Chapter 1 Sets

(i) Set A:

The set A is defined as $\{x | x \text{ is a positive integer less than 10 and } 2^x – 1 \text{ is an odd number}\}$.

The positive integers less than 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9.

We need to check the condition that $2^x - 1$ is an odd number for each of these integers.

For any positive integer x, the term $2^x$ represents a power of 2. Powers of 2 ($2^1=2$, $2^2=4$, $2^3=8$, etc.) are always even numbers.

Subtracting 1 from an even number ($2^x - 1$) always results in an odd number.

For example:

$x=1: 2^1 - 1 = 2 - 1 = 1$ (odd)

$x=2: 2^2 - 1 = 4 - 1 = 3$ (odd)

$x=3: 2^3 - 1 = 8 - 1 = 7$ (odd)

This holds true for all positive integers x.

Therefore, the condition "$2^x – 1$ is an odd number" is satisfied by all positive integers less than 10.

So, the elements of set A are all positive integers less than 10.

In roaster form, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}.

(ii) Set C:

The set C is defined as $\{x : x^2 + 7x – 8 = 0, x \in R\}$.

We need to find the values of x that satisfy the quadratic equation $x^2 + 7x – 8 = 0$.

This is a quadratic equation. We can solve it by factoring.

We look for two numbers that multiply to -8 and add up to 7. These numbers are 8 and -1.

So, the equation $x^2 + 7x – 8 = 0$ can be factored as $(x + 8)(x - 1) = 0$.

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1: $x + 8 = 0$

Subtracting 8 from both sides, we get $x = -8$.

Case 2: $x - 1 = 0$

Adding 1 to both sides, we get $x = 1$.

The solutions to the equation are $x = -8$ and $x = 1$.

The condition for the elements of C is that x must be a real number ($x \in R$). Both -8 and 1 are real numbers.

So, the elements of set C are -8 and 1.

In roaster form, C = {-8, 1}.

(i) Statement: 37 ∉ {x | x has exactly two positive factors}

The set $\{x | x \text{ has exactly two positive factors}\}$ consists of numbers which have only two positive factors: 1 and the number itself. These are known as prime numbers.

We need to determine if 37 is a prime number.

The positive factors of 37 are 1 and 37.

Since 37 has exactly two positive factors, 1 and 37, it is a prime number.

Therefore, 37 belongs to the set $\{x | x \text{ has exactly two positive factors}\}$.

The given statement says that 37 does not belong to this set, which is incorrect.

Thus, the statement is False.

(ii) Statement: 28 ∈ {y | the sum of the all positive factors of y is 2y}

The set $\{y | \text{the sum of the all positive factors of y is } 2y\}$ consists of numbers for which the sum of all positive factors (including the number itself) is twice the number. These are known as perfect numbers.

We need to find the positive factors of 28 and calculate their sum.

The positive factors of 28 are 1, 2, 4, 7, 14, and 28.

The sum of these factors is $1 + 2 + 4 + 7 + 14 + 28 = 56$.

Now, we check if this sum is equal to $2y$, where $y = 28$.

$2y = 2 \times 28 = 56$.

Since the sum of the positive factors of 28 is 56, which is equal to $2 \times 28$, 28 belongs to the given set.

Thus, the statement is True.

(iii) Statement: 7,747 ∈ {t | t is a multiple of 37}

The set $\{t | t \text{ is a multiple of } 37\}$ consists of numbers that are divisible by 37.

We need to check if 7747 is divisible by 37. We can do this by performing division.

Dividing 7747 by 37:

$7747 \div 37$

$7747 = 37 \times 209 + 14$

Since the remainder is 14 (not 0), 7747 is not exactly divisible by 37.

Therefore, 7747 is not a multiple of 37.

The given statement says that 7,747 belongs to the set of multiples of 37, which is incorrect.

Thus, the statement is False.

(i) Show that Y ⊂ X ∪ Y

To Prove:

The set Y is a subset of the set X ∪ Y.

Proof:

To prove that a set A is a subset of a set B ($A \subset B$), we must show that every element of A is also an element of B.

Let 'y' be an arbitrary element of the set Y.

The definition of the union of two sets, X ∪ Y, is the set of all elements that are in X, or in Y, or in both.

Since 'y' is an element of Y, by the definition of union, 'y' must also be an element of X ∪ Y.

Since any arbitrary element 'y' in Y is also in X ∪ Y, it follows that Y is a subset of X ∪ Y.

(ii) Show that X ∩ Y ⊂ X

To Prove:

The set X ∩ Y is a subset of the set X.

Proof:

Let 'z' be an arbitrary element of the set X ∩ Y.

The definition of the intersection of two sets, X ∩ Y, is the set of all elements that are in both X and Y.

Therefore, if 'z' is an element of X ∩ Y, it means that 'z' is an element of X and 'z' is also an element of Y.

From this, it directly follows that 'z' must be an element of X.

Since any arbitrary element 'z' in X ∩ Y is also in X, it follows that X ∩ Y is a subset of X.

(iii) Show that X ⊂ Y ⇒ X ∩ Y = X

Given:

$X \subset Y$ (X is a subset of Y).

To Prove:

$X \cap Y = X$.

Proof:

To prove that two sets are equal, we must show that each set is a subset of the other. We need to prove two things:

1. $X \cap Y \subset X$

2. $X \subset X \cap Y$

Part 1: Prove $X \cap Y \subset X$

This is a general property of sets and was proven in part (ii) above. It is always true, regardless of whether X is a subset of Y.

Part 2: Prove $X \subset X \cap Y$

For this part, we will use the given condition that $X \subset Y$.

Let 'x' be an arbitrary element of the set X.

Since we are given that $X \subset Y$, every element of X must also be an element of Y.

So, we now know that 'x' is an element of X and 'x' is also an element of Y.

By the definition of intersection, if an element belongs to both X and Y, it must belong to their intersection, $X \cap Y$.

Since any arbitrary element 'x' in X is also in X ∩ Y, it follows that $X \subset X \cap Y$.

Because we have shown that $X \cap Y \subset X$ and $X \subset X \cap Y$, we can conclude that the two sets are equal.

Given: N = {1, 2, 3, ..., 100}.

(i) Subset A:

The subset A of N consists of elements from N that are odd numbers.

An odd number is an integer that is not divisible by 2.

We need to find all numbers in the set N = {1, 2, 3, ..., 100} that are odd.

These numbers are 1, 3, 5, 7, 9, ..., up to the largest odd number less than or equal to 100, which is 99.

In set-builder form, A = $\{x \in N | x \text{ is an odd number}\}$.

In roaster form, A = {1, 3, 5, ..., 99}.

(ii) Subset B:

The subset B of N consists of elements $y$ such that $y = x + 2$ for some $x \in N$, and $y$ must also be in N.

In set-builder form, B = $\{y \in N | y = x + 2 \text{ for some } x \in N\}$.

We take each element $x$ from N = {1, 2, 3, ..., 100}, calculate $x + 2$, and check if the result is in N.

For $x = 1$, $x+2 = 1+2 = 3$. $3 \in N$.

For $x = 2$, $x+2 = 2+2 = 4$. $4 \in N$.

For $x = 3$, $x+2 = 3+2 = 5$. $5 \in N$.

...

For $x = 98$, $x+2 = 98+2 = 100$. $100 \in N$.

For $x = 99$, $x+2 = 99+2 = 101$. $101 \notin N$ (since N goes up to 100).

For $x = 100$, $x+2 = 100+2 = 102$. $102 \notin N$.

So, the elements generated by $x+2$ (where $x \in N$) that are also within N are 3, 4, 5, ..., up to 100.

In roaster form, B = {3, 4, 5, ..., 100}.

Given: Set E = {2, 4, 6, 8, 10}.

Let n represent any member of E. We need to form new sets based on the given expressions.

(i) Set of numbers represented by n + 1:

We apply the expression $n + 1$ to each element n in set E.

For n = 2: $2 + 1 = 3$

For n = 4: $4 + 1 = 5$

For n = 6: $6 + 1 = 7$

For n = 8: $8 + 1 = 9$

For n = 10: $10 + 1 = 11$

The set of numbers represented by $n + 1$ is {3, 5, 7, 9, 11}.

(ii) Set of numbers represented by n²:

We apply the expression $n^2$ to each element n in set E.

For n = 2: $2^2 = 4$

For n = 4: $4^2 = 16$

For n = 6: $6^2 = 36$

For n = 8: $8^2 = 64$

For n = 10: $10^2 = 100$

The set of numbers represented by $n^2$ is {4, 16, 36, 64, 100}.

Given: Set X = {1, 2, 3, 4, 5, 6}.

Let n be any member of X.

(i) The set of n such that $n \in X$ and $2n \notin X$:

We check the condition for each element n in X:

For $n = 1$, $2n = 2 \times 1 = 2$. Is $2 \notin X$? No, $2 \in X$. So, 1 does not satisfy the condition.

For $n = 2$, $2n = 2 \times 2 = 4$. Is $4 \notin X$? No, $4 \in X$. So, 2 does not satisfy the condition.

For $n = 3$, $2n = 2 \times 3 = 6$. Is $6 \notin X$? No, $6 \in X$. So, 3 does not satisfy the condition.

For $n = 4$, $2n = 2 \times 4 = 8$. Is $8 \notin X$? Yes, $8$ is not in the set {1, 2, 3, 4, 5, 6}. So, 4 satisfies the condition.

For $n = 5$, $2n = 2 \times 5 = 10$. Is $10 \notin X$? Yes, $10$ is not in the set {1, 2, 3, 4, 5, 6}. So, 5 satisfies the condition.

For $n = 6$, $2n = 2 \times 6 = 12$. Is $12 \notin X$? Yes, $12$ is not in the set {1, 2, 3, 4, 5, 6}. So, 6 satisfies the condition.

The elements n in X that satisfy the condition $2n \notin X$ are 4, 5, and 6.

The set is {4, 5, 6}.

(ii) The set of n such that $n \in X$ and $n + 5 = 8$:

We need to find the value of n that satisfies the equation $n + 5 = 8$ and is also in the set X.

Subtract 5 from both sides of the equation: $n = 8 - 5 = 3$.

Now check if this value of n is in set X = {1, 2, 3, 4, 5, 6}.

Yes, $3 \in X$.

The element n in X that satisfies the condition $n + 5 = 8$ is 3.

The set is {3}.

(iii) The set of n such that $n \in X$ and $n > 4$:

We need to find the elements n in X that are greater than 4.

The elements in X are 1, 2, 3, 4, 5, 6.

Check which elements are greater than 4:

Is $1 > 4$? No.

Is $2 > 4$? No.

Is $3 > 4$? No.

Is $4 > 4$? No.

Is $5 > 4$? Yes.

Is $6 > 4$? Yes.

The elements n in X that satisfy the condition $n > 4$ are 5 and 6.

The set is {5, 6}.

Given:

U = The set of all students in the school (Universal Set).

E = The set of students studying English.

M = The set of students studying Mathematics.

(i) All the students who study Mathematics study English, but some students who study English do not study Mathematics.

This statement means that the set of students who study Mathematics (M) is a subset of the set of students who study English (E). Since some students study English but not Mathematics, M is a proper subset of E.

This relationship is denoted by $M \subset E$.

The Venn diagram illustrating this relationship is shown below, where the circle representing M is completely inside the circle representing E.

(ii) There is no student who studies both Mathematics and English.

This statement means that there are no common students between the set of students studying Mathematics (M) and the set of students studying English (E). Therefore, the sets M and E are disjoint sets.

This relationship is denoted by $M \cap E = \phi$, where $\phi$ represents the empty set.

The Venn diagram illustrating this relationship is shown below, with two non-overlapping circles for M and E.

(iii) Some of the students study Mathematics but do not study English, some study English but do not study Mathematics, and some study both.

This statement indicates that the sets M and E have some students in common, and each set also has students that are not in the other. This means the sets are intersecting sets.

This implies that:

$M - E \neq \phi$ (Students studying only Mathematics)

$E - M \neq \phi$ (Students studying only English)

$M \cap E \neq \phi$ (Students studying both subjects)

The Venn diagram for this relationship shows two overlapping circles.

(iv) Not all students study Mathematics, but every student studying English studies Mathematics.

The statement "every student studying English studies Mathematics" implies that the set of students studying English (E) is a subset of the set of students studying Mathematics (M). The phrase "Not all students study Mathematics" along with the first part implies that there are students who study Mathematics but not English. Thus, E is a proper subset of M.

This relationship is denoted by $E \subset M$.

The Venn diagram for this case shows the circle for E completely inside the circle for M.

The statement $(A \cap B) \cup C = A \cap (B \cup C)$ is False.

This equality does not hold true for all sets A, B, and C.

Justification: (by Counterexample)

Let's consider specific sets for A, B, and C.

Let U = {1, 2, 3, 4, 5}

Let A = {1, 2, 3}

Let B = {2, 4}

Let C = {3, 5}

Now, let's evaluate the Left Hand Side (LHS): $(A \cap B) \cup C$

First, find the intersection of A and B:

$A \cap B = \{1, 2, 3\} \cap \{2, 4\} = \{2\}$

Next, find the union of $(A \cap B)$ and C:

$(A \cap B) \cup C = \{2\} \cup \{3, 5\} = \{2, 3, 5\}$

Now, let's evaluate the Right Hand Side (RHS): $A \cap (B \cup C)$

First, find the union of B and C:

$B \cup C = \{2, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$

Next, find the intersection of A and $(B \cup C)$:

$A \cap (B \cup C) = \{1, 2, 3\} \cap \{2, 3, 4, 5\} = \{2, 3\}$

Comparing the results of LHS and RHS:

LHS = $\{2, 3, 5\}$

RHS = $\{2, 3\}$

Since $\{2, 3, 5\} \neq \{2, 3\}$, the equality $(A \cap B) \cup C = A \cap (B \cup C)$ does not hold for these sets.

Therefore, the given statement is false for all sets A, B, and C.

To Prove: $A - (A \cap B) = A - B$ for all sets A and B.

We will start with the Left Hand Side (LHS) and transform it using set properties until we reach the Right Hand Side (RHS).

Consider the LHS: $A - (A \cap B)$.

Using the definition of set difference, $X - Y = X \cap Y'$, we can write:

$A - (A \cap B) = A \cap (A \cap B)'$

Now, we apply De Morgan's Law to $(A \cap B)'$. De Morgan's Law states that $(X \cap Y)' = X' \cup Y'$.

So, $(A \cap B)' = A' \cup B'$.

Substitute this back into the expression:

$A \cap (A' \cup B')$

Next, we use the Distributive Law, which states that $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$.

Applying this law, we distribute A over the union in the parenthesis:

$A \cap (A' \cup B') = (A \cap A') \cup (A \cap B')$

Now, we use the property of intersection of a set with its complement, which states that $A \cap A' = \emptyset$ (the empty set).

Substitute $\emptyset$ into the expression:

$(A \cap A') \cup (A \cap B') = \emptyset \cup (A \cap B')$

Finally, we use the property of the union of a set with the empty set, which states that $\emptyset \cup X = X$.

Applying this property:

$\emptyset \cup (A \cap B') = A \cap B'$

Recall the definition of set difference: $A - B = A \cap B'$.

So, we have successfully transformed the LHS to $A \cap B'$, which is equal to $A - B$, the RHS.

Therefore, $A - (A \cap B) = A - B$.

Proof Summary:

$A - (A \cap B)$

$= A \cap (A \cap B)'$ [Definition of set difference]

$= A \cap (A' \cup B')$ [De Morgan's Law]

$= (A \cap A') \cup (A \cap B')$ [Distributive Law]

$= \emptyset \cup (A \cap B')$ [$A \cap A' = \emptyset$]

$= A \cap B'$ [Union with empty set]

$= A - B$ [Definition of set difference]

Thus, $A - (A \cap B) = A - B$.

Given: A, B, and C are any sets.

To Prove: $(A – B) \cap (C – B) = (A \cap C) – B$

Proof:

We need to show that the set on the Left Hand Side (LHS) is equal to the set on the Right Hand Side (RHS).

Consider the LHS: $(A – B) \cap (C – B)$.

Using the definition of set difference, $X - Y = X \cap Y'$, we can rewrite the terms:

$A – B = A \cap B'$

$C – B = C \cap B'$

Substitute these into the LHS expression:

$(A \cap B') \cap (C \cap B')$

Using the associative property of intersection $(X \cap Y) \cap Z = X \cap (Y \cap Z)$:

$A \cap B' \cap C \cap B'$

Using the commutative property of intersection $X \cap Y = Y \cap X$, we can rearrange the terms:

$A \cap C \cap B' \cap B'$

Using the idempotent property of intersection $X \cap X = X$, we have $B' \cap B' = B'$:

$A \cap C \cap B'$

Using the associative property again, we can group the terms:

$(A \cap C) \cap B'$

Now, using the definition of set difference $X \cap Y' = X - Y$ in reverse, we can rewrite this expression:

$(A \cap C) - B$

This is the expression for the Right Hand Side (RHS).

Since we have shown that LHS can be transformed into RHS using valid set properties, the equality holds for all sets A, B, and C.

Thus, $(A – B) \cap (C – B) = (A \cap C) – B$ is a True statement.

Step-by-step Proof:

LHS = $(A – B) \cap (C – B)$

= $(A \cap B') \cap (C \cap B')$ [Definition of set difference]

= $A \cap B' \cap C \cap B'$ [Associativity of intersection]

= $A \cap C \cap B' \cap B'$ [Commutativity of intersection]

= $A \cap C \cap B'$ [Idempotent property of intersection]

= $(A \cap C) \cap B'$ [Associativity of intersection]

= $(A \cap C) – B$ [Definition of set difference]

= RHS

Therefore, $(A – B) \cap (C – B) = (A \cap C) – B$ is true for all sets A, B, and C.

Given: A, B, and C are any sets.

To Prove: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

Proof:

We need to show that the set on the Left Hand Side (LHS) is equal to the set on the Right Hand Side (RHS).

Consider the LHS: $A \cup (B \cap C)$.

We will show that an arbitrary element $x$ belongs to the LHS if and only if it belongs to the RHS.

Let $x$ be an arbitrary element.

$x \in A \cup (B \cap C)$

By the definition of union, this means $x$ is in A or $x$ is in $(B \cap C)$.

$\iff x \in A \text{ or } (x \in B \text{ and } x \in C)$

Now, we distribute the "or $x \in A$" over the "and". This is similar to the distributive law in logic or algebra.

$\iff (x \in A \text{ or } x \in B) \text{ and } (x \in A \text{ or } x \in C)$

By the definition of union, "$x \in A \text{ or } x \in B$" means $x \in A \cup B$.

By the definition of union, "$x \in A \text{ or } x \in C$" means $x \in A \cup C$.

So, the statement becomes:

$\iff (x \in A \cup B) \text{ and } (x \in A \cup C)$

By the definition of intersection, "$x \in (A \cup B) \text{ and } x \in (A \cup C)$" means $x \in (A \cup B) \cap (A \cup C)$.

$\iff x \in (A \cup B) \cap (A \cup C)$

Thus, $x \in A \cup (B \cap C)$ if and only if $x \in (A \cup B) \cap (A \cup C)$.

This shows that every element in $A \cup (B \cap C)$ is also in $(A \cup B) \cap (A \cup C)$, and every element in $(A \cup B) \cap (A \cup C)$ is also in $A \cup (B \cap C)$.

Therefore, the two sets are equal.

$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Alternatively (Using set properties directly):

Consider the RHS: $(A \cup B) \cap (A \cup C)$.

We use the distributive law of intersection over union, which states that $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$. Here we apply it in reverse, recognizing $A$ as the common set being distributed.

$(A \cup B) \cap (A \cup C) = A \cup (B \cap C)$

This directly transforms the RHS into the LHS.

Thus, $(A \cup B) \cap (A \cup C) = A \cup (B \cap C)$.

This property is known as the Distributive Law of Union over Intersection.

Given:

P is the set of prime numbers. So, $P = \{2, 3, 5, 7, 11, ... \}$.

S is a set defined as $S = \{t \mid 2^t – 1 \text{ is a prime}\}$.

To Prove:

We need to prove that S is a subset of P, i.e., $S \subset P$.

Proof:

To prove that $S \subset P$, we must show that every element of set S is also an element of set P. In other words, if an element $t$ belongs to S, then $t$ must be a prime number.

Let $t$ be an arbitrary element of S.

By the definition of set S, if $t \in S$, then $2^t - 1$ is a prime number.

We will now prove that $t$ must be a prime number by the method of contradiction.

Let us assume that $t$ is not a prime number. If $t$ is not prime, it must be either 1 or a composite number.

Case 1: $t = 1$

If $t = 1$, then the value of the expression is:

$2^t - 1 = 2^1 - 1 = 2 - 1 = 1$.

The number 1 is neither prime nor composite. Therefore, $2^1 - 1$ is not a prime number. This contradicts our initial condition that $2^t - 1$ is a prime number for $t \in S$. Thus, $t$ cannot be 1.

Case 2: $t$ is a composite number

If $t$ is a composite number, then it can be written as a product of two integers $a$ and $b$, where $a > 1$ and $b > 1$.

Let $t = ab$.

Now, let's substitute this into the expression $2^t - 1$:

$2^t - 1 = 2^{ab} - 1 = (2^a)^b - 1$.

We know the algebraic identity for the difference of powers: $x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + ... + x + 1)$.

Using this identity with $x = 2^a$ and $n = b$, we can factorize our expression:

$2^t - 1 = (2^a - 1)((2^a)^{b-1} + (2^a)^{b-2} + ... + 2^a + 1)$.

This shows that $(2^a - 1)$ is a factor of $2^t - 1$. For $2^t - 1$ to be composite, its factors must be integers greater than 1.

Let's examine the two factors:

Factor 1: $(2^a - 1)$

Since we assumed $a > 1$, it follows that $2^a > 2^1 = 2$.

Therefore, $2^a - 1 > 2 - 1 = 1$. So, the first factor is an integer greater than 1.

Factor 2: $((2^a)^{b-1} + (2^a)^{b-2} + ... + 2^a + 1)$

Since $a > 1$ and $b > 1$, both $2^a$ and the powers of $2^a$ are integers greater than 1. The sum of these positive integers is clearly an integer greater than 1.

Since $2^t - 1$ can be expressed as a product of two integers, both of which are greater than 1, $2^t - 1$ is a composite number.

This contradicts our initial condition that for $t \in S$, $2^t - 1$ is a prime number.

Both cases ($t=1$ and $t$ being composite) lead to a contradiction. Therefore, our assumption that $t$ is not a prime number must be false.

This implies that $t$ must be a prime number.

So, if $t \in S$, then $t \in P$.

Since every element of S is also an element of P, we can conclude that S is a subset of P.

Hence, $S \subset P$ is proved.

Given:

Let U be the set of all 50 students.

Let M be the set of students who passed in Mathematics.

Let P be the set of students who passed in Physics.

Let C be the set of students who passed in Chemistry.

We are given the following information:

• Total number of students = 50.
• Each student passed in at least one subject, so $n(M \cup P \cup C) = 50$.
• $n(M) = 37$
• $n(P) = 24$
• $n(C) = 43$
• At most 19 passed Mathematics and Physics, so $n(M \cap P) \le 19$.
• At most 29 passed Mathematics and Chemistry, so $n(M \cap C) \le 29$.
• At most 20 passed Physics and Chemistry, so $n(P \cap C) \le 20$.

To Find:

The largest possible number of students that could have passed all three examinations, i.e., the maximum possible value of $n(M \cap P \cap C)$.

Solution:

We use the Principle of Inclusion-Exclusion for three sets:

$n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(M \cap C) \ $$ - n(P \cap C) + n(M \cap P \cap C)$

Let $x = n(M \cap P \cap C)$. This is the number of students who passed all three subjects.

Substituting the given values into the formula:

$50 = 37 + 24 + 43 - [n(M \cap P) + n(M \cap C) + n(P \cap C)] + x$

$50 = 104 - [n(M \cap P) + n(M \cap C) + n(P \cap C)] + x$

Rearranging the equation to find the sum of the pairwise intersections:

Let's define the number of students who passed in exactly two subjects:

• Let 'a' be the number of students who passed in Mathematics and Physics only. So, $a = n(M \cap P) - n(M \cap P \cap C) = n(M \cap P) - x$.
• Let 'b' be the number of students who passed in Mathematics and Chemistry only. So, $b = n(M \cap C) - n(M \cap P \cap C) = n(M \cap C) - x$.
• Let 'c' be the number of students who passed in Physics and Chemistry only. So, $c = n(P \cap C) - n(M \cap P \cap C) = n(P \cap C) - x$.

Since the number of students cannot be negative, $a \ge 0$, $b \ge 0$, and $c \ge 0$.

From the definitions of a, b, and c, we can write:

$n(M \cap P) = a + x$

$n(M \cap C) = b + x$

$n(P \cap C) = c + x$

Substituting these into equation (i):

$(a + x) + (b + x) + (c + x) = 54 + x$

$a + b + c + 3x = 54 + x$

$a + b + c = 54 - 2x$

Now, let's use the "at most" conditions given in the problem:

$n(M \cap P) \le 19 \implies a + x \le 19 \implies a \le 19 - x$

$n(M \cap C) \le 29 \implies b + x \le 29 \implies b \le 29 - x$

$n(P \cap C) \le 20 \implies c + x \le 20 \implies c \le 20 - x$

Adding these three inequalities:

$a + b + c \le (19 - x) + (29 - x) + (20 - x)$

$a + b + c \le 68 - 3x$

We now have two expressions involving $a + b + c$:

$a + b + c = 54 - 2x$

$a + b + c \le 68 - 3x$

Substituting the first expression into the second one:

$54 - 2x \le 68 - 3x$

Now, we solve for $x$:

$3x - 2x \le 68 - 54$

$x \le 14$

This result shows that the number of students who passed all three subjects, $x$, must be less than or equal to 14.

Therefore, the largest possible number of students that could have passed all three examinations is 14.

Given:

There are 20 sets $X_r$, each with $|X_r| = 5$ elements, for $r=1, 2, \dots, 20$.

There are n sets $Y_r$, each with $|Y_r| = 2$ elements, for $r=1, 2, \dots, n$.

The union of all $X_r$ is equal to the union of all $Y_r$, denoted by S: $\bigcup\limits_{r=1}^{20} X_r = S = \bigcup\limits_{r=1}^{n} Y_r$.

Each element of S belongs to exactly 10 of the $X_r$'s.

Each element of S belongs to exactly 4 of the $Y_r$'s.

We can calculate the sum of the cardinalities of the sets $X_r$. This sum counts each element in the union S as many times as it appears in the sets $X_r$.

Since each element of S belongs to exactly 10 of the $X_r$'s, the total sum of cardinalities is 10 times the cardinality of S.

$\sum\limits_{r=1}^{20} |X_r| = 10 \times |S|$

We know $|X_r| = 5$ for each r, so:

$\sum\limits_{r=1}^{20} 5 = 20 \times 5 = 100$

Thus, $100 = 10 \times |S|$.

Dividing by 10, we find the cardinality of S:

$|S| = \frac{100}{10} = 10$

Similarly, we can calculate the sum of the cardinalities of the sets $Y_r$. This sum counts each element in the union S as many times as it appears in the sets $Y_r$.

Since each element of S belongs to exactly 4 of the $Y_r$'s, the total sum of cardinalities is 4 times the cardinality of S.

$\sum\limits_{r=1}^{n} |Y_r| = 4 \times |S|$

We know $|Y_r| = 2$ for each r, so:

$\sum\limits_{r=1}^{n} 2 = n \times 2 = 2n$

Thus, $2n = 4 \times |S|$.

Substitute the value of $|S| = 10$ we found:

$2n = 4 \times 10$

$2n = 40$

Divide by 2 to find n:

$n = \frac{40}{2} = 20$

The value of n is 20.

Comparing this with the given options:

(A) 10

(B) 20

(C) 100

(D) 50

The correct option is (B).

The final answer is (B) 20.

Let the first finite set have m elements.

The total number of subsets of the first set is $2^m$.

Let the second finite set have n elements.

The total number of subsets of the second set is $2^n$.

According to the problem statement, the total number of subsets of the first set is 56 more than the total number of subsets of the second set.

This can be written as an equation:

$2^m = 2^n + 56$

Rearranging the equation, we get:

$2^m - 2^n = 56$

Since the difference is positive, we know that $2^m > 2^n$, which implies $m > n$.

We can factor out $2^n$ from the left side of the equation:

$2^n(2^{m-n} - 1) = 56$

Now, let's find the prime factorization of 56:

$56 = 8 \times 7 = 2^3 \times 7$

So the equation becomes:

$2^n(2^{m-n} - 1) = 2^3 \times 7$

Since $m > n$, $m-n$ is a positive integer. Thus, $2^{m-n}$ is an even integer greater than or equal to $2^1 = 2$. Therefore, $2^{m-n} - 1$ is always an odd integer greater than or equal to $2^1 - 1 = 1$.

Comparing the factors on both sides of the equation $2^n(2^{m-n} - 1) = 2^3 \times 7$, we must equate the power of 2 terms and the odd terms separately.

The power of 2 term on the left is $2^n$. The power of 2 term on the right is $2^3$.

So, $2^n = 2^3$, which implies $n = 3$.

The odd term on the left is $2^{m-n} - 1$. The odd term on the right is 7.

So, $2^{m-n} - 1 = 7$.

Adding 1 to both sides, we get $2^{m-n} = 8$.

Since $8 = 2^3$, we have $2^{m-n} = 2^3$, which implies $m-n = 3$.

We have a system of two equations with m and n:

$n = 3$

$m - n = 3$

Substitute the value of n from the first equation into the second equation:

$m - 3 = 3$

Add 3 to both sides:

$m = 3 + 3 = 6$

So, the values of m and n are 6 and 3 respectively.

Let's check if these values satisfy the original equation:

Number of subsets of the first set (m=6): $2^6 = 64$.

Number of subsets of the second set (n=3): $2^3 = 8$.

$2^m = 2^n + 56$

$64 = 8 + 56$

$64 = 64$

The equation is satisfied.

The values of m and n respectively are 6 and 3.

Comparing this with the given options:

(A) 7, 6

(B) 5, 1

(C) 6, 3

(D) 8, 7

The correct option is (C).

The final answer is (C) 6, 3.

We need to simplify the given set expression using set properties.

The expression is: $(A \cup B \cup C) \cap (A \cap B' \cap C')' \cap C'$

Let's simplify the second term using De Morgan's Law: $(A \cap B' \cap C')'$

De Morgan's Law states $(X \cap Y)' = X' \cup Y'$ and $(X \cup Y)' = X' \cap Y'$.

Applying it to $(A \cap (B' \cap C'))'$: $(A)' \cup (B' \cap C')'$

Applying De Morgan's Law again to $(B' \cap C')'$: $(B')' \cup (C')' = B \cup C$

So, $(A \cap B' \cap C')' = A' \cup (B \cup C)$.

Now substitute this back into the original expression:

$(A \cup B \cup C) \cap (A' \cup B \cup C) \cap C'$

Let's rearrange the terms using the commutative property of intersection:

$((A \cup B \cup C) \cap (A' \cup B \cup C)) \cap C'$

Notice that the term $(B \cup C)$ is common in the first two sets within the outer intersection. We can use the distributive law of intersection over union in reverse: $(X \cup Y) \cap (X' \cup Y) = (X \cap X') \cup Y$.

Here, let $X = A$ and $Y = B \cup C$. Then $X' = A'$.

$(A \cup (B \cup C)) \cap (A' \cup (B \cup C)) = (A \cap A') \cup (B \cup C)$

We know that $A \cap A' = \emptyset$ (the empty set).

So, $(A \cap A') \cup (B \cup C) = \emptyset \cup (B \cup C) = B \cup C$.

Now substitute this back into the expression:

$(B \cup C) \cap C'$

Now, distribute the intersection with $C'$ over the union $(B \cup C)$:

$(B \cap C') \cup (C \cap C')$

We know that $C \cap C' = \emptyset$ (the empty set).

So, $(B \cap C') \cup (C \cap C') = (B \cap C') \cup \emptyset$

The union of any set with the empty set is the set itself:

$(B \cap C') \cup \emptyset = B \cap C'$

Thus, the simplified expression is $B \cap C'$.

Let's trace the steps:

$(A \cup B \cup C) \cap (A \cap B' \cap C')' \cap C'$

= $(A \cup B \cup C) \cap (A' \cup (B' \cap C')') \cap C'$ [De Morgan's Law]

= $(A \cup B \cup C) \cap (A' \cup (B \cup C)) \cap C'$ [De Morgan's Law and Double Complement Law]

= $((A \cup (B \cup C)) \cap (A' \cup (B \cup C))) \cap C'$ [Commutative and Associative Laws of Intersection]

= $((A \cap A') \cup (B \cup C)) \cap C'$ [Distributive Law: $(X \cup Y) \cap (X' \cup Y) = (X \cap X') \cup Y$ with $X=A, Y=B \cup C$]

= $(\emptyset \cup (B \cup C)) \cap C'$ [Property of complement]

= $(B \cup C) \cap C'$ [Property of empty set]

= $(B \cap C') \cup (C \cap C')$ [Distributive Law of Intersection over Union]

= $(B \cap C') \cup \emptyset$ [Property of complement]

= $B \cap C'$ [Property of empty set]

The simplified expression is $B \cap C'$.

Comparing this with the given options:

(A) B ∩ C'

(B) A ∩ C

(C) B ∪ C'

(D) A ∩ C'

The correct option is (A).

The final answer is (A) B ∩ C'.

Solution:

We know the Principle of Inclusion-Exclusion for two finite sets A and B, which states:

Here, $n(X)$ represents the cardinality (number of elements) of a set X.

• $n(A \cup B)$ is the number of elements in the union of A and B.
• $n(A \cap B)$ is the number of elements in the intersection of A and B.

To find an expression for $n(A) + n(B)$, we can rearrange the formula above.

By adding $n(A \cap B)$ to both sides of the equation, we get:

$n(A \cup B) + n(A \cap B) = n(A) + n(B) - n(A \cap B) + n(A \cap B)$

$n(A \cup B) + n(A \cap B) = n(A) + n(B)$

Therefore, the expression that is equal to $n(A) + n(B)$ is $n(A \cup B) + n(A \cap B)$.

Conceptually: When we add $n(A)$ and $n(B)$, we count the elements in the intersection $(A \cap B)$ twice. The term $n(A \cup B)$ counts the elements in the union, where the intersection is counted only once. By adding $n(A \cap B)$ to $n(A \cup B)$, we effectively count the intersection elements twice, which balances the equation.

The blank should be filled with $n(A \cup B) + n(A \cap B)$.

Solution:

Let A be a finite set with $n$ elements. We want to find the total number of subsets of A.

Let $A = \{a_1, a_2, a_3, ..., a_n\}$.

To construct a subset of A, we must make a decision for each element of A: either to include it in the subset or to exclude it.

• For the element $a_1$, there are 2 choices (include or exclude).
• For the element $a_2$, there are 2 choices (include or exclude).
• ...
• For the element $a_n$, there are 2 choices (include or exclude).

Since the choice for each element is independent of the others, the total number of ways to form a subset is the product of the number of choices for each element.

Total number of subsets = $\underbrace{2 \times 2 \times 2 \times \dots \times 2}_{n \text{ times}} = 2^n$.

Alternate Solution (Using Combinations):

The total number of subsets is the sum of the number of subsets with 0 elements, 1 element, 2 elements, and so on, up to $n$ elements.

The number of subsets containing exactly $k$ elements from a set of $n$ elements is given by the combination formula $\binom{n}{k}$.

So, Total number of subsets = (subsets with 0 elements) + (subsets with 1 element) + ... + (subsets with n elements)

Total number of subsets = $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}$

From the Binomial Theorem, we know that:

$(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k = \binom{n}{0}x^n y^0 + \binom{n}{1}x^{n-1}y^1 + \dots + \binom{n}{n}x^0 y^n$

By substituting $x=1$ and $y=1$ into the theorem, we get:

$(1+1)^n = \binom{n}{0}(1)^n(1)^0 + \binom{n}{1}(1)^{n-1}(1)^1 + \dots + \binom{n}{n}(1)^0(1)^n$

$2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}$

This confirms that the total number of subsets is $2^n$.

The blank should be filled with $2^n$.

Given:

R = {x ∈ Z | x is divisible by 2} = the set of all even integers.

S = {y ∈ Z | y is divisible by 3} = the set of all integers that are multiples of 3.

The statement is R ∩ S = $\emptyset$ (the empty set).

The intersection R ∩ S consists of all elements that are in both set R and set S.

An element $z \in R \cap S$ means $z \in R$ and $z \in S$.

If $z \in R$, then $z$ is an integer divisible by 2. This means $z$ is a multiple of 2, so $z = 2k$ for some integer $k$.

If $z \in S$, then $z$ is an integer divisible by 3. This means $z$ is a multiple of 3, so $z = 3m$ for some integer $m$.

So, the elements in R ∩ S are integers that are divisible by both 2 and 3.

A number that is divisible by both 2 and 3 is divisible by their least common multiple (LCM).

The LCM of 2 and 3 is 6.

So, the elements in R ∩ S are integers that are divisible by 6. These are the multiples of 6.

R ∩ S = {..., -12, -6, 0, 6, 12, ...}

This set R ∩ S contains elements such as 0, 6, -6, 12, -12, etc.

For example, $6 \in R$ because 6 is divisible by 2 (6 = 2 × 3). Also, $6 \in S$ because 6 is divisible by 3 (6 = 3 × 2). Therefore, $6 \in R \cap S$.

Since the set R ∩ S contains elements (like 6), it is not the empty set ($\emptyset$).

The statement R ∩ S = $\emptyset$ is therefore incorrect.

Thus, the statement is False.

Given:

Q is the set of rational numbers.

R is the set of real numbers.

The statement is Q ∩ R = Q.

We know the relationship between rational numbers and real numbers.

A rational number is a number that can be expressed as a fraction $\frac{p}{q}$ where p and q are integers and $q \neq 0$.

A real number is any number that can be placed on the number line. The set of real numbers R includes all rational numbers and all irrational numbers (numbers that cannot be expressed as a simple fraction, like $\sqrt{2}$ or $\pi$).

By definition, every rational number is a real number. This means that the set of rational numbers Q is a subset of the set of real numbers R.

$Q \subset R$

The intersection of two sets X and Y, denoted by $X \cap Y$, contains all elements that are common to both X and Y.

In this case, we are considering Q ∩ R.

An element $x \in Q \cap R$ means $x \in Q$ and $x \in R$.

Since every element of Q is also an element of R (because $Q \subset R$), an element $x$ is in both sets if and only if $x$ is in Q.

Therefore, the elements common to both Q and R are precisely the elements of Q.

So, $Q \cap R = Q$.

This is a general property of sets: if $A \subset B$, then $A \cap B = A$.

Since Q is a subset of R, their intersection is Q.

The statement Q ∩ R = Q is therefore correct.

Thus, the statement is True.

Solution:

(i) A = {x : x ∈ R, 2x + 11 = 15}

To find the elements of set A, we need to solve the equation $2x + 11 = 15$ for $x \in R$.

$2x = 15 - 11$

$2x = 4$

$x = \frac{4}{2}$

$x = 2$

The only real number that satisfies the condition $2x + 11 = 15$ is 2.

Therefore, the set A in roster form is:

A = {2}

(ii) B = {x | x² = x, x ∈ R}

To find the elements of set B, we need to solve the equation $x^2 = x$ for $x \in R$.

$x^2 - x = 0$

Factor out x:

$x(x - 1) = 0$

This equation is satisfied if $x = 0$ or if $x - 1 = 0$.

If $x - 1 = 0$, then $x = 1$.

The real numbers that satisfy the condition $x^2 = x$ are 0 and 1.

Therefore, the set B in roster form is:

B = {0, 1}

(iii) C = {x | x is a positive factor of a prime number p}

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let p be any prime number (e.g., 2, 3, 5, 7, 11, ...).

The positive factors of a prime number p are exactly 1 and p.

Therefore, the set C contains these two factors.

In roster form, the set C is:

C = {1, p}

Solution:

(i) D = {t | t³ = t, t ∈ R}

To find the elements of set D, we need to solve the equation $t^3 = t$ for $t \in R$.

$t^3 - t = 0$

$t(t^2 - 1) = 0$

Using the difference of squares formula, $t^2 - 1 = (t - 1)(t + 1)$.

So, the equation becomes $t(t - 1)(t + 1) = 0$.

This equation is satisfied if any of the factors are equal to zero:

$t = 0$ or $t - 1 = 0$ or $t + 1 = 0$

Solving for t:

$t = 0$, $t = 1$, or $t = -1$.

All these values are real numbers.

Therefore, the set D in roster form is:

D = {-1, 0, 1}

(ii) E = {w | $\frac{w \;-\; 2}{w \;+\; 3} = 3$, w ∈ R}

To find the elements of set E, we need to solve the equation $\frac{w - 2}{w + 3} = 3$ for $w \in R$.

First, note that the denominator cannot be zero, so $w + 3 \neq 0$, which means $w \neq -3$.

Multiply both sides by $(w + 3)$:

$w - 2 = 3(w + 3)$

$w - 2 = 3w + 9$

Subtract w from both sides and subtract 9 from both sides:

$-2 - 9 = 3w - w$

$-11 = 2w$

Divide by 2:

$w = -\frac{11}{2}$

This value is a real number and is not equal to -3.

Therefore, the set E in roster form is:

E = {$-\frac{11}{2}$}

(iii) F = {x | x⁴ – 5x² + 6 = 0, x ∈ R}

To find the elements of set F, we need to solve the equation $x^4 – 5x^2 + 6 = 0$ for $x \in R$.

This is a quadratic equation in terms of $x^2$. Let $y = x^2$. Substituting this into the equation, we get:

$y^2 - 5y + 6 = 0$

Factor the quadratic equation:

$(y - 2)(y - 3) = 0$

This gives two possible values for y:

$y - 2 = 0 \implies y = 2$

$y - 3 = 0 \implies y = 3$

Now substitute back $y = x^2$:

Case 1: $x^2 = 2$

Taking the square root of both sides, $x = \pm\sqrt{2}$. Both $\sqrt{2}$ and $-\sqrt{2}$ are real numbers.

Case 2: $x^2 = 3$

Taking the square root of both sides, $x = \pm\sqrt{3}$. Both $\sqrt{3}$ and $-\sqrt{3}$ are real numbers.

The real numbers that satisfy the condition are $-\sqrt{3}$, $-\sqrt{2}$, $\sqrt{2}$, and $\sqrt{3}$.

Therefore, the set F in roster form is:

F = {$-\sqrt{3}$, $-\sqrt{2}$, $\sqrt{2}$, $\sqrt{3}$}

Solution:

The set Y is defined as the set of positive factors of the number $N = 2^{p – 1} (2^{p} – 1)$, where $2^{p} – 1$ is a prime number.

Let $q = 2^p - 1$. We are given that $q$ is a prime number. The number is $N = 2^{p-1} \cdot q$.

The positive factors of $2^{p-1}$ are $2^0, 2^1, 2^2, \ldots, 2^{p-1}$.

Since $q$ is a prime number, its positive factors are $1$ and $q$.

The positive factors of the number $N = 2^{p-1} \cdot q$ are obtained by taking the product of each factor of $2^{p-1}$ with each factor of $q$.

The factors are of the form $2^k \cdot q^m$, where $0 \leq k \leq p-1$ and $0 \leq m \leq 1$ (since $q$ is prime, its power in the prime factorization of $N$ is 1).

For $m=0$, the factors are $2^k \cdot q^0 = 2^k \cdot 1 = 2^k$. These are $2^0, 2^1, \ldots, 2^{p-1}$, which are $1, 2, \ldots, 2^{p-1}$.

For $m=1$, the factors are $2^k \cdot q^1 = 2^k \cdot q$. These are $2^0 \cdot q, 2^1 \cdot q, \ldots, 2^{p-1} \cdot q$, which are $q, 2q, \ldots, 2^{p-1}q$.

Substituting $q = 2^p - 1$, the factors are $2^k$ for $k=0, 1, \ldots, p-1$ and $2^k (2^p - 1)$ for $k=0, 1, \ldots, p-1$.

Since $2^p - 1$ is prime, and for $p \ge 2$ (as $2^p-1$ is prime, it must be greater than 1, so $p>1$), $2^p-1 \ge 3$. Thus, none of the terms $2^k$ (which are powers of 2) can be equal to any of the terms $2^k(2^p-1)$ (which are multiples of an odd prime $\ge 3$). Therefore, all these factors are distinct.

Listing all the factors in the set Y:

Y = {$1, 2, 2^2, \ldots, 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), \ldots, 2^{p-1}(2^p-1)$}

Therefore, the set Y in roster form is:

Y = {$1, 2, 2^2, \ldots, 2^{p-1}, 2^p-1, 2(2^p-1), \ldots, 2^{p-1}(2^p-1)$}

Solution:

(i) 35 ∈ {x | x has exactly four positive factors}.

To determine if this statement is true or false, we find the positive factors of 35.

The positive factors of 35 are the positive integers that divide 35 without leaving a remainder.

$35 \div 1 = 35$

$35 \div 5 = 7$

$35 \div 7 = 5$

$35 \div 35 = 1$

The positive factors of 35 are 1, 5, 7, and 35.

There are exactly four positive factors of 35.

Thus, 35 belongs to the set {x | x has exactly four positive factors}.

Therefore, the statement is True.

(ii) 128 ∈ {y | the sum of all the positive factors of y is 2y}.

This statement claims that 128 is a perfect number (a number where the sum of its positive factors equals twice the number).

First, find the positive factors of 128.

The prime factorization of 128 is $128 = 2^7$.

The positive factors of $2^7$ are $2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7$.

These factors are 1, 2, 4, 8, 16, 32, 64, and 128.

Now, calculate the sum of these factors:

Sum = $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128$

This is a geometric series sum $\sum_{k=0}^{7} 2^k = \frac{2^{7+1} - 1}{2 - 1} = \frac{2^8 - 1}{1} = 256 - 1 = 255$.

Next, calculate $2y = 2 \times 128 = 256$.

Compare the sum of factors with 2y:

$255 \neq 256$

The sum of the positive factors of 128 is 255, which is not equal to $2 \times 128$.

Thus, 128 is not a perfect number, and it does not belong to the given set.

Therefore, the statement is False.

(iii) 3 ∉ {x | x⁴ – 5x³ + 2x² – 112x + 6 = 0}.

This statement claims that 3 is not a solution (root) to the equation $x^4 – 5x^3 + 2x^2 – 112x + 6 = 0$.

To check this, substitute $x = 3$ into the polynomial $P(x) = x^4 – 5x^3 + 2x^2 – 112x + 6$ and evaluate it.

$P(3) = (3)^4 – 5(3)^3 + 2(3)^2 – 112(3) + 6$

$P(3) = 81 – 5(27) + 2(9) – 336 + 6$

$P(3) = 81 – 135 + 18 – 336 + 6$

$P(3) = (81 + 18 + 6) – (135 + 336)$

$P(3) = 105 – 471$

$P(3) = -366$

Since $P(3) = -366 \neq 0$, $x=3$ is not a root of the equation $x^4 – 5x^3 + 2x^2 – 112x + 6 = 0$.

Thus, 3 does not belong to the set {x | x⁴ – 5x³ + 2x² – 112x + 6 = 0}.

Therefore, the statement is True.

(iv) 496 ∉ {y | the sum of all the positive factors of y is 2y}.

This statement claims that 496 is not a perfect number.

To check this, find the positive factors of 496 and sum them.

First, find the prime factorization of 496:

$\begin{array}{c|cc} 2 & 496 \\ \hline 2 & 248 \\ \hline 2 & 124 \\ \hline 2 & 62 \\ \hline 31 & 31 \\ \hline & 1 \end{array}$

So, $496 = 2^4 \times 31^1$.

The sum of positive factors $\sigma(n)$ is calculated using the formula $\sigma(p^a) = \frac{p^{a+1}-1}{p-1}$ for a prime power $p^a$, and $\sigma(n) = \prod \sigma(p_i^{a_i})$ for $n = \prod p_i^{a_i}$.

Sum of factors of 496 = $\sigma(2^4) \times \sigma(31^1)$

$\sigma(2^4) = \frac{2^{4+1} - 1}{2 - 1} = \frac{2^5 - 1}{1} = 32 - 1 = 31$.

$\sigma(31^1) = \frac{31^{1+1} - 1}{31 - 1} = \frac{31^2 - 1}{30} = \frac{961 - 1}{30} = \frac{960}{30} = 32$.

Sum of factors of 496 = $31 \times 32 = 992$.

Next, calculate $2y = 2 \times 496 = 992$.

Compare the sum of factors with 2y:

$992 = 992$

The sum of the positive factors of 496 is 992, which is equal to $2 \times 496$.

Thus, 496 is a perfect number, and it does belong to the given set.

Therefore, the statement "496 ∉ {y | the sum of all the positive factors of y is 2y}" is False.

Given:

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

Set M = {3, 4, 5, 6}

Set N = {1, 3, 5}

To Verify:

$L – (M \cup N) = (L – M) \cap (L – N)$

Solution:

We need to evaluate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately and check if they are equal.

Evaluate LHS: $L – (M \cup N)$

First, find the union of sets M and N ($M \cup N$). The union contains all elements present in either M or N or both.

$M \cup N = \{3, 4, 5, 6\} \cup \{1, 3, 5\}$

$M \cup N = \{1, 3, 4, 5, 6\}$

Next, find the difference between set L and the set $(M \cup N)$. The difference $L – (M \cup N)$ contains all elements that are in L but are not in $(M \cup N)$.

$L = \{1, 2, 3, 4\}$

$M \cup N = \{1, 3, 4, 5, 6\}$

$L – (M \cup N) = \{1, 2, 3, 4\} – \{1, 3, 4, 5, 6\}$

Elements in L are 1, 2, 3, 4.

Elements in $(M \cup N)$ are 1, 3, 4, 5, 6.

Elements in L that are not in $(M \cup N)$: Only 2 is in L but not in $\{1, 3, 4, 5, 6\}$.

So, $L – (M \cup N) = \{2\}$.

Evaluate RHS: $(L – M) \cap (L – N)$

First, find the difference between set L and set M ($L – M$). This set contains all elements that are in L but not in M.

$L = \{1, 2, 3, 4\}$

$M = \{3, 4, 5, 6\}$

$L – M = \{1, 2, 3, 4\} – \{3, 4, 5, 6\}$

Elements in L are 1, 2, 3, 4.

Elements in M are 3, 4, 5, 6.

Elements in L that are not in M: 1 and 2 are in L but not in $\{3, 4, 5, 6\}$.

$L – M = \{1, 2\}$

Next, find the difference between set L and set N ($L – N$). This set contains all elements that are in L but not in N.

$L = \{1, 2, 3, 4\}$

$N = \{1, 3, 5\}$

$L – N = \{1, 2, 3, 4\} – \{1, 3, 5\}$

Elements in L are 1, 2, 3, 4.

Elements in N are 1, 3, 5.

Elements in L that are not in N: 2 and 4 are in L but not in $\{1, 3, 5\}$.

$L – N = \{2, 4\}$

Finally, find the intersection of $(L – M)$ and $(L – N)$. The intersection contains all elements that are common to both sets.

$L – M = \{1, 2\}$

$L – N = \{2, 4\}$

$(L – M) \cap (L – N) = \{1, 2\} \cap \{2, 4\}$

The element common to both $\{1, 2\}$ and $\{2, 4\}$ is 2.

So, $(L – M) \cap (L – N) = \{2\}$.

Verification:

From (i), LHS = $\{2\}$.

From (ii), RHS = $\{2\}$.

Since LHS = RHS, the identity $L – (M \cup N) = (L – M) \cap (L – N)$ is verified for the given sets L, M, and N.

Given:

A and B are subsets of a universal set U.

(i) To prove: $A \subset A \cup B$

Proof:

To prove that A is a subset of $A \cup B$, we must show that every element of A is also an element of $A \cup B$.

Let $x$ be an arbitrary element of set A.

So, $x \in A$.

By the definition of the union of two sets, $A \cup B$ contains all elements that are in A, or in B, or in both.

The condition for an element to be in $A \cup B$ is: $x \in A$ or $x \in B$.

Since we have established that $x \in A$, this condition is satisfied.

Therefore, $x \in A \cup B$.

We have shown that for any arbitrary element $x$, if $x \in A$, then $x \in A \cup B$.

Hence, by the definition of a subset, $A \subset A \cup B$.

(ii) To prove: $A \subset B \Leftrightarrow A \cup B = B$

Proof:

This is a biconditional statement, which means we must prove it in two parts.

Part 1: Prove that if $A \subset B$, then $A \cup B = B$.

Assume that $A \subset B$.

To prove that $A \cup B = B$, we need to show that $(A \cup B) \subset B$ and $B \subset (A \cup B)$.

First, let's show that $(A \cup B) \subset B$.

Let $x$ be an arbitrary element of $A \cup B$.

By definition of union, $x \in A$ or $x \in B$.

Case 1: $x \in A$. Since we have assumed $A \subset B$, if $x \in A$, then it must be that $x \in B$.

Case 2: $x \in B$.

In both cases, if $x \in A \cup B$, then $x \in B$. Therefore, $(A \cup B) \subset B$.

Next, let's show that $B \subset (A \cup B)$.

As shown in part (i) of this question, a set is always a subset of its union with any other set. Thus, $B \subset (A \cup B)$ is always true.

Since $(A \cup B) \subset B$ and $B \subset (A \cup B)$, we can conclude that $A \cup B = B$.

Part 2: Prove that if $A \cup B = B$, then $A \subset B$.

Assume that $A \cup B = B$.

To prove that $A \subset B$, we need to show that every element of A is also an element of B.

Let $x$ be an arbitrary element of A, so $x \in A$.

From part (i), we know that for any sets A and B, $A \subset (A \cup B)$.

So, since $x \in A$, it follows that $x \in (A \cup B)$.

But we have assumed that $A \cup B = B$.

Therefore, if $x \in (A \cup B)$, it must be that $x \in B$.

We have shown that if $x \in A$, then $x \in B$.

Hence, by the definition of a subset, $A \subset B$.

Since both parts of the biconditional statement have been proven, we can conclude that $A \subset B \Leftrightarrow A \cup B = B$.

(iii) To prove: $(A \cap B) \subset A$

Proof:

To prove that $(A \cap B)$ is a subset of A, we must show that every element of $(A \cap B)$ is also an element of A.

Let $x$ be an arbitrary element of the set $(A \cap B)$.

So, $x \in (A \cap B)$.

By the definition of the intersection of two sets, $(A \cap B)$ contains all elements that are in both A and B.

This means that if $x \in (A \cap B)$, then $x \in A$ and $x \in B$.

From this compound statement, the statement "$x \in A$" must be true.

We have shown that for any arbitrary element $x$, if $x \in (A \cap B)$, then $x \in A$.

Hence, by the definition of a subset, $(A \cap B) \subset A$.

Given:

Set N = {1, 2, 3, ..., 100}

To Write:

(i) The subset of N whose elements are even numbers.

(ii) The subset of N whose elements are perfect square numbers.

Solution:

The set N contains all natural numbers from 1 to 100, i.e., N = {1, 2, 3, 4, ..., 99, 100}.

(i) Subset of N with even numbers:

An even number is an integer that is divisible by 2.

We need to list all the even numbers in the set N.

The even numbers from 1 to 100 are 2, 4, 6, 8, 10, and so on, up to 100.

Let E be the subset of N consisting of even numbers.

E = {x | x ∈ N and x is even}

Listing the elements in roster form:

E = {2, 4, 6, 8, 10, 12, ..., 100}

(ii) Subset of N with perfect square numbers:

A perfect square number is an integer that is the square of an integer. We need to find the integers $m$ such that $m^2 \in N$, i.e., $1 \leq m^2 \leq 100$.

We find the squares of integers starting from 1:

$1^2 = 1$ (which is in N)

$2^2 = 4$ (which is in N)

$3^2 = 9$ (which is in N)

$4^2 = 16$ (which is in N)

$5^2 = 25$ (which is in N)

$6^2 = 36$ (which is in N)

$7^2 = 49$ (which is in N)

$8^2 = 64$ (which is in N)

$9^2 = 81$ (which is in N)

$10^2 = 100$ (which is in N)

$11^2 = 121$ (which is not in N, as it is greater than 100)

So, the perfect square numbers in N are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

Let P be the subset of N consisting of perfect square numbers.

P = {x | x ∈ N and x is a perfect square}

Listing the elements in roster form:

P = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

Given:

Set X = {1, 2, 3}

n represents any member of X.

To Write:

The sets containing all numbers represented by:

(i) 4n

(ii) n + 6

(iii) $\frac{n}{2}$

(iv) n – 1

Solution:

We will evaluate the expression for each element in set X.

(i) Set of numbers represented by 4n:

• For $n = 1$: $4 \times 1 = 4$
• For $n = 2$: $4 \times 2 = 8$
• For $n = 3$: $4 \times 3 = 12$

The set containing these numbers is {4, 8, 12}.

Set = {4, 8, 12}

(ii) Set of numbers represented by n + 6:

• For $n = 1$: $1 + 6 = 7$
• For $n = 2$: $2 + 6 = 8$
• For $n = 3$: $3 + 6 = 9$

The set containing these numbers is {7, 8, 9}.

Set = {7, 8, 9}

(iii) Set of numbers represented by $\frac{n}{2}$:

• For $n = 1$: $\frac{1}{2}$
• For $n = 2$: $\frac{2}{2} = 1$
• For $n = 3$: $\frac{3}{2}$

The set containing these numbers is {$\frac{1}{2}$, 1, $\frac{3}{2}$}.

Set = {$\frac{1}{2}$, 1, $\frac{3}{2}$}

(iv) Set of numbers represented by n – 1:

• For $n = 1$: $1 - 1 = 0$
• For $n = 2$: $2 - 1 = 1$
• For $n = 3$: $3 - 1 = 2$

The set containing these numbers is {0, 1, 2}.

Set = {0, 1, 2}

Given:

Set Y = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

a represents any element of Y.

To Write:

The sets containing all elements satisfying the given conditions.

Solution:

We will check each condition for every element in set Y.

(i) a ∈ Y but a² ∉ Y

We need to find the elements 'a' in Y such that their square ($a^2$) is not in Y.

• For $a = 1$: $a^2 = 1^2 = 1$. $1 \in Y$. This element does not satisfy the condition.
• For $a = 2$: $a^2 = 2^2 = 4$. $4 \in Y$. This element does not satisfy the condition.
• For $a = 3$: $a^2 = 3^2 = 9$. $9 \in Y$. This element does not satisfy the condition.
• For $a = 4$: $a^2 = 4^2 = 16$. $16 \notin Y$ (since Y goes up to 10). This element satisfies the condition.
• For $a = 5$: $a^2 = 5^2 = 25$. $25 \notin Y$. This element satisfies the condition.
• For $a = 6$: $a^2 = 6^2 = 36$. $36 \notin Y$. This element satisfies the condition.
• For $a = 7$: $a^2 = 7^2 = 49$. $49 \notin Y$. This element satisfies the condition.
• For $a = 8$: $a^2 = 8^2 = 64$. $64 \notin Y$. This element satisfies the condition.
• For $a = 9$: $a^2 = 9^2 = 81$. $81 \notin Y$. This element satisfies the condition.
• For $a = 10$: $a^2 = 10^2 = 100$. $100 \notin Y$. This element satisfies the condition.

The elements 'a' in Y that satisfy the condition $a^2 \notin Y$ are 4, 5, 6, 7, 8, 9, 10.

Set = {4, 5, 6, 7, 8, 9, 10}

(ii) a + 1 = 6, a ∈ Y

We need to find the element 'a' in Y that satisfies the equation $a + 1 = 6$.

Solving the equation: $a + 1 = 6 \implies a = 6 - 1 \implies a = 5$.

We check if $a = 5$ is in the set Y = {1, 2, 3, ..., 10}. Yes, 5 ∈ Y.

The only element 'a' in Y that satisfies the condition is 5.

Set = {5}

(iii) a is less than 6 and a ∈ Y

We need to find the elements 'a' in Y that are less than 6.

The elements in Y are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

We list the elements from this set that are strictly less than 6:

1, 2, 3, 4, 5.

These elements satisfy both conditions: they are in Y and they are less than 6.

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

Given:

The Universal Set, U, is the set of all whole numbers. $U = \{0, 1, 2, 3, ... \}$.

The subsets are:

• $A = \{2, 4, 6, 8, 12, 20\}$
• $B = \{3, 6, 9, 12, 15\}$
• $C = \{5, 10, 15, 20\}$

Solution:

To draw the Venn diagram, we first need to find the elements in the intersection of these sets.

1. Intersection of A and B ($A \cap B$):

These are the elements common to both A and B.

$A \cap B = \{6, 12\}$

2. Intersection of B and C ($B \cap C$):

These are the elements common to both B and C.

$B \cap C = \{15\}$

3. Intersection of A and C ($A \cap C$):

These are the elements common to both A and C.

$A \cap C = \{20\}$

4. Intersection of A, B, and C ($A \cap B \cap C$):

These are the elements common to all three sets.

$A \cap B \cap C = \phi$ (the empty set), as there are no elements common to all three sets.

Now we can determine the elements that belong to each specific region of the Venn diagram:

• Elements in A only: $A - (B \cup C) = \{2, 4, 8\}$
• Elements in B only: $B - (A \cup C) = \{3, 9\}$
• Elements in C only: $C - (A \cup B) = \{5, 10\}$
• Elements in A and B only: $(A \cap B) - C = \{6, 12\}$
• Elements in B and C only: $(B \cap C) - A = \{15\}$
• Elements in A and C only: $(A \cap C) - B = \{20\}$
• Elements in A and B and C: $A \cap B \cap C = \phi$

Venn Diagram:

The Venn diagram is drawn with a rectangle representing the universal set U and three overlapping circles for sets A, B, and C. The elements are placed in their respective regions.

Given:

We have the following sets:

• U: The set of all boys and girls in a school (Universal Set).
• G: The set of all girls in the school.
• B: The set of all boys in the school.
• S: The set of all students who take swimming.

We are also told that some, but not all, students take swimming.

Analysis of Set Relationships:

Before drawing the diagram, let's analyze the relationships between the sets:

1. Relationship between U, G, and B:
   • The set of all students (U) is composed entirely of girls (G) and boys (B).
   • A student cannot be both a boy and a girl, so the sets G and B are disjoint. This means their intersection is the empty set: $G \cap B = \phi$.
   • The union of all girls and all boys makes up the entire school. Therefore, $G \cup B = U$.
   • This means G and B form a partition of the universal set U.
2. Relationship of S with other sets:
   • S is the set of students who swim, so it is a subset of U ($S \subset U$).
   • "Some, but not all, students... take swimming" means that S is not an empty set ($S \neq \phi$) and S is not equal to the universal set ($S \neq U$). It is a proper subset of U.
   • Since both boys and girls can swim, the set S will likely overlap with both set B and set G.
   • The intersection $S \cap G$ represents the girls who swim.
   • The intersection $S \cap B$ represents the boys who swim.

Venn Diagram:

Based on the analysis, we can draw the Venn diagram as follows:

• The universal set U is represented by a rectangle.
• This rectangle is divided into two non-overlapping regions, representing the disjoint sets B (Boys) and G (Girls). Together, these two regions fill the entire rectangle U.
• The set S (Swimmers) is represented by a circle that overlaps with both the B region and the G region. This shows that some boys are swimmers and some girls are swimmers.
• The circle for S does not cover the entire rectangle, as not all students are swimmers.

Given:

A, B, and C are any three sets.

To Show:

$(A – B) \cap (C – B) = A – (B \cup C)$

Verification of the Statement

The given identity is incorrect. We can demonstrate this with a counterexample.

Counterexample:

Let's define three sets:

• $A = \{1, 2, 3\}$
• $B = \{3, 4, 5\}$
• $C = \{1, 5, 6\}$

Now, let's calculate the Left Hand Side (LHS) and the Right Hand Side (RHS) of the given equation.

LHS = $(A – B) \cap (C – B)$

First, calculate the set differences:

$A – B = \{1, 2, 3\} – \{3, 4, 5\} = \{1, 2\}$

$C – B = \{1, 5, 6\} – \{3, 4, 5\} = \{1, 6\}$

Now, find their intersection:

$(A – B) \cap (C – B) = \{1, 2\} \cap \{1, 6\} = \{1\}$

RHS = $A – (B \cup C)$

First, calculate the union of B and C:

$B \cup C = \{3, 4, 5\} \cup \{1, 5, 6\} = \{1, 3, 4, 5, 6\}$

Now, calculate the set difference:

$A – (B \cup C) = \{1, 2, 3\} – \{1, 3, 4, 5, 6\} = \{2\}$

Conclusion:

Since LHS = $\{1\}$ and RHS = $\{2\}$, we have $LHS \neq RHS$. Therefore, the statement $(A – B) \cap (C – B) = A – (B \cup C)$ is not true for all sets A, B, and C.

Correction and Proof of a Valid Identity

It is likely that there is a typographical error in the question. A related and correct set identity is:

$(A – B) \cap (C – B) = (A \cap C) – B$

We will now prove this corrected statement.

Proof:

To prove the equality of the two sets, we will show that each set is a subset of the other.

Part 1: Show that $(A – B) \cap (C – B) \subseteq (A \cap C) – B$

Let $x$ be an arbitrary element of $(A – B) \cap (C – B)$.

$\implies x \in (A – B)$ and $x \in (C – B)$

$\implies (x \in A$ and $x \notin B$) and ($x \in C$ and $x \notin B$)

By combining these conditions, we can say:

$\implies (x \in A$ and $x \in C$) and ($x \notin B$)

$\implies x \in (A \cap C)$ and $x \notin B$

$\implies x \in (A \cap C) – B$

Thus, any element in $(A – B) \cap (C – B)$ is also in $(A \cap C) – B$.

Part 2: Show that $(A \cap C) – B \subseteq (A – B) \cap (C – B)$

Let $y$ be an arbitrary element of $(A \cap C) – B$.

$\implies y \in (A \cap C)$ and $y \notin B$

$\implies (y \in A$ and $y \in C$) and ($y \notin B$)

We can distribute the condition $y \notin B$:

$\implies (y \in A$ and $y \notin B$) and ($y \in C$ and $y \notin B$)

$\implies y \in (A – B)$ and $y \in (C – B)$

$\implies y \in (A – B) \cap (C – B)$

Thus, any element in $(A \cap C) – B$ is also in $(A – B) \cap (C – B)$.

From (i) and (ii), since both sets are subsets of each other, they must be equal.

Hence, $(A – B) \cap (C – B) = (A \cap C) – B$ is proved.

Alternate Solution (Using Properties of Sets)

We will prove the corrected identity $(A – B) \cap (C – B) = (A \cap C) – B$.

Let's start with the LHS:

LHS $= (A – B) \cap (C – B)$

$= (A \cap B') \cap (C \cap B')$

[By definition of set difference, $X – Y = X \cap Y'$]

$= A \cap B' \cap C \cap B'$

[Associative property of intersection]

$= A \cap C \cap B' \cap B'$

[Commutative property of intersection]

$= (A \cap C) \cap (B' \cap B')$

[Associative property of intersection]

$= (A \cap C) \cap B'$

[Idempotent law, $X \cap X = X$]

$= (A \cap C) – B$

[By definition of set difference]

= RHS

Hence, the identity is proved.

Statement: For all sets A and B, $(A – B) \cup (A \cap B) = A$.

Justification:

We will prove this statement using element-wise arguments.

We need to show that $(A – B) \cup (A \cap B) \subseteq A$ and $A \subseteq (A – B) \cup (A \cap B)$.

Part 1: Show that $(A – B) \cup (A \cap B) \subseteq A$

Let $x$ be an arbitrary element such that $x \in (A – B) \cup (A \cap B)$.

By the definition of union, this means $x \in (A – B)$ or $x \in (A \cap B)$.

Case 1: $x \in (A – B)$.

By the definition of set difference, $x \in A$ and $x \notin B$. If $x \in A$ and $x \notin B$, then it implies $x \in A$.

Case 2: $x \in (A \cap B)$.

By the definition of intersection, $x \in A$ and $x \in B$. If $x \in A$ and $x \in B$, then it implies $x \in A$.

In both cases, if $x \in (A – B) \cup (A \cap B)$, it follows that $x \in A$.

Thus, $(A – B) \cup (A \cap B) \subseteq A$.

Part 2: Show that $A \subseteq (A – B) \cup (A \cap B)$

Let $x$ be an arbitrary element such that $x \in A$.

For any element $x$, it either belongs to B or it does not belong to B.

Case 1: $x \in A$ and $x \notin B$.

By the definition of set difference, if $x \in A$ and $x \notin B$, then $x \in (A – B)$. If $x \in (A – B)$, then $x \in (A – B) \cup (A \cap B)$ by the definition of union.

Case 2: $x \in A$ and $x \in B$.

By the definition of intersection, if $x \in A$ and $x \in B$, then $x \in (A \cap B)$. If $x \in (A \cap B)$, then $x \in (A – B) \cup (A \cap B)$ by the definition of union.

Since any element $x \in A$ must either be in B or not in B, it must belong to either $(A – B)$ or $(A \cap B)$.

Thus, if $x \in A$, then $x \in (A – B) \cup (A \cap B)$.

Therefore, $A \subseteq (A – B) \cup (A \cap B)$.

Since we have shown both $(A – B) \cup (A \cap B) \subseteq A$ and $A \subseteq (A – B) \cup (A \cap B)$, we conclude that $(A – B) \cup (A \cap B) = A$.

This identity represents that the set A can be partitioned into two disjoint parts: the part of A that is not in B (A - B) and the part of A that is in B (A ∩ B). The union of these two parts gives the entire set A.

Therefore, the statement is True.

Statement: For all sets A, B and C, $A – (B – C) = (A – B) – C$.

Justification:

We will examine both sides of the equation using set difference properties.

Recall that for any sets X and Y, $X – Y = X \cap Y'$. Here $Y'$ denotes the complement of Y with respect to the universal set U. We assume A, B, C are subsets of U.

Consider the Left Hand Side (LHS):

$LHS = A – (B – C)$

$LHS = A \cap (B – C)'$ (Using $X - Y = X \cap Y'$)

$LHS = A \cap (B \cap C')'$ (Using $B - C = B \cap C'$)

$LHS = A \cap (B' \cup (C')')$ (Using De Morgan's Law: $(X \cap Y)' = X' \cup Y'$)

$LHS = A \cap (B' \cup C)$ (Using $(C')' = C$)

$LHS = (A \cap B') \cup (A \cap C)$ (Using Distributive Law: $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$)

$LHS = (A – B) \cup (A \cap C)$ (Using $A \cap B' = A - B$)

So, $A – (B – C) = (A – B) \cup (A \cap C)$.

Now consider the Right Hand Side (RHS):

$RHS = (A – B) – C$

$RHS = (A \cap B') – C$ (Using $X - Y = X \cap Y'$)

$RHS = (A \cap B') \cap C'$ (Using $X - Y = X \cap Y'$)

$RHS = A \cap B' \cap C'$ (Intersection is associative)

$RHS = A \cap (B' \cap C')$

$RHS = A \cap (B \cup C)'$ (Using De Morgan's Law: $(X \cup Y)' = X' \cap Y'$)

$RHS = A – (B \cup C)$ (Using $X \cap Y' = X - Y$)

So, $(A – B) – C = A – (B \cup C)$.

We need to check if $(A – B) \cup (A \cap C) = A – (B \cup C)$ for all sets A, B, and C.

Let's use a counterexample.

Let $A = \{1, 2\}$, $B = \{2, 3\}$, $C = \{1, 3\}$.

LHS: $A – (B – C)$

$B – C = \{2, 3\} – \{1, 3\} = \{2\}$ (Elements in B but not in C)

$A – (B – C) = \{1, 2\} – \{2\} = \{1\}$ (Elements in A but not in $\{2\}$)

So, LHS = $\{1\}$.

RHS: $(A – B) – C$

$A – B = \{1, 2\} – \{2, 3\} = \{1\}$ (Elements in A but not in B)

$(A – B) – C = \{1\} – \{1, 3\} = \{\}$ (Elements in $\{1\}$ but not in C)

So, RHS = $\{\}$.

Since LHS = $\{1\}$ and RHS = $\{\}$, LHS $\neq$ RHS.

This shows that the statement is not true for all sets A, B, and C.

Therefore, the statement is False.

Statement: For all sets A, B and C, if $A \subset B$, then $A \cap C \subset B \cap C$.

Justification:

We need to show that if $A \subset B$, then every element in $A \cap C$ is also an element in $B \cap C$.

Assume that $A \subset B$. This means that every element that belongs to set A also belongs to set B.

Let $x$ be an arbitrary element such that $x \in A \cap C$.

By the definition of intersection, $x \in A \cap C$ means that $x \in A$ and $x \in C$.

Since we assumed $A \subset B$ and we know that $x \in A$, it follows from the definition of a subset that $x \in B$.

So now we have two conditions: $x \in B$ and $x \in C$.

By the definition of intersection, if $x \in B$ and $x \in C$, then $x \in B \cap C$.

Thus, we have shown that if $x \in A \cap C$, then $x \in B \cap C$.

By the definition of a subset, this means that $A \cap C$ is a subset of $B \cap C$.

Therefore, if $A \subset B$, then $A \cap C \subset B \cap C$ for all sets A, B, and C.

The statement is True.

Statement: For all sets A, B and C, if $A \subset B$, then $A \cup C \subset B \cup C$.

Justification:

We need to show that if $A \subset B$, then every element in $A \cup C$ is also an element in $B \cup C$.

Assume that $A \subset B$. This means that every element that belongs to set A also belongs to set B.

Let $x$ be an arbitrary element such that $x \in A \cup C$.

By the definition of union, $x \in A \cup C$ means that $x \in A$ or $x \in C$.

We consider two cases:

Case 1: $x \in A$.

Since we are given that $A \subset B$, if $x \in A$, then by the definition of a subset, $x$ must also be an element of B.

So, if $x \in A$, then $x \in B$.

By the definition of union, if $x \in B$, then $x \in B \cup C$ (since $B \cup C$ includes all elements of B).

Case 2: $x \in C$.

By the definition of union, if $x \in C$, then $x \in B \cup C$ (since $B \cup C$ includes all elements of C).

In both cases (when $x \in A$ or when $x \in C$), we find that $x \in B \cup C$.

Thus, we have shown that if $x \in A \cup C$, then $x \in B \cup C$.

By the definition of a subset, this means that $A \cup C$ is a subset of $B \cup C$.

Therefore, if $A \subset B$, then $A \cup C \subset B \cup C$ for all sets A, B, and C.

The statement is True.

Statement: For all sets A, B and C, if $A \subset C$ and $B \subset C$, then $A \cup B \subset C$.

Justification:

We are given that $A \subset C$ and $B \subset C$.

The condition $A \subset C$ means that every element in set A is also an element in set C.

The condition $B \subset C$ means that every element in set B is also an element in set C.

To show that $A \cup B \subset C$, we need to prove that every element in the set $A \cup B$ is also an element in set C.

Let $x$ be an arbitrary element such that $x \in A \cup B$.

By the definition of the union of sets, $x \in A \cup B$ means that $x \in A$ or $x \in B$.

We consider two cases based on this condition:

Case 1: $x \in A$.

Since we are given that $A \subset C$, if $x \in A$, then by the definition of a subset, it must be true that $x \in C$.

Case 2: $x \in B$.

Since we are given that $B \subset C$, if $x \in B$, then by the definition of a subset, it must be true that $x \in C$.

In both cases, whether $x$ is in A or in B, we have shown that $x$ must be in C.

Thus, for any element $x \in A \cup B$, it follows that $x \in C$.

By the definition of a subset, this means that the set $A \cup B$ is a subset of the set C.

Therefore, if $A \subset C$ and $B \subset C$, then $A \cup B \subset C$ for all sets A, B, and C.

The statement is True.

To Prove: For all sets A and B, $A \cup (B – A) = A \cup B$

Proof:

We will use the definition of set difference, $B – A = B \cap A'$, where $A'$ is the complement of A with respect to the universal set U.

Consider the Left Hand Side (LHS):

$LHS = A \cup (B – A)$

$LHS = A \cup (B \cap A')$ (Using the definition of set difference)

Now we apply the Distributive Law for union over intersection: $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$. Here, X=A, Y=B, Z=$A'$.

$LHS = (A \cup B) \cap (A \∪ A')$

The union of a set and its complement is the universal set: $A \cup A' = U$.

$LHS = (A \cup B) \cap U$

The intersection of any set with the universal set is the set itself: $(A \cup B) \cap U = A \cup B$.

$LHS = A \cup B$

This is the Right Hand Side (RHS).

Therefore, $A \cup (B – A) = A \cup B$ is proven using set properties.

Alternatively (using element-wise argument):

To prove $A \cup (B – A) = A \cup B$, we show that $A \cup (B – A) \subseteq A \cup B$ and $A \cup B \subseteq A \cup (B – A)$.

Part 1: Show that $A \cup (B – A) \subseteq A \cup B$.

Let $x \in A \cup (B – A)$. By definition of union, $x \in A$ or $x \in (B – A)$.

If $x \in A$, then $x \in A \cup B$ (by definition of union). This case holds.

If $x \in (B – A)$, then by definition of set difference, $x \in B$ and $x \notin A$. If $x \in B$, then $x \in A \cup B$ (by definition of union). This case also holds.

Since both cases imply $x \in A \cup B$, we have $A \cup (B – A) \subseteq A \cup B$.

Part 2: Show that $A \cup B \subseteq A \cup (B – A)$.

Let $x \in A \cup B$. By definition of union, $x \in A$ or $x \in B$.

Case 1: $x \in A$. If $x \in A$, then $x \in A \cup (B – A)$ (by definition of union). This case holds.

Case 2: $x \in B$. If $x \in B$, we consider two sub-cases based on whether $x$ is in A or not.

Subcase 2a: $x \in B$ and $x \in A$. This is the same as $x \in A \cap B$. If $x \in A$, then $x \in A \cup (B – A)$ as shown in Case 1. This subcase holds.

Subcase 2b: $x \in B$ and $x \notin A$. This is the definition of $x \in (B – A)$. If $x \in (B – A)$, then $x \in A \cup (B – A)$ (by definition of union). This subcase holds.

In all cases, if $x \in A \cup B$, then $x \in A \cup (B – A)$.

Thus, $A \cup B \subseteq A \cup (B – A)$.

Since both inclusions are proven, $A \cup (B – A) = A \cup B$.

To Prove: For all sets A and B, $A – (A – B) = A \cap B$

Proof:

We will use the definition of set difference, $X – Y = X \cap Y'$, where $Y'$ is the complement of Y with respect to the universal set U.

Consider the Left Hand Side (LHS):

$LHS = A – (A – B)$

$LHS = A \cap (A – B)'$ (Using the definition of set difference, $X - Y = X \cap Y'$)

$LHS = A \cap (A \cap B')'$ (Using the definition of set difference, $A - B = A \cap B'$)

$LHS = A \cap (A' \cup (B')')$ (Using De Morgan's Law: $(X \cap Y)' = X' \cup Y'$)

$LHS = A \cap (A' \cup B)$ (Using $(B')' = B$)

Now we apply the Distributive Law for intersection over union: $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$. Here, X=A, Y=$A'$, Z=B.

$LHS = (A \cap A') \cup (A \cap B)$

The intersection of a set and its complement is the empty set: $A \cap A' = \emptyset$.

$LHS = \emptyset \cup (A \cap B)$

The union of the empty set with any set is the set itself: $\emptyset \cup X = X$.

$LHS = A \cap B$

This is the Right Hand Side (RHS).

Therefore, $A – (A – B) = A \cap B$ is proven using set properties.

Alternatively (using element-wise argument):

To prove $A – (A – B) = A \cap B$, we show that $A – (A – B) \subseteq A \cap B$ and $A \cap B \subseteq A – (A – B)$.

Part 1: Show that $A – (A – B) \subseteq A \cap B$.

Let $x \in A – (A – B)$. By definition of set difference, $x \in A$ and $x \notin (A – B)$.

The condition $x \notin (A – B)$ means that it is not the case that ($x \in A$ and $x \notin B$). Using De Morgan's Law for logical statements, this is equivalent to ($x \notin A$ or $x \in B$).

So, we have ($x \in A$) and ($x \notin A$ or $x \in B$).

Using the distributive property of logical AND over OR: ($x \in A$ and $x \notin A$) or ($x \in A$ and $x \in B$).

The condition ($x \in A$ and $x \notin A$) is a contradiction, which is always false. It represents the empty set ($\emptyset$).

So, we have False or ($x \in A$ and $x \in B$).

This simplifies to ($x \in A$ and $x \in B$).

By the definition of intersection, ($x \in A$ and $x \in B$) means $x \in A \cap B$.

Thus, if $x \in A – (A – B)$, then $x \in A \cap B$.

Therefore, $A – (A – B) \subseteq A \cap B$.

Part 2: Show that $A \cap B \subseteq A – (A – B)$.

Let $x \in A \cap B$. By definition of intersection, $x \in A$ and $x \in B$.

We need to show that $x \in A – (A – B)$, which means $x \in A$ and $x \notin (A – B)$.

We already know $x \in A$.

Now we need to show $x \notin (A – B)$. By definition of set difference, $x \in (A – B)$ means $x \in A$ and $x \notin B$.

However, we know that $x \in B$. So, the condition $x \notin B$ is false for this element $x$.

Since one of the conditions ($x \in A$ and $x \notin B$) is false, the statement "$x \in A$ and $x \notin B$" is false. This means $x \notin (A – B)$.

So, if $x \in A \cap B$, then $x \in A$ and $x \notin (A – B)$.

By the definition of set difference, this means $x \in A – (A – B)$.

Thus, if $x \in A \cap B$, then $x \in A – (A – B)$.

Therefore, $A \cap B \subseteq A – (A – B)$.

Since both inclusions are proven, $A – (A – B) = A \cap B$.

To Prove: For all sets A and B, $A – (A \cap B) = A – B$

Proof using Set Properties:

We will start with the Left Hand Side (LHS) and transform it using set properties until we reach the Right Hand Side (RHS).

Recall the definition of set difference: For any sets X and Y, $X – Y = X \cap Y'$, where $Y'$ is the complement of set Y.

Consider the LHS:

$LHS = A – (A \cap B)$

Using the definition of set difference ($X – Y = X \cap Y'$), where $X=A$ and $Y=(A \cap B)$:

$LHS = A \cap (A \cap B)'$

Now, apply De Morgan's Law for the complement of an intersection: $(X \cap Y)' = X' \cup Y'$. Here, $X=A$ and $Y=B$.

$LHS = A \cap (A' \cup B')$

Apply the Distributive Law of intersection over union: $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$. Here, $X=A$, $Y=A'$, and $Z=B'$.

$LHS = (A \cap A') \cup (A \cap B')$

The intersection of a set and its complement is the empty set: $A \cap A' = \emptyset$.

$LHS = \emptyset \cup (A \cap B')$

The union of the empty set with any set is the set itself: $\emptyset \cup X = X$. Here, $X = (A \cap B')$.

$LHS = A \cap B'$

Finally, using the definition of set difference again: $X \cap Y' = X – Y$. Here, $X=A$ and $Y=B$.

$LHS = A – B$

This is the Right Hand Side (RHS).

Therefore, $A – (A \cap B) = A – B$ is proven using set properties.

To Prove: For all sets A and B, $(A \cup B) – B = A – B$

Proof using Set Properties:

We will start with the Left Hand Side (LHS) and transform it using set properties until we reach the Right Hand Side (RHS).

Recall the definition of set difference: For any sets X and Y, $X – Y = X \cap Y'$, where $Y'$ is the complement of set Y.

Consider the LHS:

$LHS = (A \cup B) – B$

Using the definition of set difference ($X – Y = X \cap Y'$), where $X=(A \cup B)$ and $Y=B$:

$LHS = (A \cup B) \cap B'$

Apply the Distributive Law of intersection over union: $(X \cup Y) \cap Z = (X \cap Z) \cup (Y \cap Z)$. Here, $X=A$, $Y=B$, and $Z=B'$.

$LHS = (A \cap B') \cup (B \cap B')$

The intersection of a set and its complement is the empty set: $B \cap B' = \emptyset$.

$LHS = (A \cap B') \cup \emptyset$

The union of the empty set with any set is the set itself: $X \cup \emptyset = X$. Here, $X = (A \cap B')$.

$LHS = A \cap B'$

Finally, using the definition of set difference again: $X \cap Y' = X – Y$. Here, $X=A$ and $Y=B$.

$LHS = A – B$

This is the Right Hand Side (RHS).

Therefore, $(A \cup B) – B = A – B$ is proven using set properties.

Alternatively (using element-wise argument):

To prove $(A \cup B) – B = A – B$, we show that $(A \cup B) – B \subseteq A – B$ and $A – B \subseteq (A \cup B) – B$.

Part 1: Show that $(A \cup B) – B \subseteq A – B$.

Let $x \in (A \cup B) – B$. By definition of set difference, $x \in (A \cup B)$ and $x \notin B$.

The condition $x \in (A \cup B)$ means $x \in A$ or $x \in B$.

So, we have ($x \in A$ or $x \in B$) and ($x \notin B$).

Using the distributive property of logical AND over OR: ($x \in A$ and $x \notin B$) or ($x \in B$ and $x \notin B$).

The condition ($x \in B$ and $x \notin B$) is a contradiction, which is always false. It represents the empty set ($\emptyset$).

So, we have ($x \in A$ and $x \notin B$) or False.

This simplifies to ($x \in A$ and $x \notin B$).

By the definition of set difference, ($x \in A$ and $x \notin B$) means $x \in A – B$.

Thus, if $x \in (A \cup B) – B$, then $x \in A – B$.

Therefore, $(A \cup B) – B \subseteq A – B$.

Part 2: Show that $A – B \subseteq (A \cup B) – B$.

Let $x \in A – B$. By definition of set difference, $x \in A$ and $x \notin B$.

We need to show that $x \in (A \cup B) – B$, which means $x \in (A \cup B)$ and $x \notin B$.

We already know $x \notin B$.

Now we need to show $x \in (A \cup B)$. By definition of union, $x \in (A \cup B)$ means $x \in A$ or $x \in B$.

We know $x \in A$. Since $x \in A$, the condition ($x \in A$ or $x \in B$) is true.

So, if $x \in A – B$, then $x \in A$ (which implies $x \in A \cup B$) and $x \notin B$.

By the definition of set difference, this means $x \in (A \cup B) – B$.

Thus, if $x \in A – B$, then $x \in (A \cup B) – B$.

Therefore, $A – B \subseteq (A \cup B) – B$.

Since both inclusions are proven, $(A \cup B) – B = A – B$.

Statement: Let $T = \{x\;|\;\frac{x\;+\;5}{x\;-\;7} -5=\frac{4x\;-\;40}{13\;-\;x} \}$. Is T an empty set? Justify your answer.

Justification:

To determine if the set T is empty, we need to find the values of $x$ that satisfy the given equation and the conditions on the variable $x$. The set T contains all such values of $x$.

The given equation is:

$\frac{x\;+\;5}{x\;-\;7} -5=\frac{4x\;-\;40}{13\;-\;x}$

First, we must identify any restrictions on the value of $x$ due to the denominators. The denominators cannot be zero.

$x - 7 \neq 0 \implies x \neq 7$

$13 - x \neq 0 \implies x \neq 13$

Now, let's solve the equation for $x$.

Combine the terms on the Left Hand Side (LHS):

$\frac{x\;+\;5}{x\;-\;7} - \frac{5(x\;-\;7)}{x\;-\;7} = \frac{4x\;-\;40}{13\;-\;x}$

$\frac{x\;+\;5 - 5(x\;-\;7)}{x\;-\;7} = \frac{4x\;-\;40}{13\;-\;x}$

$\frac{x\;+\;5 - 5x + 35}{x\;-\;7} = \frac{4x\;-\;40}{13\;-\;x}$

$\frac{-4x + 40}{x\;-\;7} = \frac{4x\;-\;40}{13\;-\;x}$

Notice that the numerator on the LHS, $-4x + 40$, is the negative of the numerator on the RHS, $4x - 40$. Let's rewrite $-4x + 40$ as $-(4x - 40)$.

$\frac{-(4x\;-\;40)}{x\;-\;7} = \frac{4x\;-\;40}{13\;-\;x}$

Let $Z = 4x - 40$. The equation becomes:

$\frac{-Z}{x\;-\;7} = \frac{Z}{13\;-\;x}$

We can consider two cases based on the value of Z:

Case 1: $Z = 0$

If $4x - 40 = 0$, then $4x = 40$, which means $x = 10$.

We check if this value of $x$ satisfies the initial restrictions: $x = 10 \neq 7$ and $x = 10 \neq 13$. The restrictions are satisfied.

Now we verify if $x = 10$ is a solution to the original equation by substituting it:

LHS = $\frac{10\;+\;5}{10\;-\;7} - 5 = \frac{15}{3} - 5 = 5 - 5 = 0$

RHS = $\frac{4(10)\;-\;40}{13\;-\;10} = \frac{40\;-\;40}{3} = \frac{0}{3} = 0$

Since LHS = RHS, $x = 10$ is a valid solution to the equation.

Case 2: $Z \neq 0$

If $4x - 40 \neq 0$, we can divide both sides of the equation $\frac{-(4x\;-\;40)}{x\;-\;7} = \frac{4x\;-\;40}{13\;-\;x}$ by $(4x - 40)$.

$\frac{-1}{x\;-\;7} = \frac{1}{13\;-\;x}$

Cross-multiply:

$-1(13 - x) = 1(x - 7)$

$-13 + x = x - 7$

Subtract $x$ from both sides:

$-13 = -7$

This is a false statement. This means there are no values of $x$ that satisfy the equation when $4x - 40 \neq 0$.

From the two cases, the only solution to the equation is $x = 10$.

The set T is defined as the set of all $x$ that satisfy the equation and the restrictions.

Since $x = 10$ is a solution and satisfies the restrictions, it is an element of set T.

$T = \{10\}$

An empty set is a set that contains no elements. The set T contains one element, which is 10.

Therefore, T is not an empty set.

To Show: For all sets A, B, and C, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Proof (using element-wise argument):

This is the Distributive Law of intersection over union. To prove the equality of the two sets, we will show that each set is a subset of the other.

Part 1: Show that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$

Let $x$ be an arbitrary element such that $x \in A \cap (B \cup C)$.

By the definition of intersection, this means $x \in A$ and $x \in (B \cup C)$.

By the definition of union, $x \in (B \cup C)$ means $x \in B$ or $x \in C$.

So, we have the conditions: ($x \in A$) and ($x \in B$ or $x \in C$).

Using the distributive property of logical AND over logical OR, this is equivalent to:

($x \in A$ and $x \in B$) or ($x \in A$ and $x \in C$).

By the definition of intersection, ($x \in A$ and $x \in B$) means $x \in A \cap B$.

By the definition of intersection, ($x \in A$ and $x \in C$) means $x \in A \cap C$.

So, the condition becomes: $x \in (A \cap B)$ or $x \in (A \cap C)$.

By the definition of union, $x \in (A \cap B)$ or $x \in (A \cap C)$ means $x \in (A \cap B) \cup (A \cap C)$.

Thus, we have shown that if $x \in A \cap (B \cup C)$, then $x \in (A \cap B) \cup (A \cap C)$.

Therefore, $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$.

Part 2: Show that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$

Let $x$ be an arbitrary element such that $x \in (A \cap B) \cup (A \cap C)$.

By the definition of union, this means $x \in (A \cap B)$ or $x \in (A \cap C)$.

Case 1: $x \in (A \cap B)$.

By the definition of intersection, $x \in A \cap B$ means $x \in A$ and $x \in B$.

Since $x \in B$, by the definition of union, $x \in B \cup C$ (as $B \cup C$ contains all elements of B).

So, if $x \in A \cap B$, we have $x \in A$ and $x \in B \cup C$. By the definition of intersection, this means $x \in A \cap (B \cup C)$.

Case 2: $x \in (A \cap C)$.

By the definition of intersection, $x \in A \cap C$ means $x \in A$ and $x \in C$.

Since $x \in C$, by the definition of union, $x \in B \cup C$ (as $B \cup C$ contains all elements of C).

So, if $x \in A \cap C$, we have $x \in A$ and $x \in B \cup C$. By the definition of intersection, this means $x \in A \cap (B \cup C)$.

In both cases, if $x \in (A \cap B) \cup (A \cap C)$, it follows that $x \in A \cap (B \cup C)$.

Thus, $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.

Since both inclusions $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$ and $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$ are proven, we conclude that the sets are equal.

$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Given:

Let E be the set of students who passed in English.

Let M be the set of students who passed in Mathematics.

Let S be the set of students who passed in Science.

Total number of students, $n(U) = 100$.

• Number of students who passed in English, $n(E) = 15$.
• Number of students who passed in Mathematics, $n(M) = 12$.
• Number of students who passed in Science, $n(S) = 8$.
• Number of students who passed in English and Mathematics, $n(E \cap M) = 6$.
• Number of students who passed in Mathematics and Science, $n(M \cap S) = 7$.
• Number of students who passed in English and Science, $n(E \cap S) = 4$.
• Number of students who passed in all three subjects, $n(E \cap M \cap S) = 4$.

To Find:

The number of students who passed:

(i) in English and Mathematics but not in Science.

(ii) in Mathematics and Science but not in English.

(iii) in Mathematics only.

(iv) in more than one subject only.

Solution:

We will solve each part using the given data and formulas from set theory.

(i) Number of students who passed in English and Mathematics but not in Science

This is the number of students in the set $(E \cap M)$ but not in S. We can find this by subtracting the number of students who passed all three subjects from the number of students who passed English and Mathematics.

Number of students = $n(E \cap M) - n(E \cap M \cap S)$

= $6 - 4$

= $2$

So, 2 students passed in English and Mathematics but not in Science.

(ii) Number of students who passed in Mathematics and Science but not in English

This is the number of students in the set $(M \cap S)$ but not in E. We subtract the number of students who passed all three subjects from the number of students who passed Mathematics and Science.

Number of students = $n(M \cap S) - n(E \cap M \cap S)$

= $7 - 4$

= $3$

So, 3 students passed in Mathematics and Science but not in English.

(iii) Number of students who passed in Mathematics only

To find the number of students who passed only in Mathematics, we need to subtract the number of students who passed Mathematics along with other subjects from the total number of students who passed Mathematics.

Number of students = $n(M) - [(\text{passed in M and E only}) \ $$ + (\text{passed in M and S only}) + (\text{passed in all three})]$

= $n(M) - [ (n(E \cap M) - n(E \cap M \cap S)) + (n(M \cap S) - n(E \cap M \cap S)) \ $$ + n(E \cap M \cap S) ]$

= $12 - [ (6 - 4) + (7 - 4) + 4 ]$

= $12 - [ 2 + 3 + 4 ]$

= $12 - 9$

= $3$

So, 3 students passed in Mathematics only.

(iv) Number of students who passed in more than one subject only

This is the number of students who passed in at least two subjects. This is the sum of students who passed in exactly two subjects and those who passed in all three subjects.

Number of students = (passed in E and M only) + (passed in M and S only) + (passed in E and S only) + (passed in all three)

First, we find the number of students who passed in English and Science only:

$n(E \cap S \text{ only}) = n(E \cap S) - n(E \cap M \cap S) = 4 - 4 = 0$

Now, we sum up all the required parts:

Total = $(n(E \cap M) - n(E \cap M \cap S)) + (n(M \cap S) - n(E \cap M \cap S)) \ $$ + (n(E \cap S) - n(E \cap M \cap S)) + n(E \cap M \cap S)$

= $(6 - 4) + (7 - 4) + (4 - 4) + 4$

= $2 + 3 + 0 + 4$

= $9$

So, 9 students passed in more than one subject.

Given:

Total number of students = 60

Number of students who play Cricket, $|C| = 25$

Number of students who play Tennis, $|T| = 20$

Number of students who play both Cricket and Tennis, $|C \cap T| = 10$

To Find:

The number of students who play neither game.

Solution:

Let C be the set of students who play Cricket and T be the set of students who play Tennis.

We are given the following information:

$|C| = 25$

$|T| = 20$

$|C \cap T| = 10$

The number of students who play at least one of the games is given by the formula for the union of two sets:

$|C \cup T| = |C| + |T| - |C \cap T|$

Substitute the given values into the formula:

$|C \cup T| = 25 + 20 - 10$

$|C \cup T| = 45 - 10$

The number of students who play at least one game is 35.

The total number of students in the class is 60.

The students who play neither game are those in the total group who are not in the set of students who play at least one game.

Number of students who play neither = Total number of students - $|C \cup T|$

Number of students who play neither = $60 - 35$

Number of students who play neither = 25

Therefore, 25 students play neither game.

Given:

Total number of students in the survey = 200

Let M be the set of students who study Mathematics.

Let P be the set of students who study Physics.

Let C be the set of students who study Chemistry.

We are given the following information:

$|M| = 120$

$|P| = 90$

$|C| = 70$

$|M \cap P| = 40$

$|P \cap C| = 30$

$|C \cap M| = 50$

Number of students who study none of these subjects = 20

To Find:

The number of students who study all three subjects, i.e., $|M \cap P \cap C|$.

Solution:

Let U be the set of all students in the survey. The total number of students is $|U| = 200$.

The number of students who study at least one of the three subjects is the total number of students minus those who study none of the subjects.

$|M \cup P \cup C| = |U| - (\text{Number of students who study none})$

$|M \cup P \cup C| = 200 - 20$

The principle of inclusion-exclusion for three sets is given by the formula:

$|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |P \cap C| - |C \cap M| \ $$ + |M \cap P \cap C|$

We know $|M \cup P \cup C|$ from (i) and we are given the values for the individual sets and pairwise intersections. We need to find $|M \cap P \cap C|$.

Substitute the known values into the formula:

$180 = 120 + 90 + 70 - 40 - 30 - 50 + |M \cap P \cap C|$

$180 = (120 + 90 + 70) - (40 + 30 + 50) + |M \cap P \cap C|$

$180 = 280 - 120 + |M \cap P \cap C|$

$180 = 160 + |M \cap P \cap C|$

Now, solve for $|M \cap P \cap C|$:

$|M \cap P \cap C| = 180 - 160$

$|M \cap P \cap C| = 20$

Therefore, the number of students who study all three subjects is 20.

Given:

Total number of families in the town = 10,000

Let U be the set of all families in the town, so $|U| = 10000$.

Let A, B, and C be the sets of families who buy newspaper A, B, and C respectively.

We convert the given percentages into the number of families:

$|A| = 40\%$ of $10000 = \frac{40}{100} \times 10000 = 4000$

$|B| = 20\%$ of $10000 = \frac{20}{100} \times 10000 = 2000$

$|C| = 10\%$ of $10000 = \frac{10}{100} \times 10000 = 1000$

$|A \cap B| = 5\%$ of $10000 = \frac{5}{100} \times 10000 = 500$

$|B \cap C| = 3\%$ of $10000 = \frac{3}{100} \times 10000 = 300$

$|A \cap C| = 4\%$ of $10000 = \frac{4}{100} \times 10000 = 400$

$|A \cap B \cap C| = 2\%$ of $10000 = \frac{2}{100} \times 10000 = 200$

To Find:

(a) The number of families which buy newspaper A only.

(b) The number of families which buy none of A, B and C.

Solution:

(a) The number of families which buy newspaper A only:

The number of families who buy newspaper A only is the number of families in set A minus those who also buy B or C (or both). This can be calculated as $|A|$ minus the sum of the numbers in the intersections involving A, adjusted for the triple intersection.

Number of families who buy A and B but not C = $|A \cap B| - |A \cap B \cap C| = 500 - 200 = 300$.

Number of families who buy A and C but not B = $|A \cap C| - |A \cap B \cap C| = 400 - 200 = 200$.

Number of families who buy A and B and C = $|A \cap B \cap C| = 200$.

The number of families who buy newspaper A only is the total number in A minus those who are in the overlaps with B or C (or both).

$|A \text{ only}| = |A| - (\text{Number in A and B only}) \ $$ - (\text{Number in A and C only}) \ $$ - (\text{Number in A and B and C})$

$|A \text{ only}| = |A| - (|A \cap B| - |A \cap B \cap C|) - (|A \cap C| - |A \cap B \cap C|) \ $$ - |A \cap B \cap C|$

Substitute the values:

$|A \text{ only}| = 4000 - (500 - 200) - (400 - 200) - 200$

$|A \text{ only}| = 4000 - 300 - 200 - 200$

$|A \text{ only}| = 4000 - 700$

$|A \text{ only}| = 3300$

The number of families which buy newspaper A only is 3300.

(b) The number of families which buy none of A, B and C:

First, we need to find the number of families who buy at least one of the three newspapers. This is given by the union of the three sets, $|A \cup B \cup C|$.

Using the Principle of Inclusion-Exclusion for three sets:

$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| \ $$ + |A \cap B \cap C|$

Substitute the calculated values:

$|A \cup B \cup C| = 4000 + 2000 + 1000 - 500 - 300 - 400 + 200$

$|A \cup B \cup C| = 7000 - (500 + 300 + 400) + 200$

$|A \cup B \cup C| = 7000 - 1200 + 200$

$|A \cup B \cup C| = 5800 + 200$

$|A \cup B \cup C| = 6000$

The number of families who buy at least one newspaper is 6000.

The number of families who buy none of the newspapers is the total number of families minus the number of families who buy at least one newspaper.

Number who buy none = $|U| - |A \cup B \cup C|$

Number who buy none = $10000 - 6000$

Number who buy none = $4000$

The number of families which buy none of A, B and C is 4000.

Given:

Let F be the set of students studying French.

Let E be the set of students studying English.

Let S be the set of students studying Sanskrit.

Total number of students in the group, $n(U) = 50$.

• Number of students studying French, $n(F) = 17$.
• Number of students studying English, $n(E) = 13$.
• Number of students studying Sanskrit, $n(S) = 15$.
• Number of students studying French and English, $n(F \cap E) = 9$.
• Number of students studying English and Sanskrit, $n(E \cap S) = 4$.
• Number of students studying French and Sanskrit, $n(F \cap S) = 5$.
• Number of students studying all three languages, $n(F \cap E \cap S) = 3$.

To Find:

The number of students who study:

(i) French only

(ii) English only

(iii) Sanskrit only

(iv) English and Sanskrit but not French

(v) French and Sanskrit but not English

(vi) French and English but not Sanskrit

(vii) at least one of the three languages

(viii) none of the three languages

Solution:

First, let's calculate the number of students in each exclusive category based on the given data.

Number of students who study French and English only = $n(F \cap E) - n(F \cap E \cap S) = 9 - 3 = 6$.

Number of students who study English and Sanskrit only = $n(E \cap S) - n(F \cap E \cap S) = 4 - 3 = 1$.

Number of students who study French and Sanskrit only = $n(F \cap S) - n(F \cap E \cap S) = 5 - 3 = 2$.

Now we can find the answers to the questions.

(i) Number of students who study French only

This is the total number of French students minus those who study French with other languages.

Number = $n(F) - [(\text{French and English only}) \ $$ + (\text{French and Sanskrit only}) \ $$ + (\text{All three})]$

= $17 - [6 + 2 + 3]$

= $17 - 11$

= $6$

So, 6 students study French only.

(ii) Number of students who study English only

This is the total number of English students minus those who study English with other languages.

Number = $n(E) - [(\text{French and English only}) \ $$ + (\text{English and Sanskrit only}) \ $$ + (\text{All three})]$

= $13 - [6 + 1 + 3]$

= $13 - 10$

= $3$

So, 3 students study English only.

(iii) Number of students who study Sanskrit only

This is the total number of Sanskrit students minus those who study Sanskrit with other languages.

Number = $n(S) - [(\text{French and Sanskrit only}) \ $$ + (\text{English and Sanskrit only}) \ $$ + (\text{All three})]$

= $15 - [2 + 1 + 3]$

= $15 - 6$

= $9$

So, 9 students study Sanskrit only.

(iv) Number of students who study English and Sanskrit but not French

This is the number of students who study English and Sanskrit only, which we calculated earlier.

Number = $n(E \cap S) - n(F \cap E \cap S)$

= $4 - 3$

= $1$

So, 1 student studies English and Sanskrit but not French.

(v) Number of students who study French and Sanskrit but not English

This is the number of students who study French and Sanskrit only, which we calculated earlier.

Number = $n(F \cap S) - n(F \cap E \cap S)$

= $5 - 3$

= $2$

So, 2 students study French and Sanskrit but not English.

(vi) Number of students who study French and English but not Sanskrit

This is the number of students who study French and English only, which we calculated earlier.

Number = $n(F \cap E) - n(F \cap E \cap S)$

= $9 - 3$

= $6$

So, 6 students study French and English but not Sanskrit.

(vii) Number of students who study at least one of the three languages

This is the number of students in the union of the three sets, $n(F \cup E \cup S)$.

Using the Principle of Inclusion-Exclusion:

$n(F \cup E \cup S) = n(F) + n(E) + n(S) - n(F \cap E) - n(E \cap S) \ $$ - n(F \cap S) + n(F \cap E \cap S)$

= $17 + 13 + 15 - 9 - 4 - 5 + 3$

= $45 - 18 + 3$

= $30$

So, 30 students study at least one of the three languages.

(viii) Number of students who study none of the three languages

This is the total number of students minus the number of students who study at least one language.

Number = $n(U) - n(F \cup E \cup S)$

= $50 - 30$

= $20$

So, 20 students study none of the three languages.

Solution:

Let S be the union of the sets $A_i$ and also the union of the sets $B_j$. We are given $S = \bigcup\limits_{i=1}^{30} A_i = \bigcup\limits_{j=1}^{n} B_j$.

We are given that there are 30 sets $A_1, A_2, ..., A_{30}$, and each set $A_i$ has 5 elements. So, $|A_i| = 5$ for $i = 1, 2, ..., 30$.

The sum of the cardinalities of the sets $A_i$ is $\sum\limits_{i=1}^{30} |A_i| = 30 \times 5 = 150$.

We are also given that each element of S belongs to exactly 10 of the $A_i$'s.

The sum $\sum\limits_{i=1}^{30} |A_i|$ counts each element in the union S exactly 10 times. Therefore, we have the relationship:

$\sum\limits_{i=1}^{30} |A_i| = 10 \times |S|$

$150 = 10 \times |S|$

From this, we can find the cardinality of the set S:

$|S| = \frac{150}{10} = 15$

We are also given that there are n sets $B_1, B_2, ..., B_n$, and each set $B_j$ has 3 elements. So, $|B_j| = 3$ for $j = 1, 2, ..., n$.

The sum of the cardinalities of the sets $B_j$ is $\sum\limits_{j=1}^{n} |B_j| = n \times 3 = 3n$.

We are given that each element of S belongs to exactly 9 of the $B_j$'s.

The sum $\sum\limits_{j=1}^{n} |B_j|$ counts each element in the union S exactly 9 times. Therefore, we have the relationship:

$\sum\limits_{j=1}^{n} |B_j| = 9 \times |S|$

Substitute the value of $|S| = 15$ and the sum $\sum\limits_{j=1}^{n} |B_j| = 3n$ into this equation:

$3n = 9 \times 15$

$3n = 135$

Now, solve for n:

$n = \frac{135}{3}$

$n = 45$

The value of n is 45.

Comparing this result with the given options:

(A) 15

(B) 3

(C) 45

(D) 35

The calculated value $n=45$ matches option (C).

The final answer is $\boxed{45}$.

Given:

Let the two finite sets be A and B.

Number of elements in set A, $n(A) = m$.

Number of elements in set B, $n(B) = n$.

The number of subsets of a set with $k$ elements is $2^k$.

• Number of subsets of set A = $2^m$.
• Number of subsets of set B = $2^n$.

According to the question, the number of subsets of the first set is 112 more than that of the second set. This can be written as an equation:

Solution:

We need to solve the equation for $m$ and $n$.

$2^m - 2^n = 112$

Since the difference is positive, we know that $2^m > 2^n$, which implies that $m > n$.

Let's factor out the smaller power of 2, which is $2^n$:

$2^n(2^{m-n} - 1) = 112$

Now, we find the prime factorization of 112.

$\begin{array}{c|cc} 2 & 112 \\ \hline 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

So, $112 = 2 \times 2 \times 2 \times 2 \times 7 = 2^4 \times 7$.

Substitute this back into the equation:

$2^n(2^{m-n} - 1) = 2^4 \times 7$

Now, we can compare the factors on both sides of the equation. Note that $2^n$ is an even number (a power of 2), and since $m-n > 0$, $2^{m-n}$ is an even integer greater than or equal to 2. Thus, $(2^{m-n} - 1)$ is an odd integer.

By comparing the even and odd parts on both sides:

Comparing the even parts:

$2^n = 2^4$

$\implies n = 4$

Comparing the odd parts:

$2^{m-n} - 1 = 7$

$2^{m-n} = 7 + 1$

$2^{m-n} = 8$

$2^{m-n} = 2^3$

$\implies m-n = 3$

Now, substitute the value of $n=4$ into this equation to find $m$:

$m - 4 = 3$

$m = 3 + 4$

$m = 7$

The values of $m$ and $n$ are 7 and 4, respectively.

Therefore, the correct option is (B) 7, 4.

To Simplify:

The set expression $(A \cap B')' \cup (B \cap C)$.

Solution:

We need to simplify the given set expression: $(A \cap B')' \cup (B \cap C)$.

First, let's simplify the part $(A \cap B')'$ using De Morgan's Law, which states that $(X \cap Y)' = X' \cup Y'$.

Applying De Morgan's Law to $(A \cap B')'$:

$(A \cap B')' = A' \cup (B')'$

We know that $(B')' = B$ (Law of Double Complementation).

So, $(A \cap B')' = A' \cup B$.

Now, substitute this back into the original expression:

$(A \cap B')' \cup (B \cap C) = (A' \cup B) \cup (B \cap C)$

Using the Associative Law for union, $(X \cup Y) \cup Z = X \cup (Y \cup Z)$, we can write:

$A' \cup (B \cup (B \cap C))$

Now, consider the expression inside the parenthesis: $B \cup (B \cap C)$.

Using the Absorption Law, which states that $X \cup (X \cap Y) = X$, we can simplify $B \cup (B \cap C)$. Here, $X = B$ and $Y = C$.

So, $B \cup (B \cap C) = B$.

Substituting this back into the expression:

$A' \cup (B \cup (B \cap C)) = A' \cup B$.

Thus, the simplified expression is $A' \cup B$.

Comparing this result with the given options:

(A) $A' \cup B \cup C$

(B) $A' \cup B$

(C) $A' \cup C'$

(D) $A' \cap B$

The simplified expression matches option (B).

The correct answer is (B) A' ∪ B.

Given:

We are given five sets of quadrilaterals in a plane:

• $F_1$ = The set of all parallelograms.
• $F_2$ = The set of all rectangles.
• $F_3$ = The set of all rhombuses.
• $F_4$ = The set of all squares.
• $F_5$ = The set of all trapeziums.

Analysis of Relationships:

Let's establish the relationships between these sets based on their geometric definitions:

1. A rectangle is a parallelogram with four right angles. Therefore, every rectangle is a parallelogram. This means $F_2$ is a subset of $F_1$.

   $F_2 \subset F_1$

2. A rhombus is a parallelogram with four equal sides. Therefore, every rhombus is a parallelogram. This means $F_3$ is a subset of $F_1$.

   $F_3 \subset F_1$

3. A square is a parallelogram with four right angles and four equal sides. This means a square is both a rectangle and a rhombus. Therefore, every square is a parallelogram. This means $F_4$ is a subset of $F_1$.

   $F_4 \subset F_1$

4. From the definition of a square, it is the intersection of the set of rectangles and the set of rhombuses.

   $F_4 = F_2 \cap F_3$

5. A trapezium is a quadrilateral with at least one pair of parallel sides. Since a parallelogram has two pairs of parallel sides, every parallelogram is a trapezium.

   $F_1 \subset F_5$

Evaluating the Options:

Now we will check each option to see which one can be equal to $F_1$.

(A) $F_2 \cap F_3$

This is the set of figures that are both rectangles and rhombuses. As established above, this is the definition of a square.

$F_2 \cap F_3 = F_4$

Since the set of all parallelograms ($F_1$) is not the same as the set of all squares ($F_4$), this option is incorrect.

(B) $F_3 \cap F_4$

This is the set of figures that are both rhombuses and squares. Since every square is a rhombus, $F_4$ is a subset of $F_3$. The intersection of a set and its subset is the subset itself.

$F_3 \cap F_4 = F_4$

This option is also incorrect, as $F_1 \neq F_4$.

(C) $F_2 \cup F_5$

This is the union of the set of rectangles and the set of trapeziums. Since every rectangle is a parallelogram, and every parallelogram is a trapezium, it follows that $F_2 \subset F_5$. The union of a set and its subset is the superset.

$F_2 \cup F_5 = F_5$

The set of all parallelograms ($F_1$) is not the same as the set of all trapeziums ($F_5$), as some trapeziums are not parallelograms. So, this option is incorrect.

(D) $F_2 \cup F_3 \cup F_4 \cup F_1$

This is the union of the sets of rectangles, rhombuses, squares, and parallelograms.

As we have established:

• $F_2 \subset F_1$
• $F_3 \subset F_1$
• $F_4 \subset F_1$

When we take the union of a set with one or more of its subsets, the result is the set itself. In this case, $F_1$ is the largest set, containing all the others.

$F_2 \cup F_3 \cup F_4 \cup F_1 = F_1$

This statement is true.

Therefore, the correct option is (D) $F_2 \cup F_3 \cup F_4 \cup F_1$.

Given:

We are given three sets of points:

• S = The set of points inside a square.
• T = The set of points inside a triangle.
• C = The set of points inside a circle.

The conditions are:

1. The triangle and circle intersect each other.
2. The triangle is contained in the square.
3. The circle is contained in the square.

Analysis of Relationships:

Let's translate the given geometric conditions into the language of set theory:

• "The triangle is contained in the square" means that every point inside the triangle is also inside the square. Therefore, T is a subset of S.

  $T \subset S$

• "The circle is contained in the square" means that every point inside the circle is also inside the square. Therefore, C is a subset of S.

  $C \subset S$

• "The triangle and circle intersect each other" means there are points that are common to both the triangle and the circle. Therefore, their intersection is not an empty set.

  $T \cap C \neq \phi$

Evaluating the Options:

We will now examine each option based on these set relationships.

(A) S ∩ T ∩ C = φ

This statement claims that the intersection of all three sets is the empty set. We know that $T \subset S$ and $C \subset S$. The intersection of a set with its subset is the subset itself. So, $S \cap T = T$ and $S \cap C = C$. Therefore, $S \cap T \cap C = (S \cap T) \cap C = T \cap C$. The problem states that the triangle and circle intersect, which means $T \cap C \neq \phi$. Thus, $S \cap T \cap C \neq \phi$. The statement in option (A) is false.

(B) S ∪ T ∪ C = C

This statement claims that the union of all three sets is equal to set C. We know that $C \subset S$. The union of a set and its subset is the superset. So, $S \cup C = S$. Since S is the largest set (the superset), the union of all three cannot be C unless S and T are also subsets of C, which is contrary to the given information. The statement in option (B) is false.

(C) S ∪ T ∪ C = S

This statement claims that the union of all three sets is equal to set S. Let's evaluate the union. We know that $T \subset S$ and $C \subset S$. When we take the union of a set with one of its subsets, the result is the superset. So, $S \cup T = S$. And, $S \cup C = S$. Therefore, $S \cup T \cup C = (S \cup T) \cup C = S \cup C = S$. The statement in option (C) is true.

(D) S ∪ T = S ∩ C

Let's evaluate the left side (LHS) and the right side (RHS) of this equation.

LHS: $S \cup T$. Since $T \subset S$, we have $S \cup T = S$.

RHS: $S \cap C$. Since $C \subset S$, we have $S \cap C = C$.

So the equation becomes $S = C$. This is not necessarily true; the square can be much larger than the circle it contains. The statement in option (D) is false.

Therefore, the correct option is (C) S ∪ T ∪ C = S.

Given:

$R$ is the set of points inside a rectangle.

The rectangle has sides of length $a$ and $b$, where $a > 1$ and $b > 1$.

Two sides of the rectangle are along the positive direction of the x-axis and y-axis.

This means the vertices of the rectangle are at (0, 0), (a, 0), (0, b), and (a, b).

To Find:

The set notation that represents the set of points $R$ inside the rectangle.

Solution:

The rectangle is defined by the region in the Cartesian plane where the x-coordinates range from 0 to $a$ and the y-coordinates range from 0 to $b$.

Mathematically, the region covered by the rectangle (including its boundary) can be described as:

$\{(x, y) : 0 \leq x \leq a, 0 \leq y \leq b\}$.

The question asks for the set of points inside the rectangle. The term "inside" typically refers to the interior of the region, which excludes the boundary (the sides and vertices).

For a point $(x, y)$ to be strictly inside the rectangle, its x-coordinate must be greater than 0 but less than $a$, and its y-coordinate must be greater than 0 but less than $b$.

So, for points $(x, y)$ inside the rectangle, the conditions must be:

$0 < x < a$

and

$0 < y < b$.

Therefore, the set $R$ of points inside the rectangle is given by:

$R = \{(x, y) : 0 < x < a \text{ and } 0 < y < b\}$.

Let's compare this with the given options:

(A) $R = \{(x, y) : 0 \leq x \leq a, 0 \leq y \leq b\}$ - This includes the boundary (sides and vertices), representing the closed rectangle. Incorrect.

(B) $R = \{(x, y) : 0 \leq x < a, 0 \leq y \leq b\}$ - This excludes the boundary at $x=a$ but includes the boundaries at $x=0$, $y=0$, and $y=b$. Incorrect.

(C) $R = \{(x, y) : 0 \leq x \leq a, 0 < y < b\}$ - This excludes the boundaries at $y=0$ and $y=b$ but includes the boundaries at $x=0$ and $x=a$. Incorrect.

(D) $R = \{(x, y) : 0 < x < a, 0 < y < b\}$ - This excludes all boundaries (sides and vertices), representing the open rectangle, which corresponds to the points strictly inside. Correct.

The set of points inside the rectangle is the set of all points $(x, y)$ such that $x$ is strictly between 0 and $a$, and $y$ is strictly between 0 and $b$.

The correct option is (D) R = {(x, y) : 0 < x < a, 0 < y < b}.

Given:

Total number of students in the class, $|U| = 60$.

Number of students who play cricket, $|C| = 25$.

Number of students who play tennis, $|T| = 20$.

Number of students who play both games, $|C \cap T| = 10$.

To Find:

The number of students who play neither cricket nor tennis.

Solution:

Let $C$ be the set of students who play cricket and $T$ be the set of students who play tennis.

We are given:

$|U| = 60$

$|C| = 25$

$|T| = 20$

$|C \cap T| = 10$

The number of students who play at least one of the two games is given by the Principle of Inclusion-Exclusion for two sets:

$|C \cup T| = |C| + |T| - |C \cap T|$

Substitute the given values into the formula:

$|C \cup T| = 25 + 20 - 10$

$|C \cup T| = 45 - 10$

$|C \cup T| = 35$

This value, 35, represents the number of students who play at least one game (either cricket, or tennis, or both).

The number of students who play neither game is the total number of students minus the number of students who play at least one game.

Number of students who play neither = $|U| - |C \cup T|$

Number of students who play neither = $60 - 35$

Number of students who play neither = $25$

The number of students who play neither is 25.

Comparing this result with the given options:

(A) 0

(B) 25

(C) 35

(D) 45

The correct option is (B).

The number of students who play neither is 25.

Given:

Total number of persons in the town, $|U| = 840$.

Number of persons who read Hindi, $|H| = 450$.

Number of persons who read English, $|E| = 300$.

Number of persons who read both Hindi and English, $|H \cap E| = 200$.

To Find:

The number of persons who read neither Hindi nor English.

Solution:

Let $H$ be the set of persons who read Hindi and $E$ be the set of persons who read English.

We need to find the number of persons who read neither Hindi nor English, which is the number of elements outside the union of sets $H$ and $E$. This can be represented as $|(H \cup E)'|$ or $|U| - |H \cup E|$.

First, we find the number of persons who read at least one of the two languages, which is $|H \cup E|$.

Using the Principle of Inclusion-Exclusion:

$|H \cup E| = |H| + |E| - |H \cap E|$

Substitute the given values:

$|H \cup E| = 450 + 300 - 200$

$|H \cup E| = 750 - 200$

$|H \cup E| = 550$

So, 550 persons read at least one language.

The number of persons who read neither language is the total number of persons minus the number of persons who read at least one language.

Number of persons who read neither = $|U| - |H \cup E|$

Number of persons who read neither = $840 - 550$

Number of persons who read neither = $290$

The number of persons who read neither is 290.

Comparing this result with the given options:

(A) 210

(B) 290

(C) 180

(D) 260

The correct option is (B).

The number of persons who read neither is 290.

Given:

We are given two sets defined as:

$X = \{8^n – 7n – 1 \mid n \in \mathbb{N}\}$

$Y = \{49n – 49 \mid n \in \mathbb{N}\}$

where $\mathbb{N}$ is the set of natural numbers, i.e., $\mathbb{N} = \{1, 2, 3, ...\}$.

To Find:

The correct relationship between sets X and Y.

Solution:

Let's analyze the elements of each set to determine their relationship.

Step 1: Analyze the elements of Set X

The general element of set X is $x_n = 8^n – 7n – 1$.

We can analyze this expression using the Binomial Theorem by writing $8^n$ as $(1+7)^n$.

According to the Binomial Theorem:

$(1+7)^n = \binom{n}{0}1^n + \binom{n}{1}1^{n-1}(7)^1 + \binom{n}{2}1^{n-2}(7)^2 \ $$ + \binom{n}{3}1^{n-3}(7)^3 + \dots + \binom{n}{n}(7)^n$

$8^n = 1 + n \cdot 7 + \frac{n(n-1)}{2} \cdot 49 + \dots + 7^n$

Now, substitute this back into the expression for $x_n$:

$x_n = (1 + 7n + \binom{n}{2}49 + \binom{n}{3}343 + \dots) – 7n – 1$

$x_n = \binom{n}{2}49 + \binom{n}{3}343 + \dots + 7^n$

We can factor out $49$ from every term, since all terms from $\binom{n}{2}7^2$ onwards contain a factor of at least $7^2 = 49$.

$x_n = 49 \left[ \binom{n}{2} + \binom{n}{3}7 + \dots \right]$

This shows that every element of set X is a multiple of 49. Let's list the first few elements:

• For $n=1: x_1 = 8^1 - 7(1) - 1 = 0$.
• For $n=2: x_2 = 8^2 - 7(2) - 1 = 64 - 14 - 1 = 49$.
• For $n=3: x_3 = 8^3 - 7(3) - 1 = 512 - 21 - 1 = 490$.

So, $X = \{0, 49, 490, ...\}$.

Step 2: Analyze the elements of Set Y

The general element of set Y is $y_n = 49n – 49$.

We can factor out 49:

$y_n = 49(n-1)$.

Since $n \in \{1, 2, 3, ...\}$, the term $(n-1)$ will take values $\{0, 1, 2, ...\}$.

Therefore, set Y consists of all non-negative multiples of 49.

Let's list the first few elements:

• For $n=1: y_1 = 49(1-1) = 0$.
• For $n=2: y_2 = 49(2-1) = 49$.
• For $n=3: y_3 = 49(3-1) = 98$.
• For $n=4: y_4 = 49(4-1) = 147$.

So, $Y = \{0, 49, 98, 147, 490, ...\}$.

Step 3: Compare Set X and Set Y

From Step 1, we proved that every element of X is a multiple of 49. From Step 2, we found that Y is the set of all non-negative multiples of 49. Therefore, every element of X must be contained in Y. This means X is a subset of Y ($X \subseteq Y$).

Now, we must check if the sets are equal ($X=Y$) or if X is a proper subset of Y ($X \subset Y$). For the sets to be equal, every element of Y must also be in X.

Let's consider the element $98$, which is in Y (generated when $n=3$). Is $98$ in X?

If $98 \in X$, then there must be some natural number $k$ such that:

$8^k - 7k - 1 = 98$

$8^k - 7k = 99$

Let's check values of $k$:

• For $k=1: 8^1 - 7(1) = 1$, which is not 99.
• For $k=2: 8^2 - 7(2) = 64 - 14 = 50$, which is not 99.
• For $k=3: 8^3 - 7(3) = 512 - 21 = 491$, which is not 99.

Since the function $f(k) = 8^k - 7k$ is strictly increasing for $k \ge 1$, and its value is less than 99 at $k=2$ and greater than 99 at $k=3$, there is no natural number $k$ for which the value is exactly 99.

Therefore, $98 \notin X$.

Since we have found an element ($98$) that is in Y but not in X, the sets are not equal. X is a proper subset of Y.

Conclusion: $X \subset Y$.

Thus, the correct option is (A).

While option (C) suggests the sets are equal, our analysis shows that set Y contains all non-negative multiples of 49, whereas set X contains only a selection of those multiples. Therefore, the statement $X = Y$ is incorrect.

The correct option is (A) X ⊂ Y.

Given:

Let $C_1$ be the set of people who watch the first News Channel.

Let $C_2$ be the set of people who watch the another channel.

Percentage of people watching the first channel, $|C_1| = 63\%$.

Percentage of people watching the second channel, $|C_2| = 76\%$.

Percentage of people watching both channels, $|C_1 \cap C_2| = x\%$.

The total percentage of people surveyed is $100\%$. Let this be the universal set $U$, so $|U|=100\%$.

To Find:

The possible range of values for $x$.

Solution:

The percentage of people who watch at least one of the two channels is given by the Principle of Inclusion-Exclusion:

$|C_1 \cup C_2| = |C_1| + |C_2| - |C_1 \cap C_2|$

Substitute the given percentages:

$|C_1 \cup C_2| = 63\% + 76\% - x\%$

$|C_1 \cup C_2| = (139 - x)\%$.

The number of people watching at least one channel cannot exceed the total number of people. Thus, the percentage of people watching at least one channel cannot exceed $100\%$.

$|C_1 \cup C_2| \leq |U|$

$(139 - x)\% \leq 100\%$

$139 - x \leq 100$

$139 - 100 \leq x$

$39 \leq x$.

Also, the number of people watching both channels cannot be more than the number of people watching either channel individually. The intersection of two sets cannot be larger than the size of either set.

$|C_1 \cap C_2| \leq |C_1|$

$x\% \leq 63\%$

$x \leq 63$.

And,

$|C_1 \cap C_2| \leq |C_2|$

$x\% \leq 76\%$

$x \leq 76$.

The condition $x \leq 63$ is stricter than $x \leq 76$, so we take $x \leq 63$.

Combining the two conditions we found:

From $|C_1 \cup C_2| \leq 100\%$, we have $39 \leq x$.

From $|C_1 \cap C_2| \leq |C_1|$, we have $x \leq 63$.

Therefore, the possible range for $x$ is $39 \leq x \leq 63$.

Comparing this result with the given options:

(A) x = 35 (35 is not in the range $[39, 63]$)

(B) x = 63 (63 is in the range $[39, 63]$)

(C) 39 ≤ x ≤ 63 (This is the determined range)

(D) x = 39 (39 is in the range $[39, 63]$)

Option (C) represents the full range of possible values for $x$.

The correct option is (C) 39 ≤ x ≤ 63.

Given:

Set $A = \{(x,y) \mid y = \frac{1}{x}, x \in R, x \neq 0 \}$

Set $B = \{(x, y) \mid y = – x, x \in R\}$

To Find:

The intersection of sets $A$ and $B$, i.e., $A \cap B$.

Solution:

The set $A$ consists of all points $(x, y)$ in the Cartesian plane that lie on the curve $y = \frac{1}{x}$, excluding the point where $x=0$ (which is not on the curve anyway). This curve is a hyperbola.

The set $B$ consists of all points $(x, y)$ in the Cartesian plane that lie on the line $y = -x$. This line passes through the origin with a slope of -1.

The intersection of sets $A$ and $B$, denoted by $A \cap B$, is the set of all points $(x, y)$ that belong to both set $A$ and set $B$.

For a point $(x, y)$ to be in $A \cap B$, it must satisfy the conditions for both sets simultaneously. This means the coordinates $(x, y)$ must satisfy both equations:

We can find the intersection points by substituting the expression for $y$ from equation (ii) into equation (i):

Substitute $y = -x$ into $y = \frac{1}{x}$:

$-x = \frac{1}{x}$

Since set $A$ requires $x \neq 0$, we can safely multiply both sides of the equation by $x$:

$-x \times x = \frac{1}{x} \times x$

$-x^2 = 1$

$x^2 = -1$

We are looking for real values of $x$ because the sets are defined over $x \in R$. The equation $x^2 = -1$ has no real solutions. The solutions are complex numbers $x = i$ and $x = -i$.

Since there is no real number $x$ that satisfies the equation $x^2 = -1$, there is no real value of $x$ for which the point $(x, y)$ lies on both the curve $y = \frac{1}{x}$ and the line $y = -x$.

Therefore, the intersection of the two sets $A$ and $B$ is the empty set.

$A \cap B = \phi$.

Comparing this result with the given options:

(A) $A \cap B = A$ (Incorrect, as $A$ is not empty)

(B) $A \cap B = B$ (Incorrect, as $B$ is not empty)

(C) $A \cap B = \phi$ (Correct)

(D) $A \cup B = A$ (Incorrect, as this implies $B \subseteq A$, which is not true)

The correct option is (C) A ∩ B = φ.

Given:

Two sets $A$ and $B$.

The set expression to evaluate is $A \cap (A \cup B)$.

To Find:

The simplified form of the expression $A \cap (A \cup B)$.

Solution:

We are asked to simplify the expression $A \cap (A \cup B)$.

Let's consider the meaning of the union $A \cup B$. The set $A \cup B$ contains all elements that are in set $A$, or in set $B$, or in both.

So, any element that is in $A$ is definitely also in $A \cup B$.

Now consider the intersection $A \cap (A \cup B)$. This set contains all elements that are simultaneously in set $A$ AND in set $(A \cup B)$.

Since every element of $A$ is already contained in $A \cup B$, the elements that are common to both $A$ and $A \cup B$ are precisely the elements of $A$.

Therefore, $A \cap (A \cup B) = A$.

This result is a fundamental property in set theory known as the Absorption Law, which states that for any two sets $A$ and $B$, $A \cap (A \cup B) = A$ and $A \cup (A \cap B) = A$.

Using the Absorption Law, $A \cap (A \cup B) = A$.

Comparing the simplified expression with the given options:

(A) A

(B) B

(C) $\phi$

(D) A $\cap$ B

The simplified expression matches option (A).

The correct option is (A) A.

Given:

Set $A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}$. This is the set of odd natural numbers from 1 to 17.

Set $B = \{2, 4, ..., 18\}$. This is the set of even natural numbers from 2 to 18.

The universal set is $N$, the set of natural numbers, $N = \{1, 2, 3, 4, \dots\}$.

To Simplify:

The set expression $A' \cup ((A \cup B) \cap B')$.

Solution:

We need to simplify the given expression $A' \cup ((A \cup B) \cap B')$.

Let's simplify the part within the parentheses first: $(A \cup B) \cap B'$.

Using the distributive law for set operations, $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$, we can rewrite $(A \cup B) \cap B'$ as $(A \cap B') \cup (B \cap B')$.

$(A \cup B) \cap B' = (A \cap B') \cup (B \cap B')$.

Now, consider the term $(B \cap B')$. The intersection of a set $B$ with its complement $B'$ is the empty set $\phi$.

$(B \cap B') = \phi$.

Substitute this back into the expression:

$(A \cap B') \cup \phi$.

The union of any set with the empty set is the set itself. So, $(A \cap B') \cup \phi = A \cap B'$.

Thus, the expression inside the parentheses simplifies to $A \cap B'$.

Now, substitute this back into the original expression:

$A' \cup ((A \cup B) \cap B') = A' \cup (A \cap B')$.

We can use the distributive law again, this time for union over intersection, $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$. Here, $X = A'$, $Y = A$, and $Z = B'$.

$A' \cup (A \cap B') = (A' \cup A) \cap (A' \cup B')$.

Consider the term $(A' \cup A)$. The union of a set $A$ with its complement $A'$ is the universal set $N$.

$(A' \cup A) = N$.

Substitute this back:

$N \cap (A' \cup B')$.

The intersection of the universal set $N$ with any set is that set itself. So, $N \cap (A' \cup B') = A' \cup B'$.

The simplified expression is $A' \cup B'$.

Using De Morgan's Law, $A' \cup B' = (A \cap B)'$.

Now, let's determine the set $A \cap B$.

Set $A$ contains odd numbers $\{1, 3, 5, \dots, 17\}$.

Set $B$ contains even numbers $\{2, 4, \dots, 18\}$.

Since there are no common elements between a set of only odd numbers and a set of only even numbers, the intersection $A \cap B$ is the empty set $\phi$.

$A \cap B = \phi$.

Finally, we need to find the complement of the empty set, $(A \cap B)' = \phi'$.

The complement of the empty set with respect to the universal set $N$ is the universal set $N$ itself.

$\phi' = N$.

Therefore, the simplified expression $A' \cup ((A \cup B) \cap B')$ is equal to $N$.

Comparing this result with the given options:

(A) $\phi$

(B) N

(C) A

(D) B

The simplified expression matches option (B).

The correct answer is (B) N.

Given:

We are given two sets:

S = {x | x is a positive multiple of 3 less than 100}

P = {x | x is a prime number less than 20}

To Find:

The value of n(S) + n(P), where n(A) represents the number of elements in set A.

Solution:

First, we determine the number of elements in set S, denoted by n(S).

The set S contains all positive multiples of 3 that are less than 100. These elements can be represented as $3k$, where $k$ is a positive integer.

The condition is $3k < 100$.

To find the possible values of $k$, we divide by 3:

$k < \frac{100}{3}$

$k < 33.33...$

Since $k$ must be a positive integer, the possible values for $k$ are $1, 2, 3, \dots, 33$.

This means there are 33 elements in set S.

So, $n(S) = 33$.

Next, we determine the number of elements in set P, denoted by n(P).

The set P contains all prime numbers that are less than 20. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's list the prime numbers less than 20:

$P = \{2, 3, 5, 7, 11, 13, 17, 19\}$

By counting these elements, we find that there are 8 elements in set P.

So, $n(P) = 8$.

Finally, we calculate the sum n(S) + n(P).

$n(S) + n(P) = 33 + 8$

$n(S) + n(P) = 41$

Conclusion:

The calculated value of n(S) + n(P) is 41.

However, this value is not among the given options: (A) 34, (B) 31, (C) 33, (D) 30. This indicates that there is likely an error in the options provided with the question.

Given:

Two sets $X$ and $Y$.

$X'$ denotes the complement of set $X$.

The set expression to evaluate is $X \cap (X \cup Y)'$.

To Simplify:

The expression $X \cap (X \cup Y)'$.

Solution:

We need to simplify the given expression: $X \cap (X \cup Y)'$.

First, simplify the term inside the parentheses using De Morgan's Law: $(X \cup Y)' = X' \cap Y'$.

Substitute this into the expression:

$X \cap (X' \cap Y')$.

Use the Associative Law for intersection, which states that $(A \cap B) \cap C = A \cap (B \cap C)$:

$(X \cap X') \cap Y'$.

Now, consider the term $(X \cap X')$. The intersection of any set with its complement is the empty set $\phi$.

$X \cap X' = \phi$.

Substitute this back into the expression:

$\phi \cap Y'$.

The intersection of the empty set with any set ($Y'$ in this case) is always the empty set $\phi$.

$\phi \cap Y' = \phi$.

Thus, the simplified expression is $\phi$.

Comparing the simplified expression with the given options:

(A) X

(B) Y

(C) $\phi$

(D) X $\cap$ Y

The simplified expression matches option (C).

The correct answer is (C) $\phi$.

Given:

The set $\{x \in R : 1 \leq x < 2\}$.

To Write:

The set in interval notation.

Solution:

The given set is $\{x \in R : 1 \leq x < 2\}$.

This set includes all real numbers $x$ such that $x$ is greater than or equal to 1 and strictly less than 2.

In interval notation, a closed interval $[a, b]$ includes the endpoints $a$ and $b$, i.e., $a \leq x \leq b$.

An open interval $(a, b)$ excludes the endpoints $a$ and $b$, i.e., $a < x < b$.

A half-open or half-closed interval includes one endpoint but excludes the other.

The condition $1 \leq x$ means that 1 is included in the interval. This is represented by a square bracket '[' at the beginning of the interval.

The condition $x < 2$ means that 2 is not included in the interval. This is represented by a parenthesis ')' at the end of the interval.

Combining these, the interval representation of the set $\{x \in R : 1 \leq x < 2\}$ is $[1, 2)$.

The set $\{x \in R : 1 \leq x < 2\}$ can be written as $\mathbf{[1, 2)}$.

Given:

Set $A = \phi$ (the empty set).

To Find:

The number of elements in the power set of $A$, denoted as $n(P(A))$.

Solution:

The power set of a set $A$, denoted by $P(A)$, is the set of all subsets of $A$.

The number of elements in the power set of a set with $k$ elements is given by the formula $2^k$.

In this case, the set is $A = \phi$. The number of elements in the empty set is $n(A) = 0$.

Using the formula for the number of elements in the power set:

$n(P(A)) = 2^{n(A)}$

$n(P(\phi)) = 2^0$

$n(P(\phi)) = 1$.

Alternatively, let's list the subsets of the empty set $\phi$.

The only subset of the empty set is the empty set itself, $\phi$.

So, the power set of $\phi$ is $P(\phi) = \{\phi\}$.

The number of elements in the set $\{\phi\}$ is 1.

Thus, $n(P(\phi)) = 1$.

When $A = \phi$, then the number of elements in $P(A)$ is $\mathbf{1}$.

Given:

$A$ and $B$ are finite sets.

$A \subset B$ (This means $A$ is a proper subset of $B$, implying every element in $A$ is in $B$, and there is at least one element in $B$ not in $A$. However, for the union, the property $A \subseteq B$ is sufficient).

To Find:

$n(A \cup B)$.

Solution:

We are given that $A \subset B$. This means that every element of set $A$ is also an element of set $B$.

The union of two sets $A$ and $B$, denoted by $A \cup B$, is the set containing all elements that are in $A$ or in $B$ (or both).

Since every element in $A$ is already in $B$, when we form the union $A \cup B$, we are combining all elements of $B$ with all elements of $A$. Because all elements of $A$ are already present in $B$, the union $A \cup B$ will contain exactly the same elements as set $B$.

Therefore, if $A \subseteq B$ (which includes the case $A \subset B$), then the union of $A$ and $B$ is equal to $B$.

$A \cup B = B$.

The number of elements in the union $n(A \cup B)$ is equal to the number of elements in set $B$, which is $n(B)$.

$n(A \cup B) = n(B)$.

If $A$ and $B$ are finite sets such that $A \subset B$, then $n (A \cup B) = \mathbf{n(B)}$.

Given:

Two sets $A$ and $B$.

The set difference $A - B$.

To Write:

An equivalent expression for $A - B$.

Solution:

The set difference $A - B$ (read as "A minus B") is the set of elements which are in $A$ but not in $B$.

In set-builder notation, $A - B = \{x \mid x \in A \text{ and } x \notin B\}$.

The condition $x \notin B$ means that $x$ is in the complement of $B$, which is denoted by $B'$ (or $B^c$). Assuming a universal set $U$, $B' = \{x \mid x \in U \text{ and } x \notin B\}$.

So, $x \in A - B \iff x \in A \text{ and } x \in B'$.

The set of elements that are in both $A$ and $B'$ is the intersection of $A$ and $B'$.

Thus, $A - B = A \cap B'$.

If $A$ and $B$ are any two sets, then $A – B$ is equal to $\mathbf{A \cap B'}$.

Given:

Set $A = \{1, 2\}$.

To Find:

The power set of $A$, denoted as $P(A)$.

Solution:

The power set of a set $A$, denoted by $P(A)$, is the set of all possible subsets of $A$.

The subsets of the set $A = \{1, 2\}$ are:

1. The empty set ($\phi$ or \{\})

2. Subsets containing one element: $\{1\}$, $\{2\}$

3. The subset containing all elements (the set itself): $\{1, 2\}$

Listing all these subsets as elements of a new set gives the power set $P(A)$.

$P(A) = \{\phi, \{1\}, \{2\}, \{1, 2\}\}$.

Note that the number of elements in set $A$ is $n(A) = 2$. The number of elements in its power set is $n(P(A)) = 2^{n(A)} = 2^2 = 4$. Our list contains 4 subsets, which is consistent.

Power set of the set $A = \{1, 2\}$ is $\mathbf{\{\phi, \{1\}, \{2\}, \{1, 2\}\}}$.

Given:

Set $A = \{1, 3, 5\}$

Set $B = \{2, 4, 6\}$

Set $C = \{0, 2, 4, 6, 8\}$

To Find:

A possible universal set for the sets A, B, and C.

Solution:

A universal set ($U$) for a collection of sets is a set that contains all the elements of all the sets in the collection.

In other words, for sets A, B, and C, a universal set $U$ must satisfy the condition that $A \subseteq U$, $B \subseteq U$, and $C \subseteq U$.

This means that $U$ must contain every element that is in A, or in B, or in C. The smallest such set is the union of A, B, and C.

$A \cup B \cup C = \{1, 3, 5\} \cup \{2, 4, 6\} \cup \{0, 2, 4, 6, 8\}$

To find the union, we list all unique elements from the three sets:

Elements from A: 1, 3, 5

Elements from B: 2, 4, 6

Elements from C: 0, 2, 4, 6, 8

Combining these unique elements: 0, 1, 2, 3, 4, 5, 6, 8.

$A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$.

Any set that contains all the elements in $A \cup B \cup C$ can be a universal set for A, B, and C.

The simplest (and smallest) choice for a universal set is the union itself.

$U = A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$.

Other valid universal sets could be larger sets that contain these elements, such as the set of all integers, the set of all real numbers, or even the set of all single-digit non-negative integers $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Since the question asks for "can be", the union of the sets is a valid answer.

The universal set of all the three sets A, B, and C can be $\mathbf{\{0, 1, 2, 3, 4, 5, 6, 8\}}$.

Given:

$U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

$A = \{1, 2, 3, 5\}$

$B = \{2, 4, 6, 7\}$

$C = \{2, 3, 4, 8\}$

To Find:

(i) $(B \cup C)'$

(ii) $(C – A)'$

Solution:

(i) First, find the union of sets B and C: $B \cup C$.

$B \cup C = \{2, 4, 6, 7\} \cup \{2, 3, 4, 8\}$

$B \cup C = \{2, 3, 4, 6, 7, 8\}$.

Now, find the complement of $(B \cup C)$ with respect to the universal set $U$.

$(B \cup C)' = U - (B \cup C)$

$(B \cup C)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} - \{2, 3, 4, 6, 7, 8\}$

$(B \cup C)'$ contains elements in $U$ that are not in $B \cup C$.

$(B \cup C)' = \mathbf{\{1, 5, 9, 10\}}$.

(ii) First, find the set difference $C – A$.

$C – A$ contains elements in $C$ that are not in $A$.

$C = \{2, 3, 4, 8\}$

$A = \{1, 2, 3, 5\}$

Elements in C which are not in A are 4 and 8.

$C – A = \{4, 8\}$.

Now, find the complement of $(C – A)$ with respect to the universal set $U$.

$(C – A)' = U - (C – A)$

$(C – A)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} - \{4, 8\}$

$(C – A)'$ contains elements in $U$ that are not in $C – A$.

$(C – A)' = \mathbf{\{1, 2, 3, 5, 6, 7, 9, 10\}}$.

Given:

Two sets $A$ and $B$.

The set expression to simplify is $A - (A \cap B)$.

To Simplify:

The expression $A - (A \cap B)$.

Solution:

We need to simplify the given expression: $A - (A \cap B)$.

Recall the definition of set difference: For any two sets $X$ and $Y$, $X - Y$ is the set of elements that are in $X$ but not in $Y$. This can also be written as $X \cap Y'$, where $Y'$ is the complement of $Y$.

Applying this definition to $A - (A \cap B)$, where $X = A$ and $Y = A \cap B$:

$A - (A \cap B) = A \cap (A \cap B)'$.

Now, use De Morgan's Law to simplify $(A \cap B)'$. De Morgan's Law states that $(X \cap Y)' = X' \cup Y'$.

$(A \cap B)' = A' \cup B'$.

Substitute this back into the expression:

$A \cap (A' \cup B')$.

Next, use the Distributive Law for intersection over union, which states that $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$. Here, $X=A$, $Y=A'$, and $Z=B'$.

$A \cap (A' \cup B') = (A \cap A') \cup (A \cap B')$.

Consider the term $(A \cap A')$. The intersection of a set $A$ with its complement $A'$ is the empty set $\phi$.

$A \cap A' = \phi$.

Substitute this back into the expression:

$\phi \cup (A \cap B')$.

The union of the empty set $\phi$ with any set (in this case, $A \cap B'$) is the set itself.

$\phi \cup (A \cap B') = A \cap B'$.

Finally, recall the definition of set difference again: $A \cap B'$ is equal to $A - B$.

$A \cap B' = A - B$.

Thus, the simplified expression is $A - B$.

For all sets A and B, $A – (A \cap B)$ is equal to $\mathbf{A - B}$.

Given:

We need to match the set expressions in the left column with their simplified equivalents in the right column for any sets A, B, and C.

Solution:

We will simplify each expression from the left column one by one.

(i) $((A' \cup B') - A)'$

Using the definition of set difference, $X - Y = X \cap Y'$:

$((A' \cup B') - A)' = ((A' \cup B') \cap A')'$

By the Absorption Law, $X \cup (X \cap Y) = X$ and $(X \cap Y) \cup X = X$, a similar law is $(X \cup Y) \cap X = X$. Here, let $X=A'$, the expression inside becomes:

$(A' \cup B') \cap A' = A'$

So, the original expression simplifies to:

$(A')' = A$ (by the Law of Double Complementation)

Thus, (i) matches with (b).

(ii) $[B' \cup (B' - A)]'$

Using the definition of set difference, $B' - A = B' \cap A'$:

$[B' \cup (B' \cap A')]'$

By the Absorption Law, $X \cup (X \cap Y) = X$. Here, let $X=B'$, the expression inside the brackets becomes:

$B' \cup (B' \cap A') = B'$

So, the original expression simplifies to:

$(B')' = B$ (by the Law of Double Complementation)

Thus, (ii) matches with (c).

(iii) $(A - B) - (B - C)$

Using the definition of set difference, $X - Y = X \cap Y'$:

$(A - B) - (B - C) = (A \cap B') - (B \cap C')$

$= (A \cap B') \cap (B \cap C')'$

Using De Morgan's Law on the second part, $(B \cap C')' = B' \cup (C')' = B' \cup C$:

$= (A \cap B') \cap (B' \cup C)$

Using the Distributive Law, $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$:

$= ((A \cap B') \cap B') \cup ((A \cap B') \cap C)$

Using the Idempotent Law, $B' \cap B' = B'$:

$= (A \cap B') \cup (A \cap B' \cap C)$

By the Absorption Law, $X \cup (X \cap Z) = X$. Let $X = A \cap B'$ and $Z=C$:

$= A \cap B'$

$= A - B$

Thus, (iii) matches with (a).

(iv) $(A - B) \cap (C - B)$

Using the definition of set difference:

$(A - B) \cap (C - B) = (A \cap B') \cap (C \cap B')$

Using Commutative and Associative laws for intersection:

$= A \cap B' \cap C \cap B'$

$= A \cap C \cap B' \cap B'$

Using the Idempotent Law, $B' \cap B' = B'$:

$= (A \cap C) \cap B'$

$= (A \cap C) - B$

Thus, (iv) matches with (f).

(v) $A \times (B \cap C)$

This is a distributive property of the Cartesian product over set intersection.

$A \times (B \cap C) = (A \times B) \cap (A \times C)$

Thus, (v) matches with (d).

(vi) $A \times (B \cup C)$

This is a distributive property of the Cartesian product over set union.

$A \times (B \cup C) = (A \times B) \cup (A \times C)$

Thus, (vi) matches with (e).

Final Matching:

Given Statement:

If $A$ is any set, then $A \subset A$.

To Determine:

Whether the statement is True or False.

Solution:

The symbol $\subset$ denotes a proper subset.

A set $X$ is a proper subset of a set $Y$ (denoted by $X \subset Y$) if and only if $X$ is a subset of $Y$ ($X \subseteq Y$) and $X$ is not equal to $Y$ ($X \neq Y$).

The statement $A \subset A$ claims that $A$ is a proper subset of itself.

For $A$ to be a proper subset of $A$, two conditions must be met:

1. $A \subseteq A$ (A is a subset of A)

2. $A \neq A$ (A is not equal to A)

The first condition, $A \subseteq A$, is always true because every element of $A$ is indeed an element of $A$.

However, the second condition, $A \neq A$, is always false. A set is always equal to itself.

Since the second condition is false, the statement $A \subset A$ is false. A set cannot be a proper subset of itself.

Note that the statement $A \subseteq A$ (A is a subset of A) is always true. The symbol $\subseteq$ denotes a subset, which can be either proper or improper (when the sets are equal).

The statement "If A is any set, then A ⊂ A" is False.

Given:

Set $M = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

Set $B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

The statement is $B \not\subset M$.

To Determine:

Whether the statement $B \not\subset M$ is True or False.

Solution:

The symbol $\not\subset$ means "is not a proper subset of".

A set $X$ is a proper subset of a set $Y$, denoted by $X \subset Y$, if $X$ is a subset of $Y$ ($X \subseteq Y$) and $X$ is not equal to $Y$ ($X \neq Y$).

The notation $B \not\subset M$ means that $B$ is not a proper subset of $M$. This statement is true if either $B$ is not a subset of $M$ ($B \not\subseteq M$), or if $B$ is a subset of $M$ but $B$ is equal to $M$ ($B \subseteq M$ and $B=M$).

Let's compare the sets $M$ and $B$.

$M = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

$B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

We can see that set $B$ contains exactly the same elements as set $M$.

Therefore, $B = M$.

Now let's check if $B$ is a proper subset of $M$.

Condition 1: Is $B \subseteq M$? Yes, because every element in $B$ is also in $M$ (since $B=M$).

Condition 2: Is $B \neq M$? No, because $B$ and $M$ have the same elements, so $B = M$.

Since the second condition ($B \neq M$) is false, $B$ is not a proper subset of $M$.

The statement $B \not\subset M$ means "B is not a proper subset of M". Since we concluded that B is indeed not a proper subset of M, the statement is true.

The statement "Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then B ⊄ M" is True.

Given Statement:

The sets $\{1, 2, 3, 4\}$ and $\{3, 4, 5, 6\}$ are equal.

To Determine:

Whether the statement is True or False.

Solution:

Two sets are said to be equal if and only if they have exactly the same elements.

Let $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$.

For sets A and B to be equal, every element in A must be in B, and every element in B must be in A.

Let's examine the elements:

The element 1 is in set A, but it is not in set B.

The element 2 is in set A, but it is not in set B.

Since there are elements in set A that are not in set B (namely 1 and 2), set A is not a subset of set B ($A \not\subseteq B$).

For completeness, let's also check the other direction:

The element 5 is in set B, but it is not in set A.

The element 6 is in set B, but it is not in set A.

Since there are elements in set B that are not in set A (namely 5 and 6), set B is not a subset of set A ($B \not\subseteq A$).

Because sets A and B do not contain the same elements, they are not equal.

The statement "The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal" is False.

Given:

$Q$ is the set of rational numbers.

$Z$ is the set of integers.

The statement is $Q \cup Z = Q$.

To Determine:

Whether the statement $Q \cup Z = Q$ is True or False.

Solution:

The set of integers $Z$ is defined as $\{ \dots, -3, -2, -1, 0, 1, 2, 3, \dots \}$.

The set of rational numbers $Q$ is defined as $\{ \frac{p}{q} \mid p \in Z, q \in Z, q \neq 0 \}$.

Consider any integer $z \in Z$. Any integer $z$ can be written in the form $\frac{z}{1}$, where $z \in Z$ and $1 \in Z$ with $1 \neq 0$.

Since every integer can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$, every integer is a rational number.

Thus, the set of integers $Z$ is a subset of the set of rational numbers $Q$.

$Z \subseteq Q$.

The union of two sets $X$ and $Y$, denoted by $X \cup Y$, is the set of all elements that are in $X$ or in $Y$ (or both).

We want to evaluate $Q \cup Z$.

Since $Z \subseteq Q$, every element in $Z$ is also in $Q$.

The union $Q \cup Z$ contains all elements that are in $Q$ or in $Z$. Since all elements of $Z$ are already in $Q$, the set of elements in $Q \cup Z$ is the same as the set of elements in $Q$.

Therefore, if $Z \subseteq Q$, then $Q \cup Z = Q$.

Since integers are indeed a subset of rational numbers, the statement $Q \cup Z = Q$ is true.

The statement "Q ∪ Z = Q, where Q is the set of rational numbers and Z is the set of integers" is True.

Given:

Set $R = \{x \in Z \mid x \text{ is divisible by 2}\}$

Set $T = \{x \in Z \mid x \text{ is divisible by 6}\}$

The statement is $T \subset R$.

To Determine:

Whether the statement $T \subset R$ is True or False.

Solution:

The set $R$ consists of all integers that are multiples of 2 (even integers). $R = \{\dots, -4, -2, 0, 2, 4, 6, \dots\}$

The set $T$ consists of all integers that are multiples of 6. $T = \{\dots, -12, -6, 0, 6, 12, 18, \dots\}$

The statement $T \subset R$ means that $T$ is a proper subset of $R$. This requires two conditions to be met:

1. $T \subseteq R$ (Every element in $T$ is also in $R$).

2. $T \neq R$ (There is at least one element in $R$ that is not in $T$).

Let's check Condition 1: $T \subseteq R$.

If an integer $x$ is in $T$, it means $x$ is divisible by 6. So $x = 6k$ for some integer $k$.

We can write $x = 2 \times (3k)$. Since $k$ is an integer, $3k$ is also an integer.

This shows that $x$ is divisible by 2.

By the definition of set $R$, if an integer is divisible by 2, it belongs to $R$.

Therefore, every element in $T$ is also in $R$. This means $T$ is a subset of $R$, so $T \subseteq R$. Condition 1 is True.

Let's check Condition 2: $T \neq R$.

We need to find if there is any element in $R$ that is not in $T$.

An element in $R$ is an integer divisible by 2. For example, consider the integer 2.

Is $2 \in R$? Yes, because 2 is divisible by 2.

Is $2 \in T$? $T$ contains integers divisible by 6. 2 is not divisible by 6.

Therefore, $2 \notin T$.

Since we found an element (2) that is in $R$ but not in $T$, the sets $R$ and $T$ are not equal ($R \neq T$). This also implies $T \neq R$. Condition 2 is True.

Because both $T \subseteq R$ and $T \neq R$ are true, the statement $T \subset R$ is true.

The statement "Let sets R and T be defined as R = {x ∈ Z | x is divisible by 2} and T = {x ∈ Z | x is divisible by 6}. Then T ⊂ R" is True.

Given:

Set $A = \{0, 1, 2\}$

Set $B = \{x \in R \mid 0 \leq x \leq 2\}$

The statement is $A = B$.

To Determine:

Whether the statement $A = B$ is True or False.

Solution:

Two sets are equal if and only if they contain exactly the same elements.

Set $A$ is a finite set containing three specific integer elements: 0, 1, and 2.

Set $B$ is the set of all real numbers $x$ such that $x$ is greater than or equal to 0 and less than or equal to 2. This can be written in interval notation as $[0, 2]$.

Set $B$ contains not only the integers 0, 1, and 2, but also all the real numbers in between, such as 0.5, $\sqrt{2}$ (approximately 1.414), 1.5, etc.

For example, consider the number 0.5.

Is $0.5 \in B$? Yes, because 0.5 is a real number and $0 \leq 0.5 \leq 2$.

Is $0.5 \in A$? $A = \{0, 1, 2\}$. The number 0.5 is not in set A.

Since there is at least one element (like 0.5) that is in set B but not in set A, the two sets do not contain the same elements.

Therefore, sets A and B are not equal.

The statement "Given A = {0, 1, 2}, B = {x ∈ R | 0 ≤ x ≤ 2}. Then A = B" is False.