.

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Check if f is Onto:

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

The domain of f is $\mathbf{R}_*$. The codomain of f is $\mathbf{R}_*$.

We need to check if for every $y \in \mathbf{R}_*$ (codomain), there exists an $x \in \mathbf{R}_*$ (domain) such that $f(x) = y$.

Let $y$ be an arbitrary non-zero real number in the codomain $\mathbf{R}_*$.

We want to find an $x \in \mathbf{R}_*$ such that $f(x) = y$.

Using the definition of f(x), we have $\frac{1}{x} = y$.

Solving for $x$, we take the reciprocal of both sides (since $y \ne 0$): $x = \frac{1}{y}$.

Since $y$ is a non-zero real number, its reciprocal $\frac{1}{y}$ is also a non-zero real number. Thus, $x = \frac{1}{y} \in \mathbf{R}_*$.

So, for every non-zero real number $y$ in the codomain, we found a non-zero real number $x = \frac{1}{y}$ in the domain such that $f(x) = f(\frac{1}{y}) = \frac{1}{1/y} = y$.

Therefore, for every $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$.

Hence, the function f is onto.

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Conclusion (Part 1):

The function $f : \mathbf{R}_* \to \mathbf{R}_*$ defined by $f(x) = \frac{1}{x}$ is both one-one and onto.

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Solution (Part 2):

Consider the case where the domain is $\mathbf{N}$ (set of natural numbers, {1, 2, 3, ...}) and the codomain is $\mathbf{R}_*$ (set of non-zero real numbers).

Let the new function be $g : \mathbf{N} \to \mathbf{R}_*$ defined by $g(x) = \frac{1}{x}$.

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Check if g is One-one:

Let $x_1, x_2 \in \mathbf{N}$ such that $g(x_1) = g(x_2)$.

By the definition of g, $g(x_1) = \frac{1}{x_1}$ and $g(x_2) = \frac{1}{x_2}$.

So, $\frac{1}{x_1} = \frac{1}{x_2}$.

Since $x_1, x_2 \in \mathbf{N}$, they are non-zero positive integers. Taking reciprocals, we get $x_1 = x_2$.

Thus, $g(x_1) = g(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in \mathbf{N}$.

Therefore, the function g is one-one.

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Check if g is Onto:

The domain of g is $\mathbf{N} = \{1, 2, 3, ...\}$. The codomain of g is $\mathbf{R}_*$.

We need to check if for every $y \in \mathbf{R}_*$ (codomain), there exists an $x \in \mathbf{N}$ (domain) such that $g(x) = y$.

Set $g(x) = y$: $\frac{1}{x} = y$.

Solving for $x$, we get $x = \frac{1}{y}$.

Let's consider an element in the codomain $\mathbf{R}_*$, for example, $y = 2$.

We need to find if there exists an $x \in \mathbf{N}$ such that $g(x) = 2$, which means $\frac{1}{x} = 2$.

Solving for $x$, we get $x = \frac{1}{2}$.

However, $1/2$ is not a natural number ($1/2 \notin \mathbf{N}$).

So, for the element $y=2$ in the codomain $\mathbf{R}_*$, there is no element $x$ in the domain $\mathbf{N}$ such that $g(x) = 2$.

As another example, consider $y = \sqrt{2}$. We need $x = \frac{1}{\sqrt{2}}$, which is also not a natural number.

The range of g is the set of reciprocals of natural numbers: Range(g) = $\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...\}$.

The codomain is $\mathbf{R}_*$, which includes numbers like 2, $\sqrt{2}$, $-1$, etc., that are not in the range of g.

Range(g) $\neq$ $\mathbf{R}_*$.

Therefore, the function g is not onto.

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Conclusion (Part 2):

If the domain $\mathbf{R}_*$ is replaced by $\mathbf{N}$ and the codomain remains $\mathbf{R}_*$, the function $g(x) = \frac{1}{x}$ is one-one but not onto.

So, the result (that the function is both one-one and onto) is not true in this case.

Let's check the injectivity and surjectivity for each function separately.

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(i) f : N $\to$ N given by f(x) = $x^2$

Domain (N) = {1, 2, 3, ...}, Codomain (N) = {1, 2, 3, ...}

Injectivity:

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

$x_1^2 = x_2^2$

Since $x_1, x_2 \in$ N, they are positive integers. Taking square root on both sides, we get:

$x_1 = x_2$

Thus, for different elements in the domain, the images are different.

Therefore, f is injective.

Surjectivity:

For surjectivity, for every $y$ in the codomain N, there must exist an $x$ in the domain N such that $f(x) = y$.

$x^2 = y$

This means $x = \sqrt{y}$. For $x$ to be in N, $\sqrt{y}$ must be a natural number.

Consider $y = 2$, which is in the codomain N. $x = \sqrt{2}$. This is not a natural number. There is no $x$ in N such that $f(x) = 2$.

Consider $y = 3$, which is in the codomain N. $x = \sqrt{3}$. This is not a natural number. There is no $x$ in N such that $f(x) = 3$.

The range of f is the set of perfect squares {1, 4, 9, 16, ...}, which is a proper subset of the codomain N.

Therefore, f is not surjective.

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(ii) f : Z $\to$ Z given by f(x) = $x^2$

Domain (Z) = {..., -2, -1, 0, 1, 2, ...}, Codomain (Z) = {..., -2, -1, 0, 1, 2, ...}

Injectivity:

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

$x_1^2 = x_2^2$

This implies $x_1 = \pm x_2$.

For example, consider $x_1 = 2$ and $x_2 = -2$. Both are in the domain Z.

$f(2) = 2^2 = 4$

$f(-2) = (-2)^2 = 4$

Here, $f(2) = f(-2)$ but $2 \neq -2$.

Therefore, f is not injective.

Surjectivity:

For surjectivity, for every $y$ in the codomain Z, there must exist an $x$ in the domain Z such that $f(x) = y$.

$x^2 = y$

This means $x = \pm\sqrt{y}$. For $x$ to be in Z, $y$ must be a non-negative perfect square.

Consider $y = -1$, which is in the codomain Z. $x^2 = -1$. There is no real $x$ satisfying this, so no integer $x$.

Consider $y = 2$, which is in the codomain Z. $x^2 = 2$. $x = \pm\sqrt{2}$, which are not integers.

Consider $y = 3$, which is in the codomain Z. $x^2 = 3$. $x = \pm\sqrt{3}$, which are not integers.

The range of f is the set of non-negative perfect squares {0, 1, 4, 9, ...}, which is a proper subset of the codomain Z.

Therefore, f is not surjective.

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(iii) f : R $\to$ R given by f(x) = $x^2$

Domain (R) = Set of all real numbers, Codomain (R) = Set of all real numbers.

Injectivity:

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

$x_1^2 = x_2^2$

This implies $x_1 = \pm x_2$.

For example, consider $x_1 = 2$ and $x_2 = -2$. Both are in the domain R.

$f(2) = 2^2 = 4$

$f(-2) = (-2)^2 = 4$

Here, $f(2) = f(-2)$ but $2 \neq -2$.

Therefore, f is not injective.

Surjectivity:

For surjectivity, for every $y$ in the codomain R, there must exist an $x$ in the domain R such that $f(x) = y$.

$x^2 = y$

This means $x = \pm\sqrt{y}$. For $x$ to be a real number, $\sqrt{y}$ must be a real number, which means $y$ must be non-negative ($y \geq 0$).

Consider $y = -1$, which is in the codomain R. $x^2 = -1$. There is no real $x$ satisfying this.

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

Therefore, f is not surjective.

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(iv) f : N $\to$ N given by f(x) = $x^3$

Domain (N) = {1, 2, 3, ...}, Codomain (N) = {1, 2, 3, ...}

Injectivity:

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

$x_1^3 = x_2^3$

Taking the cube root on both sides, we get:

$x_1 = x_2$

Thus, for different elements in the domain, the images are different.

Therefore, f is injective.

Surjectivity:

For surjectivity, for every $y$ in the codomain N, there must exist an $x$ in the domain N such that $f(x) = y$.

$x^3 = y$

This means $x = \sqrt[3]{y}$. For $x$ to be in N, $\sqrt[3]{y}$ must be a natural number.

Consider $y = 2$, which is in the codomain N. $x = \sqrt[3]{2}$. This is not a natural number.

Consider $y = 3$, which is in the codomain N. $x = \sqrt[3]{3}$. This is not a natural number.

Consider $y = 4$, which is in the codomain N. $x = \sqrt[3]{4}$. This is not a natural number.

The range of f is the set of perfect cubes {1, 8, 27, 64, ...}, which is a proper subset of the codomain N.

Therefore, f is not surjective.

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(v) f : Z $\to$ Z given by f(x) = $x^3$

Domain (Z) = {..., -2, -1, 0, 1, 2, ...}, Codomain (Z) = {..., -2, -1, 0, 1, 2, ...}

Injectivity:

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

$x_1^3 = x_2^3$

Taking the cube root on both sides, we get:

$x_1 = x_2$

Thus, for different elements in the domain, the images are different.

Therefore, f is injective.

Surjectivity:

For surjectivity, for every $y$ in the codomain Z, there must exist an $x$ in the domain Z such that $f(x) = y$.

$x^3 = y$

This means $x = \sqrt[3]{y}$. For $x$ to be in Z, $y$ must be a perfect cube.

Consider $y = 2$, which is in the codomain Z. $x = \sqrt[3]{2}$, which is not an integer.

Consider $y = -2$, which is in the codomain Z. $x = \sqrt[3]{-2}$, which is not an integer.

Consider $y = 3$, which is in the codomain Z. $x = \sqrt[3]{3}$, which is not an integer.

The range of f is the set of perfect cubes {..., -27, -8, -1, 0, 1, 8, 27, ...}, which is a proper subset of the codomain Z.

Therefore, f is not surjective.

Solution:

The given function is f : R $\to$ R, defined by f(x) = [x].

Here, [x] denotes the greatest integer less than or equal to x.

Domain (R) = Set of all real numbers.

Codomain (R) = Set of all real numbers.

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Injectivity (One-one):

For a function to be injective, different elements in the domain must have different images in the codomain.

We need to check if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in$ R.

Consider $x_1 = 1.5$ and $x_2 = 1.9$. Both are in the domain R.

$f(x_1) = f(1.5) = [1.5] = 1$

$f(x_2) = f(1.9) = [1.9] = 1$

Here, we have $f(1.5) = f(1.9) = 1$, but $1.5 \neq 1.9$.

Since there exist different elements in the domain that have the same image in the codomain, the function is not injective.

Therefore, f is not one-one.

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Surjectivity (Onto):

For a function to be surjective, every element in the codomain must be the image of at least one element in the domain.

We need to check if for every $y \in$ R (codomain), there exists an $x \in$ R (domain) such that $f(x) = y$, i.e., $[x] = y$.

The range of the greatest integer function $f(x) = [x]$ is the set of all integers, i.e., Z = {..., -2, -1, 0, 1, 2, ...}.

The codomain of the function is R, the set of all real numbers.

The range of f (Z) is a proper subset of the codomain (R).

For example, consider $y = 0.5$, which is in the codomain R.

We need to find an $x \in$ R such that $f(x) = [x] = 0.5$.

By the definition of the greatest integer function, [x] is always an integer. It cannot be a non-integer value like 0.5.

Thus, there is no real number $x$ such that $[x] = 0.5$.

Since there are elements in the codomain (e.g., 0.5, $\sqrt{2}$, $\pi$, etc.) that are not the image of any element in the domain, the function is not surjective.

Therefore, f is not onto.

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Conclusion:

The Greatest Integer Function f : R $\to$ R, given by f(x) = [x], is neither one-one nor onto.

Solution:

The given function is f : R $\to$ R, defined by f(x) = |x|.

The modulus function is defined as:

$|x| = \begin{cases} x & , & \text{if } x \ge 0 \\ -x & , & \text{if } x < 0 \end{cases}$

Domain (R) = Set of all real numbers.

Codomain (R) = Set of all real numbers.

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Injectivity (One-one):

For a function to be injective, if $f(x_1) = f(x_2)$, then it must imply $x_1 = x_2$ for all $x_1, x_2 \in$ R.

Let's check if this holds true for the modulus function.

Consider $x_1 = 2$ and $x_2 = -2$. Both are real numbers and belong to the domain R.

$f(x_1) = f(2) = |2| = 2$

$f(x_2) = f(-2) = |-2| = -(-2) = 2$

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

However, the inputs are different: $2 \neq -2$.

Since there exist distinct elements in the domain (2 and -2) that map to the same image in the codomain (2), the function is not injective.

Therefore, f is neither one-one.

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Surjectivity (Onto):

For a function to be surjective, every element in the codomain must have at least one pre-image in the domain. That is, for every $y \in$ R (codomain), there must exist an $x \in$ R (domain) such that $f(x) = y$, i.e., $|x| = y$.

The definition of the modulus function tells us that $|x|$ is always non-negative for any real number $x$.

$|x| \geq 0$ for all $x \in$ R.

This means the range of the function f(x) = |x| is the set of all non-negative real numbers, which is the interval $[0, \infty)$.

The codomain of the function is R, the set of all real numbers.

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

Consider an element from the codomain that is not in the range. For example, consider $y = -3$. $-3$ is a real number and is in the codomain R.

We need to find an $x \in$ R such that $f(x) = |x| = -3$.

According to the definition of the modulus, $|x|$ cannot be a negative number like -3 for any real $x$.

There is no real number $x$ whose modulus is -3.

Since there are elements in the codomain (all negative real numbers) that are not the image of any element in the domain, the function is not surjective.

Therefore, f is neither onto.

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Conclusion:

The modulus function f : R $\to$ R, given by f(x) = |x|, is neither one-one nor onto.

Solution:

The given function is f : R $\to$ R, defined by the Signum function:

$f(x) = \begin{cases} 1 & , & x > 0 \\ 0 & , & x = 0 \\ -1 & , & x < 0 \end{cases}$

Domain (R) = Set of all real numbers.

Codomain (R) = Set of all real numbers.

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Injectivity (One-one):

For a function to be injective, if $f(x_1) = f(x_2)$, then it must imply $x_1 = x_2$ for all $x_1, x_2 \in$ R.

Let's check if this holds for the Signum function.

Consider $x_1 = 5$ and $x_2 = 10$. Both are real numbers in the domain R, and both are greater than 0.

$f(x_1) = f(5) = 1$ (since $5 > 0$)

$f(x_2) = f(10) = 1$ (since $10 > 0$)

We have $f(5) = f(10) = 1$.

However, the inputs are different: $5 \neq 10$.

Since different elements in the domain (like 5 and 10) map to the same image (1) in the codomain, the function is not injective.

Similarly, consider $x_1 = -5$ and $x_2 = -10$. Both are real numbers in the domain R, and both are less than 0.

$f(x_1) = f(-5) = -1$ (since $-5 < 0$)

$f(x_2) = f(-10) = -1$ (since $-10 < 0$)

We have $f(-5) = f(-10) = -1$.

However, the inputs are different: $-5 \neq -10$.

This again shows the function is not injective.

Therefore, f is neither one-one.

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Surjectivity (Onto):

For a function to be surjective, every element in the codomain must be the image of at least one element in the domain. That is, for every $y \in$ R (codomain), there must exist an $x \in$ R (domain) such that $f(x) = y$.

Looking at the definition of the Signum function, the possible output values are limited to only three distinct values:

$f(x) = 1$ if $x > 0$

$f(x) = 0$ if $x = 0$

$f(x) = -1$ if $x < 0$

The range of the function f(x) is the set { -1, 0, 1 }.

The codomain of the function is R, the set of all real numbers.

The range { -1, 0, 1 } is a finite set and is a proper subset of the infinite set R.

Consider an element from the codomain that is not in the range. For example, consider $y = 2$. $2$ is a real number and is in the codomain R.

We need to find an $x \in$ R such that $f(x) = 2$.

Based on the definition of $f(x)$, the output can only be 1, 0, or -1. It can never be 2.

Thus, there is no real number $x$ such that $f(x) = 2$.

Similarly, any real number in the codomain R, other than -1, 0, or 1 (e.g., 0.5, -10, $\sqrt{3}$, $\pi$), does not have a pre-image in the domain R.

Since there are elements in the codomain that are not the image of any element in the domain, the function is not surjective.

Therefore, f is neither onto.

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Conclusion:

The Signum Function f : R $\to$ R, given by the definition provided, is neither one-one nor onto.

Given:

Set A = {1, 2, 3}

Set B = {4, 5, 6, 7}

Function f : A $\to$ B given by f = {(1, 4), (2, 5), (3, 6)}

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To Show:

The function f is one-one (injective).

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Solution:

A function f : A $\to$ B is said to be one-one or injective if different elements of set A have different images in set B. In other words, for any two elements $x_1, x_2 \in$ A, if $f(x_1) = f(x_2)$, then it must imply that $x_1 = x_2$.

The function f is given as the set of ordered pairs {(1, 4), (2, 5), (3, 6)}.

From the definition of f, the images of the elements in the domain A are:

$f(1) = 4$

$f(2) = 5$

$f(3) = 6$

Now, let's consider any two distinct elements from the domain A and check their images:

For elements 1 and 2 in A ($1 \neq 2$), their images are $f(1) = 4$ and $f(2) = 5$. Clearly, $4 \neq 5$.

For elements 1 and 3 in A ($1 \neq 3$), their images are $f(1) = 4$ and $f(3) = 6$. Clearly, $4 \neq 6$.

For elements 2 and 3 in A ($2 \neq 3$), their images are $f(2) = 5$ and $f(3) = 6$. Clearly, $5 \neq 6$.

In every case where the elements in the domain A are distinct (1, 2, and 3), their corresponding images in the codomain B (4, 5, and 6, respectively) are also distinct.

Alternatively, let's use the condition: if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

The possible images are 4, 5, and 6.

If $f(x_1) = f(x_2) = 4$, then from the definition of f, both $x_1$ and $x_2$ must be equal to 1. Thus, $x_1 = x_2 = 1$.

If $f(x_1) = f(x_2) = 5$, then both $x_1$ and $x_2$ must be equal to 2. Thus, $x_1 = x_2 = 2$.

If $f(x_1) = f(x_2) = 6$, then both $x_1$ and $x_2$ must be equal to 3. Thus, $x_1 = x_2 = 3$.

In all cases where the images are equal, the elements in the domain are also equal.

---

Conclusion:

Since every distinct element in set A maps to a distinct image in set B under the function f, the function f is one-one.

Solution:

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(i) f : R $\to$ R defined by f(x) = 3 – 4x

Domain (R) = Set of all real numbers.

Codomain (R) = Set of all real numbers.

Injectivity (One-one):

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

$3 - 4x_1 = 3 - 4x_2$

Subtracting 3 from both sides:

$-4x_1 = -4x_2$

Dividing by -4:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in$ R, the function f is injective (one-one).

Surjectivity (Onto):

For surjectivity, for every $y$ in the codomain R, there must exist an $x$ in the domain R such that $f(x) = y$.

Let $y \in$ R be any element in the codomain.

We set $f(x) = y$ and solve for $x$:

$3 - 4x = y$

$-4x = y - 3$

$x = \frac{y - 3}{-4}$

$x = \frac{3 - y}{4}$

Since $y$ is a real number, $\frac{3 - y}{4}$ is also a real number. Thus, for every $y \in$ R, there exists a real number $x = \frac{3 - y}{4}$ in the domain R such that $f(x) = y$.

Therefore, f is surjective (onto).

Conclusion:

Since the function f is both injective (one-one) and surjective (onto), it is bijective.

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(ii) f : R $\to$ R defined by f(x) = 1 + $x^2$

Domain (R) = Set of all real numbers.

Codomain (R) = Set of all real numbers.

Injectivity (One-one):

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

$1 + x_1^2 = 1 + x_2^2$

Subtracting 1 from both sides:

$x_1^2 = x_2^2$

Taking the square root of both sides:

$x_1 = \pm x_2$

This implies that $x_1$ is either equal to $x_2$ or $x_1$ is equal to $-x_2$. For the function to be injective, we must have $x_1 = x_2$ only.

Consider a counterexample: Let $x_1 = 2$ and $x_2 = -2$. Both are in the domain R.

$f(2) = 1 + (2)^2 = 1 + 4 = 5$

$f(-2) = 1 + (-2)^2 = 1 + 4 = 5$

Here, $f(2) = f(-2) = 5$, but $2 \neq -2$.

Since there exist distinct elements in the domain (2 and -2) that have the same image (5), the function is not injective (not one-one).

Surjectivity (Onto):

For surjectivity, for every $y$ in the codomain R, there must exist an $x$ in the domain R such that $f(x) = y$.

Let $y \in$ R be any element in the codomain.

We set $f(x) = y$ and solve for $x$:

$1 + x^2 = y$

$x^2 = y - 1$

For $x$ to be a real number, $x^2$ must be non-negative. Therefore, we must have $y - 1 \geq 0$, which means $y \geq 1$.

This shows that only real numbers $y$ that are greater than or equal to 1 can be images under this function.

The range of the function is $[1, \infty)$.

The codomain is R, which includes numbers less than 1.

Consider a value in the codomain that is less than 1, for example, $y = 0$. $0 \in$ R (codomain).

We need to find an $x \in$ R such that $f(x) = 1 + x^2 = 0$. This means $x^2 = -1$.

There is no real number $x$ whose square is -1.

Since there are elements in the codomain (e.g., all real numbers less than 1) that do not have a pre-image in the domain R, the function is not surjective (not onto).

Conclusion:

Since the function f is neither injective nor surjective, it is not bijective.

Given:

Sets A and B.

Function f : A $\times$ B $\to$ B $\times$ A defined by f(a, b) = (b, a) for every (a, b) $\in$ A $\times$ B.

---

To Show:

The function f is a bijective function.

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Solution:

To show that f is a bijective function, we need to prove that it is both injective (one-one) and surjective (onto).

Injectivity (One-one):

A function f is injective if for any two elements in the domain, their images are equal only if the elements themselves are equal.

Let $(a_1, b_1)$ and $(a_2, b_2)$ be any two elements in the domain A $\times$ B such that $f((a_1, b_1)) = f((a_2, b_2))$.

By the definition of the function f, $f((a_1, b_1)) = (b_1, a_1)$ and $f((a_2, b_2)) = (b_2, a_2)$.

So, we have:

$(b_1, a_1) = (b_2, a_2)$

By the definition of equality of ordered pairs, this means the corresponding components must be equal:

$b_1 = b_2$

$a_1 = a_2$

Since $a_1 = a_2$ and $b_1 = b_2$, it follows that the ordered pairs $(a_1, b_1)$ and $(a_2, b_2)$ are equal.

$(a_1, b_1) = (a_2, b_2)$

Therefore, $f((a_1, b_1)) = f((a_2, b_2))$ implies $(a_1, b_1) = (a_2, b_2)$.

Hence, the function f is injective (one-one).

Surjectivity (Onto):

A function f is surjective if for every element in the codomain, there exists at least one element in the domain that maps to it.

Let $(y, x)$ be an arbitrary element in the codomain B $\times$ A. By definition, $y \in$ B and $x \in$ A.

We need to find an element $(a, b)$ in the domain A $\times$ B such that $f((a, b)) = (y, x)$.

By the definition of f, $f((a, b)) = (b, a)$.

So, we set the image equal to the target element:

$(b, a) = (y, x)$

By the definition of equality of ordered pairs, this implies:

$b = y$

$a = x$

Since $x \in$ A and $y \in$ B, the ordered pair $(x, y)$ is an element of the domain A $\times$ B.

Let's check the image of $(x, y)$ under f:

$f((x, y)) = (y, x)$ (by definition of f)

So, for any element $(y, x)$ in the codomain B $\times$ A, we found an element $(x, y)$ in the domain A $\times$ B such that $f((x, y)) = (y, x)$.

Hence, every element in the codomain has a pre-image in the domain.

Therefore, the function f is surjective (onto).

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Conclusion:

Since the function f : A $\times$ B $\to$ B $\times$ A defined by f(a, b) = (b, a) is both injective (one-one) and surjective (onto), it is a bijective function.

Given:

The function $f : N \to N$, where N is the set of natural numbers ($N = \{1, 2, 3, ...\}$).

The function is defined by $f(n) = \begin{cases} \frac{n + 1}{2} & , & \text{if } n \text{ is odd} \\ \frac{n}{2} & , & \text{if } n \text{ is even} \end{cases}$ for all $n \in N$.

---

To State and Justify:

Whether the function f is bijective.

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Solution:

A function is bijective if and only if it is both injective (one-one) and surjective (onto).

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Check for Injectivity (One-one):

A function is injective if distinct elements in the domain map to distinct elements in the codomain. That is, if $n_1, n_2 \in N$ and $n_1 \neq n_2$, then $f(n_1) \neq f(n_2)$. Equivalently, if $f(n_1) = f(n_2)$, then $n_1 = n_2$.

Let's evaluate the function for the first few natural numbers:

$f(1) = \frac{1+1}{2} = \frac{2}{2} = 1$ (since 1 is odd)

$f(2) = \frac{2}{2} = 1$ (since 2 is even)

We observe that $f(1) = 1$ and $f(2) = 1$.

Here, we have two distinct elements in the domain, $n_1 = 1$ and $n_2 = 2$, such that $n_1 \neq n_2$, but their images are equal, $f(n_1) = f(n_2)$.

Since the function is not one-one, it is not injective.

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Check for Surjectivity (Onto):

A function is surjective if for every element $y$ in the codomain N, there exists at least one element $n$ in the domain N such that $f(n) = y$.

Let $y$ be an arbitrary natural number in the codomain N.

We need to find an $n \in N$ such that $f(n) = y$.

Consider $n_1 = 2y - 1$. If $y \in N$, then $y \ge 1$, so $2y \ge 2$, and $2y - 1 \ge 1$. Thus, $n_1 = 2y - 1$ is a natural number. Also, $2y - 1$ is always an odd number.

Let's find the image of $n_1 = 2y - 1$ under f:

$f(n_1) = f(2y - 1) = \frac{(2y - 1) + 1}{2} = \frac{2y}{2} = y$ (since $2y-1$ is odd).

Consider $n_2 = 2y$. If $y \in N$, then $y \ge 1$, so $2y \ge 2$. Thus, $n_2 = 2y$ is an even natural number.

Let's find the image of $n_2 = 2y$ under f:

$f(n_2) = f(2y) = \frac{2y}{2} = y$ (since $2y$ is even).

So, for any natural number $y$ in the codomain, we can find at least one natural number $n$ in the domain (in fact, we found two: $2y-1$ and $2y$) such that $f(n) = y$.

For example, if $y=3$, we can take $n=2(3)-1=5$ (odd), $f(5) = (5+1)/2=3$. Or we can take $n=2(3)=6$ (even), $f(6)=6/2=3$. Both 5 and 6 are in the domain N.

Since every element in the codomain N has at least one pre-image in the domain N, the function f is surjective (onto).

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Conclusion:

The function f is surjective (onto) but not injective (one-one).

For a function to be bijective, it must be both injective and surjective.

Since f is not injective, it is not bijective.

Given:

Set A = R – {3}

Set B = R – {1}

Function f : A $\to$ B defined by $f(x) = \frac{x - 2}{x - 3}$ for all $x \in$ A.

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To Determine:

Whether the function f is one-one (injective) and onto (surjective).

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Solution:

To justify whether f is one-one and onto, we check for injectivity and surjectivity.

Injectivity (One-one):

A function f : A $\to$ B is injective if for any two elements $x_1, x_2 \in$ A, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

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

$\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}$

Since $x_1, x_2 \in$ A, $x_1 \neq 3$ and $x_2 \neq 3$. Thus, $x_1 - 3 \neq 0$ and $x_2 - 3 \neq 0$. We can cross-multiply:

$(x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)$

Expand both sides:

$x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6$

Subtract $x_1x_2$ and 6 from both sides:

$-3x_1 - 2x_2 = -3x_2 - 2x_1$

Rearrange the terms to group $x_1$ and $x_2$:

$-3x_1 + 2x_1 = -3x_2 + 2x_2$

$-x_1 = -x_2$

Multiply by -1:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in$ A, the function f is injective (one-one).

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Surjectivity (Onto):

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

Let $y \in$ B be an arbitrary element in the codomain. By definition of B, $y$ is a real number and $y \neq 1$.

We set $f(x) = y$ and solve for $x$ in terms of $y$:

$y = \frac{x - 2}{x - 3}$

Since $x \in$ A, $x \neq 3$, so $x - 3 \neq 0$. We can multiply both sides by $(x - 3)$:

$y(x - 3) = x - 2$

Distribute y on the left side:

$yx - 3y = x - 2$

Collect all terms containing $x$ on one side and the remaining terms on the other side:

$yx - x = 3y - 2$

Factor out $x$ from the left side:

$x(y - 1) = 3y - 2$

Since $y \in$ B, we know that $y \neq 1$. Therefore, $y - 1 \neq 0$. We can divide both sides by $(y - 1)$:

$x = \frac{3y - 2}{y - 1}$

Now, we need to verify if this value of $x$ is always in the domain A for every $y \in$ B.

For $x$ to be in A, $x$ must be a real number and $x \neq 3$.

Since $y$ is a real number ($y \in$ B $\subset$ R) and $y - 1 \neq 0$, the expression $x = \frac{3y - 2}{y - 1}$ results in a real number for any $y \in$ B.

Next, we check if $x$ can be equal to 3 for any $y \in$ B. Assume $x = 3$ and substitute it into the expression for $x$ in terms of $y$:

$3 = \frac{3y - 2}{y - 1}$

Multiply by $(y - 1)$ (which is non-zero since $y \in$ B):

$3(y - 1) = 3y - 2$

$3y - 3 = 3y - 2$

Subtract $3y$ from both sides:

$-3 = -2$

This is a contradiction. This means our assumption that $x$ can be equal to 3 for some $y \in$ B is false. Therefore, for every $y \in$ B, the calculated value of $x$ is never equal to 3.

Since $x$ is always a real number and $x \neq 3$, the calculated $x$ is always in the domain A = R – {3}.

Thus, for every $y \in$ B, there exists an $x \in$ A such that $f(x) = y$.

Therefore, the function f is surjective (onto).

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Conclusion:

Since the function f is both injective (one-one) and surjective (onto), it is a bijective function.

p>The given function is f : R $\to$ R, defined by f(x) = $x^4$.

Domain (R) = Set of all real numbers.

Codomain (R) = Set of all real numbers.

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Injectivity (One-one):

For a function to be injective, if $f(x_1) = f(x_2)$ for $x_1, x_2 \in$ R, then it must imply $x_1 = x_2$.

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

$x_1^4 = x_2^4$

Taking the fourth root on both sides, we get:

$x_1 = \pm x_2$

This means that $x_1$ can be equal to $x_2$ or $x_1$ can be equal to $-x_2$. For the function to be one-one, we require $x_1 = x_2$ always.

Consider a counterexample: Let $x_1 = 2$ and $x_2 = -2$. Both are in the domain R.

$f(2) = 2^4 = 16$

$f(-2) = (-2)^4 = 16$

Here, $f(2) = f(-2) = 16$, but $2 \neq -2$.

Since there exist distinct elements in the domain (2 and -2) that have the same image (16) in the codomain, the function is not injective (not one-one).

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Surjectivity (Onto):

For a function to be surjective, every element in the codomain must be the image of at least one element in the domain. That is, for every $y \in$ R (codomain), there must exist an $x \in$ R (domain) such that $f(x) = y$, i.e., $x^4 = y$.

The fourth power of any real number is always non-negative.

$x^4 \ge 0$ for all $x \in$ R.

This means the range of the function f(x) = $x^4$ is the set of all non-negative real numbers, which is the interval $[0, \infty)$.

The codomain of the function is R, the set of all real numbers.

The range $[0, \infty)$ is a proper subset of the codomain R (which includes negative numbers).

Consider a value in the codomain that is not in the range. For example, let $y = -5$. $-5$ is a real number and is in the codomain R.

We need to find an $x \in$ R such that $f(x) = x^4 = -5$.

There is no real number $x$ whose fourth power is a negative number.

Since there are elements in the codomain (all negative real numbers) that do not have a pre-image in the domain R, the function is not surjective (not onto).

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Conclusion:

The function f is neither injective nor surjective.

Based on the analysis:

(A) f is one-one onto - False (not one-one, not onto)

(B) f is many-one onto - False (many-one, but 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.

Solution:

The given function is f : R $\to$ R, defined by f(x) = 3x.

Domain (R) = Set of all real numbers.

Codomain (R) = Set of all real numbers.

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Injectivity (One-one):

For a function to be injective, if $f(x_1) = f(x_2)$ for any $x_1, x_2 \in$ R, then it must imply $x_1 = x_2$.

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

$3x_1 = 3x_2$

Divide both sides by 3:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in$ R, the function f is injective (one-one).

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Surjectivity (Onto):

For a function to be surjective, for every element $y$ in the codomain R, there must exist an element $x$ in the domain R such that $f(x) = y$.

Let $y \in$ R be any element in the codomain.

We set $f(x) = y$ and solve for $x$:

$3x = y$

Divide both sides by 3:

$x = \frac{y}{3}$

Since $y$ is a real number, $\frac{y}{3}$ is also a real number. Thus, for every $y \in$ R, there exists a real number $x = \frac{y}{3}$ in the domain R such that $f(x) = 3 \left( \frac{y}{3} \right) = y$.

Therefore, f is surjective (onto).

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Conclusion:

Since the function f is both injective (one-one) and surjective (onto), it is bijective.

Based on the analysis:

(A) f is one-one onto - True

(B) f is many-one onto - False (it is one-one)

(C) f is one-one but not onto - False (it is onto)

(D) f is neither one-one nor onto - False (it is both)

The correct answer is (A) f is one-one onto.

Given:

Function f : {2, 3, 4, 5} $\to$ {3, 4, 5, 9} defined by:

$f(2) = 3$

$f(3) = 4$

$f(4) = 5$

$f(5) = 5$

Function g : {3, 4, 5, 9} $\to$ {7, 11, 15} defined by:

$g(3) = 7$

$g(4) = 7$

$g(5) = 11$

$g(9) = 11$

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To Find:

The composite function gof.

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Solution:

The composite function gof is defined as $(g \circ f)(x) = g(f(x))$ for all $x$ in the domain of f.

The domain of f is {2, 3, 4, 5}. We need to find the value of gof for each element in this domain.

For $x = 2$:

$(g \circ f)(2) = g(f(2))$

Since $f(2) = 3$, we have:

$(g \circ f)(2) = g(3)$

Since $g(3) = 7$, we have:

$(g \circ f)(2) = 7$

For $x = 3$:

$(g \circ f)(3) = g(f(3))$

Since $f(3) = 4$, we have:

$(g \circ f)(3) = g(4)$

Since $g(4) = 7$, we have:

$(g \circ f)(3) = 7$

For $x = 4$:

$(g \circ f)(4) = g(f(4))$

Since $f(4) = 5$, we have:

$(g \circ f)(4) = g(5)$

Since $g(5) = 11$, we have:

$(g \circ f)(4) = 11$

For $x = 5$:

$(g \circ f)(5) = g(f(5))$

Since $f(5) = 5$, we have:

$(g \circ f)(5) = g(5)$

Since $g(5) = 11$, we have:

$(g \circ f)(5) = 11$

The composite function gof maps elements from {2, 3, 4, 5} to {7, 11}.

We can represent gof as a set of ordered pairs:

gof = {(2, 7), (3, 7), (4, 11), (5, 11)}

Given:

Function f : R $\to$ R defined by f(x) = cos x

Function g : R $\to$ R defined by g(x) = $3x^2$

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To Find:

The composite functions gof and fog.

To Show:

gof $\neq$ fog.

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Solution:

Finding gof:

The composite function gof is defined as $(g \circ f)(x) = g(f(x))$.

Substitute $f(x) = \cos x$ into g(x):

$(g \circ f)(x) = g(\cos x)$

Using the definition of g(x) = $3x^2$, where the input is now $\cos x$:

$(g \circ f)(x) = 3(\cos x)^2$

$(g \circ f)(x) = 3\cos^2 x$

So, gof : R $\to$ R is defined by $(g \circ f)(x) = 3\cos^2 x$.

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Finding fog:

The composite function fog is defined as $(f \circ g)(x) = f(g(x))$.

Substitute $g(x) = 3x^2$ into f(x):

$(f \circ g)(x) = f(3x^2)$

Using the definition of f(x) = $\cos x$, where the input is now $3x^2$:

$(f \circ g)(x) = \cos(3x^2)$

So, fog : R $\to$ R is defined by $(f \circ g)(x) = \cos(3x^2)$.

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Showing gof $\neq$ fog:

To show that gof $\neq$ fog, we need to find at least one value of x in the domain (R) for which $(g \circ f)(x) \neq (f \circ g)(x)$.

We have found that:

$(g \circ f)(x) = 3\cos^2 x$

$(f \circ g)(x) = \cos(3x^2)$

These two expressions are generally not equal for all values of x.

Consider a specific value for x, for example, $x = \frac{\pi}{2}$.

For gof:

$(g \circ f)\left(\frac{\pi}{2}\right) = 3\cos^2 \left(\frac{\pi}{2}\right)$

Since $\cos \left(\frac{\pi}{2}\right) = 0$, we have:

$(g \circ f)\left(\frac{\pi}{2}\right) = 3(0)^2 = 3 \times 0 = 0$

For fog:

$(f \circ g)\left(\frac{\pi}{2}\right) = \cos\left(3 \left(\frac{\pi}{2}\right)^2\right)$

$(f \circ g)\left(\frac{\pi}{2}\right) = \cos\left(3 \left(\frac{\pi^2}{4}\right)\right)$

$(f \circ g)\left(\frac{\pi}{2}\right) = \cos\left(\frac{3\pi^2}{4}\right)$

The value $\cos\left(\frac{3\pi^2}{4}\right)$ is approximately $\cos\left(\frac{3 \times (9.87)}{4}\right) = \cos(7.4025)$. This is a value between -1 and 1, but it is not 0.

Since $0 \neq \cos\left(\frac{3\pi^2}{4}\right)$, we have shown that $(g \circ f)\left(\frac{\pi}{2}\right) \neq (f \circ g)\left(\frac{\pi}{2}\right)$.

Therefore, gof $\neq$ fog.

Given:

Function f : N $\to$ Y defined by f(x) = 4x + 3.

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

Codomain is Y = {y $\in$ N: y = 4x + 3 for some x $\in$ N}.

The set Y is the range of the function f.

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To Show:

The function f(x) is invertible.

To Find:

The inverse function, $f^{-1}$.

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Solution:

A function is invertible if and only if it is bijective (both injective and surjective).

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:

$4x_1 = 4x_2$

Divide by 4:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in$ N, the function f is injective (one-one).

Surjectivity (Onto):

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

In this case, the codomain Y is defined as the set of all values $y$ in N that can be expressed as $4x + 3$ for some $x \in$ N. This is precisely the definition of the range of the function f.

Since the codomain Y is equal to the range of the function f, for every element $y$ in the codomain Y, there exists an element $x$ in the domain N such that $f(x) = y$.

Therefore, f is surjective (onto) by the definition of the set Y.

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Invertibility:

Since the function f is both injective (one-one) and surjective (onto), it is bijective. Therefore, the function f is invertible.

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Finding the inverse function $f^{-1}$:

The inverse function $f^{-1}$ maps elements from the codomain Y back to the domain N. So, $f^{-1}$ : Y $\to$ N.

Let $y$ be an element in the codomain Y. We have $y = f(x)$ for some $x \in$ N.

$y = 4x + 3$

To find the inverse function, we solve this equation for $x$ in terms of $y$:

$y - 3 = 4x$

$x = \frac{y - 3}{4}$

The inverse function $f^{-1}$ maps $y$ to $x$. So, we can write $f^{-1}(y) = x$.

$f^{-1}(y) = \frac{y - 3}{4}$

The domain of $f^{-1}$ is Y, and the codomain of $f^{-1}$ is N.

We can write the inverse function using the variable $x$ instead of $y$, by replacing $y$ with $x$ in the expression for $f^{-1}(y)$:

$f^{-1}(x) = \frac{x - 3}{4}$

The domain of $f^{-1}$ is Y, and the codomain of $f^{-1}$ is N.

Note that for $f^{-1}(x)$ to be in N, $x$ must be in Y, meaning $x$ must be of the form $4k+3$ for some $k \in$ N. If $x = 4k+3$, then $f^{-1}(x) = \frac{(4k+3) - 3}{4} = \frac{4k}{4} = k$, which is in N.

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Conclusion:

The function f is invertible because it is bijective. The inverse function is $f^{-1}$ : Y $\to$ N defined by $f^{-1}(y) = \frac{y - 3}{4}$ (or $f^{-1}(x) = \frac{x - 3}{4}$).

Given:

R$_1$ and R$_2$ are equivalence relations in a set A.

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To Show:

R$_1 \cap$ R$_2$ is also an equivalence relation in A.

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Solution:

To show that R$_1 \cap$ R$_2$ is an equivalence relation, we need to prove that it is reflexive, symmetric, and transitive.

Since R$_1$ and R$_2$ are equivalence relations on A, they satisfy the following properties:

For R$_1$:

1. Reflexive: $(a, a) \in$ R$_1$ for all $a \in$ A.

2. Symmetric: If $(a, b) \in$ R$_1$, then $(b, a) \in$ R$_1$ for all $a, b \in$ A.

3. Transitive: If $(a, b) \in$ R$_1$ and $(b, c) \in$ R$_1$, then $(a, c) \in$ R$_1$ for all $a, b, c \in$ A.

For R$_2$:

1. Reflexive: $(a, a) \in$ R$_2$ for all $a \in$ A.

2. Symmetric: If $(a, b) \in$ R$_2$, then $(b, a) \in$ R$_2$ for all $a, b \in$ A.

3. Transitive: If $(a, b) \in$ R$_2$ and $(b, c) \in$ R$_2$, then $(a, c) \in$ R$_2$ for all $a, b, c \in$ A.

Let R = R$_1 \cap$ R$_2$. An ordered pair $(a, b)$ is in R if and only if $(a, b) \in$ R$_1$ and $(a, b) \in$ R$_2$.

Now, let's check the three properties for R = R$_1 \cap$ R$_2$:

1. Reflexivity:

We need to show that $(a, a) \in$ R for all $a \in$ A.

Since R$_1$ is reflexive, $(a, a) \in$ R$_1$ for all $a \in$ A.

Since R$_2$ is reflexive, $(a, a) \in$ R$_2$ for all $a \in$ A.

Since $(a, a)$ is in both R$_1$ and R$_2$, by the definition of intersection, $(a, a) \in$ R$_1 \cap$ R$_2$ for all $a \in$ A.

Therefore, R is reflexive.

2. Symmetry:

We need to show that if $(a, b) \in$ R, then $(b, a) \in$ R for all $a, b \in$ A.

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

By the definition of R = R$_1 \cap$ R$_2$, this means $(a, b) \in$ 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$_2$.

Since $(b, a)$ is in both R$_1$ and R$_2$, by the definition of intersection, $(b, a) \in$ R$_1 \cap$ R$_2$.

Therefore, if $(a, b) \in$ R, then $(b, a) \in$ R. So, 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 $a, b, c \in$ A.

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

Since $(a, b) \in$ R, by the definition of R = R$_1 \cap$ R$_2$, this means $(a, b) \in$ R$_1$ and $(a, b) \in$ R$_2$.

Since $(b, c) \in$ R, by the definition of R = R$_1 \cap$ R$_2$, this means $(b, c) \in$ 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$_2$.

Since $(a, c)$ is in both R$_1$ and R$_2$, by the definition of intersection, $(a, c) \in$ R$_1 \cap$ R$_2$.

Therefore, if $(a, b) \in$ R and $(b, c) \in$ R, then $(a, c) \in$ R. So, R is transitive.

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Conclusion:

Since R = R$_1 \cap$ R$_2$ is reflexive, symmetric, and transitive, it is an equivalence relation on the set A.

Given:

A set A of ordered pairs of positive integers, i.e., A = N $\times$ N.

A relation R on A defined by $(x, y) \text{ R } (u, v)$ if and only if $xv = yu$, where $(x, y), (u, v) \in$ A.

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To Show:

R is an equivalence relation on A.

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Solution:

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

Let $(x, y), (u, v), (a, b)$ be arbitrary elements in A. Since A is the set of ordered pairs of positive integers, we have $x, y, u, v, a, b \in$ N.

1. Reflexivity:

We need to show that $(x, y) \text{ R } (x, y)$ for every $(x, y) \in$ A.

According to the definition of R, $(x, y) \text{ R } (x, y)$ if and only if the product of the first component of the first pair and the second component of the second pair is equal to the product of the second component of the first pair and the first component of the second pair.

So, we need to check if $x y = y x$.

Since $x$ and $y$ are positive integers, the multiplication of $x$ and $y$ is commutative in the set of integers (and thus in positive integers). That is, $xy = yx$.

Thus, $(x, y) \text{ R } (x, y)$ for all $(x, y) \in$ A.

Therefore, R is reflexive.

2. Symmetry:

We need to show that if $(x, y) \text{ R } (u, v)$, then $(u, v) \text{ R } (x, y)$ for all $(x, y), (u, v) \in$ A.

Assume $(x, y) \text{ R } (u, v)$.

By the definition of R, this means $xv = yu$.

We need to show that $(u, v) \text{ R } (x, y)$, which means checking if $uy = vx$.

Given $xv = yu$.

Since the multiplication of integers is commutative, we can write $yu = xv$ as $uy = vx$.

This is exactly the condition for $(u, v) \text{ R } (x, y)$.

Therefore, if $(x, y) \text{ R } (u, v)$, then $(u, v) \text{ R } (x, y)$. So, R is symmetric.

3. Transitivity:

We need to show that if $(x, y) \text{ R } (u, v)$ and $(u, v) \text{ R } (a, b)$, then $(x, y) \text{ R } (a, b)$ for all $(x, y), (u, v), (a, b) \in$ A.

Assume $(x, y) \text{ R } (u, v)$ and $(u, v) \text{ R } (a, b)$.

From $(x, y) \text{ R } (u, v)$, we have $xv = yu$ ... (1)

From $(u, v) \text{ R } (a, b)$, we have $ub = va$ ... (2)

We need to show that $(x, y) \text{ R } (a, b)$, which means we need to show $xb = ya$.

From equation (1), $xv = yu$. Multiply both sides by $b$:

$xvb = yub$ ... (3)

From equation (2), $ub = va$. Multiply both sides by $y$:

$uby = vay$ ... (4)

The left side of (3) is $xvb$. The right side of (4) is $vay$.

The term $ub$ appears in the right side of (3) ($yub$) and the left side of (4) ($uby$). Since multiplication is commutative, $yub = uby$.

So, from (3) and (4), we have $xvb = yub$ and $uby = vay$. Since $yub = uby$, we can write $xvb = uby$ and $uby = vay$.

By transitivity of equality, $xvb = vay$.

So, $xvb = vay$.

We want to show $xb = ya$. We can divide both sides of $xvb = vay$ by $v$. Since $v$ is a positive integer (part of an ordered pair of positive integers), $v \neq 0$.

$\frac{xvb}{v} = \frac{vay}{v}$

$\frac{xb\cancel{v}}{\cancel{v}} = \frac{ya\cancel{v}}{\cancel{v}}$

$xb = ya$

This is the condition for $(x, y) \text{ R } (a, b)$.

Therefore, if $(x, y) \text{ R } (u, v)$ and $(u, v) \text{ R } (a, b)$, then $(x, y) \text{ R } (a, b)$. So, R is transitive.

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Conclusion:

Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation on the set A of ordered pairs of positive integers.

Note: This relation R essentially establishes that two ordered pairs $(x, y)$ and $(u, v)$ are related if the fractions $\frac{x}{y}$ and $\frac{u}{v}$ are equivalent, given that $x, y, u, v$ are positive integers.

Given:

Set X = {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Relation R$_1$ on X: R$_1$ = {(x, y) : x, y $\in$ X and x – y is divisible by 3}.

Relation R$_2$ on X: R$_2$ = {(x, y) : x, y $\in$ X and ({x, y} $\subset$ {1, 4, 7} or {x, y} $\subset$ {2, 5, 8} or {x, y} $\subset$ {3, 6, 9})}.

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To Show:

R$_1$ = R$_2$

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Solution (Proof):

To show that R$_1$ = R$_2$, we must show that every ordered pair in R$_1$ is also in R$_2$ (i.e., R$_1 \subseteq$ R$_2$) and every ordered pair in R$_2$ is also in R$_1$ (i.e., R$_2 \subseteq$ R$_1$).

Let's first analyze the condition for R$_1$. An ordered pair $(x, y) \in$ R$_1$ if and only if $x - y$ is divisible by 3. This is equivalent to saying that $x \equiv y \pmod{3}$. This means that $x$ and $y$ have the same remainder when divided by 3.

Let's partition the set X based on the remainder when divided by 3:

Set of elements with remainder 1: $\{z \in X : z \equiv 1 \pmod{3}\} = \{1, 4, 7\}$

Set of elements with remainder 2: $\{z \in X : z \equiv 2 \pmod{3}\} = \{2, 5, 8\}$

Set of elements with remainder 0: $\{z \in X : z \equiv 0 \pmod{3}\} = \{3, 6, 9\}$

So, $(x, y) \in$ R$_1$ if and only if $x$ and $y$ belong to the same set among $\{1, 4, 7\}$, $\{2, 5, 8\}$, or $\{3, 6, 9\}$.

Now, let's analyze the condition for R$_2$. An ordered pair $(x, y) \in$ R$_2$ if and only if $\{x, y\} \subset \{1, 4, 7\}$ or $\{x, y\} \subset \{2, 5, 8\}$ or $\{x, y\} \subset \{3, 6, 9\}$.

The notation $\{x, y\} \subset S$ means that both elements $x$ and $y$ are members of the set $S$. So, the condition for R$_2$ means that $x$ and $y$ are both in $\{1, 4, 7\}$, or $x$ and $y$ are both in $\{2, 5, 8\}$, or $x$ and $y$ are both in $\{3, 6, 9\}$.

This is precisely the same condition as derived for R$_1$. The condition "$x$ and $y$ belong to the same set among $\{1, 4, 7\}$, $\{2, 5, 8\}$, or $\{3, 6, 9\}$" is equivalent to the condition "$x$ and $y$ have the same remainder when divided by 3".

Let's demonstrate the equivalence formally:

Part 1: Show R$_1 \subseteq$ R$_2$

Let $(x, y) \in$ R$_1$. By definition of R$_1$, $x - y$ is divisible by 3. This implies $x$ and $y$ have the same remainder when divided by 3. Since the sets $\{1, 4, 7\}$, $\{2, 5, 8\}$, and $\{3, 6, 9\}$ partition X based on remainders modulo 3, both $x$ and $y$ must belong to the same set from this partition. If $x$ and $y$ are both in $\{1, 4, 7\}$, then $\{x, y\} \subset \{1, 4, 7\}$. If $x$ and $y$ are both in $\{2, 5, 8\}$, then $\{x, y\} \subset \{2, 5, 8\}$. If $x$ and $y$ are both in $\{3, 6, 9\}$, then $\{x, y\} \subset \{3, 6, 9\}$. In any of these cases, the condition for $(x, y) \in$ R$_2$ is satisfied. Therefore, R$_1 \subseteq$ R$_2$.

Part 2: Show R$_2 \subseteq$ R$_1$

Let $(x, y) \in$ R$_2$. By definition of R$_2$, one of the following conditions must be true:

(a) $\{x, y\} \subset \{1, 4, 7\}$. This means both $x$ and $y$ are in $\{1, 4, 7\}$. Any two numbers from this set have a difference that is a multiple of 3 (e.g., $4-1=3$, $7-1=6$, $7-4=3$). Therefore, $x - y$ is divisible by 3, which means $(x, y) \in$ R$_1$.

(b) $\{x, y\} \subset \{2, 5, 8\}$. This means both $x$ and $y$ are in $\{2, 5, 8\}$. Any two numbers from this set have a difference that is a multiple of 3 (e.g., $5-2=3$, $8-2=6$, $8-5=3$). Therefore, $x - y$ is divisible by 3, which means $(x, y) \in$ R$_1$.

(c) $\{x, y\} \subset \{3, 6, 9\}$. This means both $x$ and $y$ are in $\{3, 6, 9\}$. Any two numbers from this set have a difference that is a multiple of 3 (e.g., $6-3=3$, $9-3=6$, $9-6=3$). Therefore, $x - y$ is divisible by 3, which means $(x, y) \in$ R$_1$.

In all possible cases where $(x, y) \in$ R$_2$, we have shown that $(x, y) \in$ R$_1$. Therefore, R$_2 \subseteq$ R$_1$.

Since R$_1 \subseteq$ R$_2$ and R$_2 \subseteq$ R$_1$, it follows that R$_1$ and R$_2$ contain exactly the same ordered pairs.

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Conclusion:

Thus, R$_1$ = R$_2$.

Given:

A function f : X $\to$ Y.

A relation R in X defined by R = {(a, b) : f(a) = f(b)}.

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To Examine:

Whether R is an equivalence relation.

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Solution:

To determine if R is an equivalence relation on set X, we need to check if it satisfies the three properties: reflexivity, symmetry, and transitivity.

Let a, b, and c be arbitrary elements in the set X.

1. Reflexivity:

For R to be reflexive, for every element $a \in$ X, the ordered pair $(a, a)$ must be in R.

According to the definition of R, $(a, a) \in$ R if and only if $f(a) = f(a)$.

The equality $f(a) = f(a)$ is always true for any element $a$ in the domain X, as a function maps an element to exactly one element in the codomain.

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

Therefore, R is reflexive.

2. Symmetry:

For R to be symmetric, if $(a, b) \in$ R for any $a, b \in$ X, then $(b, a)$ must also be in R.

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

By the definition of R, this means $f(a) = f(b)$.

We need to check if $(b, a) \in$ R, which means checking if $f(b) = f(a)$.

Since the equality relation is symmetric, if $f(a) = f(b)$, it is also true that $f(b) = f(a)$.

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

Therefore, R is symmetric.

3. Transitivity:

For R to be transitive, if $(a, b) \in$ R and $(b, c) \in$ R for any $a, b, c \in$ X, then $(a, c)$ must also be in R.

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

From $(a, b) \in$ R, by the definition of R, we have $f(a) = f(b)$.

From $(b, c) \in$ R, by the definition of R, we have $f(b) = f(c)$.

We have $f(a) = f(b)$ and $f(b) = f(c)$. By the transitive property of equality, if $f(a)$ is equal to $f(b)$, and $f(b)$ is equal to $f(c)$, then $f(a)$ must be equal to $f(c)$.

So, $f(a) = f(c)$.

By the definition of R, $f(a) = f(c)$ means 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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Conclusion:

Since the relation R satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on the set X.

This relation R partitions the set X into disjoint equivalence classes, where each class consists of all elements in X that are mapped to the same element in Y by the function f. The equivalence classes correspond to the pre-images of the elements in the range of f.

Given:

Set A = {1, 2, 3}.

We are considering functions f : A $\to$ A.

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To Find:

The number of all one-one functions from set A to itself.

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Solution:

Let the elements of set A be $a_1 = 1$, $a_2 = 2$, and $a_3 = 3$. The codomain is also set A, with elements 1, 2, and 3.

A function f : A $\to$ A is one-one (injective) if every distinct element in the domain A has a distinct image in the codomain A.

Let's determine the possible images for each element in the domain to form a one-one function:

Consider the element 1 from the domain. It can be mapped to any of the 3 elements in the codomain {1, 2, 3}. So, there are 3 choices for $f(1)$.

Consider the element 2 from the domain. Since the function must be one-one, the image of 2, $f(2)$, cannot be the same as $f(1)$. Since one element from the codomain has already been used as the image of 1, there are only $3 - 1 = 2$ remaining elements in the codomain that 2 can be mapped to. So, there are 2 choices for $f(2)$.

Consider the element 3 from the domain. Since the function must be one-one, the image of 3, $f(3)$, cannot be the same as $f(1)$ or $f(2)$. Since two elements from the codomain have already been used as the images of 1 and 2, there is only $3 - 2 = 1$ remaining element in the codomain that 3 can be mapped to. So, there is 1 choice for $f(3)$.

To find the total number of one-one functions, we multiply the number of choices for the image of each element in the domain:

Total number of one-one functions = (Number of choices for $f(1)$) $\times$ (Number of choices for $f(2)$) $\times$ (Number of choices for $f(3)$)

Total number of one-one functions = $3 \times 2 \times 1 = 6$

This is equivalent to finding the number of permutations of 3 elements taken 3 at a time, denoted as P(3, 3) or $3!$.

For a set A with n elements, the number of one-one functions from A to itself is n!.

In this case, n = 3. So, the number of one-one functions is $3! = 3 \times 2 \times 1 = 6$.

These 6 one-one functions are also onto (surjective) because the domain and codomain have the same finite number of elements. A function from a finite set to itself is one-one if and only if it is onto. Therefore, these 6 functions are also bijective.

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Conclusion:

The number of all one-one functions from set A = {1, 2, 3} to itself is 6.

The possible one-one functions are:

1. f = {(1, 1), (2, 2), (3, 3)}

2. f = {(1, 1), (2, 3), (3, 2)}

3. f = {(1, 2), (2, 1), (3, 3)}

4. f = {(1, 2), (2, 3), (3, 1)}

5. f = {(1, 3), (2, 1), (3, 2)}

6. f = {(1, 3), (2, 2), (3, 1)}

Given:

Set A = $\{1, 2, 3\}$.

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To Show:

The number of relations on A containing (1, 2) and (2, 3) which are reflexive and transitive but not symmetric is three.

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Solution:

Let R be a relation on the set $A = \{1, 2, 3\}$. The relation R must satisfy the following properties:

1. Reflexive: $(a, a) \in R$ for all $a \in A$. Thus, R must contain $\{(1, 1), (2, 2), (3, 3)\}$.

2. Contains (1, 2) and (2, 3): $(1, 2) \in R$ and $(2, 3) \in R$.

3. Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

4. Not Symmetric: There exists $(a, b) \in R$ such that $(b, a) \notin R$.

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From the required properties, R must contain $\{(1, 1), (2, 2), (3, 3)\}$ (for reflexivity) and $\{(1, 2), (2, 3)\}$ (given). For transitivity, since $(1, 2) \in R$ and $(2, 3) \in R$, the pair $(1, 3)$ must also be in R.

The smallest relation satisfying these conditions is $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$.

We verify that $R_1$ is reflexive, contains (1, 2) and (2, 3), and is transitive. We check for symmetry: $(1, 2) \in R_1$ but $(2, 1) \notin R_1$. Therefore, $R_1$ is not symmetric.

So, $R_1$ is one such relation.

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Now, we consider adding other pairs from $A \times A \setminus R_1 = \{(2, 1), (3, 1), (3, 2)\}$ to $R_1$ such that the resulting relation remains transitive and not symmetric.

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1. Adding the pair $(2, 1)$:

Consider $R_2 = R_1 \cup \{(2, 1)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$.

This relation is reflexive and contains the required pairs.

We check for transitivity. The addition of $(2, 1)$ requires checking new paths involving 2 and 1. All such paths result in pairs already present in $R_2$. So, $R_2$ is transitive.

$R_2$ is not symmetric since, for example, $(1, 3) \in R_2$ but $(3, 1) \notin R_2$.

Thus, $R_2$ is a second such relation.

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2. Adding the pair $(3, 2)$:

Consider $R_3 = R_1 \cup \{(3, 2)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$.

This relation is reflexive and contains the required pairs.

We check for transitivity. The addition of $(3, 2)$ requires checking new paths involving 3 and 2. All such paths result in pairs already present in $R_3$. So, $R_3$ is transitive.

$R_3$ is not symmetric since, for example, $(1, 2) \in R_3$ but $(2, 1) \notin R_3$.

Thus, $R_3$ is a third such relation.

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3. Adding the pair $(3, 1)$ or any combination that forces $(3, 1)$:

If a relation R contains $(3, 1)$, then for transitivity to hold:

Since $(2, 3) \in R_1 \subseteq R$ and $(3, 1) \in R$, it must be that $(2, 1) \in R$.

Similarly, since $(3, 1) \in R$ and $(1, 2) \in R_1 \subseteq R$, it must be that $(3, 2) \in R$.

Thus, any transitive relation satisfying the conditions and containing $(3, 1)$ must also contain $(2, 1)$ and $(3, 2)$.

The relation $R_1 \cup \{(3, 1), (2, 1), (3, 2)\}$ is the complete set $A \times A = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\}$.

The relation $A \times A$ is a symmetric relation.

Since the required relations must *not* be symmetric, no relation containing $(3, 1)$ satisfies the condition.

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4. Adding both $(2, 1)$ and $(3, 2)$:

Consider $R_4 = R_1 \cup \{(2, 1), (3, 2)\}$. This relation is reflexive and contains the required pairs.

However, for $R_4$ to be transitive, since $(3, 2) \in R_4$ and $(2, 1) \in R_4$, the pair $(3, 1)$ must be in $R_4$.

But $(3, 1) \notin R_4$. Thus, $R_4$ is not transitive.

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The only relations satisfying all the given conditions (containing (1, 2) and (2, 3), reflexive, transitive, and not symmetric) are $R_1$, $R_2$, and $R_3$. There are exactly three such relations.

Thus, 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 considering equivalence relations R on A that contain the pairs $(1, 2)$ and $(2, 1)$.

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To Show:

The number of such equivalence relations on A is two.

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Solution:

An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive.

There is a fundamental connection between equivalence relations on a set and partitions of that set. Every equivalence relation on a set corresponds to a unique partition of the set, and every partition of a set corresponds to a unique equivalence relation.

The condition that a relation R contains the pair $(a, b)$ means that the elements $a$ and $b$ are related under R. In the context of equivalence relations and partitions, if $(a, b) \in R$, then $a$ and $b$ belong to the same part (subset) in the corresponding partition of the set.

We are given that the equivalence relation R must contain $(1, 2)$ and $(2, 1)$. According to the correspondence, this means that the elements 1 and 2 must belong to the same part in the partition of the set $A = \{1, 2, 3\}$.

Let's list all possible partitions of the set $A = \{1, 2, 3\}$ where the elements 1 and 2 are grouped together in the same subset (part):

Possibility 1: The elements 1 and 2 are in one part, and the element 3 is in a different part.

This gives the partition $P_1 = \{\{1, 2\}, \{3\}\}$.

The equivalence relation $R_1$ corresponding to this partition consists of all ordered pairs $(a, b)$ where $a$ and $b$ are in the same part of $P_1$.

Pairs from $\{1, 2\}$ are $(1, 1), (1, 2), (2, 1), (2, 2)$.

Pairs from $\{3\}$ are $(3, 3)$.

So, $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. This relation contains $(1, 2)$ and $(2, 1)$. We can verify that $R_1$ is reflexive, symmetric, and transitive, so it is an equivalence relation.

Possibility 2: The elements 1, 2, and 3 are all in the same part.

This gives the partition $P_2 = \{\{1, 2, 3\}\}$.

The equivalence relation $R_2$ corresponding to this partition consists of all ordered pairs $(a, b)$ where $a$ and $b$ are in the same part of $P_2$. Since all elements are in the same part, every ordered pair from $A \times A$ is included.

So, $R_2 = A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$. This relation clearly contains $(1, 2)$ and $(2, 1)$. The relation $A \times A$ is always reflexive, symmetric, and transitive, so it is an equivalence relation.

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These are the only two possible partitions of the set $\{1, 2, 3\}$ where the elements 1 and 2 belong to the same part. Since there is a one-to-one correspondence between such partitions and the desired equivalence relations, there are exactly two such equivalence relations.

The two equivalence relations containing (1, 2) and (2, 1) are:

$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$

$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$

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Thus, the number of equivalence relations in the set $\{1, 2, 3\}$ containing (1, 2) and (2, 1) is two.

Given:

The identity function $I_N : N \to N$ defined by $I_N(x) = x$ for all $x \in N$.

A function $f : N \to N$ defined by $f(x) = (I_N + I_N)(x) = I_N(x) + I_N(x) = x + x = 2x$.

The domain and codomain for both functions are the set of natural numbers, $N = \{1, 2, 3, ...\}$.

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To Show:

1. $I_N$ is an onto function.

2. $I_N + I_N$ (i.e., the function $f(x) = 2x$) is not an onto function.

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Solution:

A function $g : A \to B$ is said to be onto (or surjective) if for every element $y$ in the codomain $B$, there exists at least one element $x$ in the domain $A$ such that $g(x) = y$.

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Part 1: Showing $I_N$ is onto.

Consider the function $I_N : N \to N$ defined by $I_N(x) = x$.

The domain is $N$ and the codomain is $N$.

Let $y$ be any arbitrary element in the codomain $N$.

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

From the definition of $I_N$, we have $I_N(x) = x$.

So, we need $x = y$.

Since $y$ is an element of the codomain $N$ (which contains all natural numbers), the value $x = y$ is also a natural number, meaning $x \in N$ (the domain).

Thus, for every $y$ in the codomain $N$, there exists an element $x = y$ in the domain $N$ such that $I_N(x) = y$.

Therefore, the identity function $I_N : N \to N$ is onto.

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Part 2: Showing $I_N + I_N$ is not onto.

Consider the function $f : N \to N$ defined by $f(x) = (I_N + I_N)(x) = 2x$.

The domain is $N$ and the codomain is $N$.

For $f$ to be onto, for every element $y$ in the codomain $N$, there must exist an element $x$ in the domain $N$ such that $f(x) = y$.

The equation $f(x) = y$ translates to $2x = y$.

Solving for $x$, we get $x = \frac{y}{2}$.

For $f$ to be onto, for every $y \in N$ (codomain), the value $x = \frac{y}{2}$ must be an element of $N$ (domain).

Let's consider an element in the codomain $N$. Let $y = 3$.

For $y=3$ to be in the range of $f$, there must exist an $x \in N$ such that $f(x) = 3$, i.e., $2x = 3$.

This implies $x = \frac{3}{2}$.

However, $\frac{3}{2}$ is not a natural number ($x \notin N$).

This means that there is no element in the domain $N$ that maps to the element $3$ in the codomain $N$.

Since there exists at least one element in the codomain ($y=3$) that is not the image of any element in the domain, the function $f(x) = 2x$ is not onto.

Given:

Function f : $\left[ 0, \frac{π}{2} \right]$ $\to$ R given by f(x) = sin x.

Function g : $\left[ 0, \frac{π}{2} \right]$ $\to$ R given by g(x) = cos x.

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To Show:

1. f is one-one on $\left[ 0, \frac{π}{2} \right]$.

2. g is one-one on $\left[ 0, \frac{π}{2} \right]$.

3. f + g is not one-one on $\left[ 0, \frac{π}{2} \right]$.

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Solution:

A function h : A $\to$ B is said to be one-one (or injective) if for any $x_1, x_2 \in A$, h($x_1$) = h($x_2$) implies $x_1 = x_2$. Equivalently, if $x_1 \neq x_2$, then h($x_1$) $\neq$ h($x_2$).

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1. To show f(x) = sin x is one-one on $\left[ 0, \frac{π}{2} \right]$.

Let $x_1, x_2 \in \left[ 0, \frac{π}{2} \right]$ such that f($x_1$) = f($x_2$).

This means sin $x_1$ = sin $x_2$.

On the interval $\left[ 0, \frac{π}{2} \right]$, the sine function is strictly increasing. This means that if sin $x_1$ = sin $x_2$ for $x_1, x_2$ in this interval, then $x_1$ must be equal to $x_2$.

Alternatively, consider sin $x_1$ - sin $x_2$ = 0.

Using the formula sin A - sin B = $2 \text{ cos } \left( \frac{A+B}{2} \right) \text{ sin } \left( \frac{A-B}{2} \right)$, we get:

$2 \text{ cos } \left( \frac{x_1+x_2}{2} \right) \text{ sin } \left( \frac{x_1-x_2}{2} \right) = 0$

Since $x_1, x_2 \in \left[ 0, \frac{π}{2} \right]$, we have $0 \leq x_1 \leq \frac{π}{2}$ and $0 \leq x_2 \leq \frac{π}{2}$.

Thus, $0 \leq x_1+x_2 \leq π$, which means $0 \leq \frac{x_1+x_2}{2} \leq \frac{π}{2}$.

In the interval $\left[ 0, \frac{π}{2} \right]$, cos $\left( \frac{x_1+x_2}{2} \right)$ is non-negative. cos $\left( \frac{x_1+x_2}{2} \right) = 0$ only if $\frac{x_1+x_2}{2} = \frac{π}{2}$, which implies $x_1+x_2 = π$. This can only happen if $x_1 = x_2 = \frac{π}{2}$. In this specific case, $x_1-x_2 = 0$, and sin $\left( \frac{x_1-x_2}{2} \right) = \text{sin } 0 = 0$.

If $x_1+x_2 < π$, then cos $\left( \frac{x_1+x_2}{2} \right) > 0$.

For the product to be zero, we must have sin $\left( \frac{x_1-x_2}{2} \right) = 0$.

Since $0 \leq x_1, x_2 \leq \frac{π}{2}$, we have $-\frac{π}{2} \leq x_1-x_2 \leq \frac{π}{2}$.

Thus, $-\frac{π}{4} \leq \frac{x_1-x_2}{2} \leq \frac{π}{4}$.

In the interval $\left[ -\frac{π}{4}, \frac{π}{4} \right]$, sin $\theta = 0$ only if $\theta = 0$.

Therefore, $\frac{x_1-x_2}{2} = 0$, which implies $x_1-x_2 = 0$, so $x_1 = x_2$.

Since f($x_1$) = f($x_2$) implies $x_1 = x_2$ for all $x_1, x_2 \in \left[ 0, \frac{π}{2} \right]$, the function f is one-one on this interval.

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2. To show g(x) = cos x is one-one on $\left[ 0, \frac{π}{2} \right]$.

Let $x_1, x_2 \in \left[ 0, \frac{π}{2} \right]$ such that g($x_1$) = g($x_2$).

This means cos $x_1$ = cos $x_2$.

On the interval $\left[ 0, \frac{π}{2} \right]$, the cosine function is strictly decreasing. This means that if cos $x_1$ = cos $x_2$ for $x_1, x_2$ in this interval, then $x_1$ must be equal to $x_2$.

Alternatively, consider cos $x_1$ - cos $x_2$ = 0.

Using the formula cos A - cos B = $-2 \text{ sin } \left( \frac{A+B}{2} \right) \text{ sin } \left( \frac{A-B}{2} \right)$, we get:

$-2 \text{ sin } \left( \frac{x_1+x_2}{2} \right) \text{ sin } \left( \frac{x_1-x_2}{2} \right) = 0$

Since $x_1, x_2 \in \left[ 0, \frac{π}{2} \right]$, we have $0 \leq x_1+x_2 \leq π$, which means $0 \leq \frac{x_1+x_2}{2} \leq \frac{π}{2}$.

In the interval $\left[ 0, \frac{π}{2} \right]$, sin $\left( \frac{x_1+x_2}{2} \right)$ is non-negative. sin $\left( \frac{x_1+x_2}{2} \right) = 0$ only if $\frac{x_1+x_2}{2} = 0$, which implies $x_1+x_2 = 0$. This can only happen if $x_1 = x_2 = 0$. In this specific case, $x_1-x_2 = 0$, and sin $\left( \frac{x_1-x_2}{2} \right) = \text{sin } 0 = 0$.

If $x_1+x_2 > 0$, then sin $\left( \frac{x_1+x_2}{2} \right) > 0$ (unless $x_1+x_2 = π$, i.e., $x_1=x_2=\frac{\pi}{2}$, where sin $\frac{\pi}{2}=1$).

For the product to be zero, we must have sin $\left( \frac{x_1-x_2}{2} \right) = 0$.

As shown before, $-\frac{π}{4} \leq \frac{x_1-x_2}{2} \leq \frac{π}{4}$.

In this interval, sin $\left( \frac{x_1-x_2}{2} \right) = 0$ implies $\frac{x_1-x_2}{2} = 0$, which means $x_1 = x_2$.

Since g($x_1$) = g($x_2$) implies $x_1 = x_2$ for all $x_1, x_2 \in \left[ 0, \frac{π}{2} \right]$, the function g is one-one on this interval.

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3. To show f + g is not one-one on $\left[ 0, \frac{π}{2} \right]$.

Let h(x) = (f + g)(x) = f(x) + g(x) = sin x + cos x.

To show that h(x) is not one-one on $\left[ 0, \frac{π}{2} \right]$, we need to find two distinct values $x_1, x_2 \in \left[ 0, \frac{π}{2} \right]$ such that h($x_1$) = h($x_2$).

Consider the values $x_1 = 0$ and $x_2 = \frac{π}{2}$, both of which are in the domain $\left[ 0, \frac{π}{2} \right]$. Clearly, $x_1 \neq x_2$.

Let's evaluate h(0):

h(0) = sin 0 + cos 0 = $0 + 1 = 1$.

Let's evaluate h($\frac{π}{2}$):

h($\frac{π}{2}$) = sin $\frac{π}{2}$ + cos $\frac{π}{2}$ = $1 + 0 = 1$.

We have h(0) = 1 and h($\frac{π}{2}$) = 1, but $0 \neq \frac{π}{2}$.

Since there exist distinct values in the domain that map to the same value in the codomain, the function h(x) = f(x) + g(x) is not one-one on $\left[ 0, \frac{π}{2} \right]$.

This completes the proof.

Given:

The function f : R $\to$ {x ∈ R : – 1 < x < 1} defined by f(x) = $\frac{x}{1 + |x|}$, x ∈ R.

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To Prove:

The function f is both one-one and onto.

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Proof (One-one):

To show that f is one-one, we must prove that for any $x_1, x_2 \in$ R, if f($x_1$) = f($x_2$), then $x_1 = x_2$.

Let f($x_1$) = f($x_2$). This means $\frac{x_1}{1 + |x_1|} = \frac{x_2}{1 + |x_2|}$.

We consider different cases based on the signs of $x_1$ and $x_2$.

Case 1: $x_1 \ge 0$ and $x_2 \ge 0$.

In this case, $|x_1| = x_1$ and $|x_2| = x_2$. The equation becomes:

$\frac{x_1}{1 + x_1} = \frac{x_2}{1 + x_2}$

$x_1(1 + x_2) = x_2(1 + x_1)$

$x_1 + x_1x_2 = x_2 + x_1x_2$

$x_1 = x_2$

So, if $x_1 \ge 0$ and $x_2 \ge 0$, f($x_1$) = f($x_2$) implies $x_1 = x_2$.

Case 2: $x_1 < 0$ and $x_2 < 0$.

In this case, $|x_1| = -x_1$ and $|x_2| = -x_2$. The equation becomes:

$\frac{x_1}{1 + (-x_1)} = \frac{x_2}{1 + (-x_2)}$

$\frac{x_1}{1 - x_1} = \frac{x_2}{1 - x_2}$

$x_1(1 - x_2) = x_2(1 - x_1)$

$x_1 - x_1x_2 = x_2 - x_1x_2$

$x_1 = x_2$

So, if $x_1 < 0$ and $x_2 < 0$, f($x_1$) = f($x_2$) implies $x_1 = x_2$.

Case 3: $x_1 \ge 0$ and $x_2 < 0$.

If $x_1 \ge 0$, then $|x_1| = x_1$, so f($x_1$) = $\frac{x_1}{1+x_1}$. Since $x_1 \ge 0$, $1+x_1 \ge 1$. If $x_1 = 0$, f(0) = 0. If $x_1 > 0$, then $1+x_1 > 1$, so $0 < \frac{x_1}{1+x_1} = 1 - \frac{1}{1+x_1} < 1$. Thus, for $x_1 \ge 0$, f($x_1$) $\ge 0$.

If $x_2 < 0$, then $|x_2| = -x_2$, so f($x_2$) = $\frac{x_2}{1+(-x_2)} = \frac{x_2}{1-x_2}$. Since $x_2 < 0$ and $1-x_2 > 1$, the numerator is negative and the denominator is positive. Thus, for $x_2 < 0$, f($x_2$) < 0.

If f($x_1$) = f($x_2$), then f($x_1$) must be $\ge 0$ and f($x_2$) must be < 0, which is a contradiction unless both are 0. But f($x_1$) = 0 only if $x_1 = 0$, and f($x_2$) can never be 0 if $x_2 < 0$. Therefore, f($x_1$) $\neq$ f($x_2$) in this case.

Case 4: $x_1 < 0$ and $x_2 \ge 0$.

This case is symmetric to Case 3. f($x_1$) < 0 and f($x_2$) $\ge 0$. Thus, f($x_1$) $\neq$ f($x_2$).

From all cases, we conclude that f($x_1$) = f($x_2$) implies $x_1$ and $x_2$ must have the same sign (or both zero), and in those situations, we proved $x_1 = x_2$.

Therefore, the function f is one-one.

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Proof (Onto):

To show that f is onto the set {y ∈ R : – 1 < y < 1}, we must prove that for every y in the codomain (-1, 1), there exists an x in the domain R such that f(x) = y.

Let y ∈ (-1, 1). We need to find x such that f(x) = y, i.e., $\frac{x}{1 + |x|} = y$.

We consider two cases for y based on its sign.

Case 1: $0 \le y < 1$.

If $y=0$, we need f(x) = 0, which means $\frac{x}{1+|x|} = 0$. This implies $x = 0$. For $x=0$, $|x|=0$, and f(0) = $\frac{0}{1+0} = 0$. So x=0 is the preimage of y=0.

If $0 < y < 1$, since f(x) = y > 0, it implies $\frac{x}{1+|x|} > 0$. This can only happen if x and $1+|x|$ have the same sign. Since $1+|x|$ is always positive, x must be positive. So, we assume x > 0. In this case, $|x| = x$.

The equation becomes $\frac{x}{1 + x} = y$.

$x = y(1 + x)$

$x = y + yx$

$x - yx = y$

$x(1 - y) = y$

Since $y < 1$, $1 - y \neq 0$. So, $x = \frac{y}{1 - y}$.

Since $0 < y < 1$, the numerator y is positive and the denominator $1-y$ is positive. Thus, $x = \frac{y}{1-y} > 0$. This satisfies our assumption that x > 0. Also, since $0 < y < 1$, $1-y < 1$, so $\frac{1}{1-y} > 1$. Thus $x = y \cdot \frac{1}{1-y} > y$. For example, if $y=0.5$, $x=\frac{0.5}{0.5}=1$. If $y=0.9$, $x=\frac{0.9}{0.1}=9$. As y approaches 1 from below, x approaches $\infty$. This value of x is in R.

Let's verify f(x) for $x = \frac{y}{1-y}$ where $0 < y < 1$. Since $x > 0$, $|x| = x$.

$f(x) = \frac{x}{1+|x|} = \frac{\frac{y}{1-y}}{1+\frac{y}{1-y}} = \frac{\frac{y}{1-y}}{\frac{1-y+y}{1-y}} = \frac{\frac{y}{1-y}}{\frac{1}{1-y}} = y$.

So, for any $y \in [0, 1)$, there exists $x = \frac{y}{1-y} \in R$ such that f(x) = y.

Case 2: $-1 < y < 0$.

If $-1 < y < 0$, since f(x) = y < 0, it implies $\frac{x}{1+|x|} < 0$. Since $1+|x|$ is always positive, x must be negative. So, we assume x < 0. In this case, $|x| = -x$.

The equation becomes $\frac{x}{1 + (-x)} = y$, i.e., $\frac{x}{1 - x} = y$.

$x = y(1 - x)$

$x = y - yx$

$x + yx = y$

$x(1 + y) = y$

Since $y > -1$, $1 + y \neq 0$. So, $x = \frac{y}{1 + y}$.

Since $-1 < y < 0$, the numerator y is negative and the denominator $1+y$ is positive. Thus, $x = \frac{y}{1+y} < 0$. This satisfies our assumption that x < 0. For example, if $y=-0.5$, $x=\frac{-0.5}{0.5}=-1$. If $y=-0.9$, $x=\frac{-0.9}{0.1}=-9$. As y approaches -1 from above, x approaches $-\infty$. This value of x is in R.

Let's verify f(x) for $x = \frac{y}{1+y}$ where $-1 < y < 0$. Since $x < 0$, $|x| = -x$.

$f(x) = \frac{x}{1+|x|} = \frac{\frac{y}{1+y}}{1+(-\frac{y}{1+y})} = \frac{\frac{y}{1+y}}{1-\frac{y}{1+y}} = \frac{\frac{y}{1+y}}{\frac{1+y-y}{1+y}} = \frac{\frac{y}{1+y}}{\frac{1}{1+y}} = y$.

So, for any $y \in (-1, 0)$, there exists $x = \frac{y}{1+y} \in R$ such that f(x) = y.

Combining Case 1 and Case 2, for every y ∈ (-1, 1), there exists an x ∈ R (either $x=\frac{y}{1-y}$ for $y \ge 0$ or $x=\frac{y}{1+y}$ for $y < 0$) such that f(x) = y.

Therefore, the function f is onto.

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

Given:

The function f : R $\to$ R defined by f(x) = $x^3$, x ∈ R.

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To Prove:

The function f is injective (one-one).

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Proof:

To show that the function f is injective, we need to prove that for any two elements $x_1, x_2$ in the domain R, if f($x_1$) = f($x_2$), then $x_1$ must be equal to $x_2$.

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

By the definition of the function f, this means:

$x_1^3 = x_2^3$

We can rearrange this equation:

$x_1^3 - x_2^3 = 0$

We use the algebraic identity for the difference of cubes, which states that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Applying this identity, we get:

$(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0$

For the product of two real numbers to be zero, at least one of the numbers must be zero. Therefore, either $(x_1 - x_2) = 0$ or $(x_1^2 + x_1x_2 + x_2^2) = 0$.

Case 1: $x_1 - x_2 = 0$.

This directly implies $x_1 = x_2$. This is what we want to conclude for injectivity.

Case 2: $x_1^2 + x_1x_2 + x_2^2 = 0$.

We need to determine when this quadratic expression in terms of $x_1$ (or $x_2$) is zero for real values of $x_1$ and $x_2$. We can complete the square for this expression with respect to $x_1$:

$x_1^2 + x_1x_2 + x_2^2 = \left( x_1^2 + x_1x_2 + \frac{x_2^2}{4} \right) - \frac{x_2^2}{4} + x_2^2$

$x_1^2 + x_1x_2 + x_2^2 = \left( x_1 + \frac{x_2}{2} \right)^2 + \frac{3x_2^2}{4}$

For this sum to be zero, since $(x_1 + \frac{x_2}{2})^2$ is a square of a real number, it is always $\ge 0$. Similarly, $\frac{3x_2^2}{4}$ is also always $\ge 0$ for real $x_2$.

The sum of two non-negative numbers is zero if and only if both numbers are zero.

So, we must have $\left( x_1 + \frac{x_2}{2} \right)^2 = 0$ and $\frac{3x_2^2}{4} = 0$.

From $\frac{3x_2^2}{4} = 0$, we get $x_2^2 = 0$, which implies $x_2 = 0$.

Substituting $x_2 = 0$ into $\left( x_1 + \frac{x_2}{2} \right)^2 = 0$, we get $\left( x_1 + \frac{0}{2} \right)^2 = x_1^2 = 0$, which implies $x_1 = 0$.

Thus, the expression $x_1^2 + x_1x_2 + x_2^2 = 0$ only when $x_1 = 0$ and $x_2 = 0$. In this specific scenario, $x_1 = x_2$ (both are 0).

Combining both cases, we see that if $(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0$, then either $x_1 = x_2$ directly from the first factor, or $x_1 = x_2 = 0$ from the second factor being zero. In all scenarios where f($x_1$) = f($x_2$), we conclude that $x_1 = x_2$.

Therefore, the function f : R $\to$ R given by f(x) = $x^3$ is injective.

Given:

A non-empty set X.

P(X) is the power set of X (the set of all subsets of X).

A relation R on P(X) is defined for A, B $\in$ P(X) as A R B if and only if A $\subset$ B (A is a subset of B).

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To Determine:

Is the relation R an equivalence relation on P(X)? Justify the answer.

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Solution:

For a relation R on a set S to be an equivalence relation, it must satisfy the following three properties for all elements A, B, C $\in$ S:

1. Reflexivity: A R A must be true.

2. Symmetry: If A R B is true, then B R A must also be true.

3. Transitivity: If A R B is true and B R C is true, then A R C must also be true.

Let's examine the given relation R on P(X) where A R B means A $\subset$ B.

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1. Reflexivity:

For any subset A $\in$ P(X), is A $\subset$ A?

By the definition of a subset, a set A is a subset of a set B if every element of A is also an element of B. Since every element in set A is indeed an element in set A, the condition A $\subset$ A is always true for any set A.

Thus, the relation R is reflexive.

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2. Symmetry:

For any subsets A, B $\in$ P(X), if A R B (i.e., A $\subset$ B) is true, is B R A (i.e., B $\subset$ A) also true?

Consider a non-empty set X. Let's take a specific example. Since X is non-empty, let x be an element in X. Consider the subsets A = {x} and B = X. Both A and B are in P(X).

Is A $\subset$ B? Yes, if A = {x} and B = X, then A is a subset of B.

Is B $\subset$ A? Is X $\subset$ {x}?

If X has more than one element (or if X has exactly one element x, but we chose A to be the empty set $\emptyset$), then X is not a subset of {x}. For instance, if X = {x, y} with $x \neq y$, then y $\in$ X but y $\notin$ {x}. So X is not a subset of {x}.

Therefore, A $\subset$ B does not necessarily imply B $\subset$ A.

The relation R is not symmetric.

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3. Transitivity:

For any subsets A, B, C $\in$ P(X), if A R B (i.e., A $\subset$ B) is true and B R C (i.e., B $\subset$ C) is true, is A R C (i.e., A $\subset$ C) also true?

Suppose A $\subset$ B and B $\subset$ C.

A $\subset$ B means every element in A is also in B.

B $\subset$ C means every element in B is also in C.

Let a be an arbitrary element in A. Since A $\subset$ B, a must also be an element in B. Now, since a is in B and B $\subset$ C, a must also be an element in C.

Since every element in A is also in C, it follows that A $\subset$ C.

Thus, if A $\subset$ B and B $\subset$ C, then A $\subset$ C is always true.

The relation R is transitive.

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Justification:

The relation R defined by A R B if A $\subset$ B is reflexive and transitive, but it is not symmetric. For a relation to be an equivalence relation, it must satisfy all three properties (reflexivity, symmetry, and transitivity).

Since the symmetry property is not satisfied by the relation R on P(X), R is not an equivalence relation on P(X).

Given:

The set A = {1, 2, 3, ..., n}.

We are considering functions f : A $\to$ A.

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To Find:

The number of all onto functions from the set {1, 2, 3, ..., n} to itself.

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Solution:

Let A be a finite set with n elements, i.e., $|A| = n$. We are looking for the number of onto functions from A to A.

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

When considering functions between two finite sets with the same number of elements, there is a special relationship between one-one (injective) functions, onto (surjective) functions, and bijective functions.

For a function f : A $\to$ B where $|A| = |B| = n$ (finite sets), the following statements are equivalent:

1. f is one-one (injective).

2. f is onto (surjective).

3. f is bijective (both one-one and onto).

This means that a function from a finite set to itself is onto if and only if it is one-one. A function that is both one-one and onto is called a bijection.

A bijection from a set to itself is also known as a permutation of the set.

Therefore, the number of onto functions from the set {1, 2, ..., n} to itself is equal to the number of bijections from {1, 2, ..., n} to {1, 2, ..., n}.

The number of bijections from a set of n elements to itself is the number of ways to arrange the n elements, which is given by n factorial ($n!$).

The factorial of a non-negative integer n, denoted by $n!$, is the product of all positive integers less than or equal to n. $n! = n \times (n-1) \times (n-2) \times \ ... \ \times 2 \times 1$.

Thus, the number of onto functions from the set {1, 2, 3, ..., n} to itself is $n!$.

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Conclusion:

The number of all onto functions from the set {1, 2, 3, ..., n} to itself is $n!$.

Given:

Set A = {– 1, 0, 1, 2}

Set B = {– 4, – 2, 0, 2}

Function f : A $\to$ B defined by f(x) = $x^2 – x$, for x ∈ A.

Function g : A $\to$ B defined by g(x) = $2\left|x- \frac{1}{2} \right|-1$, for x ∈ A.

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To Determine:

Are functions f and g equal? Justify the answer.

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Solution:

Two functions f : A $\to$ B and g : A $\to$ B are said to be equal if and only if f(x) = g(x) for every element x in their common domain A.

We need to evaluate f(x) and g(x) for each element in the set A = {– 1, 0, 1, 2} and compare the results.

Let's calculate the values of f(x) for each x ∈ A:

For x = – 1:

f(– 1) = $(-1)^2 - (-1) = 1 + 1 = 2$

For x = 0:

f(0) = $0^2 - 0 = 0 - 0 = 0$

For x = 1:

f(1) = $1^2 - 1 = 1 - 1 = 0$

For x = 2:

f(2) = $2^2 - 2 = 4 - 2 = 2$

The values of f(x) for x ∈ A are: f(– 1) = 2, f(0) = 0, f(1) = 0, f(2) = 2. All these values {2, 0} are elements of the codomain B = {– 4, – 2, 0, 2}.

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Now, let's calculate the values of g(x) for each x ∈ A:

For x = – 1:

g(– 1) = $2\left|-1- \frac{1}{2} \right|-1 = 2\left|-\frac{3}{2}\right|-1 = 2 \times \frac{3}{2} - 1 = 3 - 1 = 2$

For x = 0:

g(0) = $2\left|0- \frac{1}{2} \right|-1 = 2\left|-\frac{1}{2}\right|-1 = 2 \times \frac{1}{2} - 1 = 1 - 1 = 0$

For x = 1:

g(1) = $2\left|1- \frac{1}{2} \right|-1 = 2\left|\frac{1}{2}\right|-1 = 2 \times \frac{1}{2} - 1 = 1 - 1 = 0$

For x = 2:

g(2) = $2\left|2- \frac{1}{2} \right|-1 = 2\left|\frac{4}{2}- \frac{1}{2}\right|-1 = 2\left|\frac{3}{2}\right|-1 = 2 \times \frac{3}{2} - 1 = 3 - 1 = 2$

The values of g(x) for x ∈ A are: g(– 1) = 2, g(0) = 0, g(1) = 0, g(2) = 2. All these values {2, 0} are elements of the codomain B = {– 4, – 2, 0, 2}.

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Now, let's compare the values of f(x) and g(x) for each element in A:

f(– 1) = 2 and g(– 1) = 2. So, f(– 1) = g(– 1).

f(0) = 0 and g(0) = 0. So, f(0) = g(0).

f(1) = 0 and g(1) = 0. So, f(1) = g(1).

f(2) = 2 and g(2) = 2. So, f(2) = g(2).

Since f(x) = g(x) for all x ∈ A, the functions f and g are equal.

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Justification:

We have shown by evaluating the function values for each element in the domain A that f(x) is equal to g(x) for every x ∈ A. According to the definition of equal functions, if the domain and codomain are the same for two functions, and their values are equal for every element in the domain, then the functions are equal.

Thus, functions f and g are equal.

Given:

The set A = {1, 2, 3}.

A relation R on A such that R $\subseteq$ A $\times$ A.

The relation R must contain the ordered pairs (1, 2) and (1, 3).

The relation R must be reflexive and symmetric.

The relation R must not be transitive.

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To Find:

The number of such relations R.

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Solution:

Let R be a relation on the set A = {1, 2, 3}. R is a subset of A $\times$ A.

A $\times$ A = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)}.

We are given that R must satisfy the following conditions:

1. R contains (1, 2) and (1, 3).

2. R is reflexive: For all a $\in$ A, (a, a) $\in$ R.

3. R is symmetric: If (a, b) $\in$ R, then (b, a) $\in$ R.

4. R is not transitive: There exist a, b, c $\in$ A such that (a, b) $\in$ R and (b, c) $\in$ R, but (a, c) $\notin$ R.

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First, let's determine the minimum set of ordered pairs that must be in R based on conditions 1, 2, and 3.

From condition 2 (reflexivity), R must contain:

(1, 1), (2, 2), (3, 3)

From condition 1, R must contain:

(1, 2), (1, 3)

From condition 3 (symmetry), for every pair (a, b) in R, the pair (b, a) must also be in R. Since (1, 2) $\in$ R, its symmetric pair (2, 1) must be in R. Since (1, 3) $\in$ R, its symmetric pair (3, 1) must be in R.

So, any relation R satisfying conditions 1, 2, and 3 must contain at least the following set of ordered pairs:

$S_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)\}$

The remaining ordered pairs in A $\times$ A are (2, 3) and (3, 2). To maintain symmetry, if (2, 3) is included in R, then (3, 2) must also be included, and vice versa. Thus, these two pairs must either both be in R or neither be in R.

Let $T = \{(2, 3), (3, 2)\}$. The possible relations R that are reflexive, symmetric, and contain (1, 2) and (1, 3) are of the form $R = S_0 \cup X$, where X is a subset of T such that if (a,b) $\in$ X, then (b,a) $\in$ X. The only possible choices for X are $\emptyset$ or T.

So, there are two possible relations that satisfy conditions 1, 2, and 3:

Relation $R_1 = S_0 \cup \emptyset = \{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)\}$.

Relation $R_2 = S_0 \cup T = \{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1), (2, 3), (3, 2)\}$.

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Now, we need to check which of these relations is not transitive (condition 4).

Recall that a relation R is transitive if for all a, b, c $\in$ A, whenever (a, b) $\in$ R and (b, c) $\in$ R, it follows that (a, c) $\in$ R.

Let's check transitivity for $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)\}$.

Consider the pairs (2, 1) $\in R_1$ and (1, 3) $\in R_1$. According to transitivity, (2, 3) must be in $R_1$. However, (2, 3) is not in $R_1$.

Therefore, $R_1$ is not transitive.

Relation $R_1$ satisfies all four conditions (contains (1,2) and (1,3), reflexive, symmetric, and not transitive). So, $R_1$ is one such relation.

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Let's check transitivity for $R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1), (2, 3), (3, 2)\}$.

This relation $R_2$ includes all possible symmetric pairs based on the initial conditions and reflexivity. This relation represents the situation where elements {1, 2, 3} are all related to each other. This is the universal relation $A \times A$, which is always an equivalence relation. An equivalence relation is transitive by definition.

Alternatively, we can check some paths:

(1, 2) $\in R_2$ and (2, 3) $\in R_2 \implies$ (1, 3) $\in R_2$. Yes.

(2, 1) $\in R_2$ and (1, 3) $\in R_2 \implies$ (2, 3) $\in R_2$. Yes.

(1, 3) $\in R_2$ and (3, 2) $\in R_2 \implies$ (1, 2) $\in R_2$. Yes.

... and so on for all possible combinations. All required resulting pairs are present in $R_2$.

Therefore, $R_2$ is transitive.

Relation $R_2$ satisfies conditions 1, 2, and 3 but fails condition 4 (it is transitive, not non-transitive). So, $R_2$ is not a required relation.

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We found only one relation ($R_1$) that satisfies all the given conditions.

The number of such relations is 1.

Comparing this with the given options:

(A) 1

(B) 2

(C) 3

(D) 4

The correct option is (A).

Given:

The set A = {1, 2, 3}.

We are looking for equivalence relations R on A such that R contains the ordered pair (1, 2).

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To Find:

The number of such equivalence relations.

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Solution:

An equivalence relation R on a set A is a relation that is:

1. Reflexive: (a, a) $\in$ R for all a $\in$ A.

2. Symmetric: If (a, b) $\in$ R, then (b, a) $\in$ R for all a, b $\in$ A.

3. Transitive: If (a, b) $\in$ R and (b, c) $\in$ R, then (a, c) $\in$ R for all a, b, c $\in$ A.

An equivalence relation on a set A partitions the set into disjoint non-empty subsets called equivalence classes. The relation R is given by the union of the Cartesian products of these equivalence classes with themselves. That is, if $\{A_1, A_2, \ldots, A_k\}$ is a partition of A, the corresponding equivalence relation is $R = (A_1 \times A_1) \cup (A_2 \times A_2) \cup \ldots \cup (A_k \times A_k)$. The condition (a, b) $\in$ R means that a and b belong to the same subset (equivalence class) in the partition.

We are given that the equivalence relation R must contain the pair (1, 2). This implies that the elements 1 and 2 must belong to the same equivalence class in the partition of A.

The set A = {1, 2, 3}. We need to find partitions of A where 1 and 2 are in the same part (subset).

Let's list the possible partitions of A = {1, 2, 3} where 1 and 2 are together:

Partition 1:

The elements 1 and 2 are in one subset, and the element 3 is in a separate subset.

The partition is P$_1$ = {{1, 2}, {3}}.

This is a valid partition because the subsets {1, 2} and {3} are non-empty, disjoint, and their union is {1, 2, 3}.

The equivalence relation corresponding to this partition is $R_1 = (\{1, 2\} \times \{1, 2\}) \cup (\{3\} \times \{3\})$.

$R_1 = \{(1, 1), (1, 2), (2, 1), (2, 2)\} \cup \{(3, 3)\}$.

$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$.

This relation contains (1, 2). Let's verify if it is an equivalence relation:

• Reflexive: Yes, (1,1), (2,2), (3,3) are in $R_1$.
• Symmetric: Yes, for every (a,b) $\in R_1$, (b,a) $\in R_1$. (1,2) $\leftrightarrow$ (2,1).
• Transitive: Yes, for example, (1,2) $\in R_1$ and (2,1) $\in R_1 \implies$ (1,1) $\in R_1$. (2,1) $\in R_1$ and (1,2) $\in R_1 \implies$ (2,2) $\in R_1$. There are no pairs that would require (1,3), (3,1), (2,3), or (3,2) to be in $R_1$.

So, $R_1$ is an equivalence relation containing (1, 2).

Partition 2:

The elements 1, 2, and 3 are all in the same subset.

The partition is P$_2$ = {{1, 2, 3}}.

This is a valid partition because the subset {1, 2, 3} is non-empty and its union is {1, 2, 3}.

The equivalence relation corresponding to this partition is $R_2 = (\{1, 2, 3\} \times \{1, 2, 3\})$.

$R_2 = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$.

This is the universal relation on A. This relation contains (1, 2).

Let's verify if it is an equivalence relation:

• Reflexive: Yes, all (a,a) are included.
• Symmetric: Yes, if (a,b) is in $R_2$, then (b,a) is in $R_2$.
• Transitive: Yes, the universal relation is always transitive.

So, $R_2$ is an equivalence relation containing (1, 2).

These are the only two partitions of {1, 2, 3} where 1 and 2 are in the same subset. Each partition corresponds to exactly one equivalence relation.

Thus, there are exactly 2 equivalence relations on A that contain the pair (1, 2).

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Conclusion:

The number of equivalence relations containing (1, 2) on the set {1, 2, 3} is 2.

The correct option is (B).