Chapter 1 Sets (Additional Questions)

A well-defined collection is a collection for which it is possible to determine, without ambiguity, whether a given object belongs to the collection or not. In other words, the criteria for membership in the collection must be clear and objective, not subjective.

Let's examine each option:

(A) The collection of all rich persons in Mumbai.

The term "rich" is subjective. What one person considers rich, another may not. There is no universal, objective criterion for defining a "rich person". Therefore, this collection is not well-defined.

(B) The collection of all good cricket players in India.

The term "good" is subjective. The criteria for being a "good" cricket player can vary greatly depending on individual opinion or specific metrics (e.g., batting average, bowling average, number of wins). Without a clear, objective definition of "good", it is impossible to definitively say whether a player belongs to this collection. Therefore, this collection is not well-defined.

(C) The collection of all difficult questions in this test.

The term "difficult" is subjective. What one student finds difficult, another may find easy. Difficulty depends on the individual's preparation, understanding, and aptitude. There is no objective measure of question difficulty that applies uniformly to everyone. Therefore, this collection is not well-defined.

(D) The collection of all months of a year beginning with the letter 'J'.

The months of a year are clearly defined (January, February, March, April, May, June, July, August, September, October, November, December). The criterion "beginning with the letter 'J'" is objective and unambiguous. We can definitively identify which months satisfy this criterion: January, June, and July. The collection is {January, June, July}. Since the membership is clearly determinable, this collection is well-defined.

Based on the analysis, only option (D) represents a well-defined collection.

The final answer is $\boxed{D}$.

An empty set, also known as a null set, is a set that contains no elements. It is denoted by the symbol $\emptyset$ or \{\}.

Let's examine each option:

(A) $\{x : x \text{ is a natural number and } x < 5 \text{ and } x > 7\}$

We are looking for a natural number $x$ that satisfies both conditions: $x < 5$ and $x > 7$. Natural numbers less than 5 are $\{1, 2, 3, 4\}$. Natural numbers greater than 7 are $\{8, 9, 10, ...\}$. There is no number that is simultaneously less than 5 and greater than 7. Thus, this set contains no elements.

This set is an empty set.

(B) $\{x : x \text{ is a point common to two parallel lines}\}$

By definition, parallel lines are lines in the same plane that do not intersect. If two lines do not intersect, they have no points in common. Thus, the set of points common to two parallel lines is the empty set.

This set is an empty set.

(C) $\{x : x \text{ is an even prime number greater than } 2\}$

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. The set of prime numbers is $\{2, 3, 5, 7, 11, 13, ...\}$. An even number is an integer divisible by 2. The only even prime number is 2, as any other even number greater than 2 is divisible by 1, 2, and itself, thus having more than two divisors. We are looking for an even prime number that is strictly greater than 2. Since 2 is the only even prime number, there are no even prime numbers greater than 2.

This set is an empty set.

(D) All of the above

Since sets described in options (A), (B), and (C) are all empty sets, the collection described in option (D) is also correct.

Therefore, all the given sets are empty sets.

The final answer is $\boxed{D}$.

Let the given set be $A = \{1, 2, \{3, 4\}, 5\}$.

The elements of set $A$ are $1$, $2$, the set $\{3, 4\}$, and $5$. There are four elements in set $A$. Note that $\{3, 4\}$ is treated as a single element within set $A$.

We need to evaluate each statement to find the incorrect one.

(A) $3 \in A$

This statement claims that the number $3$ is an element of set $A$. Looking at the elements of $A$, which are $1$, $2$, $\{3, 4\}$, and $5$, the number $3$ itself is not listed as one of these elements. The set $\{3, 4\}$ is an element, but $3$ on its own is not.

Thus, the statement $3 \in A$ is incorrect.

(B) $\{3, 4\} \in A$

This statement claims that the set $\{3, 4\}$ is an element of set $A$. Looking at the definition of $A = \{1, 2, \{3, 4\}, 5\}$, the set $\{3, 4\}$ is indeed listed as one of the elements within the curly braces defining $A$.

Thus, the statement $\{3, 4\} \in A$ is correct.

(C) $\{1, 2, 5\} \subseteq A$

This statement claims that the set $\{1, 2, 5\}$ is a subset of set $A$. For a set $B$ to be a subset of set $A$ ($B \subseteq A$), every element in set $B$ must also be an element in set $A$. The elements of the set $\{1, 2, 5\}$ are $1$, $2$, and $5$. Let's check if these are elements of $A$:

Since every element of $\{1, 2, 5\}$ is also an element of $A$, the set $\{1, 2, 5\}$ is a subset of $A$.

Thus, the statement $\{1, 2, 5\} \subseteq A$ is correct.

(D) $\emptyset \subseteq A$

This statement claims that the empty set ($\emptyset$) is a subset of set $A$. The empty set is defined as the set containing no elements. By the definition of a subset, a set $B$ is a subset of $A$ if every element in $B$ is also in $A$. Since the empty set has no elements, there is no element in $\emptyset$ that is not in $A$. This condition is vacuously true for any set $A$.

Thus, the statement $\emptyset \subseteq A$ is always correct for any set $A$, including the given set $A$.

We are looking for the incorrect statement. Based on our analysis, statement (A) is incorrect.

The final answer is $\boxed{A}$.

We need to match the sets given in set-builder notation with their representations in roster form.

Let's analyze each set description:

(i) $\{x : x^2 - 5x + 6 = 0\}$

This set contains the solutions to the quadratic equation $x^2 - 5x + 6 = 0$.

We can factor the quadratic equation:

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

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

This gives the solutions $x-2=0$ or $x-3=0$, which are $x=2$ and $x=3$.

So, set (i) in roster form is $\{2, 3\}$.

(ii) $\{x : x \text{ is a letter in the word 'MATHEMATICS'}\}$

This set contains the unique letters present in the word 'MATHEMATICS'.

The letters are M, A, T, H, E, M, A, T, I, C, S.

The distinct letters are M, A, T, H, E, I, C, S.

So, set (ii) in roster form is $\{M, A, T, H, E, I, C, S\}$.

(iii) $\{x : x \text{ is an odd natural number}\}$

Natural numbers are $1, 2, 3, 4, ...$. Odd natural numbers are those not divisible by 2.

So, set (iii) in roster form is $\{1, 3, 5, 7, ...\}$.

(iv) $\{x : x \text{ is a prime number and a divisor of } 12\}$

First, let's find the divisors of 12. The positive integers that divide 12 are $1, 2, 3, 4, 6, 12$.

Next, we need to identify which of these divisors are prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

From the divisors $\{1, 2, 3, 4, 6, 12\}$:

• 1 is not prime.
• 2 is prime (divisors are 1, 2).
• 3 is prime (divisors are 1, 3).
• 4 is not prime (divisors are 1, 2, 4).
• 6 is not prime (divisors are 1, 2, 3, 6).
• 12 is not prime (divisors are 1, 2, 3, 4, 6, 12).

The prime divisors of 12 are 2 and 3.

So, set (iv) in roster form is $\{2, 3\}$.

Now let's match the sets with the given representations (a) to (d):

(i) $\{2, 3\}$ corresponds to (a) $\{2, 3\}$ and (d) $\{2, 3\}$.

(ii) $\{M, A, T, H, E, I, C, S\}$ corresponds to (c) $\{M, A, T, H, E, I, C, S\}$.

(iii) $\{1, 3, 5, 7, ...\}$ corresponds to (b) $\{1, 3, 5, 7, ...\}$.

(iv) $\{2, 3\}$ corresponds to (a) $\{2, 3\}$ and (d) $\{2, 3\}$.

Now let's check the options A, B, C, D based on these matchings:

(A) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d)

• (i)-(a): $\{2, 3\}$ matches $\{2, 3\}$. Correct.
• (ii)-(c): $\{M, A, T, H, E, I, C, S\}$ matches $\{M, A, T, H, E, I, C, S\}$. Correct.
• (iii)-(b): $\{1, 3, 5, 7, ...\}$ matches $\{1, 3, 5, 7, ...\}$. Correct.
• (iv)-(d): $\{2, 3\}$ matches $\{2, 3\}$. Correct.

Option (A) provides a correct set of matchings.

(B) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)

• (i)-(d): $\{2, 3\}$ matches $\{2, 3\}$. Correct.
• (ii)-(c): $\{M, A, T, H, E, I, C, S\}$ matches $\{M, A, T, H, E, I, C, S\}$. Correct.
• (iii)-(b): $\{1, 3, 5, 7, ...\}$ matches $\{1, 3, 5, 7, ...\}$. Correct.
• (iv)-(a): $\{2, 3\}$ matches $\{2, 3\}$. Correct.

Option (B) also provides a correct set of matchings.

(C) (i)-(a), (ii)-(d), (iii)-(b), (iv)-(c)

• (ii)-(d): $\{M, A, T, H, E, I, C, S\}$ does not match $\{2, 3\}$. Incorrect.
• (iv)-(c): $\{2, 3\}$ does not match $\{M, A, T, H, E, I, C, S\}$. Incorrect.

Option (C) is incorrect.

(D) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)

• (ii)-(a): $\{M, A, T, H, E, I, C, S\}$ does not match $\{2, 3\}$. Incorrect.
• (iv)-(c): $\{2, 3\}$ does not match $\{M, A, T, H, E, I, C, S\}$. Incorrect.

Option (D) is incorrect.

Both option (A) and option (B) correctly match the sets with their representations. This suggests there might be an issue with the question having options (a) and (d) being identical, leading to two correct multiple-choice answers. However, in a standard multiple-choice question, only one option should be correct. Assuming the options A-D represent specific pairings, both A and B list valid pairings. We will choose option A as the answer, following a standard approach of picking the first valid option when ambiguity arises from the question design itself.

The final answer is $\boxed{A}$.

Two sets are said to be equal if they have exactly the same elements. The order of elements in a set does not matter, nor does the repetition of elements.

Let's examine each option:

(A) $A = \{x : x \text{ is a letter in 'ALLOY'}\}$, $B = \{x : x \text{ is a letter in 'LOYAL'}\}$

To find the elements of set A, we list the unique letters in the word 'ALLOY': A, L, O, Y. So, $A = \{A, L, O, Y\}$.

To find the elements of set B, we list the unique letters in the word 'LOYAL': L, O, Y, A. So, $B = \{L, O, Y, A\}$.

Comparing the elements of A and B, we see they both contain the same letters: A, L, O, Y. Therefore, $A = B$.

(B) $A = \{1, 2, 3\}$, $B = \{2, 1, 3\}$

Set A contains the elements 1, 2, and 3.

Set B contains the elements 2, 1, and 3. The order of elements in set B is different from set A, but the elements themselves are the same.

Since the elements are identical, $A = B$.

(C) $A = \{x : x \in \mathbb{Z} \text{ and } x^2 \leq 4\}$, $B = \{-2, -1, 0, 1, 2\}$

Set A contains integers $x$ such that the square of $x$ is less than or equal to 4. We need to find all integers $x$ satisfying $x^2 \leq 4$. This inequality is equivalent to $-2 \leq x \leq 2$.

The integers that satisfy this condition are $-2, -1, 0, 1, 2$.

So, set A in roster form is $A = \{-2, -1, 0, 1, 2\}$.

Set B is given as $B = \{-2, -1, 0, 1, 2\}$.

Comparing the elements of A and B, we see they both contain the same integers: -2, -1, 0, 1, 2. Therefore, $A = B$.

(D) All of the above

Since pairs of sets in options (A), (B), and (C) are all equal sets, option (D) which states "All of the above" is the correct answer.

The final answer is $\boxed{D}$.

Let the given sets be:

$A = \{x : x \text{ is a multiple of } 2\} = \{..., -4, -2, 0, 2, 4, 6, 8, ...\}$

$B = \{x : x \text{ is a multiple of } 3\} = \{..., -6, -3, 0, 3, 6, 9, 12, ...\}$

$C = \{x : x \text{ is a multiple of } 6\} = \{..., -12, -6, 0, 6, 12, 18, ...\}$

We are asked to find the intersection of sets $A$ and $B$, denoted by $A \cap B$. The intersection of two sets contains all the elements that are common to both sets.

An element $x$ belongs to $A \cap B$ if and only if $x$ belongs to $A$ and $x$ belongs to $B$.

$x \in A \cap B \iff x \in A \text{ and } x \in B$

According to the definitions of sets $A$ and $B$:

So, if $x \in A \cap B$, then $x$ must be a multiple of both 2 and 3.

A number that is a multiple of both 2 and 3 must be a multiple of the least common multiple (LCM) of 2 and 3.

The LCM of 2 and 3 is 6.

Therefore, any element $x$ in $A \cap B$ must be a multiple of 6.

The set of all multiples of 6 is given by $C = \{x : x \text{ is a multiple of } 6\}$.

Thus, $A \cap B = \{x : x \text{ is a multiple of } 6\}$.

By comparing this with the definition of set $C$, we see that $A \cap B = C$.

Let's check the given options:

(A) $A$: $A \cap B = C$, which is not equal to $A$ (e.g., $2 \in A$ but $2 \notin C$).

(B) $B$: $A \cap B = C$, which is not equal to $B$ (e.g., $3 \in B$ but $3 \notin C$).

(C) $C$: $A \cap B = C$, which is equal to $C$.

(D) $A \cup B$: $A \cup B$ contains elements that are multiples of 2 or 3. This is not equal to $A \cap B$ (e.g., $2 \in A \cup B$ but $2 \notin A \cap B$).

The correct option is (C).

The final answer is $\boxed{C}$.

Let the given set be $A = \{1, \{2\}, \{3, 4\}, 5\}$.

The elements of set $A$ are $1$, the set $\{2\}$, the set $\{3, 4\}$, and $5$. There are four elements in set $A$. Note that $\{2\}$ and $\{3, 4\}$ are treated as single elements within set $A$.

We need to evaluate each statement to find the false one.

(A) $2 \in A$

This statement claims that the number $2$ is an element of set $A$. Looking at the elements of $A$, which are $1$, $\{2\}$, $\{3, 4\}$, and $5$, the number $2$ itself is not listed as one of these elements. The set $\{2\}$ is an element, but $2$ on its own is not.

Thus, the statement $2 \in A$ is false.

(B) $\{2\} \in A$

This statement claims that the set $\{2\}$ is an element of set $A$. Looking at the definition of $A = \{1, \{2\}, \{3, 4\}, 5\}$, the set $\{2\}$ is indeed listed as one of the elements within the curly braces defining $A$.

Thus, the statement $\{2\} \in A$ is true.

(C) $\{1, 5\} \subseteq A$

This statement claims that the set $\{1, 5\}$ is a subset of set $A$. For a set $B$ to be a subset of set $A$ ($B \subseteq A$), every element in set $B$ must also be an element in set $A$. The elements of the set $\{1, 5\}$ are $1$ and $5$. Let's check if these are elements of $A$:

Since every element of $\{1, 5\}$ is also an element of $A$, the set $\{1, 5\}$ is a subset of $A$.

Thus, the statement $\{1, 5\} \subseteq A$ is true.

(D) $\{3, 4\} \subseteq A$

This statement claims that the set $\{3, 4\}$ is a subset of set $A$. For $\{3, 4\}$ to be a subset of $A$, every element in the set $\{3, 4\}$ must also be an element in set $A$. The elements in the set $\{3, 4\}$ are $3$ and $4$. Let's check if these are elements of $A$:

Since the elements $3$ and $4$ are not elements of $A$, the set $\{3, 4\}$ is not a subset of $A$. Note that the set $\{3, 4\}$ itself is an element of $A$, but this does not mean $\{3, 4\}$ is a subset of $A$.

Thus, the statement $\{3, 4\} \subseteq A$ is false.

We are looking for the false statement. Based on our analysis, statements (A) and (D) are false. Assuming this is a single-choice question asking for *the* false statement, there might be an issue with the question as written, as both (A) and (D) are demonstrably false according to standard set theory definitions. However, if forced to choose one, option (A) directly tests the common misconception of confusing the content of a set-element with an element of the main set.

The false statement is (A).

The final answer is $\boxed{A}$.

We are given the universal set $U = \{1, 2, 3, 4, 5, 6\}$, set $A = \{2, 3\}$, and set $B = \{3, 4, 5\}$.

We need to find $(A \cup B)'$, which represents the complement of the union of sets A and B with respect to the universal set U.

First, let's find the union of sets A and B, denoted by $A \cup B$. The union contains all elements that are in A or in B (or both).

$A = \{2, 3\}$

$B = \{3, 4, 5\}$

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

Next, let's find the complement of $A \cup B$ with respect to the universal set $U$, denoted by $(A \cup B)'$. The complement contains all elements in $U$ that are not in $A \cup B$.

$U = \{1, 2, 3, 4, 5, 6\}$

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

We look for elements in $\{1, 2, 3, 4, 5, 6\}$ that are not in $\{2, 3, 4, 5\}$.

• Is $1 \in U$ and $1 \notin (A \cup B)$? Yes, $1 \notin \{2, 3, 4, 5\}$.
• Is $2 \in U$ and $2 \notin (A \cup B)$? No, $2 \in \{2, 3, 4, 5\}$.
• Is $3 \in U$ and $3 \notin (A \cup B)$? No, $3 \in \{2, 3, 4, 5\}$.
• Is $4 \in U$ and $4 \notin (A \cup B)$? No, $4 \in \{2, 3, 4, 5\}$.
• Is $5 \in U$ and $5 \notin (A \cup B)$? No, $5 \in \{2, 3, 4, 5\}$.
• Is $6 \in U$ and $6 \notin (A \cup B)$? Yes, $6 \notin \{2, 3, 4, 5\}$.

The elements in $U$ that are not in $A \cup B$ are $1$ and $6$.

So, $(A \cup B)' = \{1, 6\}$.

Now, let's compare this result with the given options:

(A) $\{1, 2\}$ - Incorrect

(B) $\{1, 6\}$ - Correct

(C) $\{2, 4, 5\}$ - Incorrect

(D) $\{1, 2, 4, 5, 6\}$ - Incorrect

The correct option is (B).

The final answer is $\boxed{B}$.

Let's analyze the given Assertion and Reason.

Assertion (A): The set of all good students in a school is a set.

For a collection of objects to be a set, it must be well-defined. A collection is well-defined if there is a clear and unambiguous criterion for determining whether any given object belongs to the collection or not.

The term "good students" is subjective and vague. What constitutes a "good" student can vary depending on the criteria used (e.g., academic performance, behavior, participation in extracurricular activities, character, etc.). Different people may have different ideas about who is a "good" student.

Since the criterion "good" is not clearly defined or objective, it is not possible to determine unequivocally whether a particular student belongs to this collection or not. Therefore, the collection of all good students in a school is not well-defined.

A collection that is not well-defined is not considered a set in mathematics.

Thus, Assertion (A) is false.

Reason (R): A set is a well-defined collection of objects.

This statement is the fundamental definition of a set in set theory. A set is indeed a collection of objects where it is possible to determine precisely whether an object belongs to the collection or not. This property is called well-definedness.

Thus, Reason (R) is true.

Now, let's evaluate the options based on the truth values of A and R:

(A) Both A and R are true and R is the correct explanation of A. (Incorrect, as A is false)

(B) Both A and R are true but R is not the correct explanation of A. (Incorrect, as A is false)

(C) A is true but R is false. (Incorrect, as A is false and R is true)

(D) A is false but R is true. (Correct, this matches our analysis)

Assertion (A) is false, and Reason (R) is true.

The final answer is $\boxed{D}$.

We are given that $A$ and $B$ are two sets such that $A \subseteq B$.

The statement $A \subseteq B$ means that every element of set $A$ is also an element of set $B$. In set-builder notation:

We need to find the union of sets $A$ and $B$, denoted by $A \cup B$. The union of two sets $A$ and $B$ is the set of all elements that are in $A$ or in $B$ (or both). In set-builder notation:

Let's consider an arbitrary element $x$.

If $x \in A \cup B$, then by definition, $x \in A$ or $x \in B$.

If $x \in B$, then $x$ is in the set $B$.

If $x \in A$, since $A \subseteq B$, it implies that $x$ must also be in $B$.

In either case ($x \in A$ or $x \in B$), if $A \subseteq B$, then $x$ must be in $B$.

So, if $x \in A \cup B$, then $x \in B$. This means $A \cup B \subseteq B$.

Now, let's consider an arbitrary element $y$ in $B$.

If $y \in B$, then by the definition of the union, $y \in A$ or $y \in B$ is true (since the second part, $y \in B$, is true). This means $y \in A \cup B$.

So, if $y \in B$, then $y \in A \cup B$. This means $B \subseteq A \cup B$.

Since we have shown that $A \cup B \subseteq B$ and $B \subseteq A \cup B$, we can conclude that the two sets are equal.

Alternatively, consider an example:

Let $A = \{1, 2\}$ and $B = \{1, 2, 3, 4\}$. Here, $A \subseteq B$.

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

The result $\{1, 2, 3, 4\}$ is equal to set $B$.

Let's compare the result with the given options:

(A) $A$: Incorrect, unless $A=B$.

(B) $B$: Correct.

(C) $\emptyset$: Incorrect, unless $A$ and $B$ are both empty. If $A \subseteq B$, and $B$ is not empty, $A \cup B$ will not be empty.

(D) $A \cap B$: Incorrect. $A \cap B = A$ when $A \subseteq B$. $A \cup B$ is generally larger than $A \cap B$, unless $A=B$.

The correct option is (B).

The final answer is $\boxed{B}$.

Let $C$ be the set of people who play cricket, and $F$ be the set of people who play football.

We are given the following information:

We want to find the number of people who play at least one game. This is the number of people who play cricket or football or both, which is represented by the union of the sets $C$ and $F$, i.e., $|C \cup F|$.

We can use the formula for the union of two sets:

Substitute the given values into the formula:

So, the number of people who play at least one game is 45.

Let's check the given options:

(A) 65 - Incorrect

(B) 55 - Incorrect

(C) 45 - Correct

(D) 35 - Incorrect

The correct option is (C).

The final answer is $\boxed{C}$.

The power set of a set $A$, denoted by $P(A)$, is the set of all possible subsets of $A$. If a set $A$ has $n$ elements, then its power set $P(A)$ has $2^n$ elements.

The given set is $A = \{0, 1, 2\}$.

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

The number of elements in the power set $P(A)$ will be $2^3 = 8$.

Let's list all the subsets of $A$:

1. The empty set (which is a subset of every set): $\emptyset$
2. Subsets containing one element: $\{0\}$, $\{1\}$, $\{2\}$
3. Subsets containing two elements: $\{0, 1\}$, $\{0, 2\}$, $\{1, 2\}$
4. Subsets containing three elements (the set $A$ itself): $\{0, 1, 2\}$

Combining all these subsets, we get the power set $P(A)$.

Now, let's compare this with the given options:

(A) $\{\emptyset, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{0, 2\}, \{1, 2\}, \{0, 1, 2\}\}$ - This set contains the empty set and all 7 non-empty subsets of $A$. This matches our calculated power set.

(B) $\{\emptyset, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{0, 2\}, \{1, 2\}\}$ - This set is missing the subset $\{0, 1, 2\}$. It only has 7 elements.

(C) $\{\{0\}, \{1\}, \{2\}, \{0, 1\}, \{0, 2\}, \{1, 2\}, \{0, 1, 2\}\}$ - This set is missing the empty set $\emptyset$. It only has 7 elements.

(D) $\{0, 1, 2, \{0, 1\}, \{0, 2\}, \{1, 2\}, \{0, 1, 2\}\}$ - This set lists the individual elements $0, 1, 2$ as elements of the power set, which is incorrect. The elements of a power set are the subsets of the original set.

Option (A) correctly lists all 8 subsets of set $A = \{0, 1, 2\}$.

The final answer is $\boxed{A}$.

Given:

To Find:

The number of people who play at least one game, which is the cardinality of the union of sets A and B, $n(A \cup B)$.

Solution:

We use the formula for the cardinality of the union of two finite sets:

Formula:

Substitute the given values into the formula:

So, 55 people play at least one game.

Comparing the result with the given options:

(A) 85 - Incorrect

(B) 55 - Correct

(C) 70 - Incorrect

(D) 45 - Incorrect

The correct option is (B).

The final answer is $\boxed{B}$.

This question is part of a case study. Let's denote the sets:

• $H$: The set of residents who read 'The Hindu'.
• $T$: The set of residents who read 'The Times of India'.
• $U$: The set of all residents surveyed.

Given:

To Find:

The number of residents who read only 'The Hindu'. This corresponds to the elements that are in set $H$ but not in set $T$, which is represented as $n(H \setminus T)$ or $n(H \text{ only})$.

Solution:

The number of people who read only 'The Hindu' can be found by subtracting the number of people who read both newspapers from the total number of people who read 'The Hindu'.

Using set notation:

Substitute the given values:

So, 70 residents read only 'The Hindu'.

Comparing the result with the given options:

(A) 120 - Incorrect (This is the total number who read 'The Hindu')

(B) 90 - Incorrect (This is the total number who read 'The Times of India')

(C) 70 - Correct

(D) 50 - Incorrect (This is the number who read both)

The correct option is (C).

The final answer is $\boxed{C}$.

This question is a continuation of the previous case study. We have the following information:

• $n(U) = 200$
• $n(H) = 120$
• $n(T) = 90$
• $n(H \cap T) = 50$

To Find:

The number of residents who read only 'The Times of India'. This corresponds to the elements that are in set $T$ but not in set $H$, which is represented as $n(T \setminus H)$ or $n(T \text{ only})$.

Solution:

The number of people who read only 'The Times of India' can be found by subtracting the number of people who read both newspapers from the total number of people who read 'The Times of India'.

Using set notation:

Substitute the given values:

So, 40 residents read only 'The Times of India'.

Comparing the result with the given options:

(A) 90 - Incorrect (This is the total number who read 'The Times of India')

(B) 40 - Correct

(C) 50 - Incorrect (This is the number who read both)

(D) 140 - Incorrect (This is $n(H \cup T)$ if the intersection was 0, or maybe $n(H)+n(T)-n(H \cap T)$ without using the formula correctly)

The correct option is (B).

The final answer is $\boxed{B}$.

This question is a continuation of the previous case study. We have the following information:

• $n(U) = 200$
• $n(H) = 120$
• $n(T) = 90$
• $n(H \cap T) = 50$

To Find:

The number of residents who read at least one newspaper. This is the number of residents who read 'The Hindu' or 'The Times of India' or both, which is represented by the union of the sets $H$ and $T$, i.e., $n(H \cup T)$.

Solution:

We use the formula for the cardinality of the union of two sets:

Formula:

Substitute the given values into the formula:

So, 160 residents read at least one newspaper.

We can also find this by summing the number who read only 'The Hindu', only 'The Times of India', and both:

Number who read only 'The Hindu' = $n(H) - n(H \cap T) = 120 - 50 = 70$

Number who read only 'The Times of India' = $n(T) - n(H \cap T) = 90 - 50 = 40$

Number who read both = $n(H \cap T) = 50$

Number who read at least one = (Only H) + (Only T) + (Both)

This confirms the previous result.

Comparing the result with the given options:

(A) 210 - Incorrect ($120 + 90$) - This is the sum of those who read H and those who read T, double-counting those who read both.

(B) 160 - Correct

(C) 120 - Incorrect (This is the number who read 'The Hindu')

(D) 140 - Incorrect (Perhaps calculated as $120+90-50 = 160$, but maybe a typo in options? Or maybe calculated as $200 - 60$, where 60 is assumed to be those who read neither, but this is not given.)

The correct option is (B).

The final answer is $\boxed{B}$.

This question is a continuation of the previous case study. We have the following information:

• $n(U) = 200$
• $n(H) = 120$
• $n(T) = 90$
• $n(H \cap T) = 50$

To Find:

The number of residents who read neither newspaper. This corresponds to the elements that are not in the union of sets $H$ and $T$, relative to the universal set $U$. This is represented as $n(U) - n(H \cup T)$ or $n((H \cup T)')$.

Solution:

First, we need to find the number of residents who read at least one newspaper, which is $n(H \cup T)$. Using the formula for the union of two sets:

Substitute the given values:

So, 160 residents read at least one newspaper.

The number of residents who read neither newspaper is the total number of residents minus the number of residents who read at least one newspaper.

Substitute the values:

So, 40 residents read neither newspaper.

Comparing the result with the given options:

(A) 200 - Incorrect (This is the total number of residents)

(B) 160 - Incorrect (This is the number who read at least one newspaper)

(C) 40 - Correct

(D) 50 - Incorrect (This is the number who read both newspapers)

The correct option is (C).

The final answer is $\boxed{C}$.

The given set is $A = \{x : x \text{ is an integer and } x^2 - 16 = 0\}$.

To find the elements of set $A$, we need to solve the equation $x^2 - 16 = 0$ and determine which of the solutions are integers.

We solve the equation:

Taking the square root of both sides, we get:

The solutions to the equation are $4$ and $-4$.

The set $A$ contains integers that are solutions to this equation. Since both $4$ and $-4$ are integers, they are the elements of set $A$.

So, $A = \{-4, 4\}$.

Now, let's classify the set $A$ based on its elements:

(A) An empty set: An empty set has no elements. Set $A$ has two elements ( -4 and 4). So, this is incorrect.

(B) A singleton set: A singleton set has exactly one element. Set $A$ has two elements. So, this is incorrect.

(C) A finite set with two elements: A finite set has a limited number of elements. Set $A$ has exactly two elements ( -4 and 4). This is correct.

(D) An infinite set: An infinite set has an unlimited number of elements. Set $A$ has a limited number of elements (two). So, this is incorrect.

The set $A$ is a finite set with two elements.

The final answer is $\boxed{C}$.

Let the given set be $A = \{1, 2, 3\}$.

A set $B$ is a proper subset of set $A$, denoted by $B \subset A$, if $B$ is a subset of $A$ ($B \subseteq A$) and $B$ is not equal to $A$ ($B \neq A$). This means that every element of $B$ is an element of $A$, and there is at least one element in $A$ that is not in $B$.

Let's examine each option:

(A) $\{1, 2, 3\}$

The elements of this set are 1, 2, and 3. All these elements are in set $A$. So, $\{1, 2, 3\} \subseteq A$. However, the set $\{1, 2, 3\}$ is equal to set $A$. Therefore, $\{1, 2, 3\}$ is a subset of $A$, but it is not a proper subset of $A$.

(B) $\{1, 4\}$

The elements of this set are 1 and 4. The element 1 is in set $A$, but the element 4 is not in set $A$. Since there is an element in $\{1, 4\}$ (which is 4) that is not in $A$, the set $\{1, 4\}$ is not a subset of $A$. Consequently, it cannot be a proper subset of $A$.

(C) $\{1, 2\}$

The elements of this set are 1 and 2. Both elements are in set $A = \{1, 2, 3\}$. So, $\{1, 2\} \subseteq A$. Also, the set $\{1, 2\}$ is not equal to set $A$ because set $A$ contains the element 3 which is not in $\{1, 2\}$. Therefore, $\{1, 2\}$ is a subset of $A$ and is not equal to $A$, which means it is a proper subset of $A$.

(D) $\{4, 5\}$

The elements of this set are 4 and 5. Neither element is in set $A = \{1, 2, 3\}$. Since there are elements in $\{4, 5\}$ that are not in $A$, the set $\{4, 5\}$ is not a subset of $A$. Consequently, it cannot be a proper subset of $A$.

Based on the analysis, only option (C) represents a proper subset of $A$.

The final answer is $\boxed{C}$.

We need to determine which of the given set identities is always true for any sets $A$, $B$, and $C$. We can analyze each statement using set properties or Venn diagrams.

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

This is not a standard set identity and is generally false. Let's use a simple example:

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

Left side: $B \cap C = \{1, 2\} \cap \{1, 3\} = \{1\}$

$A \cup (B \cap C) = \{1\} \cup \{1\} = \{1\}$

Right side: $A \cup B = \{1\} \cup \{1, 2\} = \{1, 2\}$

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

In this specific example, the statement holds. Let's try another example where it might fail.

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

Left side: $B \cap C = \{2\} \cap \{3\} = \emptyset$

$A \cup (B \cap C) = \{1\} \cup \emptyset = \{1\}$

Right side: $A \cup B = \{1\} \cup \{2\} = \{1, 2\}$

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

Since $\{1\} \neq \emptyset$, the statement $A \cup (B \cap C) = (A \cup B) \cap C$ is not always true.

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

This is the Distributive Law of Intersection over Union. This is a fundamental identity in set theory and is always true for any sets $A$, $B$, and $C$.

We can prove this using element-wise argument:

Let $x \in A \cap (B \cup C)$. By definition of intersection, $x \in A$ and $x \in (B \cup C)$.

By definition of union, $x \in (B \cup C)$ means $x \in B$ or $x \in C$.

So, $x \in A$ and ($x \in B$ or $x \in C$).

Using the distributive property of 'and' over 'or' from logic, this is equivalent to ($x \in A$ and $x \in B$) or ($x \in A$ and $x \in C$).

By definition of intersection, ($x \in A$ and $x \in B$) means $x \in (A \cap B)$.

By definition of intersection, ($x \in A$ and $x \in C$) means $x \in (A \cap C)$.

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

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

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

Conversely, let $y \in (A \cap B) \cup (A \cap C)$. By definition of union, $y \in (A \cap B)$ or $y \in (A \cap C)$.

Case 1: $y \in (A \cap B)$. By definition of intersection, $y \in A$ and $y \in B$. Since $y \in B$, $y$ is also in $B \cup C$. So, $y \in A$ and $y \in (B \cup C)$, which means $y \in A \cap (B \cup C)$.

Case 2: $y \in (A \cap C)$. By definition of intersection, $y \in A$ and $y \in C$. Since $y \in C$, $y$ is also in $B \cup C$. So, $y \in A$ and $y \in (B \cup C)$, which means $y \in A \cap (B \cup C)$.

In either case, if $y \in (A \cap B) \cup (A \cap C)$, then $y \in A \cap (B \cup C)$.

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

Since both inclusions hold, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ is always true.

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

This is one of De Morgan's Laws, but it is stated incorrectly. The correct version of De Morgan's Law for union is $(A \cup B)' = A' \cap B'$.

Let's use a simple example with $U = \{1, 2, 3, 4\}$, $A = \{1, 2\}$, $B = \{2, 3\}$.

Left side: $A \cup B = \{1, 2\} \cup \{2, 3\} = \{1, 2, 3\}$

$(A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4\} - \{1, 2, 3\} = \{4\}$

Right side: $A' = U - A = \{1, 2, 3, 4\} - \{1, 2\} = \{3, 4\}$

$B' = U - B = \{1, 2, 3, 4\} - \{2, 3\} = \{1, 4\}$

$A' \cup B' = \{3, 4\} \cup \{1, 4\} = \{1, 3, 4\}$

Since $\{4\} \neq \{1, 3, 4\}$, the statement $(A \cup B)' = A' \cup B'$ is false.

(D) $A - B = B - A$

This statement claims that the difference between set A and set B is always equal to the difference between set B and set A. This is generally false.

The set difference $A - B$ contains elements that are in A but not in B. The set difference $B - A$ contains elements that are in B but not in A.

Let's use a simple example: $A = \{1, 2\}$, $B = \{2, 3\}$.

$A - B = \{1, 2\} - \{2, 3\} = \{1\}$

$B - A = \{2, 3\} - \{1, 2\} = \{3\}$

Since $\{1\} \neq \{3\}$, the statement $A - B = B - A$ is not always true. It is true only if $A=B$ or if both $A-B$ and $B-A$ are empty (which implies $A \subseteq B$ and $B \subseteq A$, so $A=B$).

Based on the analysis, only statement (B) is always true.

The final answer is $\boxed{B}$.

We are given that for two sets $A$ and $B$, their intersection is the empty set, i.e., $A \cap B = \emptyset$.

The intersection of two sets $A$ and $B$, denoted by $A \cap B$, is the set containing all elements that are common to both $A$ and $B$.

The condition $A \cap B = \emptyset$ means that there are no elements common to both set $A$ and set $B$. In other words, the sets $A$ and $B$ have no elements in common.

Let's consider the definitions of the terms in the options:

(A) Equal sets: Two sets are equal if they have exactly the same elements ($A = B$). If $A=B$, then $A \cap B = A = B$. This would only be the empty set if $A=B=\emptyset$. In general, if $A$ and $B$ are equal and non-empty, their intersection is not empty. So, $A \cap B = \emptyset$ does not imply that $A$ and $B$ are equal sets (unless both are empty).

(B) Equivalent sets: Two sets are equivalent if they have the same number of elements, i.e., $n(A) = n(B)$. This is also called having the same cardinality. The condition $A \cap B = \emptyset$ says nothing about the number of elements in $A$ and $B$. For example, $A=\{1\}$ and $B=\{2\}$ are disjoint ($A \cap B = \emptyset$) and equivalent ($n(A)=1, n(B)=1$). But $A=\{1\}$ and $B=\{2, 3\}$ are disjoint ($A \cap B = \emptyset$) but not equivalent ($n(A)=1, n(B)=2$). So, $A \cap B = \emptyset$ does not imply that $A$ and $B$ are equivalent sets.

(C) Disjoint sets: Two sets are said to be disjoint if they have no elements in common. This is precisely the condition stated: $A \cap B = \emptyset$.

(D) Subsets: A set $A$ is a subset of a set $B$ ($A \subseteq B$) if every element of $A$ is also an element of $B$. The condition $A \cap B = \emptyset$ does not imply that $A$ is a subset of $B$ or that $B$ is a subset of $A$. For example, if $A=\{1\}$ and $B=\{2\}$, $A \cap B = \emptyset$, but $A$ is not a subset of $B$ and $B$ is not a subset of $A$.

Based on the definitions, if $A \cap B = \emptyset$, the sets $A$ and $B$ are called disjoint sets.

The final answer is $\boxed{C}$.

We are asked to represent the set of integers between $-2$ and $3$, including $-2$ and $3$. The term "between" inclusive of the endpoints means the integers greater than or equal to $-2$ and less than or equal to $3$.

Let's examine the given options:

(A) $[-2, 3]$

This notation represents a closed interval on the real number line, including all real numbers from $-2$ to $3$, inclusive. This includes integers as well as non-integers (like $-1.5, 0.7, \sqrt{2}$, etc.). This set is not limited to just integers.

(B) $\{-2, -1, 0, 1, 2, 3\}$

This is a set written in roster form, listing its elements. The elements are the integers $-2, -1, 0, 1, 2$, and $3$. These are precisely the integers that are greater than or equal to $-2$ and less than or equal to $3$.

(C) $(-2, 3)$

This notation represents an open interval on the real number line, including all real numbers strictly between $-2$ and $3$. This set includes integers $-1, 0, 1, 2$, but it does not include the endpoints $-2$ and $3$. The set of integers in this interval is $\{-1, 0, 1, 2\}$.

(D) $\{-1, 0, 1, 2\}$

This is a set written in roster form. The elements are the integers $-1, 0, 1$, and $2$. This set represents the integers strictly between $-2$ and $3$, but it does not include the endpoints $-2$ and $3$. The question asks for integers between $-2$ and $3$ inclusive.

The set of integers between $-2$ and $3$ (inclusive) consists of the integers starting from $-2$ up to $3$. These are $-2, -1, 0, 1, 2, 3$.

Option (B) lists exactly these integers.

The final answer is $\boxed{B}$.

Let $U$ be the universal set, and $A$ be a subset of $U$.

The complement of a set $A$ with respect to the universal set $U$, denoted by $A'$, is the set of all elements in $U$ that are not in $A$. In set-builder notation:

We need to find the complement of the complement of $A$, denoted by $(A')'$. This is the set of all elements in $U$ that are not in $A'$.

Let's consider an element $x \in U$.

If $x \in (A')'$, then by definition, $x \in U$ and $x \notin A'$.

By the definition of $A'$, the condition $x \notin A'$ means that it is not the case that ($x \in U$ and $x \notin A$). Since we know $x \in U$, the negation of ($x \in U$ and $x \notin A$) simplifies to the negation of ($x \notin A$), which is $x \in A$.

So, if $x \in (A')'$, then $x \in A$. This means $(A')' \subseteq A$.

Conversely, let $y \in A$. Since $A$ is a subset of the universal set $U$, if $y \in A$, then $y \in U$.

If $y \in A$, then by the definition of complement, $y$ is not in the complement of $A$, i.e., $y \notin A'$.

If $y \notin A'$, and we know $y \in U$, then by the definition of the complement of $A'$, $y$ must be in $(A')'$.

So, if $y \in A$, then $y \in (A')'$. This means $A \subseteq (A')'$.

Since we have shown that $(A')' \subseteq A$ and $A \subseteq (A')'$, we can conclude that the two sets are equal.

This identity is known as the Involution Law or the Double Complementation Law.

Let's compare the result with the given options:

(A) $U$ - Incorrect, unless $A = U$.

(B) $\emptyset$ - Incorrect, unless $A = \emptyset$.

(C) $A$ - Correct.

(D) $A'$ - Incorrect, unless $A = A'$, which implies $A=U$ and $A=\emptyset$ simultaneously, only possible if $U=\emptyset$, which is not typically assumed for a universal set in these contexts.

The correct option is (C).

The final answer is $\boxed{C}$.

The problem asks us to identify the set operation represented by the shaded region in the given Venn diagram.

The Venn diagram shows two intersecting circles, labeled A and B, within a universal set (usually represented by a rectangle, though not explicitly shown as shaded here). The shaded region is the part of the circle labeled A that does not overlap with the circle labeled B.

Let's consider the definitions of the given set operations and how they are represented in a Venn diagram:

(A) $A \cup B$ (A union B): This set contains all elements that are in set A or in set B (or in both). In a Venn diagram, this represents the entire area covered by both circles A and B.

(B) $A \cap B$ (A intersection B): This set contains all elements that are common to both set A and set B. In a Venn diagram, this represents the overlapping region where the circles A and B intersect.

(C) $A - B$ (A minus B or Difference of A and B): This set contains all elements that are in set A but not in set B. In a Venn diagram, this represents the part of circle A that is outside of circle B. This is often referred to as the region of "A only".

(D) $B - A$ (B minus A or Difference of B and A): This set contains all elements that are in set B but not in set A. In a Venn diagram, this represents the part of circle B that is outside of circle A. This is often referred to as the region of "B only".

The shaded region in the provided Venn diagram corresponds exactly to the description of $A - B$: the part of circle A that does not overlap with circle B.

Comparing the shaded region with the descriptions of the options, we find that the shaded region represents $A - B$.

The final answer is $\boxed{C}$.

We are given the condition $A \cup B = A \cap B$ for two sets $A$ and $B$. We need to determine the relationship between $A$ and $B$ based on this condition.

Recall the definitions of union and intersection:

• $A \cup B = \{x : x \in A \text{ or } x \in B\}$ (The set of elements that are in $A$ or $B$ or both)
• $A \cap B = \{x : x \in A \text{ and } x \in B\}$ (The set of elements that are in both $A$ and $B$)

The condition $A \cup B = A \cap B$ means that the set of elements that are in $A$ or $B$ is exactly the same as the set of elements that are in both $A$ and $B$.

Let's consider an arbitrary element $x$.

If $x \in A$, then by the definition of union, $x \in A \cup B$.

Since $A \cup B = A \cap B$, it must be true that $x \in A \cap B$.

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

So, if $x \in A$, we conclude that $x \in B$. This means that every element of $A$ is also an element of $B$. Therefore, $A \subseteq B$.

Similarly, let's consider an arbitrary element $y$.

If $y \in B$, then by the definition of union, $y \in A \cup B$.

Since $A \cup B = A \cap B$, it must be true that $y \in A \cap B$.

By the definition of intersection, if $y \in A \cap B$, then $y \in A$ and $y \in B$.

So, if $y \in B$, we conclude that $y \in A$. This means that every element of $B$ is also an element of $A$. Therefore, $B \subseteq A$.

We have shown that if $A \cup B = A \cap B$, then $A \subseteq B$ and $B \subseteq A$.

When two sets $A$ and $B$ satisfy both $A \subseteq B$ and $B \subseteq A$, it implies that the sets have exactly the same elements. By the definition of set equality, this means $A = B$.

Alternatively, consider that $A \cap B$ is always a subset of both $A$ and $B$ (i.e., $A \cap B \subseteq A$ and $A \cap B \subseteq B$), and $A \cup B$ is always a superset of both $A$ and $B$ (i.e., $A \subseteq A \cup B$ and $B \subseteq A \cup B$).

If $A \cup B = A \cap B$, let this common set be $S$. So, $S = A \cup B$ and $S = A \cap B$.

Since $A \subseteq A \cup B$ and $A \cup B = S$, we have $A \subseteq S$. Since $S = A \cap B$ and $A \cap B \subseteq B$, we have $S \subseteq B$. Combining these, $A \subseteq S \subseteq B$, which implies $A \subseteq B$.

Similarly, since $B \subseteq A \cup B$ and $A \cup B = S$, we have $B \subseteq S$. Since $S = A \cap B$ and $A \cap B \subseteq A$, we have $S \subseteq A$. Combining these, $B \subseteq S \subseteq A$, which implies $B \subseteq A$.

Thus, $A \subseteq B$ and $B \subseteq A$, which implies $A=B$.

Let's compare the result with the given options:

(A) $A \subset B$: This means $A$ is a proper subset of $B$, i.e., $A \subseteq B$ and $A \neq B$. This is not necessarily true; equality ($A=B$) is possible.

(B) $B \subset A$: This means $B$ is a proper subset of $A$, i.e., $B \subseteq A$ and $B \neq A$. This is not necessarily true; equality ($A=B$) is possible.

(C) $A = B$: Our derivation shows that this is the required relationship.

(D) $A \cap B = \emptyset$: The condition is $A \cup B = A \cap B$. If $A \cap B = \emptyset$, then the condition becomes $A \cup B = \emptyset$. The union of two sets is the empty set only if both sets are empty. If $A=\emptyset$ and $B=\emptyset$, then $A \cup B = \emptyset$ and $A \cap B = \emptyset$, so $A \cup B = A \cap B$ holds. In this case, $A=B$. However, the condition $A \cup B = A \cap B$ does not imply that $A \cap B$ must be the empty set. For example, if $A=\{1\}$ and $B=\{1\}$, then $A \cup B = \{1\}$ and $A \cap B = \{1\}$. $A \cup B = A \cap B$ holds, and $A=B$, but $A \cap B = \{1\}$, which is not empty.

The correct conclusion from $A \cup B = A \cap B$ is that $A$ and $B$ must be the same set.

The final answer is $\boxed{C}$.

Let $H$ be the set of people who speak Hindi.

Let $E$ be the set of people who speak English.

Given:

Number of people who speak Hindi, $n(H) = 50$

Number of people who speak English, $n(E) = 20$

Number of people who speak both Hindi and English, $n(H \cap E) = 10$

To Find:

The number of people who speak at least one of these two languages, which is $n(H \cup E)$.

Solution:

We use the formula for the cardinality of the union of two finite sets:

Formula:

$n(H \cup E) = n(H) + n(E) - n(H \cap E)$

Substitute the given values into the formula:

$n(H \cup E) = 50 + 20 - 10$

$n(H \cup E) = 70 - 10$

$n(H \cup E) = 60$

So, 60 people speak at least one of these two languages.

Comparing the result with the given options:

(A) 60 - Correct

(B) 70 - Incorrect

(C) 80 - Incorrect

(D) 40 - Incorrect

The correct option is (A).

The final answer is $\boxed{A}$.

Let $A$ be a set with $n(A) = m$, where $n(A)$ represents the number of elements in set $A$.

The power set of a set $A$, denoted by $P(A)$, is the set of all possible subsets of $A$.

There is a fundamental theorem in set theory that states if a finite set $A$ has $m$ elements, then the number of subsets of $A$ (i.e., the number of elements in the power set of $A$) is $2$ raised to the power of $m$.

Mathematically, this can be expressed as:

Given that $n(A) = m$, we can substitute this value into the formula:

So, the number of elements in the power set of $A$ is $2^m$.

Let's compare this result with the given options:

(A) $m$ - Incorrect (This is the number of elements in the original set A)

(B) $m^2$ - Incorrect

(C) $2^m$ - Correct

(D) $2m$ - Incorrect

The correct option is (C).

The final answer is $\boxed{C}$.

An infinite set is a set that contains an endless number of elements, meaning its elements cannot be listed in a sequence that comes to an end. A finite set is one that has a countable number of elements, including the empty set.

Let's examine each option:

(A) The set of all days in a week.

The days in a week are Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday. This is a specific list of 7 elements.

This set is finite.

(B) The set of all citizens of India.

While the number of citizens of India is very large, it is a specific, countable number at any given point in time. The population is a finite number.

This set is finite.

(C) The set of all prime numbers.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The sequence of prime numbers begins $2, 3, 5, 7, 11, 13, 17, 19, 23, ...$. It has been proven that there are infinitely many prime numbers (Euclid's theorem).

This set is infinite.

(D) The set of all solutions to $x^2 = 1$ in $\mathbb{Z}$.

We need to find the integer values of $x$ that satisfy the equation $x^2 = 1$.

$x^2 = 1$

Taking the square root of both sides:

$x = \pm \sqrt{1}$

$x = 1$ or $x = -1$

Both $1$ and $-1$ are integers. So, the set of integer solutions is $\{-1, 1\}$. This set contains exactly two elements.

This set is finite.

Based on the analysis, the set of all prime numbers is an infinite set.

The final answer is $\boxed{C}$.

We are given two sets, $A = \{1, 2, 3\}$ and $B = \{x : x \text{ is a solution to } x^2 - 4x + 3 = 0\}$.

To compare the sets, we need to find the elements of set $B$. Set $B$ contains the solutions to the quadratic equation $x^2 - 4x + 3 = 0$.

We solve the equation $x^2 - 4x + 3 = 0$. We can factor the quadratic expression:

$x^2 - 4x + 3 = 0$

$(x-1)(x-3) = 0$

This gives the solutions $x-1=0$ or $x-3=0$, which are $x=1$ and $x=3$.

So, the set $B$ in roster form is $\{1, 3\}$.

Now we have set $A = \{1, 2, 3\}$ and set $B = \{1, 3\}$. Let's compare these two sets and evaluate the given statements.

(A) $A = B$

For two sets to be equal, they must have exactly the same elements. Set $A$ has elements 1, 2, 3. Set $B$ has elements 1, 3. Since set $A$ contains the element 2, which is not in set $B$, $A \neq B$. So, this statement is false.

(B) $A \subset B$

This statement claims that $A$ is a proper subset of $B$. For $A$ to be a proper subset of $B$, every element of $A$ must be in $B$, and $A$ must not be equal to $B$. The elements of $A$ are 1, 2, 3. The element 2 is in $A$ but is not in $B$. Therefore, $A$ is not a subset of $B$, and thus not a proper subset. So, this statement is false.

(C) $B \subset A$

This statement claims that $B$ is a proper subset of $A$. For $B$ to be a proper subset of $A$, every element of $B$ must be in $A$, and $B$ must not be equal to $A$. The elements of $B$ are 1, 3. Let's check if these are in $A = \{1, 2, 3\}$:

Since every element of $B$ is in $A$, $B \subseteq A$. Also, we already determined that $B \neq A$. Therefore, $B$ is a proper subset of $A$. So, this statement is true.

(D) $A \cap B = \emptyset$

This statement claims that the intersection of $A$ and $B$ is the empty set, meaning they have no elements in common. Let's find the intersection of $A = \{1, 2, 3\}$ and $B = \{1, 3\}$. The common elements are 1 and 3.

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

Since $A \cap B = \{1, 3\}$, which is not the empty set, this statement is false.

Based on the analysis, the statement $B \subset A$ is true.

The final answer is $\boxed{C}$.

We are asked to simplify the set expression $A \cap (A \cup B)$.

Let's consider the sets involved:

• $A$ is a set.
• $B$ is a set.
• $A \cup B$ is the union of sets $A$ and $B$, consisting of all elements that are in $A$ or $B$ (or both).

We know that for any sets $A$ and $B$, set $A$ is always a subset of the union of $A$ and $B$. This is because every element in $A$ is, by definition, also in $A \cup B$.

Now, consider the intersection $A \cap (A \cup B)$. The intersection of two sets consists of the elements that are common to both sets.

The elements common to set $A$ and set $(A \cup B)$ are precisely the elements that are in $A$ and also in $A \cup B$.

Since every element of $A$ is already in $A \cup B$ (because $A \subseteq A \cup B$), the elements common to both sets are simply all the elements of $A$.

Therefore, the intersection of $A$ with $A \cup B$ is equal to $A$.

This is a property often referred to as the Absorption Law for sets, specifically $A \cap (A \cup B) = A$ and its dual $A \cup (A \cap B) = A$.

Comparing the result with the given options:

(A) $A \cup B$ - Incorrect

(B) $A \cap B$ - Incorrect

(C) $A$ - Correct

(D) $B$ - Incorrect

The correct option is (C).

The final answer is $\boxed{C}$.

We need to identify which of the given notations does not represent a standard set of numbers commonly used in mathematics.

Let's examine each option and its standard notation:

(A) $\mathbb{N}$ (Natural Numbers)

$\mathbb{N}$ is the standard symbol used to represent the set of natural numbers. There are two common conventions for the definition of natural numbers: some include 0 ($\{0, 1, 2, 3, ...\}$), while others do not ($\{1, 2, 3, ...\}$). In either case, $\mathbb{N}$ is a standard notation for this set.

(B) $\mathbb{I}$ (Irrational Numbers)

Irrational numbers are real numbers that cannot be expressed as a simple fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Examples include $\sqrt{2}$, $\pi$, $e$. While the set of irrational numbers is a standard concept, the notation $\mathbb{I}$ is not universally recognized as the standard symbol for this set. The set of irrational numbers is often denoted as $\mathbb{R} \setminus \mathbb{Q}$ (the set of real numbers minus the set of rational numbers).

(C) $\mathbb{Z}$ (Integers)

$\mathbb{Z}$ is the standard symbol used to represent the set of integers. The set of integers includes all whole numbers (positive and negative) and zero: $\{..., -3, -2, -1, 0, 1, 2, 3, ...\}$.

(D) $\mathbb{F}$ (Fractions)

A fraction is a number that represents a part of a whole. While the concept of fractions is standard, the notation $\mathbb{F}$ is not a standard symbol for the set of all fractions. The set of all numbers that can be expressed as a fraction $\frac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q \neq 0$ is the set of rational numbers, which is denoted by the standard symbol $\mathbb{Q}$. The term 'fractions' can sometimes refer specifically to non-integer rational numbers or positive rational numbers, but $\mathbb{F}$ is not a standard symbol for any of these sets.

Among the given options, $\mathbb{N}$ and $\mathbb{Z}$ are standard notations. $\mathbb{I}$ is sometimes used for irrational numbers but is not as universally standard as $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$. $\mathbb{F}$ for the set of fractions is definitely not a standard notation; the standard notation for rational numbers is $\mathbb{Q}$.

The question asks which is NOT a standard set of numbers, implying its representation is not standard or the concept itself isn't a primary classification. The set of fractions is a concept, but $\mathbb{F}$ is not its standard symbol, unlike $\mathbb{N}$ and $\mathbb{Z}$. $\mathbb{I}$ is less standard than $\mathbb{N}$ and $\mathbb{Z}$, but more recognized than $\mathbb{F}$. Given the options, $\mathbb{F}$ is the least standard notation for a set of numbers.

The final answer is $\boxed{D}$.

We are given sets $A = \{a, b, c, d\}$ and $B = \{c, d, e, f\}$. We need to find the symmetric difference of A and B, denoted by $A \Delta B$, which is defined as $(A-B) \cup (B-A)$.

The universal set $U$ is the set of all letters in the English alphabet, but it is not directly needed for calculating $A \Delta B$ from $A$ and $B$.

First, let's find the set difference $A - B$. This set contains all elements that are in $A$ but not in $B$.

$A = \{a, b, c, d\}$

$B = \{c, d, e, f\}$

The elements in A that are not in B are 'a' and 'b'.

Next, let's find the set difference $B - A$. This set contains all elements that are in $B$ but not in $A$.

$A = \{a, b, c, d\}$

$B = \{c, d, e, f\}$

The elements in B that are not in A are 'e' and 'f'.

Finally, the symmetric difference $A \Delta B$ is the union of the set differences $(A-B)$ and $(B-A)$.

Substitute the results from the previous steps:

The symmetric difference of A and B is $\{a, b, e, f\}$.

Comparing the result with the given options:

(A) $\{a, b, e, f\}$ - Correct

(B) $\{c, d\}$ - Incorrect (This is $A \cap B$)

(C) $\{a, b, c, d, e, f\}$ - Incorrect (This is $A \cup B$)

(D) $\{g, h, i, ..., z\}$ - Incorrect (These are elements outside $A \cup B$, relative to U)

The correct option is (A).

The final answer is $\boxed{A}$.

The empty set is denoted by $\emptyset$ or \{\}. It is a set that contains no elements.

A subset of a set $A$ is a set $B$ such that every element in $B$ is also in $A$. Mathematically, $B \subseteq A$ if for all $x$, if $x \in B$, then $x \in A$.

Let's consider the empty set $\emptyset$. We want to find all sets $B$ such that $B \subseteq \emptyset$.

According to the definition of a subset, $B \subseteq \emptyset$ if for all $x$, if $x \in B$, then $x \in \emptyset$.

The statement "$x \in \emptyset$" is always false, because the empty set contains no elements.

For the implication "if $x \in B$, then $x \in \emptyset$" to be true, the premise "$x \in B$" must be false whenever the conclusion "$x \in \emptyset$" is false. Since "$x \in \emptyset$" is always false, "$x \in B$" must always be false for any element $x$. This means that there is no element $x$ such that $x \in B$. A set with no elements is the empty set itself.

Thus, the only set $B$ for which "$x \in B$ implies $x \in \emptyset$" is true is when $B$ is the empty set $\emptyset$.

The only subset of the empty set $\emptyset$ is the empty set itself.

Alternatively, we can use the formula for the number of subsets of a finite set. If a set $A$ has $m$ elements ($n(A)=m$), then the number of subsets of $A$ is $2^m$.

The empty set has 0 elements, so $n(\emptyset) = 0$.

Using the formula, the number of subsets of the empty set is $2^{n(\emptyset)} = 2^0$.

$2^0 = 1$.

So, the empty set has exactly 1 subset, which is the empty set itself.

Comparing the result with the given options:

(A) 0 - Incorrect

(B) 1 - Correct

(C) 2 - Incorrect

(D) Infinite - Incorrect

The correct option is (B).

The final answer is $\boxed{B}$.

We are given the following information about two finite sets $A$ and $B$:

To Find:

The cardinality of the intersection of A and B, $n(A \cap B)$.

Solution:

We use the formula relating the cardinalities of union and intersection of two sets:

Formula:

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

We need to find $n(A \cap B)$. We can rearrange the formula to solve for $n(A \cap B)$:

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

Substitute the given values into the formula:

$n(A \cap B) = 20 + 30 - 50$

$n(A \cap B) = 50 - 50$

$n(A \cap B) = 0$

So, the number of elements in the intersection of A and B is 0. This means that sets A and B are disjoint ($A \cap B = \emptyset$).

Comparing the result with the given options:

(A) 0 - Correct

(B) 10 - Incorrect

(C) 20 - Incorrect

(D) 50 - Incorrect

The correct option is (A).

The final answer is $\boxed{A}$.

A finite set is a set that has a countable number of elements. An infinite set is a set that contains an endless number of elements.

Let's examine each option:

(A) The set of all points on a line segment.

A line segment consists of infinitely many points. For any two distinct points on a line segment, there are infinitely many points between them. Therefore, the number of points on a line segment is uncountable and infinite.

This set is infinite.

(B) $\{x : x \in \mathbb{Z}, x < 100\}$

This set contains all integers $x$ that are less than 100. The integers less than 100 are $\{..., -3, -2, -1, 0, 1, 2, ..., 98, 99\}$. This set extends infinitely in the negative direction.

This set is infinite.

(C) The set of all rivers in India.

While the number of rivers in India is a large number and might change slightly over time due to geographical changes or definitions, it is a specific, countable number. We can, in principle, list all the rivers in India. Therefore, the number of rivers is finite.

This set is finite.

(D) $\{x : x \in \mathbb{R}, 0 < x < 1\}$

This set represents the open interval $(0, 1)$ on the real number line. It contains all real numbers strictly between 0 and 1. Between any two distinct real numbers, there are infinitely many other real numbers. Therefore, the number of real numbers in this interval is infinite and uncountable.

This set is infinite.

Based on the analysis, only the set of all rivers in India is a finite set.

The final answer is $\boxed{C}$.

The given set is $A = \{1, 2, 3\}$. The total number of elements in set $A$ is $n(A) = 3$.

The total number of subsets of a set with $m$ elements is $2^m$. So, the total number of subsets of $A$ is $2^3 = 8$.

These subsets are:

$\emptyset$

$\{1\}$, $\{2\}$, $\{3\}$

$\{1, 2\}$, $\{1, 3\}$, $\{2, 3\}$

$\{1, 2, 3\}$

We need to find the number of subsets of $A$ that contain the element $1$. Let's look at the list of all subsets and identify those that include the element $1$:

• $\emptyset$ - Does not contain 1.
• $\{1\}$ - Contains 1.
• $\{2\}$ - Does not contain 1.
• $\{3\}$ - Does not contain 1.
• $\{1, 2\}$ - Contains 1.
• $\{1, 3\}$ - Contains 1.
• $\{2, 3\}$ - Does not contain 1.
• $\{1, 2, 3\}$ - Contains 1.

The subsets of $A$ that contain the element 1 are $\{1\}$, $\{1, 2\}$, $\{1, 3\}$, and $\{1, 2, 3\}$.

There are 4 such subsets.

Alternatively, a subset of $A$ that contains the element 1 must include 1. The other elements of the subset can be any combination of the remaining elements in $A$, which are $\{2, 3\}$. The subsets of $\{2, 3\}$ are $\emptyset$, $\{2\}$, $\{3\}$, and $\{2, 3\}$. To form a subset of $A$ containing 1, we take 1 and unite it with each subset of $\{2, 3\}$.

• $1 \cup \emptyset = \{1\}$
• $1 \cup \{2\} = \{1, 2\}$
• $1 \cup \{3\} = \{1, 3\}$
• $1 \cup \{2, 3\} = \{1, 2, 3\}$

The number of subsets of $\{2, 3\}$ is $2^{n(\{2, 3\})} = 2^2 = 4$. Each of these subsets, when combined with the element 1, forms a unique subset of $A$ containing 1. Thus, there are 4 such subsets.

In general, for a set with $n$ elements, the number of subsets containing a specific element is $2^{n-1}$. In this case, $n=3$, and the specific element is 1. The number of subsets containing 1 is $2^{3-1} = 2^2 = 4$.

Comparing the result with the given options:

(A) 2 - Incorrect

(B) 3 - Incorrect

(C) 4 - Correct

(D) 8 - Incorrect (This is the total number of subsets)

The correct option is (C).

The final answer is $\boxed{C}$.

We are given data about the number of students in different marks ranges, represented by tally marks.

Let's first determine the number of students in each range:

• Marks Range 0 - 20: $||||$ represents 4 students.
• Marks Range 20 - 40: $\bcancel{||||}$ $|||$ represents $5 + 3 = 8$ students.
• Marks Range 40 - 60: $\bcancel{||||}$ $\bcancel{||||}$ $||$ represents $5 + 5 + 2 = 12$ students.
• Marks Range 60 - 80: $\bcancel{||||}$ represents 5 students.
• Marks Range 80 - 100: $|||$ represents 3 students.

We are given that $A$ is the set of students who scored marks in the range $40 - 60$. From the table, the number of students in this range is 12.

We are given that $B$ is the set of students who scored marks in the range $60 - 80$. From the table, the number of students in this range is 5.

We need to find $n(A \cup B)$, which is the number of students who scored marks in the range $40 - 60$ or in the range $60 - 80$.

Since the marks ranges $40 - 60$ and $60 - 80$ are mutually exclusive (a student cannot score marks in both ranges simultaneously), the sets $A$ and $B$ are disjoint, meaning their intersection is the empty set ($A \cap B = \emptyset$).

The number of students who scored in both ranges is 0.

We can use the formula for the cardinality of the union of two sets:

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

Substitute the values:

$n(A \cup B) = 12 + 5 - 0$

$n(A \cup B) = 17$

Alternatively, since the sets A and B are disjoint, the number of elements in their union is simply the sum of the number of elements in each set:

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

$n(A \cup B) = 12 + 5 = 17$

Both methods give the same result.

Comparing the result with the given options:

(A) 17 - Correct

(B) 12 - Incorrect

(C) 15 - Incorrect

(D) 22 - Incorrect

The correct option is (A).

The final answer is $\boxed{A}$.

We need to identify the incorrect statement about the standard interval notation for subsets of real numbers.

Let's review the standard definitions of interval notation for real numbers, where $a$ and $b$ are real numbers with $a < b$:

• Closed Interval: $[a, b] = \{x \in \mathbb{R} : a \leq x \leq b\}$. This interval includes both endpoints $a$ and $b$.
• Open Interval: $(a, b) = \{x \in \mathbb{R} : a < x < b\}$. This interval excludes both endpoints $a$ and $b$.
• Half-Open (or Half-Closed) Intervals:
  • $[a, b) = \{x \in \mathbb{R} : a \leq x < b\}$. This interval includes the left endpoint $a$ but excludes the right endpoint $b$.
  • $(a, b] = \{x \in \mathbb{R} : a < x \leq b\}$. This interval excludes the left endpoint $a$ but includes the right endpoint $b$.

Now let's examine each given statement based on these standard definitions:

(A) $\{x \in \mathbb{R} : a \leq x \leq b\}$ is $[a, b]$.

This matches the definition of a closed interval. This statement is correct.

(B) $\{x \in \mathbb{R} : a < x < b\}$ is $(a, b)$.

This matches the definition of an open interval. This statement is correct.

(C) $\{x \in \mathbb{R} : a \leq x < b\}$ is $[a, b)$.

This matches the definition of a half-open interval where the left endpoint is included and the right endpoint is excluded. This statement is correct.

(D) $\{x \in \mathbb{R} : a < x \leq b\}$ is $(a, b]$.

This matches the definition of a half-open interval where the left endpoint is excluded and the right endpoint is included. This statement is correct.

All four statements appear to be correct according to standard interval notation. There might be an error in the question or options as presented if only one is intended to be incorrect.

Let's re-examine the question and options carefully to see if there is any subtle detail or a common misunderstanding being tested.

The question asks which statement is incorrect. Since all provided definitions seem standard and correct, let's double-check if there's any ambiguity in the notation itself.

Assuming the standard definitions as listed above, all four options are correct statements about interval notation.

However, given that this is a multiple-choice question asking for the *incorrect* statement, and all statements seem correct, there is likely an error in the question or the intended options. Without a flawed statement among the choices provided based on standard mathematical definitions, it's impossible to select an incorrect one.

Let's assume there might be a typo in one of the options, or a non-standard definition being used in the source material this question is from. If we are forced to choose, and assuming the question is valid, there must be a mistake in one of the provided notations or set descriptions. Since the provided statements (A), (B), (C), and (D) are all standard and correct definitions of intervals, there seems to be an issue with the question itself.

Assuming there's no typo and the question is valid, this would imply a different set of interval notation rules is being used or one of the descriptions is fundamentally wrong in context not provided. Based on standard mathematics, all statements are correct.

Given the need to provide an answer based on the provided format, and acknowledging the potential issue with the question, I cannot definitively identify an incorrect statement from the given options as they are written according to standard interval notation.

However, if we assume a potential typo or non-standard usage, one might speculate. For instance, if option (C) was written as $[a, b)$ but described $\{x \in \mathbb{R} : a < x < b\}$, that would be incorrect. Or if option (B) used square brackets, e.g., $\{x \in \mathbb{R} : a < x < b\}$ is $[a, b]$, that would be incorrect.

Since no such obvious error is present, and assuming standard notation, I cannot provide an incorrect option. I will state that based on standard notation, all are correct.

Let's assume there must be an error in the question and one of the options is indeed incorrect. Without knowing the intended error, I cannot proceed further to select the designated answer.

If we were forced to guess which one might be intended as incorrect in a poorly constructed question, sometimes confusion arises with half-open intervals, but options C and D are standard. Option A and B are also standard. There is no mathematical basis to select any of these as incorrect based on their description and notation.

Since I am required to provide an answer from the options, and all appear correct, I cannot fulfill the prompt's requirement to identify the *incorrect* one. This indicates a problem with the input question itself.

Given the constraints, I cannot provide a valid answer by selecting an incorrect option from the provided list, as all appear correct in standard set theory and interval notation.

I am unable to proceed with selecting an incorrect option as per the standard definitions of interval notation.

I must return one of the options A, B, C, or D. Without a genuine incorrect statement, I cannot confidently choose. Let me reconsider if there's any way to interpret the question or options differently.

The question asks "Which of the following interval notations is incorrect?". It is pairing a set-builder notation definition with an interval notation. We have checked that the pairings are correct according to standard definitions.

It seems there is an error in the question itself, making it impossible to select an incorrect statement from the given correct statements.

Assuming there is a single correct answer among the options provided in the source material (meaning one of the statements is indeed incorrect in that context), I cannot determine it without the correct version of the question or options.

However, if I must select one option, and given the standard nature of A, B, C, and D, there is no logical choice for an incorrect one. I cannot proceed to definitively answer based on the provided input.

I cannot answer this question as all listed statements appear to be correct representations of interval notation.

Note: As all options (A), (B), (C), and (D) appear to correctly match the set-builder notation with the standard interval notation, there seems to be an error in the question which asks for the "incorrect" statement. Based on standard mathematical definitions, none of the provided statements are incorrect.

Unable to provide a definitive answer as the question appears flawed.

We are given the condition $A \cup B = A$ for two sets $A$ and $B$. We need to determine which of the given statements is necessarily true based on this condition.

Recall the definition of union:

$A \cup B = \{x : x \in A \text{ or } x \in B\}$

The condition $A \cup B = A$ means that the set of elements that are in $A$ or $B$ is exactly the same as the set of elements in $A$.

Let's consider an arbitrary element $x$.

If $x \in B$, then by the definition of union, $x \in A \cup B$.

Since $A \cup B = A$, it must be true that $x \in A$.

So, if $x \in B$, then $x \in A$. This means that every element of $B$ is also an element of $A$.

By the definition of a subset, this implies that $B$ is a subset of $A$.

Let's verify this by checking the other direction. If $x \in A$, then $x \in A$ is true. By the definition of union, if $x \in A$, then "$x \in A$ or $x \in B$" is true (as the first part is true). So, $x \in A \cup B$. Since we are given $A \cup B = A$, $x \in A \cup B$ implies $x \in A$, which is consistent.

The condition $A \cup B = A$ necessarily implies that $B \subseteq A$.

Let's compare this result with the given options:

(A) $B \subseteq A$

Our derivation shows that this is necessarily true. For example, if $A=\{1, 2, 3\}$ and $B=\{1, 2\}$, then $A \cup B = \{1, 2, 3\} \cup \{1, 2\} = \{1, 2, 3\} = A$, and $B \subseteq A$ holds.

(B) $A \subseteq B$

This is not necessarily true. In the example above, $A=\{1, 2, 3\}$ and $B=\{1, 2\}$, $A \cup B = A$, but $A$ is not a subset of $B$.

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

This is not necessarily true. In the example above, $A=\{1, 2, 3\}$ and $B=\{1, 2\}$, $A \cup B = A$, but $A \cap B = \{1, 2\} \neq \emptyset$. This is only true if $B = \emptyset$, because if $B=\emptyset$, $A \cup \emptyset = A$, and $A \cap \emptyset = \emptyset$. So $A = \emptyset$.

(D) $A = \emptyset$

This is not necessarily true. In the example above, $A=\{1, 2, 3\}$, which is not empty, and $A \cup B = A$ holds. If $A = \emptyset$, then $\emptyset \cup B = \emptyset$, which implies $B = \emptyset$. So if $A = \emptyset$, then $B$ must also be $\emptyset$, and the condition $A \cup B = A$ holds ($\emptyset \cup \emptyset = \emptyset$). So, $A=\emptyset$ is a possible case, but not necessarily true for all cases where $A \cup B = A$.

The only statement that is necessarily true when $A \cup B = A$ is $B \subseteq A$.

The final answer is $\boxed{A}$.

We are given the universal set $U = \{1, 2, 3, 4, 5\}$, set $A = \{1, 2\}$, and set $B = \{2, 3\}$.

We need to find $(A \cap B)'$, which represents the complement of the intersection of sets A and B with respect to the universal set U.

First, let's find the intersection of sets A and B, denoted by $A \cap B$. The intersection contains all elements that are common to both sets.

$A = \{1, 2\}$

$B = \{2, 3\}$

The element common to both A and B is 2.

Next, let's find the complement of $A \cap B$ with respect to the universal set $U$, denoted by $(A \cap B)'$. The complement contains all elements in $U$ that are not in $A \cap B$.

$U = \{1, 2, 3, 4, 5\}$

$A \cap B = \{2\}$

We look for elements in $\{1, 2, 3, 4, 5\}$ that are not in $\{2\}$.

• Is $1 \in U$ and $1 \notin \{2\}$? Yes.
• Is $2 \in U$ and $2 \notin \{2\}$? No, $2 \in \{2\}$.
• Is $3 \in U$ and $3 \notin \{2\}$? Yes.
• Is $4 \in U$ and $4 \notin \{2\}$? Yes.
• Is $5 \in U$ and $5 \notin \{2\}$? Yes.

The elements in $U$ that are not in $A \cap B$ are $1, 3, 4, 5$.

Now, let's compare this result with the given options:

(A) $\{2\}$ - Incorrect (This is $A \cap B$)

(B) $\{1, 3, 4, 5\}$ - Correct

(C) $\{1, 2, 3\}$ - Incorrect (This is $A \cup B$)

(D) $\{4, 5\}$ - Incorrect

The correct option is (B).

The final answer is $\boxed{B}$.

We are given the set identity $A \cup B = B \cup A$. We need to identify the property illustrated by this identity.

Let's consider the definitions of the properties mentioned in the options:

(A) Associative Law: The associative law deals with grouping of operations involving three or more sets. For union, the associative law states $(A \cup B) \cup C = A \cup (B \cup C)$. For intersection, it states $(A \cap B) \cap C = A \cap (B \cap C)$. The given identity involves only two sets and changes the order of the sets in the union operation, not the grouping.

(B) Commutative Law: The commutative law states that the order of the operands does not affect the result of the operation. For set union, the commutative law states $A \cup B = B \cup A$. For set intersection, it states $A \cap B = B \cap A$. The given identity shows that the order of sets in the union operation can be swapped without changing the resulting set.

(C) Distributive Law: The distributive law relates two different operations. For sets, there are two distributive laws: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ (union over intersection) and $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ (intersection over union). The given identity only involves the union operation and does not relate two different operations.

(D) Idempotent Law: The idempotent law states that performing the operation on a set with itself yields the same set. For union, the idempotent law states $A \cup A = A$. For intersection, it states $A \cap A = A$. The given identity involves two potentially different sets $A$ and $B$, not a set with itself.

The identity $A \cup B = B \cup A$ shows that the union operation is commutative, meaning the order of the sets being united can be switched.

Therefore, the property illustrated by $A \cup B = B \cup A$ is the Commutative Law for Union.

The final answer is $\boxed{B}$.

We are given two sets, $A = \{x : x^2 = 4, x \in \mathbb{Z}\}$ and $B = \{2\}$.

To compare the sets, we need to find the elements of set $A$. Set $A$ contains integer solutions to the equation $x^2 = 4$.

We solve the equation $x^2 = 4$ for integers $x$.

$x^2 = 4$

Taking the square root of both sides, we get:

$x = \pm \sqrt{4}$

$x = 2$ or $x = -2$

Both $2$ and $-2$ are integers. So, the set $A$ in roster form is $\{-2, 2\}$.

Now we have set $A = \{-2, 2\}$ and set $B = \{2\}$. Let's compare these two sets and evaluate the given statements.

(A) $A = B$

For two sets to be equal, they must have exactly the same elements. Set $A$ has elements -2 and 2. Set $B$ has element 2. Since set $A$ contains the element -2, which is not in set $B$, $A \neq B$. So, this statement is false.

(B) $A \subset B$

This statement claims that $A$ is a proper subset of $B$. For $A$ to be a proper subset of $B$, every element of $A$ must be in $B$, and $A$ must not be equal to $B$. The elements of $A$ are -2 and 2. The element -2 is in $A$ but is not in $B$. Therefore, $A$ is not a subset of $B$, and thus not a proper subset. So, this statement is false.

(C) $B \subset A$

This statement claims that $B$ is a proper subset of $A$. For $B$ to be a proper subset of $A$, every element of $B$ must be in $A$, and $B$ must not be equal to $A$. The element of $B$ is 2. Let's check if 2 is in $A = \{-2, 2\}$:

Since every element of $B$ is in $A$, $B \subseteq A$. Also, we already determined that $B \neq A$. Therefore, $B$ is a proper subset of $A$. So, this statement is true.

(D) $A$ and $B$ are disjoint sets

Two sets are disjoint if their intersection is the empty set ($A \cap B = \emptyset$). Let's find the intersection of $A = \{-2, 2\}$ and $B = \{2\}$. The common element is 2.

$A \cap B = \{-2, 2\} \cap \{2\} = \{2\}$

Since $A \cap B = \{2\}$, which is not the empty set, $A$ and $B$ are not disjoint sets. So, this statement is false.

Based on the analysis, the statement $B \subset A$ is true.

The final answer is $\boxed{C}$.

We are given the amounts deposited by 25 students. The amounts are $\textsf{₹} 5$, $\textsf{₹} 10$, and $\textsf{₹} 20$.

Let $A$ be the set of students who deposited $\textsf{₹} 5$. The number of elements in set $A$, $n(A)$, is the count of times $\textsf{₹} 5$ appears in the data.

Let $B$ be the set of students who deposited $\textsf{₹} 10$. The number of elements in set $B$, $n(B)$, is the count of times $\textsf{₹} 10$ appears in the data.

Let $C$ be the set of students who deposited $\textsf{₹} 20$. The number of elements in set $C$, $n(C)$, is the count of times $\textsf{₹} 20$ appears in the data.

Let's count the occurrences of each amount in the table:

Amount $\textsf{₹} 5$: Count the number of 5s.

Table data: 10, 5, 20, 10, 5, 10, 20, 5, 10, 5, 20, 10, 5, 10, 20, 5, 10, 5, 20, 10, 5, 10, 20, 5, 10

Count of 5s: 5, 5, 5, 5, 5, 5, 5, 5, 5, 5. There are 10 occurrences of 5.

Amount $\textsf{₹} 10$: Count the number of 10s.

Table data: 10, 5, 20, 10, 5, 10, 20, 5, 10, 5, 20, 10, 5, 10, 20, 5, 10, 5, 20, 10, 5, 10, 20, 5, 10

Count of 10s: 10, 10, 10, 10, 10, 10, 10, 10, 10, 10. There are 10 occurrences of 10.

Amount $\textsf{₹} 20$: Count the number of 20s.

Table data: 10, 5, 20, 10, 5, 10, 20, 5, 10, 5, 20, 10, 5, 10, 20, 5, 10, 5, 20, 10, 5, 10, 20, 5, 10

Count of 20s: 20, 20, 20, 20, 20. There are 5 occurrences of 20.

We are asked to find $n(A) + n(B) + n(C)$.

The sum of the number of students who deposited each specific amount is 25. This makes sense because each of the 25 students deposited exactly one of these amounts, and the categories (depositing $\textsf{₹} 5$, $\textsf{₹} 10$, or $\textsf{₹} 20$) are mutually exclusive for a single deposit. The sets $A$, $B$, and $C$ are disjoint, and their union covers all 25 students, so $n(A \cup B \cup C) = n(A) + n(B) + n(C) = 25$.

Comparing the result with the given options:

(A) 25 - Correct

(B) 30 - Incorrect

(C) 20 - Incorrect

(D) 15 - Incorrect

The correct option is (A).

The final answer is $\boxed{A}$.

We are given the condition $A - B = A$ for two sets $A$ and $B$. We need to determine which of the given statements is necessarily true based on this condition.

Recall the definition of set difference:

$A - B = \{x : x \in A \text{ and } x \notin B\}$

The set $A - B$ consists of elements that are in $A$ but not in $B$.

The condition $A - B = A$ means that the set of elements that are in $A$ but not in $B$ is exactly the same as the set of elements in $A$.

Let's consider an arbitrary element $x$.

If $x \in A$, then since $A = A - B$, it must be true that $x \in A - B$.

By the definition of set difference, if $x \in A - B$, then $x \in A$ and $x \notin B$.

So, if $x \in A$, we conclude that $x \notin B$. This means that no element of $A$ is also an element of $B$.

By the definition of intersection, if there are no elements common to $A$ and $B$, their intersection is the empty set.

Conversely, if $A \cap B = \emptyset$, let's check if $A - B = A$.

$A - B = \{x : x \in A \text{ and } x \notin B\}$.

Since $A \cap B = \emptyset$, there is no element $x$ such that $x \in A$ and $x \in B$. This means that if $x \in A$, then it is automatically true that $x \notin B$.

So, the condition "$x \in A$ and $x \notin B$" is equivalent to just "$x \in A$".

Therefore, $A - B = \{x : x \in A\} = A$.

The condition $A - B = A$ is equivalent to $A \cap B = \emptyset$.

Let's compare this result with the given options:

(A) $A \subseteq B$

This is not necessarily true. If $A=\{1\}$ and $B=\{2\}$, then $A-B = \{1\}=A$. Here $A \cap B = \emptyset$, and $A \subseteq B$ is false.

(B) $B \subseteq A$

This is not necessarily true. If $A=\{1\}$ and $B=\{2\}$, then $A-B = \{1\}=A$. Here $A \cap B = \emptyset$, and $B \subseteq A$ is false.

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

Our derivation shows that this is necessarily true.

(D) $B = \emptyset$

This is not necessarily true. If $A=\{1\}$ and $B=\{2\}$, then $A-B = \{1\}=A$. Here $A \cap B = \emptyset$, but $B = \{2\} \neq \emptyset$. However, if $B = \emptyset$, then $A - \emptyset = A$, so the condition holds. So, $B=\emptyset$ is a possible case, but not necessarily true for all cases where $A - B = A$.

The only statement that is necessarily true when $A - B = A$ is $A \cap B = \emptyset$.

The final answer is $\boxed{C}$.

We need to evaluate the correctness of each given statement about sets.

(I) Every set is a subset of itself.

Let $A$ be any set. A set $B$ is a subset of $A$ ($B \subseteq A$) if every element in $B$ is also in $A$. To check if $A$ is a subset of itself ($A \subseteq A$), we need to check if every element in $A$ is also in $A$. This is clearly true: if $x \in A$, then $x \in A$. Therefore, every set is a subset of itself.

Statement (I) is correct.

(II) The empty set is a subset of every set.

The empty set is denoted by $\emptyset$. We need to check if $\emptyset \subseteq A$ for any set $A$. According to the definition of a subset, $\emptyset \subseteq A$ if for all $x$, if $x \in \emptyset$, then $x \in A$. The statement "$x \in \emptyset$" is always false, because the empty set contains no elements. In logic, a conditional statement "if P, then Q" is true whenever the premise P is false, regardless of the truth value of Q. Since the premise "$x \in \emptyset$" is always false, the implication "if $x \in \emptyset$, then $x \in A$" is always true for any element $x$ and any set $A$. Therefore, the empty set is a subset of every set.

Statement (II) is correct.

(III) If $A \subset B$ and $B \subset C$, then $A \subset C$.

The notation $A \subset B$ means that $A$ is a proper subset of $B$. By definition, $A \subset B$ means $A \subseteq B$ and $A \neq B$.

We are given $A \subset B$, which implies $A \subseteq B$.

We are given $B \subset C$, which implies $B \subseteq C$.

Since $A \subseteq B$ and $B \subseteq C$, by the transitivity property of subsets, it follows that $A \subseteq C$.

Now we need to check if $A \neq C$.

Since $A \subset B$, there exists at least one element in $B$ that is not in $A$. Let this element be $y$, so $y \in B$ and $y \notin A$.

Since $B \subset C$, there exists at least one element in $C$ that is not in $B$. Let this element be $z$, so $z \in C$ and $z \notin B$.

Consider the relationship between $A$ and $C$. We know $A \subseteq C$. Could $A = C$? If $A=C$, then every element of $C$ must be an element of $A$. However, since $B \subset C$, there is an element $z \in C$ such that $z \notin B$. Since $A \subseteq B$, if $z \in A$, it must also be in $B$. But $z \notin B$. Therefore, $z \notin A$. So, there is an element $z \in C$ such that $z \notin A$. This means $C$ is not a subset of $A$. Since $A \subseteq C$ and $C \not\subseteq A$, $A$ cannot be equal to $C$.

Alternatively, since $A \subset B$, $A$ has strictly fewer elements than $B$ if the sets are finite, or there's an element in B not in A if infinite. Since $B \subset C$, $B$ has strictly fewer elements than $C$ (or an element in C not in B). If $A \subset B$, then $n(A) < n(B)$ for finite sets. If $B \subset C$, then $n(B) < n(C)$ for finite sets. Thus, $n(A) < n(B) < n(C)$, which implies $n(A) < n(C)$, so $A \neq C$. Since $A \subseteq C$ and $A \neq C$, we have $A \subset C$. This transitive property holds for proper subsets.

Statement (III) is correct.

All three statements (I), (II), and (III) are correct statements about sets.

Comparing the result with the given options:

(A) (I) and (II) only - Incorrect

(B) (I) and (III) only - Incorrect

(C) (II) and (III) only - Incorrect

(D) (I), (II), and (III) - Correct

The correct option is (D).

The final answer is $\boxed{D}$.

We are given two intervals on the real number line:

We need to find the intersection of these two intervals, $A \cap B$. The intersection contains all elements that are common to both sets $A$ and $B$.

Substituting the definitions of $A$ and $B$:

For a real number $x$ to be in the intersection, it must satisfy both inequalities simultaneously:

Combining these two inequalities, we need to find the values of $x$ that satisfy both $2 \leq x$, $x < 5$, $4 < x$, and $x \leq 7$.

From $2 \leq x$ and $4 < x$, the stronger condition is $4 < x$.

From $x < 5$ and $x \leq 7$, the stronger condition is $x < 5$.

So, the combined inequality is $4 < x < 5$.

The set of real numbers $x$ such that $4 < x < 5$ is represented by the open interval $(4, 5)$.

We can visualize this on a number line:

Interval A: [---) from 2 up to 5 (5 is not included)

Interval B: (---] from 4 up to 7 (4 is not included)

The overlap is the region where both intervals exist. This region starts just after 4 and ends just before 5.

It includes numbers greater than 4 and less than 5.

For example, $4.1$ is in A and in B. $4.9$ is in A and in B.

The endpoint 4 is not in B, so it cannot be in the intersection. The endpoint 5 is not in A, so it cannot be in the intersection.

Thus, the intersection is the open interval $(4, 5)$.

Comparing the result with the given options:

(A) $[2, 7]$ - Incorrect (This is the union $A \cup B$ if adjusted for endpoints)

(B) $(4, 5)$ - Correct

(C) $[2, 4]$ - Incorrect

(D) $[5, 7]$ - Incorrect

The correct option is (B).

The final answer is $\boxed{B}$.

We are given the cardinalities of sets $A$, $B$, and their intersection:

We need to find the cardinality of the symmetric difference $A \Delta B$, which is defined as $(A-B) \cup (B-A)$.

The symmetric difference $A \Delta B$ consists of elements that are in $A$ or in $B$, but not in their intersection. In other words, it is the set of elements that belong to exactly one of the two sets $A$ and $B$.

We can find the cardinality of the set differences $A-B$ and $B-A$.

$A - B$ contains elements in $A$ that are not in $B$. The number of such elements is the total number of elements in $A$ minus the number of elements that are in both $A$ and $B$.

$B - A$ contains elements in $B$ that are not in $A$. The number of such elements is the total number of elements in $B$ minus the number of elements that are in both $A$ and $B$.

The symmetric difference is the union of $A-B$ and $B-A$. The sets $A-B$ and $B-A$ are always disjoint (an element cannot be in $A$ and not in $B$ and also in $B$ and not in $A$). Therefore, the cardinality of their union is simply the sum of their cardinalities.

Substitute the expressions for $n(A-B)$ and $n(B-A)$:

Alternatively, the symmetric difference can also be expressed as the union minus the intersection:

The sets $(A \cup B)$ and $(A \cap B)$ are not necessarily disjoint. However, $A \cap B$ is always a subset of $A \cup B$. The cardinality of the difference of a set $X$ and its subset $Y$ ($Y \subseteq X$) is $n(X - Y) = n(X) - n(Y)$.

We know the formula for the union: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

Substitute the given values:

Now substitute this into the formula for $n(A \Delta B)$:

Both methods yield the same result.

Comparing the result with the given options:

(A) $p + q - r$ - Incorrect (This is $n(A \cup B)$)

(B) $p + q - 2r$ - Correct

(C) $p + q + r$ - Incorrect

(D) $p - q + r$ - Incorrect

The correct option is (B).

The final answer is $\boxed{B}$.

We are given a set of real numbers described in set-builder notation: $\{x \in \mathbb{R} : 5 < x \leq 10\}$.

This set includes all real numbers $x$ such that $x$ is strictly greater than 5 and less than or equal to 10.

We need to represent this set using interval notation.

The standard interval notations are as follows, for real numbers $a < b$:

• $[a, b]$: includes $a$ and $b$, i.e., $a \leq x \leq b$.
• $(a, b)$: excludes $a$ and $b$, i.e., $a < x < b$.
• $[a, b)$: includes $a$ but excludes $b$, i.e., $a \leq x < b$.
• $(a, b]$: excludes $a$ but includes $b$, i.e., $a < x \leq b$.

The given set is $\{x \in \mathbb{R} : 5 < x \leq 10\}$.

Comparing the inequality $5 < x \leq 10$ with the forms above:

• The left boundary is 5, and the inequality $5 < x$ means 5 is excluded. An excluded endpoint is indicated by a parenthesis '('.
• The right boundary is 10, and the inequality $x \leq 10$ means 10 is included. An included endpoint is indicated by a square bracket ']'.

Combining these, the interval starts with an excluded 5 and ends with an included 10.

The interval notation is $(5, 10]$.

Comparing the result with the given options:

(A) $[5, 10]$ - Incorrect (Includes 5, which is excluded in the set)

(B) $(5, 10)$ - Incorrect (Excludes 10, which is included in the set)

(C) $[5, 10)$ - Incorrect (Includes 5 and excludes 10)

(D) $(5, 10]$ - Correct (Excludes 5 and includes 10)

The correct option is (D).

The final answer is $\boxed{D}$.

We are given the set $S = \{a, b, c, d\}$.

The number of elements in set $S$ is $n(S) = 4$.

Two sets are equivalent if they have the same number of elements (same cardinality). So, a set equivalent to $S$ must also have 4 elements.

Two sets are equal if they have exactly the same elements. If a set is equal to $S$, it must contain exactly the elements $a, b, c, d$ and no others.

We are looking for a set that is equivalent to $S$ ($n(\text{set})=4$) but not equal to $S$ (does not contain exactly $a, b, c, d$).

Let's examine each option:

(A) $\{a, b, c, d\}$

This set has elements $a, b, c, d$. It has 4 elements, so it is equivalent to $S$. The elements are exactly the same as in $S$. So, this set is equal to $S$. This option is not equivalent but not equal; it is equivalent and equal.

(B) $\{1, 2, 3, 4\}$

This set has elements 1, 2, 3, 4. It has 4 elements, so $n(\{1, 2, 3, 4\}) = 4$. Thus, this set is equivalent to $S$. The elements of this set are numbers, while the elements of $S$ are letters. The sets do not have the same elements. So, this set is not equal to $S$. This option is equivalent but not equal, which matches the requirement.

(C) $\{a, a, b, c, d\}$

In set notation, repeated elements are counted only once. The unique elements in this set are $a, b, c, d$. So, this set is $\{a, b, c, d\}$. It has 4 unique elements, so $n(\{a, a, b, c, d\}) = 4$. Thus, this set is equivalent to $S$. The elements are exactly the same as in $S$. So, this set is equal to $S$. This option is not equivalent but not equal; it is equivalent and equal.

(D) $\{a, b, c\}$

This set has elements $a, b, c$. It has 3 elements, so $n(\{a, b, c\}) = 3$. The number of elements (3) is not equal to $n(S)$ (4). So, this set is not equivalent to $S$. Consequently, it cannot be equal to $S$ either. This option is neither equivalent nor equal to $S$.

Based on the analysis, the set $\{1, 2, 3, 4\}$ is equivalent to $\{a, b, c, d\}$ (both have 4 elements) but is not equal to $\{a, b, c, d\}$ (they do not have the same elements).

The correct option is (B).

The final answer is $\boxed{B}$.

Let $U$ be the universal set and $A$ be any set.

The complement of set $A$ with respect to the universal set $U$, denoted by $A'$, is the set of all elements in $U$ that are not in $A$. In set-builder notation:

We are asked to find the union of set $A$ and its complement $A'$, denoted by $A \cup A'$. The union of two sets contains all elements that are in either set (or both).

Substituting the definition of $A'$:

Consider an arbitrary element $x$ from the universal set $U$. For any element $x \in U$, there are only two possibilities regarding set $A$: either $x$ is in $A$ (i.e., $x \in A$) or $x$ is not in $A$ (i.e., $x \notin A$).

• If $x \in A$, then $x$ is in the first part of the union ($x \in A$).
• If $x \notin A$, and assuming $x \in U$, then $x \in A'$ by definition ($x \in U$ and $x \notin A$). In this case, $x$ is in the second part of the union ($x \in A'$).

Since any element $x$ in the universal set $U$ must either be in $A$ or in $A'$, every element of $U$ is in $A \cup A'$. This means $U \subseteq A \cup A'$.

Also, by definition, both $A$ and $A'$ are subsets of the universal set $U$. The union of subsets of $U$ must also be a subset of $U$. So, $A \cup A' \subseteq U$.

Since $U \subseteq A \cup A'$ and $A \cup A' \subseteq U$, we can conclude that $A \cup A' = U$.

This is a fundamental property of set complements: the union of a set and its complement is the universal set. This is sometimes called the Complement Law or the Law of the Excluded Middle (in set theory context).

Comparing the result with the given options:

(A) $\emptyset$ - Incorrect (The empty set contains no elements)

(B) $A$ - Incorrect (Unless $A = U$, which is not always true)

(C) $A'$ - Incorrect (Unless $A = \emptyset$, which is not always true)

(D) $U$ - Correct

The correct option is (D).

The final answer is $\boxed{D}$.

Let $U$ be the set of all students in the class.

Let $P$ be the set of students who like Physics.

Let $C$ be the set of students who like Chemistry.

Given:

Total number of students, $n(U) = 50$

Number of students who like Physics, $n(P) = 30$

Number of students who like Chemistry, $n(C) = 25$

Number of students who like both Physics and Chemistry, $n(P \cap C) = 10$

To Find:

The number of students who like neither Physics nor Chemistry.

This is represented by $n((P \cup C)')$, which is the number of elements in the universal set that are not in the union of $P$ and $C$.

Solution:

First, we find the number of students who like at least one of the two languages, which is the cardinality of the union of sets $P$ and $C$, i.e., $n(P \cup C)$.

We use the formula for the cardinality of the union of two finite sets:

Formula:

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

Substitute the given values into the formula:

So, 45 students like at least one of the two subjects (Physics or Chemistry).

The number of students who like neither subject is the total number of students minus the number of students who like at least one subject.

Substitute the values:

So, 5 students like neither Physics nor Chemistry.

Comparing the result with the given options:

(A) 5 - Correct

(B) 10 - Incorrect

(C) 15 - Incorrect

(D) 20 - Incorrect

The correct option is (A).

The final answer is $\boxed{A}$.

The given equation is $\sin^2 \theta + \cos^2 \theta = 1$.

This equation is a fundamental trigonometric identity that holds true for all real values of the angle $\theta$.

The equation $\sin^2 \theta + \cos^2 \theta = 1$ is true for every angle $\theta \in \mathbb{R}$.

The set of all solutions to this equation is the set of all possible values of $\theta$ for which the equation is true. Since the equation is true for all real numbers $\theta$, the set of all solutions is the set of all real numbers, $\mathbb{R}$.

The set of real numbers $\mathbb{R}$ contains infinitely many elements. It is not possible to list all the real numbers in a finite sequence.

Let's classify the set of all solutions based on its nature:

(A) A finite set: A finite set has a countable number of elements. The set of real numbers is infinite. So, this is incorrect.

(B) An infinite set: An infinite set has an endless number of elements. The set of real numbers contains infinitely many elements. So, this is correct.

(C) An empty set: An empty set contains no elements. The equation $\sin^2 \theta + \cos^2 \theta = 1$ has infinitely many solutions (all real numbers), so the set of solutions is not empty. So, this is incorrect.

(D) A singleton set: A singleton set has exactly one element. The equation has infinitely many solutions, not just one. So, this is incorrect.

The set of all solutions to the equation $\sin^2 \theta + \cos^2 \theta = 1$ is the set of all real numbers, which is an infinite set.

The final answer is $\boxed{B}$.

We are given the condition $A - B = A$ for two sets $A$ and $B$. We need to determine which of the given statements is true based on this condition.

Recall the definition of set difference:

$A - B = \{x : x \in A \text{ and } x \notin B\}$

The set $A - B$ consists of elements that are in $A$ but not in $B$.

The condition $A - B = A$ means that the set of elements that are in $A$ but not in $B$ is exactly the same as the set of elements in $A$.

This implies that for any element $x$, if $x \in A$, then $x$ must satisfy the condition for being in $A - B$. That is, if $x \in A$, then $x \in A$ and $x \notin B$.

For the statement "$x \in A$ and $x \notin B$" to be true whenever "$x \in A$" is true, it must be the case that the second part "$x \notin B$" is true for all $x$ in $A$.

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

This means that there is no element that is in both $A$ and $B$. If an element is in $A$, it cannot be in $B$.

This is the definition of disjoint sets. The intersection of two sets is the set of elements common to both. If no element of $A$ is in $B$, then there are no elements common to $A$ and $B$.

Conversely, if $A \cap B = \emptyset$, let's check if $A - B = A$.

$A - B = \{x : x \in A \text{ and } x \notin B\}$.

Since $A \cap B = \emptyset$, there is no element $x$ such that $x \in A$ and $x \in B$. This means that for any element $x$, if $x \in A$, it is not possible for $x$ to also be in $B$, i.e., if $x \in A$, then $x \notin B$.

So, the condition "$x \in A$ and $x \notin B$" is equivalent to just "$x \in A$" (because if $x \in A$, the part "$x \notin B$" is automatically true due to $A \cap B = \emptyset$).

Therefore, $A - B = \{x : x \in A\} = A$.

The condition $A - B = A$ is equivalent to $A \cap B = \emptyset$.

Let's examine the given options:

(A) $A \subseteq B$: This would mean every element of A is in B. This contradicts $A \cap B = \emptyset$ unless $A = \emptyset$. Not necessarily true.

(B) $B \subseteq A$: This is not implied. For example, $A=\{1\}$, $B=\{2\}$. $A-B=\{1\}=A$, and $A \cap B=\emptyset$. But $B \not\subseteq A$. Not necessarily true.

(C) $A \cap B = \emptyset$: Our derivation shows that this is equivalent to the given condition, and therefore necessarily true.

(D) $A \cup B = U$: This statement involves the universal set $U$, which was not mentioned in the premise $A - B = A$. The condition $A - B = A$ does not provide any information about the union of $A$ and $B$ relative to a universal set. Not necessarily true.

The only statement that is necessarily true when $A - B = A$ is $A \cap B = \emptyset$.

The final answer is $\boxed{C}$.

We are given two sets defined in set-builder notation:

Here $\mathbb{N}$ represents the set of natural numbers, which we will assume to be $\{1, 2, 3, ...\}$. Prime numbers are natural numbers greater than 1 with no positive divisors other than 1 and themselves. Odd numbers are integers not divisible by 2.

First, let's list the elements of set $A$. We need prime numbers that are natural numbers and less than 10.

Prime numbers are 2, 3, 5, 7, 11, 13, ...

Prime numbers less than 10 are 2, 3, 5, 7.

Next, let's list the elements of set $B$. We need odd numbers that are natural numbers and less than 10.

Natural numbers less than 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9.

Odd numbers from this list are 1, 3, 5, 7, 9.

We need to find the union of sets $A$ and $B$, denoted by $A \cup B$. The union contains all elements that are in $A$ or in $B$ (or both).

$A = \{2, 3, 5, 7\}$

$B = \{1, 3, 5, 7, 9\}$

Combining the elements from both sets, without repetition:

Elements from A: 2, 3, 5, 7

Elements from B: 1, 3 (already listed), 5 (already listed), 7 (already listed), 9

The union is $\{1, 2, 3, 5, 7, 9\}$.

Comparing the result with the given options:

(A) $\{1, 2, 3, 5, 7, 9\}$ - Correct

(B) $\{2, 3, 5, 7\}$ - Incorrect (This is set A)

(C) $\{1, 3, 5, 7, 9\}$ - Incorrect (This is set B)

(D) $\{1, 2, 3, 5, 7\}$ - Incorrect (Missing element 9)

The correct option is (A).

The final answer is $\boxed{A}$.

We are given the following information about finite sets $A$ and $B$ within a universal set $U$:

To Find:

The cardinality of the intersection of the complements of A and B, $n(A' \cap B')$.

Solution:

We can use De Morgan's Law for sets, which states that the complement of the union of two sets is equal to the intersection of their complements:

De Morgan's Law:

Therefore, finding $n(A' \cap B')$ is equivalent to finding $n((A \cup B)')$.

The complement of the set $A \cup B$ with respect to the universal set $U$ is the set of elements in $U$ that are not in $A \cup B$. The cardinality of the complement of a set is given by the cardinality of the universal set minus the cardinality of the set.

Substitute the given values:

Since $(A \cup B)' = A' \cap B'$, we have:

The number of elements in the intersection of the complements of A and B is 20. This represents the number of elements that are neither in A nor in B.

Comparing the result with the given options:

(A) 20 - Correct

(B) 40 - Incorrect

(C) 60 - Incorrect

(D) 100 - Incorrect

The correct option is (A).

The final answer is $\boxed{A}$.

We are given the set $\{x \in \mathbb{Z} : -5 \leq x < 2\}$.

This set contains all integers $x$ such that $x$ is greater than or equal to $-5$ and strictly less than $2$.

Let's list the integers that satisfy the inequality $-5 \leq x < 2$.

The integers greater than or equal to $-5$ are $-5, -4, -3, -2, -1, 0, 1, 2, 3, ...$

The integers strictly less than $2$ are ..., $-3, -2, -1, 0, 1$.

The integers that satisfy both conditions are those from $-5$ up to, but not including, $2$.

The integers are $-5, -4, -3, -2, -1, 0, 1$.

Let the given set be $S$. In roster form, $S = \{-5, -4, -3, -2, -1, 0, 1\}$.

The cardinality of a set is the number of elements it contains. We need to count the number of elements in set $S$.

Counting the elements in $S = \{-5, -4, -3, -2, -1, 0, 1\}$, we have 7 elements.

Alternatively, for a set of integers $\{x \in \mathbb{Z} : a \leq x < b\}$, the number of integers is $b - a$. Here, $a = -5$ and $b = 2$.

Number of elements = $2 - (-5) = 2 + 5 = 7$.

Comparing the result with the given options:

(A) 5 - Incorrect

(B) 6 - Incorrect

(C) 7 - Correct

(D) 8 - Incorrect

The correct option is (C).

The final answer is $\boxed{7}$. The corresponding option is (C).

We are asked to represent the set of real numbers strictly greater than 3 using interval notation.

The set can be described in set-builder notation as $\{x \in \mathbb{R} : x > 3\}$.

We need to use interval notation to represent this set. The notation involves specifying the boundaries and whether they are included or excluded.

• The lower boundary is 3. The inequality $x > 3$ means 3 is excluded. An excluded endpoint is indicated by a parenthesis '('.
• The numbers are greater than 3, so they extend infinitely in the positive direction. This is represented by $\infty$. Infinity is always treated as an excluded endpoint, indicated by a parenthesis ')'.

Combining these, the interval starts from an excluded 3 and extends to positive infinity.

The interval notation is $(3, \infty)$.

Let's examine the given options:

(A) $(3, \infty)$

This interval represents $\{x \in \mathbb{R} : 3 < x < \infty\}$, which simplifies to $\{x \in \mathbb{R} : x > 3\}$. This matches the required set.

(B) $[3, \infty)$

This interval represents $\{x \in \mathbb{R} : 3 \leq x < \infty\}$, which simplifies to $\{x \in \mathbb{R} : x \geq 3\}$. This includes the number 3, while the question asks for numbers strictly greater than 3. This is incorrect.

(C) $(-\infty, 3)$

This interval represents $\{x \in \mathbb{R} : -\infty < x < 3\}$, which simplifies to $\{x \in \mathbb{R} : x < 3\}$. This represents real numbers strictly less than 3. This is incorrect.

(D) $(-\infty, 3]$

This interval represents $\{x \in \mathbb{R} : -\infty < x \leq 3\}$, which simplifies to $\{x \in \mathbb{R} : x \leq 3\}$. This represents real numbers less than or equal to 3. This is incorrect.

The set of real numbers greater than 3 is represented by the open interval $(3, \infty)$.

The final answer is $\boxed{A}$.

We are given that $A \cup B = B$.

The union $A \cup B$ contains all elements in $A$ or $B$. The condition $A \cup B = B$ means that any element in $A$ or $B$ must already be in $B$.

Consider an element $x \in A$. By definition of union, $x \in A \cup B$. Since $A \cup B = B$, we must have $x \in B$.

Thus, if $x \in A$, then $x \in B$. This means every element of $A$ is also an element of $B$.

This is the definition of a subset: $A \subseteq B$.

The condition $A \cup B = B$ is equivalent to $A \subseteq B$.

Now we check which of the given options is necessarily true if $A \subseteq B$ holds:

• (A) $A \subset B$: This means $A \subseteq B$ and $A \neq B$. This is not necessarily true, as $A$ could be equal to $B$.
• (B) $B \subset A$: This means $B \subseteq A$ and $B \neq A$. This contradicts $A \subseteq B$ unless $A=B=\emptyset$, which is not always the case.
• (C) $A = B$: This means $A \subseteq B$ and $B \subseteq A$. This is not necessarily true, as $A$ could be a proper subset of $B$.
• (D) $A \cap B = \emptyset$: This is not necessarily true. If $A \subseteq B$, then $A \cap B = A$. This is $\emptyset$ only if $A = \emptyset$.

Based on standard set theory, if $A \cup B = B$, then $A \subseteq B$ is necessarily true. However, this statement is not listed as an option.

Given the options, none of the statements (A), (B), (C), or (D) is necessarily true for all cases where $A \cup B = B$. This indicates a likely error in the question's options.

However, if forced to choose the most related option, $A \subseteq B$ is the direct consequence. Options (A) and (C) are specific cases of $A \subseteq B$. Neither is necessary.

Based on the provided options and standard mathematical definitions, there is no necessarily true statement among the choices. There appears to be an error in the question.

We are asked to find an equivalent expression for $A - (B \cap C)$.

Let's use the definition of set difference. The set $A - X$ consists of elements that are in $A$ but not in $X$. In this case, $X = B \cap C$.

Using De Morgan's Law for set operations (or the equivalent logical statement), we know that an element $x$ is not in the intersection of $B$ and $C$ if and only if $x$ is not in $B$ or $x$ is not in $C$.

So, the expression becomes:

Now, we apply the distributive property of 'and' over 'or' from logic:

So, the set can be written as:

Recognize that "$x \in A$ and $x \notin B$" is the definition of $x \in A - B$, and "$x \in A$ and $x \notin C$" is the definition of $x \in A - C$. The 'or' condition between these two corresponds to the union of the sets $A-B$ and $A-C$.

This is a standard identity for set difference, sometimes considered a form of De Morgan's Law applied to set difference or a distributive property of difference over intersection from the left.

Let's compare this result with the given options:

(A) $(A - B) \cap (A - C)$ - Incorrect (This is equivalent to $A - (B \cup C)$)

(B) $(A - B) \cup (A - C)$ - Correct

(C) $(A \cap B) - (A \cap C)$ - Incorrect

(D) $(A \cup B) \cap (A \cup C)$ - Incorrect (This is equivalent to $A \cup (B \cap C)$)

The correct option is (B).

The final answer is $\boxed{B}$.

We are given the set $\{x : x \in \mathbb{R}, x^2 + 1 = 0\}$.

This set contains all real numbers $x$ that are solutions to the equation $x^2 + 1 = 0$.

We need to solve the equation $x^2 + 1 = 0$ for real values of $x$.

$x^2 + 1 = 0$

$x^2 = -1$

We are looking for real numbers $x$ whose square is $-1$. In the system of real numbers, the square of any real number is always non-negative (greater than or equal to 0). There is no real number whose square is equal to $-1$.

The solutions to $x^2 = -1$ in the complex number system are $x = \pm i$, where $i$ is the imaginary unit ($i^2 = -1$). However, the set is defined for $x \in \mathbb{R}$, meaning we are only considering real number solutions.

Since there is no real number $x$ that satisfies the equation $x^2 + 1 = 0$, the set of real solutions is empty.

The set $\{x \in \mathbb{R} : x^2 + 1 = 0\}$ contains no elements.

A set that contains no elements is called the empty set, denoted by $\emptyset$ or \{\}.

Let's classify the set based on its nature:

(A) A singleton set: A singleton set has exactly one element. This set has zero elements. So, this is incorrect.

(B) An infinite set: An infinite set has an endless number of elements. This set has zero elements. So, this is incorrect.

(C) The empty set: An empty set contains no elements. This matches our finding. So, this is correct.

(D) The set of complex numbers: The set of complex numbers contains numbers of the form $a+bi$. While the solutions to $x^2+1=0$ are complex numbers ($i$ and $-i$), the set is defined to contain only real numbers ($x \in \mathbb{R}$). The set of real numbers is a subset of the set of complex numbers, but the set defined here is a subset of the real numbers satisfying the condition. Since no real number satisfies the condition, the set is the empty set, not the set of complex numbers.

The set $\{x \in \mathbb{R} : x^2 + 1 = 0\}$ is the empty set.

The final answer is $\boxed{C}$.

Let $U$ be the set of all students in the group.

Let $C$ be the set of students who like Coffee.

Let $T$ be the set of students who like Tea.

Given:

Total number of students, $n(U) = 100$

Number of students who like Coffee, $n(C) = 60$

Number of students who like Tea, $n(T) = 40$

Number of students who like neither Coffee nor Tea, $n((C \cup T)') = 20$

To Find:

The number of students who like both Coffee and Tea, which is $n(C \cap T)$.

Solution:

The number of students who like neither Coffee nor Tea is the total number of students minus the number of students who like at least one of the beverages.

Substitute the given values:

Rearrange the equation to find $n(C \cup T)$:

So, 80 students like at least one of the two beverages.

Now, we use the formula for the cardinality of the union of two finite sets:

Formula:

$n(C \cup T) = n(C) + n(T) - n(C \cap T)$

We need to find $n(C \cap T)$. We can rearrange the formula:

Substitute the values we know:

So, 20 students like both Coffee and Tea.

Comparing the result with the given options:

(A) 0 - Incorrect

(B) 10 - Incorrect

(C) 20 - Correct

(D) 40 - Incorrect

The correct option is (C).

A set is a well-defined collection of distinct objects. The objects in a set are called its elements or members.

To find the set of all positive integers which are factors of 24, we need to identify all positive integers that divide 24 without leaving a remainder.

The factors of 24 are the numbers that, when multiplied together, give 24.

These are:

$1 \times 24 = 24$

$2 \times 12 = 24$

$3 \times 8 = 24$

$4 \times 6 = 24$

The positive integers that are factors of 24 are $1, 2, 3, 4, 6, 8, 12, 24$.

Roster form (also called tabular form) is a way of describing a set by listing all its elements, separated by commas, within curly braces $\{ \}$.

Therefore, the set of all positive integers which are factors of 24 in roster form is:

$\{1, 2, 3, 4, 6, 8, 12, 24\}$

The given set is $\{1, 4, 9, 16, 25, \dots\}$.

We observe that the elements of the set are the squares of successive positive integers:

$1 = 1^2$

$4 = 2^2$

$9 = 3^2$

$16 = 4^2$

$25 = 5^2$

and so on.

Set-builder form is a notation used to describe a set by stating the properties that its elements must satisfy.

The general element of this set is the square of a positive integer. Let's represent a positive integer by $n$. Then the elements are of the form $n^2$, where $n$ belongs to the set of positive integers ($\mathbb{Z}^+$ or $\{1, 2, 3, \dots\}$).

In set-builder form, the set can be written as:

$\{x \mid x \text{ is the square of a positive integer}\}$

or using mathematical notation:

$\{n^2 \mid n \in \mathbb{Z}^+\}$

or

$\{n^2 \mid n \text{ is a positive integer}\}$

A finite set is a set that contains a finite number of elements. An infinite set is a set that contains an infinite number of elements.

(i) The set of all lines parallel to the x-axis.

In a plane, there are infinitely many distinct lines parallel to the x-axis (one for every possible y-intercept value).

Therefore, this set is an infinite set.

(ii) The set of letters in the English alphabet.

The English alphabet has a definite number of letters (26).

Therefore, this set is a finite set.

(iii) The set of all numbers which are multiples of 5.

Multiples of 5 are $5, 10, 15, 20, \dots$. This sequence continues indefinitely.

Therefore, this set is an infinite set.

(iv) The set of animals living on the Earth.

Although the number of animals is very large and constantly changing, it is a definite number at any given point in time. It is possible, in principle, to count them (even if practically impossible).

Therefore, this set is a finite set.

Two sets are said to be equal if they have exactly the same elements, regardless of their order.

(i) A = $\{2, 3\}$, B = $\{x : x \text{ is a solution of } x^2 + 5x + 6 = 0\}$

Set A is given as $\{2, 3\}$.

Set B contains the solutions of the quadratic equation $x^2 + 5x + 6 = 0$.

To find the elements of B, we solve the equation:

$x^2 + 5x + 6 = 0$

Factoring the quadratic equation, we get:

$(x+2)(x+3) = 0$

The solutions are $x+2=0$ or $x+3=0$.

So, $x = -2$ or $x = -3$.

Therefore, set B in roster form is $\{-2, -3\}$.

Comparing set A $\{2, 3\}$ and set B $\{-2, -3\}$, we see that the elements are different. The number 2 and 3 are in A, while -2 and -3 are in B. Set A contains positive numbers, while Set B contains negative numbers.

Thus, set A and set B do not have the same elements.

Hence, the pairs of sets are not equal.

(ii) A = $\{x : x \text{ is a letter in the word FOLLOW}\}$, B = $\{y : y \text{ is a letter in the word FLOW}\}$

Set A consists of the distinct letters in the word FOLLOW. The letters are F, O, L, L, O, W. The distinct letters are F, O, L, W.

So, set A in roster form is $\{F, O, L, W\}$.

Set B consists of the distinct letters in the word FLOW. The letters are F, L, O, W. The distinct letters are F, L, O, W.

So, set B in roster form is $\{F, L, O, W\}$.

Comparing set A $\{F, O, L, W\}$ and set B $\{F, L, O, W\}$, we see that both sets contain exactly the same distinct elements: F, O, L, and W. The order of elements in a set does not matter.

Thus, set A and set B have the same elements.

Hence, the pairs of sets are equal.

The cardinal number of a set is the number of distinct elements in the set. It is usually denoted by $n(S)$ for a set S.

(i) A = $\{x : x \in \mathbb{N}, x \leq 5\}$

This set contains natural numbers ($1, 2, 3, \dots$) that are less than or equal to 5.

The elements of set A are $\{1, 2, 3, 4, 5\}$.

There are 5 distinct elements in set A.

Therefore, the cardinal number of set A is $n(A) = 5$.

(ii) B = $\{\text{a, b, c, \{a, b\}}\}$

This set explicitly lists its elements. The elements are 'a', 'b', 'c', and the set '{a, b}'. Note that '{a, b}' is considered a single element of set B.

The distinct elements of set B are a, b, c, and $\{a, b\}$.

There are 4 distinct elements in set B.

Therefore, the cardinal number of set B is $n(B) = 4$.

(iii) C = $\emptyset$

The symbol $\emptyset$ represents the empty set, which is a set containing no elements.

There are 0 elements in set C.

Therefore, the cardinal number of set C is $n(C) = 0$.

(iv) D = $\{x : x \in \mathbb{Z}, x^2 = 9\}$

This set contains integers ($\dots, -3, -2, -1, 0, 1, 2, 3, \dots$) whose square is 9.

We need to solve the equation $x^2 = 9$ for integers $x$.

Taking the square root of both sides, we get $x = \pm\sqrt{9}$.

So, $x = 3$ or $x = -3$.

Both 3 and -3 are integers.

The elements of set D are $\{-3, 3\}$.

There are 2 distinct elements in set D.

Therefore, the cardinal number of set D is $n(D) = 2$.

A subset of a set A is a set containing some or all of the elements of A. Every set is a subset of itself, and the empty set is a subset of every set.

The given set is A = $\{1, 2, 3\}$.

We need to list all possible subsets of A.

Subsets with 0 elements (the empty set):

$\emptyset$ or {}

Subsets with 1 element:

$\{1\}$

$\{2\}$

$\{3\}$

Subsets with 2 elements:

$\{1, 2\}$

$\{1, 3\}$

$\{2, 3\}$

Subsets with 3 elements:

$\{1, 2, 3\}$

Listing all the subsets of A = $\{1, 2, 3\}$:

$\emptyset$, $\{1\}$, $\{2\}$, $\{3\}$, $\{1, 2\}$, $\{1, 3\}$, $\{2, 3\}$, $\{1, 2, 3\}$.

The power set P(A) of a set A is the set of all subsets of A.

P(A) = $\{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$.

The cardinality of the power set P(A) is the number of elements in P(A).

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

The number of subsets of a set with $m$ elements is given by $2^m$.

So, the cardinality of the power set P(A) is $2^{n(A)}$.

$n(P(A)) = 2^3 = 2 \times 2 \times 2 = 8$.

The cardinality of the power set P(A) is 8.

Given:

Universal set U = $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

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

Set B = $\{2, 4, 6, 8\}$

(i) Find $A' \cap B'$

First, we find the complement of set A ($A'$). The complement of A contains all elements in the universal set U that are not in A.

$A' = U \setminus A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \setminus \{1, 2, 3, 4\}$

$A' = \{5, 6, 7, 8, 9\}$

Next, we find the complement of set B ($B'$). The complement of B contains all elements in the universal set U that are not in B.

$B' = U \setminus B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \setminus \{2, 4, 6, 8\}$

$B' = \{1, 3, 5, 7, 9\}$

Now, we find the intersection of $A'$ and $B'$ ($A' \cap B'$). The intersection contains elements that are common to both $A'$ and $B'$.

$A' \cap B' = \{5, 6, 7, 8, 9\} \cap \{1, 3, 5, 7, 9\}$

The common elements are 5, 7, and 9.

$A' \cap B' = \{5, 7, 9\}$

(ii) Find $(A \cup B)'$

First, we find the union of set A and set B ($A \cup B$). The union contains all elements that are in A or in B or in both.

$A \cup B = \{1, 2, 3, 4\} \cup \{2, 4, 6, 8\}$

$A \cup B = \{1, 2, 3, 4, 6, 8\}$

Next, we find the complement of $(A \cup B)$, denoted by $(A \cup B)'$. The complement of $(A \cup B)$ contains all elements in the universal set U that are not in $(A \cup B)$.

$(A \cup B)' = U \setminus (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \setminus \{1, 2, 3, 4, 6, 8\}$

The elements in U that are not in $\{1, 2, 3, 4, 6, 8\}$ are 5, 7, and 9.

$(A \cup B)' = \{5, 7, 9\}$

Observation:

From the calculations above, we found that:

$A' \cap B' = \{5, 7, 9\}$

$(A \cup B)' = \{5, 7, 9\}$

We observe that $A' \cap B'$ is equal to $(A \cup B)'$. This is an illustration of one of De Morgan's Laws for sets, which states that for any two sets A and B, the complement of their union is equal to the intersection of their complements.

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

Given sets:

A = $\{1, 3, 5, 7, 9\}$

B = $\{2, 3, 5, 7, 11\}$

(i) Find $A \cup B$

The union of sets A and B, denoted by $A \cup B$, is the set of all elements that are in A or in B or in both.

$A \cup B = \{x \mid x \in A \text{ or } x \in B\}$

Elements in A: 1, 3, 5, 7, 9

Elements in B: 2, 3, 5, 7, 11

Combining all distinct elements from both sets:

$A \cup B = \{1, 2, 3, 5, 7, 9, 11\}$

(ii) Find $A \cap B$

The intersection of sets A and B, denoted by $A \cap B$, is the set of all elements that are common to both A and B.

$A \cap B = \{x \mid x \in A \text{ and } x \in B\}$

Elements common to A and B are 3, 5, and 7.

$A \cap B = \{3, 5, 7\}$

(iii) Find $A - B$

The set difference A minus B, denoted by $A - B$, is the set of all elements that are in A but not in B.

$A - B = \{x \mid x \in A \text{ and } x \notin B\}$

Elements in A are $\{1, 3, 5, 7, 9\}$.

Elements in B are $\{2, 3, 5, 7, 11\}$.

Elements in A that are not in B are 1 and 9.

$A - B = \{1, 9\}$

(iv) Find $B - A$

The set difference B minus A, denoted by $B - A$, is the set of all elements that are in B but not in A.

$B - A = \{x \mid x \in B \text{ and } x \notin A\}$

Elements in B are $\{2, 3, 5, 7, 11\}$.

Elements in A are $\{1, 3, 5, 7, 9\}$.

Elements in B that are not in A are 2 and 11.

$B - A = \{2, 11\}$

To represent the set $A \cap B'$ using a Venn diagram, we first understand what $B'$ and $A \cap B'$ mean.

$B'$ (the complement of B) is the set of all elements in the universal set U that are not in set B.

$A \cap B'$ (the intersection of A and B') is the set of all elements that are in set A AND are in the complement of set B.

This means $A \cap B'$ contains elements that are in A but not in B. This is equivalent to the set difference $A - B$ (or $A \setminus B$).

In a Venn diagram, we typically represent the universal set U by a rectangle and sets A and B by overlapping circles within the rectangle.

The region representing set A is the area enclosed by circle A.

The region representing set B is the area enclosed by circle B.

The region representing $B'$ is the area within the rectangle (U) but outside circle B.

The region representing $A \cap B'$ is the area that is both within circle A and within the region outside circle B. This is the portion of circle A that does not overlap with circle B.

Below is a description of how the Venn diagram would appear, with the required region shaded:

Draw a rectangle to represent the universal set U.

Inside the rectangle, draw two overlapping circles, one representing set A and the other representing set B.

The set $A \cap B'$ corresponds to the area inside the circle for set A but outside the circle for set B. This region should be shaded to represent the set $A \cap B'$.

The overlapping region of A and B represents $A \cap B$. The region outside both circles but inside the rectangle represents $(A \cup B)'$. The region inside B but outside A represents $B \cap A'$ or $B - A$.

The shaded region for $A \cap B'$ is the crescent-shaped part of circle A that does not include the intersection area ($A \cap B$).

Given:

$n(X) = 17$

$n(Y) = 23$

$n(X \cup Y) = 38$

We need to find $n(X \cap Y)$.

We use the formula for the union of two finite sets:

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$

We can rearrange this formula to find $n(X \cap Y)$:

$n(X \cap Y) = n(X) + n(Y) - n(X \cup Y)$

Substitute the given values into the formula:

$n(X \cap Y) = 17 + 23 - 38$

Perform the calculation:

$n(X \cap Y) = 40 - 38$

$n(X \cap Y) = 2$

Thus, the cardinal number of the intersection of sets X and Y is 2.

The final answer is $n(X \cap Y) = 2$.

Let H be the set of people who speak Hindi.

Let E be the set of people who speak English.

We are given the following information:

Number of people who speak Hindi, $n(H) = 25$

Number of people who speak English, $n(E) = 30$

Number of people who speak both Hindi and English, $n(H \cap E) = 15$

The question asks for the number of people who speak at least one of the two languages. This is represented by the union of the two sets, $n(H \cup E)$.

We use the formula for the number of elements in the union of two finite sets:

$n(H \cup E) = n(H) + n(E) - n(H \cap E)$

Substitute the given values into the formula:

$n(H \cup E) = 25 + 30 - 15$

Perform the calculation:

$n(H \cup E) = 55 - 15$

$n(H \cup E) = 40$

So, 40 people speak at least one of the two languages.

The number of people in the group (60) is the size of the universal set, but is not directly needed to answer the specific question asked here.

The final answer is 40.

We want to prove the identity $A \cup (A \cap B) = A$ using properties of sets.

We will start with the left-hand side (LHS) of the identity and use known set properties to transform it into the right-hand side (RHS).

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

We know that for any set A, $A = A \cap U$, where U is the universal set. This is the Identity Law for Intersection.

So, we can rewrite A as $A \cap U$.

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

Now, we can apply the Distributive Law for union over intersection, which states that for any sets X, Y, and Z, $ (X \cap Y) \cup (X \cap Z) = X \cap (Y \cup Z)$.

In our case, let X = A, Y = U, and Z = B.

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

Next, we use the property of the universal set U, which states that for any set B, $U \cup B = U$. This is the Identity Law for Union (with the universal set).

So, $A \cap (U \cup B) = A \cap U$

Finally, we again use the Identity Law for Intersection, which states that $A \cap U = A$.

So, $A \cap U = A$

Putting all the steps together:

$A \cup (A \cap B)$

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

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

$= A \cap U$

$= A$

Since we have transformed the LHS into the RHS using valid set properties, the identity is proven.

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

Given:

$n(A \cup B) = 70$

$n(A \cap B) = 20$

$n(B) = 40$

We need to find $n(A)$.

We use the formula relating the cardinalities of the union and intersection of two finite sets:

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

We can rearrange this formula to solve for $n(A)$:

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

Substitute the given values into the rearranged formula:

$n(A) = 70 - 40 + 20$}

Perform the calculation:

$n(A) = 30 + 20$

$n(A) = 50$

Thus, the cardinal number of set A is 50.

The final answer is $n(A) = 50$.

We need to prove two fundamental identities involving a set A and its complement $A'$ with respect to a universal set U.

Proof of $A \cup A' = U$:

Let U be the universal set and A be any set.

By the definition of the complement of a set, $A'$ is the set of all elements in the universal set U that are not in A.

$A' = \{x \mid x \in U \text{ and } x \notin A\}$.

Consider the union of set A and its complement $A'$, denoted by $A \cup A'$. By the definition of union, $A \cup A'$ is the set of all elements that are in A or in $A'$ (or in both).

$A \cup A' = \{x \mid x \in A \text{ or } x \in A'\}$

Substituting the definition of $A'$:

$A \cup A' = \{x \mid x \in A \text{ or } (x \in U \text{ and } x \notin A)\}$

Let's consider an arbitrary element $x \in U$. For any element $x$ in the universal set, it must either belong to set A or not belong to set A. These are the only two possibilities within U.

Case 1: If $x \in A$. Then, by the definition of union, $x \in A \cup A'$.

Case 2: If $x \notin A$. Since we are considering elements within the universal set U, this means $x \in U$ and $x \notin A$. By the definition of the complement, this is exactly the condition for $x$ to be in $A'$. So, if $x \notin A$ (within U), then $x \in A'$. By the definition of union, $x \in A \cup A'$.

In both cases, if $x \in U$, then $x \in A \cup A'$. This shows that every element in U is also in $A \cup A'$, so $U \subseteq A \cup A'$.

Conversely, let's consider an arbitrary element $y \in A \cup A'$. By the definition of union, $y \in A$ or $y \in A'$.

If $y \in A$, then since A is a set within the universal set U, it must be that $y \in U$.

If $y \in A'$, then by the definition of the complement, $y$ must be an element of the universal set U that is not in A. So, $y \in U$ and $y \notin A$. The crucial part is $y \in U$.

In both cases, if $y \in A \cup A'$, then $y \in U$. This shows that every element in $A \cup A'$ is also in U, so $A \cup A' \subseteq U$.

Since $U \subseteq A \cup A'$ and $A \cup A' \subseteq U$, we conclude that $A \cup A' = U$.

Proof of $A \cap A' = \emptyset$:

Let U be the universal set and A be any set.

By the definition of the complement of a set, $A' = \{x \mid x \in U \text{ and } x \notin A\}$.

Consider the intersection of set A and its complement $A'$, denoted by $A \cap A'$. By the definition of intersection, $A \cap A'$ is the set of all elements that are in A and in $A'$.

$A \cap A' = \{x \mid x \in A \text{ and } x \in A'\}$

Substituting the definition of $A'$:

$A \cap A' = \{x \mid x \in A \text{ and } (x \in U \text{ and } x \notin A)\}$

This simplifies to:

$A \cap A' = \{x \mid x \in A \text{ and } x \notin A \text{ and } x \in U\}$

Let's examine the condition for an element $x$ to be in the intersection: $x \in A$ and $x \notin A$. It is logically impossible for an element to be both a member of a set A and not a member of the same set A simultaneously. The statement "$x \in A$ and $x \notin A$" is a contradiction.

Therefore, there are no elements $x$ that can satisfy the condition to be in the set $A \cap A'$.

A set that contains no elements is called the empty set, denoted by $\emptyset$.

Thus, $A \cap A' = \emptyset$.

First, we write the set-builder form for the interval $[-5, 3)$.

The interval notation $[-5, 3)$ represents the set of all real numbers $x$ that are greater than or equal to $-5$ and strictly less than $3$.

The square bracket '[' at $-5$ indicates that $-5$ is included in the set.

The round bracket ')' at $3$ indicates that $3$ is not included in the set.

The numbers are specified as real numbers, denoted by $x \in \mathbb{R}$.

The conditions on $x$ are $x \ge -5$ and $x < 3$. We can combine these inequalities as $-5 \le x < 3$.

Therefore, the set-builder form for the interval $[-5, 3)$ is:

$\{x : x \in \mathbb{R}, -5 \leq x < 3\}$

Next, we write the interval notation for $\{x : x \in \mathbb{R}, x > 7\}$.

The set-builder form $\{x : x \in \mathbb{R}, x > 7\}$ represents the set of all real numbers $x$ that are strictly greater than $7$.

The condition $x > 7$ means that $x$ can be any real number larger than $7$. The number $7$ itself is not included.

Since there is no upper limit mentioned, the values of $x$ extend infinitely in the positive direction.

An endpoint that is not included is represented by a round bracket '('. For values extending infinitely, we use the symbol for infinity ($\infty$) and a round bracket ')'.

The interval starts just after $7$ and goes towards positive infinity.

Therefore, the interval notation for $\{x : x \in \mathbb{R}, x > 7\}$ is:

$(7, \infty)$

If A is a subset of B (denoted as $A \subseteq B$), it means that every element in set A is also an element in set B.

Intersection ($A \cap B$):

The intersection of two sets A and B, $A \cap B$, is the set containing all elements that are in both A and B.

Since every element of A is also in B (because $A \subseteq B$), the elements that are common to both A and B are exactly the elements of A.

Therefore, if $A \subseteq B$, then $A \cap B = A$.

Union ($A \cup B$):

The union of two sets A and B, $A \cup B$, is the set containing all elements that are in A or in B or in both.

Since every element of A is already in B (because $A \subseteq B$), when we combine all elements from A and B, we simply get all the elements of B (as A's elements are already included in B).

Therefore, if $A \subseteq B$, then $A \cup B = B$.

Justification with a Venn Diagram:

In a Venn diagram, we represent sets as circles within a rectangle (the universal set U).

To represent $A \subseteq B$, we draw the circle for set B, and then draw the circle for set A entirely inside the circle for set B. The circle for A does not extend outside the boundary of the circle for B.

For $A \cap B$, the intersection is the region where the circles overlap. Since the entire circle A is contained within circle B, the overlap region is the entire circle A. Thus, the shaded region representing $A \cap B$ is the same as the region representing A, illustrating that $A \cap B = A$.

For $A \cup B$, the union is the total region covered by both circles. Since circle A is inside circle B, the combined area of A and B is simply the area of circle B. Thus, the shaded region representing $A \cup B$ is the same as the region representing B, illustrating that $A \cup B = B$.

A Venn diagram for $A \subseteq B$ shows a smaller circle A completely enclosed within a larger circle B.

A singleton set (or unit set) is a set that contains exactly one element.

Its cardinality (number of elements) is 1.

Example of a singleton set in set-builder form:

Consider the set containing only the number 5. In roster form, this set is $\{5\}$.

We can describe this set in set-builder form by stating a property that only the number 5 satisfies.

For example, the set of all integers $x$ such that $x - 5 = 0$. The only solution to the equation $x - 5 = 0$ is $x = 5$, which is an integer.

This set can be written in set-builder form as:

$\{x \mid x \in \mathbb{Z}, x - 5 = 0\}$

Another example:

The set of all prime numbers $p$ such that $p$ is an even number. The only even prime number is 2.

This set can be written in set-builder form as:

$\{p \mid p \text{ is a prime number and } p \text{ is even}\}$

Both $\{5\}$ and $\{2\}$ are singleton sets.

The symmetric difference of two sets A and B, denoted by $A \Delta B$, is the set of elements which are in either of the sets but not in their intersection.

It can be defined in two equivalent ways:

$A \Delta B = (A \cup B) \setminus (A \cap B)$

or

$A \Delta B = (A \setminus B) \cup (B \setminus A)$

Given sets:

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

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

Method 1: Using $A \Delta B = (A \cup B) \setminus (A \cap B)$

First, find the union $A \cup B$:

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

Next, find the intersection $A \cap B$:

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

Now, find the difference $(A \cup B) \setminus (A \cap B)$:

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

Removing the elements $\{3, 4\}$ from $\{1, 2, 3, 4, 5, 6\}$, we get:

$A \Delta B = \{1, 2, 5, 6\}$

Method 2: Using $A \Delta B = (A \setminus B) \cup (B \setminus A)$

First, find the difference $A \setminus B$:

$A \setminus B = \{x \mid x \in A \text{ and } x \notin B\}$

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

Next, find the difference $B \setminus A$:

$B \setminus A = \{x \mid x \in B \text{ and } x \notin A\}$

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

Now, find the union $(A \setminus B) \cup (B \setminus A)$:

$(A \setminus B) \cup (B \setminus A) = \{1, 2\} \cup \{5, 6\}$

$A \Delta B = \{1, 2, 5, 6\}$

Both methods yield the same result.

The symmetric difference of sets A and B is $\{1, 2, 5, 6\}$.

The final answer is $A \Delta B = \{1, 2, 5, 6\}$.

The symbol $\subset$ denotes a proper subset. A set A is a proper subset of a set B if A is a subset of B (i.e., every element of A is also an element of B) and A is not equal to B (i.e., B contains at least one element not in A).

(i) $\{a, b\} \subset \{b, c, a\}$

Let A = $\{a, b\}$ and B = $\{b, c, a\}$. The elements of B are $b, c,$ and $a$. Since the order of elements in a set does not matter, B can also be written as $\{a, b, c\}$.

Check if A is a subset of B: Is $a \in B$? Yes. Is $b \in B$? Yes. Since every element of A is also in B, A is a subset of B ($A \subseteq B$).

Check if A is equal to B: A = $\{a, b\}$, B = $\{a, b, c\}$. Set B contains the element $c$, which is not in set A. Therefore, A is not equal to B ($A \neq B$).

Since A is a subset of B and A is not equal to B, A is a proper subset of B.

Therefore, the statement $\{a, b\} \subset \{b, c, a\}$ is True.

Justification: Every element of $\{a, b\}$ is present in $\{b, c, a\}$, and $\{b, c, a\}$ contains an element ($c$) which is not in $\{a, b\}$.

(ii) $\{a, e\} \subset \{\text{vowels in the English alphabet}\}$

Let A = $\{a, e\}$. Let B be the set of vowels in the English alphabet. B = $\{a, e, i, o, u\}$.

Check if A is a subset of B: Is $a \in B$? Yes. Is $e \in B$? Yes. Since every element of A is also in B, A is a subset of B ($A \subseteq B$).

Check if A is equal to B: A = $\{a, e\}$, B = $\{a, e, i, o, u\}$. Set B contains the elements $i, o, u$, which are not in set A. Therefore, A is not equal to B ($A \neq B$).

Since A is a subset of B and A is not equal to B, A is a proper subset of B.

Therefore, the statement $\{a, e\} \subset \{\text{vowels in the English alphabet}\}$ is True.

Justification: The elements $a$ and $e$ are vowels, so $\{a, e\}$ is a subset of the set of vowels. The set of vowels $\{a, e, i, o, u\}$ contains elements ($i, o, u$) which are not in $\{a, e\}$, so it is a proper subset.

(iii) $\{1, 2, 3\} \subset \{1, 3, 5\}$

Let A = $\{1, 2, 3\}$ and B = $\{1, 3, 5\}$.

Check if A is a subset of B: Is $1 \in B$? Yes. Is $2 \in B$? No. Is $3 \in B$? Yes. Since the element 2 is in A but not in B, A is not a subset of B ($A \not\subseteq B$).

If A is not a subset of B, it cannot be a proper subset of B.

Therefore, the statement $\{1, 2, 3\} \subset \{1, 3, 5\}$ is False.

Justification: The element 2 is an element of the set $\{1, 2, 3\}$ but it is not an element of the set $\{1, 3, 5\}$. For $\{1, 2, 3\}$ to be a proper subset of $\{1, 3, 5\}$, every element of $\{1, 2, 3\}$ must be in $\{1, 3, 5\}$.

To show that $(A \cap B)' = A' \cup B'$ using a Venn diagram, we will represent the regions corresponding to both sides of the identity and show that they are the same.

Draw a rectangle representing the universal set U. Inside this rectangle, draw two overlapping circles representing set A and set B.

Representing $(A \cap B)'$:

First, identify the region representing $A \cap B$. This is the overlapping area common to both circle A and circle B.

The set $(A \cap B)'$ is the complement of $A \cap B$. This is the region containing all elements in the universal set U that are not in $A \cap B$.

In the Venn diagram, shade the entire area within the rectangle U, except for the central overlapping region of A and B.

This shaded region represents $(A \cap B)'$.

Representing $A' \cup B'$:

Identify the region representing $A'$. This is the area within the rectangle U that is outside of circle A.

Identify the region representing $B'$. This is the area within the rectangle U that is outside of circle B.

The set $A' \cup B'$ is the union of $A'$ and $B'$. This is the region containing all elements that are in $A'$ or in $B'$ (or in both).

In the Venn diagram, shade the region outside circle A. Then, shade the region outside circle B. The union $A' \cup B'$ is the total shaded area from both steps.

This shaded region represents $A' \cup B'$.

Comparison:

Upon comparing the shaded region for $(A \cap B)'$ and the shaded region for $A' \cup B'$, we observe that both regions cover the exact same area in the Venn diagram: the entire universal set U excluding the intersection of A and B ($A \cap B$).

Since the regions represented by $(A \cap B)'$ and $A' \cup B'$ are identical in the Venn diagram, this visually demonstrates the identity $(A \cap B)' = A' \cup B'$. This is known as De Morgan's Second Law.

Let U be the universal set representing the total number of people in the town. Let H be the set of people who read Hindi newspaper. Let E be the set of people who read English newspaper.

We are given the following information:

Total number of people, $n(U) = 10000$

Number of people who read Hindi newspaper, $n(H) = 6000$

Number of people who read English newspaper, $n(E) = 4000$

Number of people who read both Hindi and English newspaper, $n(H \cap E) = 2500$

We need to find the number of people who read neither Hindi nor English newspaper. This is the number of people in the universal set U who are not in the union of H and E. This can be represented as $n((H \cup E)')$.

The number of people who read neither newspaper is given by the formula:

$n(\text{neither}) = n(U) - n(H \cup E)$

First, we need to find the number of people who read at least one of the two languages, which is $n(H \cup E)$. We use the formula for the union of two finite sets:

$n(H \cup E) = n(H) + n(E) - n(H \cap E)$

Substitute the given values into the formula for the union:

$n(H \cup E) = 6000 + 4000 - 2500$

$n(H \cup E) = 10000 - 2500$

$n(H \cup E) = 7500$

So, 7500 people read at least one of the two newspapers.

Now, we find the number of people who read neither newspaper using the complement formula:

$n(\text{neither}) = n(U) - n(H \cup E)$

$n(\text{neither}) = 10000 - 7500$

$n(\text{neither}) = 2500$

Thus, 2500 people read neither Hindi nor English newspaper.

The final answer is 2500.

Given that A and B are two sets such that $A \subset B$. This means that A is a proper subset of B. Consequently, every element of A is an element of B, and there is at least one element in B that is not in A.

Let U be the universal set containing A and B.

$A'$ is the complement of A with respect to U, i.e., $A' = \{x \mid x \in U \text{ and } x \notin A\}$.

$B'$ is the complement of B with respect to U, i.e., $B' = \{x \mid x \in U \text{ and } x \notin B\}$.

Consider the relationship between $A'$ and $B'$ when $A \subset B$.

If an element $x$ is not in B ($x \in B'$), then since A is contained within B, $x$ cannot be in A either (because if $x$ were in A, it would have to be in B). So, if $x \notin B$, then $x \notin A$.

This implies that every element outside B is also an element outside A. Thus, $B' \subseteq A'$.

We need to find $A' \cup B'$. Since we have established that $B' \subseteq A'$, the union of $A'$ and $B'$ will simply be the larger set, which is $A'$.

Using the property: If S is a subset of T ($S \subseteq T$), then $S \cup T = T$.

Here, $B' \subseteq A'$, so $B' \cup A' = A'$.

Since union is commutative ($A' \cup B' = B' \cup A'$), we have $A' \cup B' = A'$.

Therefore, if $A \subset B$, then $A' \cup B' = A'$.

Example:

Let the universal set be $U = \{1, 2, 3, 4, 5, 6\}$.

Let set A = $\{1, 2\}$ and set B = $\{1, 2, 3, 4\}$.

Here, $A \subset B$ because every element of A (1, 2) is in B, and B contains elements (3, 4) not in A.

Find the complements $A'$ and $B'$ with respect to U:

$A' = U \setminus A = \{1, 2, 3, 4, 5, 6\} \setminus \{1, 2\} = \{3, 4, 5, 6\}$

$B' = U \setminus B = \{1, 2, 3, 4, 5, 6\} \setminus \{1, 2, 3, 4\} = \{5, 6\}$

Now, find the union $A' \cup B'$:

$A' \cup B' = \{3, 4, 5, 6\} \cup \{5, 6\}$

$A' \cup B' = \{3, 4, 5, 6\}$

Comparing $A' \cup B'$ with $A'$, we see that:

$A' \cup B' = \{3, 4, 5, 6\}$

$A' = \{3, 4, 5, 6\}$

Thus, in this example, $A' \cup B' = A'$.

This example supports the conclusion that if $A \subset B$, then $A' \cup B' = A'$.

We need to prove De Morgan's Laws using properties of sets. These laws relate the operations of union, intersection, and complement.

De Morgan's First Law: $(A \cup B)' = A' \cap B'$

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

Part 1: Prove that $(A \cup B)' \subseteq A' \cap B'$.

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

By the definition of the complement, $x \in U$ and $x \notin (A \cup B)$.

By the definition of the union, $x \notin (A \cup B)$ means that $x$ is not in A and $x$ is not in B.

So, we have $x \in U$ and ($x \notin A$ and $x \notin B$).

Since $x \in U$ and $x \notin A$, by the definition of the complement, $x \in A'$.

Since $x \in U$ and $x \notin B$, by the definition of the complement, $x \in B'$.

Since $x \in A'$ and $x \in B'$, by the definition of the intersection, $x \in A' \cap B'$.

Thus, any element in $(A \cup B)'$ is also in $A' \cap B'$. Hence, $(A \cup B)' \subseteq A' \cap B'$.

Part 2: Prove that $A' \cap B' \subseteq (A \cup B)'$.

Let $y$ be an arbitrary element of $A' \cap B'$.

By the definition of the intersection, $y \in A'$ and $y \in B'$.

By the definition of the complement, $y \in A'$ means $y \in U$ and $y \notin A$.

And $y \in B'$ means $y \in U$ and $y \notin B$.

So, we have $y \in U$, and ($y \notin A$ and $y \notin B$).

The statement ($y \notin A$ and $y \notin B$) means that $y$ is not in A and $y$ is not in B. By the definition of the union, this is equivalent to saying that $y$ is not in the union of A and B, i.e., $y \notin (A \cup B)$.

So, we have $y \in U$ and $y \notin (A \cup B)$.

By the definition of the complement, this means $y \in (A \cup B)'$.

Thus, any element in $A' \cap B'$ is also in $(A \cup B)'$. Hence, $A' \cap B' \subseteq (A \cup B)'$.

Since $(A \cup B)' \subseteq A' \cap B'$ and $A' \cap B' \subseteq (A \cup B)'$, we conclude that $(A \cup B)' = A' \cap B'$.

De Morgan's Second Law: $(A \cap B)' = A' \cup B'$

To prove this identity, we again show that each set is a subset of the other.

Part 1: Prove that $(A \cap B)' \subseteq A' \cup B'$.

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

By the definition of the complement, $x \in U$ and $x \notin (A \cap B)$.

By the definition of the intersection, $x \notin (A \cap B)$ means that $x$ is not in the intersection of A and B. This means $x$ is not in A OR $x$ is not in B.

So, we have $x \in U$ and ($x \notin A$ or $x \notin B$).

This implies that either ($x \in U$ and $x \notin A$) or ($x \in U$ and $x \notin B$).

By the definition of the complement, ($x \in U$ and $x \notin A$) means $x \in A'$.

By the definition of the complement, ($x \in U$ and $x \notin B$) means $x \in B'$.

So, we have $x \in A'$ or $x \in B'$.

By the definition of the union, this means $x \in A' \cup B'$.

Thus, any element in $(A \cap B)'$ is also in $A' \cup B'$. Hence, $(A \cap B)' \subseteq A' \cup B'$.

Part 2: Prove that $A' \cup B' \subseteq (A \cap B)'$.

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

By the definition of the union, $y \in A'$ or $y \in B'$.

Case 1: $y \in A'$. By the definition of the complement, $y \in U$ and $y \notin A$. If $y \notin A$, then $y$ cannot be in the intersection $A \cap B$. So $y \notin (A \cap B)$. Since $y \in U$ and $y \notin (A \cap B)$, by the definition of complement, $y \in (A \cap B)'$.

Case 2: $y \in B'$. By the definition of the complement, $y \in U$ and $y \notin B$. If $y \notin B$, then $y$ cannot be in the intersection $A \cap B$. So $y \notin (A \cap B)$. Since $y \in U$ and $y \notin (A \cap B)$, by the definition of complement, $y \in (A \cap B)'$.

In both cases, if $y \in A' \cup B'$, then $y \in (A \cap B)'$.

Thus, any element in $A' \cup B'$ is also in $(A \cap B)'$. Hence, $A' \cup B' \subseteq (A \cap B)'$.

Since $(A \cap B)' \subseteq A' \cup B'$ and $A' \cup B' \subseteq (A \cap B)'$, we conclude that $(A \cap B)' = A' \cup B'$.

Given:

Total number of students in the school (Universal set U), $n(U) = 700$

Number of students playing cricket, $n(C) = 180$

Number of students playing football, $n(F) = 275$

Number of students playing both cricket and football, $n(C \cap F) = 95$

To Find:

(i) The number of students who play neither cricket nor football, $n((C \cup F)')$

(ii) The number of students who play exactly one of the two games

Solution:

To find the number of students who play neither game, we first need to find the number of students who play at least one of the two games. This is represented by the union of the two sets, $n(C \cup F)$.

Using the formula for the number of elements in the union of two finite sets:

$n(C \cup F) = n(C) + n(F) - n(C \cap F)$

Substitute the given values into the formula:

$n(C \cup F) = 180 + 275 - 95$

$n(C \cup F) = 455 - 95$}

$n(C \cup F) = 360$

So, 360 students play at least one of the two games.

(i) Number of students who play neither cricket nor football:

This is the number of students in the universal set U who are not in the set of students who play at least one game ($C \cup F$).

$n(\text{neither}) = n(U) - n(C \cup F)$

$n(\text{neither}) = 700 - 360$

$n(\text{neither}) = 340$

The number of students who play neither cricket nor football is 340.

(ii) Number of students who play exactly one of the two games:

The number of students who play exactly one game is the sum of students who play only cricket and students who play only football. This can be calculated as the total number of students playing at least one game minus the number of students playing both games.

$n(\text{exactly one}) = n(C \cup F) - n(C \cap F)$

$n(\text{exactly one}) = 360 - 95$

$n(\text{exactly one}) = 265$

Alternatively, we can calculate the number of students playing only cricket ($n(C \setminus F)$) and only football ($n(F \setminus C)$) and add them:

$n(C \setminus F) = n(C) - n(C \cap F) = 180 - 95 = 85$

$n(F \setminus C) = n(F) - n(C \cap F) = 275 - 95 = 180$

$n(\text{exactly one}) = n(C \setminus F) + n(F \setminus C) = 85 + 180 = 265$

Both methods give the same result.

The number of students who play exactly one of the two games is 265.

We need to prove the distributive law for sets: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

We can prove this identity using an element-wise argument.

Proof using Element-wise Argument:

To prove that two sets are equal, we must show that every element of the first set is an element of the second set, and vice versa.

Part 1: Prove that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.

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

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

We consider the two cases:

Case 1: $x \in A$.

If $x \in A$, then by the definition of union, $x \in A \cup B$ (since $x$ is in A) and $x \in A \cup C$ (since $x$ is in A).

Since $x \in (A \cup B)$ and $x \in (A \cup C)$, by the definition of intersection, $x \in (A \cup B) \cap (A \cup C)$.

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

If $x \in (B \cap C)$, then by the definition of intersection, $x \in B$ and $x \in C$.

Since $x \in B$, by the definition of union, $x \in A \cup B$.

Since $x \in C$, by the definition of union, $x \in A \cup C$.

Since $x \in (A \cup B)$ and $x \in (A \cup C)$, by the definition of intersection, $x \in (A \cup B) \cap (A \cup C)$.

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

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

Part 2: Prove that $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.

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

By the definition of intersection, this means $y \in (A \cup B)$ and $y \in (A \cup C)$.

By the definition of union, $y \in (A \cup B)$ means $y \in A$ or $y \in B$.

And $y \in (A \cup C)$ means $y \in A$ or $y \in C$.

So, we have ($y \in A$ or $y \in B$) and ($y \in A$ or $y \in C$).

We can use the distributive property of logical disjunction over conjunction: $(P \lor Q) \land (P \lor R) \iff P \lor (Q \land R)$.

Let P be the statement $y \in A$, Q be $y \in B$, and R be $y \in C$.

($y \in A$ or $y \in B$) and ($y \in A$ or $y \in C$) is logically equivalent to $y \in A$ or ($y \in B$ and $y \in C$).

By the definition of intersection, ($y \in B$ and $y \in C$) means $y \in (B \cap C)$.

So, we have $y \in A$ or $y \in (B \cap C)$.

By the definition of union, this means $y \in A \cup (B \cap C)$.

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

Hence, $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.

Since $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$ and $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$, we conclude that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Proof using Properties of Sets:

Consider the left-hand side (LHS): $A \cup (B \cap C)$.

We can use the identity property of intersection: $X = X \cap U$, where U is the universal set.

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

This step isn't immediately helpful for this specific identity using standard properties. Let's try working with the RHS.

Consider the right-hand side (RHS): $(A \cup B) \cap (A \cup C)$.

We can use the distributive law for intersection over union, which is the dual of the law we are trying to prove: $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$. However, applying that here would go in the wrong direction.

Let's use the element-wise definitions implicitly through the properties.

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

We know that $X \cup Y = X \cup (Y \setminus X)$. So, $A \cup B = A \cup (B \setminus A)$ and $A \cup C = A \cup (C \setminus A)$.

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

Using the property $(X \cup Y) \cap (X \cup Z) = X \cup (Y \cap Z)$ (This is the distributive law we are proving, so we cannot use it directly as a property unless it's a known axiom or previously proven lemma. Let's rethink the property based proof. A more common way uses the complement representation, but the question asks for "properties of sets", which might refer to algebraic properties).

Alternative Proof using Properties (focusing on basic identities):

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

We can use the identity $X = X \cup \emptyset$ and the complement properties. However, this becomes complex quickly.

A standard way to prove this using properties is to show that any element $x$ in LHS is in RHS and vice versa, but frame it using set operations.

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

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

Using the distributive property of logic (as used in the element-wise proof):

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

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

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

Since $x$ is in the LHS if and only if it is in the RHS, the sets are equal.

This essentially mirrors the element-wise proof but explicitly uses logical equivalences corresponding to set operations.

The fundamental properties used here are the definitions of $\cup$ and $\cap$ in terms of "or" and "and", and the logical distributive property $(P \lor Q) \land (P \lor R) \iff P \lor (Q \land R)$.

Let U be the universal set representing the total number of families surveyed. Let A, B, and C be the sets of families who buy newspaper A, newspaper B, and newspaper C, respectively.

Given:

$n(U) = 100$

$n(A) = 40$

$n(B) = 20$

$n(C) = 10$

$n(A \cap B) = 5$}

$n(B \cap C) = 3$}

$n(A \cap C) = 4$}

$n(A \cap B \cap C) = 2$

We can first find the number of families buying exactly two newspapers (i.e., the intersection of two sets but not the third).

Families buying A and B only = $n(A \cap B) - n(A \cap B \cap C) = 5 - 2 = 3$

Families buying B and C only = $n(B \cap C) - n(A \cap B \cap C) = 3 - 2 = 1$

Families buying A and C only = $n(A \cap C) - n(A \cap B \cap C) = 4 - 2 = 2$

Now, we can find the number of families buying only one newspaper.

Families buying only A = $n(A) - [\text{families buying A and B only} + \text{families buying A and C only} + \text{families buying A, B, and C}]$

$n(A \text{ only}) = n(A) - [n(A \cap B \cap C') + n(A \cap C \cap B') + n(A \cap B \cap C)]$

$n(A \text{ only}) = 40 - [3 + 2 + 2] = 40 - 7 = 33$

Families buying only B = $n(B) - [\text{families buying A and B only} + \text{families buying B and C only} + \text{families buying A, B, and C}]$

$n(B \text{ only}) = n(B) - [n(A \cap B \cap C') + n(B \cap C \cap A') + n(A \cap B \cap C)]$

$n(B \text{ only}) = 20 - [3 + 1 + 2] = 20 - 6 = 14$

Families buying only C = $n(C) - [\text{families buying A and C only} + \text{families buying B and C only} + \text{families buying A, B, and C}]$

$n(C \text{ only}) = n(C) - [n(A \cap C \cap B') + n(B \cap C \cap A') + n(A \cap B \cap C)]$

$n(C \text{ only}) = 10 - [2 + 1 + 2] = 10 - 5 = 5$

(i) Number of families who buy only newspaper A:

From the calculation above, the number of families buying only newspaper A is 33.

(ii) Number of families who buy exactly two newspapers:

This is the sum of families buying A and B only, B and C only, and A and C only.

$n(\text{exactly two}) = n(A \cap B \cap C') + n(B \cap C \cap A') + n(A \cap C \cap B')$

$n(\text{exactly two}) = 3 + 1 + 2 = 6$

The number of families who buy exactly two newspapers is 6.

(iii) Number of families who buy at least one newspaper:

This is the number of families in the union of sets A, B, and C, i.e., $n(A \cup B \cup C)$.

Using the Principle of Inclusion-Exclusion for three sets:

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

$n(A \cup B \cup C) = 40 + 20 + 10 - 5 - 3 - 4 + 2$

$n(A \cup B \cup C) = 70 - 12 + 2$

$n(A \cup B \cup C) = 60$

Alternatively, using the sum of disjoint regions:

$n(A \cup B \cup C) = n(A \text{ only}) + n(B \text{ only}) + n(C \text{ only}) + n(A \cap B \text{ only}) + n(B \cap C \text{ only}) + n(A \cap C \text{ only}) + n(A \cap B \cap C)$

$n(A \cup B \cup C) = 33 + 14 + 5 + 3 + 1 + 2 + 2 = 60$

The number of families who buy at least one newspaper is 60.

(iv) Number of families who buy none of the newspapers:

This is the number of families in the universal set who are not in the union of A, B, and C.

$n(\text{none}) = n(U) - n(A \cup B \cup C)$

$n(\text{none}) = 100 - 60$

$n(\text{none}) = 40$

The number of families who buy none of the newspapers is 40.

Let U be the universal set of all students in the school.

Let A be the set of students who play cricket.

Let B be the set of students who play football.

Let C be the set of students who play hockey.

(i) Students who play exactly one of the games:

This set consists of students who are in A but not in B and C ($A \cap B' \cap C'$), OR students who are in B but not in A and C ($B \cap A' \cap C'$), OR students who are in C but not in A and B ($C \cap A' \cap B'$).

These three groups are disjoint, so we take their union.

Set notation: $(A \cap B' \cap C') \cup (B \cap A' \cap C') \cup (C \cap A' \cap B')$

Alternatively, this can be expressed using set difference based on unions and intersections, but the intersection of sets and complements is typically clearer for specifying "only in A". Another common notation for "A only" is $A \setminus (B \cup C)$.

Set notation (Alternative): $(A \setminus (B \cup C)) \cup (B \setminus (A \cup C)) \cup (C \setminus (A \cup B))$

The most standard representation of the "only A" region is $A \cap B' \cap C'$. Thus, the first notation is preferred.

(ii) Students who play exactly two of the games:

This set consists of students who play A and B but not C ($A \cap B \cap C'$), OR students who play B and C but not A ($B \cap C \cap A'$), OR students who play A and C but not B ($A \cap C \cap B'$).

These three groups are disjoint, so we take their union.

Set notation: $(A \cap B \cap C') \cup (B \cap C \cap A') \cup (A \cap C \cap B')$

Alternatively, using set difference on intersections:

Set notation (Alternative): $((A \cap B) \setminus C) \cup ((B \cap C) \setminus A) \cup ((A \cap C) \setminus B)$

The first notation using intersection of sets and complements is preferred.

(iii) Students who play at least one of the games:

This set consists of all students who play A, or B, or C, or any combination of these. This is the definition of the union of the three sets.

Set notation: $A \cup B \cup C$

(iv) Students who play none of the games:

This set consists of students in the universal set U who do not play any of the three games. This is the complement of the set of students who play at least one game ($A \cup B \cup C$).

Set notation: $(A \cup B \cup C)'$

Using De Morgan's Law, this is equivalent to the intersection of the complements of the individual sets.

Set notation (Alternative): $A' \cap B' \cap C'$

Both notations are valid and represent the same set. The first notation is more directly the complement of "at least one".

We will prove the two statements separately.

Proof 1: $A \setminus (A \cap B) = A \setminus B$

We will use properties of sets to prove this identity.

Recall the definition of set difference: $X \setminus Y = X \cap Y'$.

Using this definition on the left-hand side (LHS):

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

Now, we apply De Morgan's Law for intersection, which states that $(X \cap Y)' = X' \cup Y'$.

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

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

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

We use the Complement Law, which states that the intersection of a set and its complement is the empty set: $A \cap A' = \emptyset$.

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

Finally, we use the Identity Law for Union, which states that the union of the empty set with any set is the set itself: $\emptyset \cup X = X$.

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

By the definition of set difference, $A \cap B'$ is equal to $A \setminus B$, which is the right-hand side (RHS).

$A \cap B' = A \setminus B$

Thus, we have shown that $A \setminus (A \cap B) = A \setminus B$.

Proof 2: $A \cup B = A \cap B \iff A = B$

This is a biconditional statement. We need to prove two implications:

Part 1: If $A = B$, then $A \cup B = A \cap B$.

Assume $A = B$.

Consider the union $A \cup B$. Since $B = A$, this is $A \cup A$. By the Idempotent Law for union, $A \cup A = A$.

Consider the intersection $A \cap B$. Since $B = A$, this is $A \cap A$. By the Idempotent Law for intersection, $A \cap A = A$.

Since both $A \cup B$ and $A \cap B$ are equal to A, we have $A \cup B = A \cap B$.

Thus, if $A = B$, then $A \cup B = A \cap B$.

Part 2: If $A \cup B = A \cap B$, then $A = B$.

Assume $A \cup B = A \cap B$.

To prove that $A = B$, we must show that $A \subseteq B$ and $B \subseteq A$.

First, let's prove $A \subseteq B$.

Let $x$ be an arbitrary element of set A ($x \in A$).

By the definition of union, if $x \in A$, then $x \in A \cup B$.

Given that $A \cup B = A \cap B$, it follows that if $x \in A \cup B$, then $x$ must also be in $A \cap B$.

By the definition of intersection, if $x \in A \cap B$, then $x$ is in both A and B ($x \in A$ and $x \in B$).

Therefore, if $x \in A$, we have shown that $x \in B$. This proves that $A \subseteq B$.

Next, let's prove $B \subseteq A$.

Let $y$ be an arbitrary element of set B ($y \in B$).

By the definition of union, if $y \in B$, then $y \in A \cup B$.

Given that $A \cup B = A \cap B$, it follows that if $y \in A \cup B$, then $y$ must also be in $A \cap B$.

By the definition of intersection, if $y \in A \cap B$, then $y$ is in both A and B ($y \in A$ and $y \in B$).

Therefore, if $y \in B$, we have shown that $y \in A$. This proves that $B \subseteq A$.

Since we have shown that $A \subseteq B$ and $B \subseteq A$, by the definition of set equality, $A = B$.

Thus, if $A \cup B = A \cap B$, then $A = B$.

Since both implications have been proven, the biconditional statement $A \cup B = A \cap B \iff A = B$ is true.

Let C be the set of students enrolled in the Computer Science program.

Let P be the set of students enrolled in the Physics program.

Let H be the set of students enrolled in the Chemistry program.

Given:

$n(C) = 200$

$n(P) = 150$

$n(H) = 100$

$n(C \cap P) = 60$

$n(P \cap H) = 40$

$n(C \cap H) = 30$

$n(C \cap P \cap H) = 10$

Calculation of $n(C \cup P \cup H)$:

The number of students enrolled in at least one program is given by the union of the three sets. Using the Principle of Inclusion-Exclusion for three sets:

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

$n(C \cup P \cup H) = 200 + 150 + 100 - 60 - 40 - 30 + 10$

$n(C \cup P \cup H) = 450 - (60 + 40 + 30) + 10$

$n(C \cup P \cup H) = 450 - 130 + 10$

$n(C \cup P \cup H) = 320 + 10$

$n(C \cup P \cup H) = 330$

(i) Number of students who are enrolled in at least one program:

This is $n(C \cup P \cup H)$.

$n(\text{at least one}) = 330$

The number of students enrolled in at least one program is 330.

(ii) Number of students who are enrolled in exactly two programs:

This is the sum of the number of students in (C and P only) + (P and H only) + (C and H only).

Number in C and P only = $n(C \cap P) - n(C \cap P \cap H) = 60 - 10 = 50$

Number in P and H only = $n(P \cap H) - n(C \cap P \cap H) = 40 - 10 = 30$

Number in C and H only = $n(C \cap H) - n(C \cap P \cap H) = 30 - 10 = 20$

$n(\text{exactly two}) = (n(C \cap P) - n(C \cap P \cap H)) + (n(P \cap H) - n(C \cap P \cap H)) + (n(C \cap H) - n(C \cap P \cap H))$

$n(\text{exactly two}) = 50 + 30 + 20 = 100$

The number of students enrolled in exactly two programs is 100.

(iii) Number of students who are enrolled in only the computer science program:

This is the number of students in C who are not in P or H. This can be calculated as $n(C) - n(C \cap P \text{ only}) - n(C \cap H \text{ only}) - n(C \cap P \cap H)$.

$n(C \text{ only}) = n(C) - [ (n(C \cap P) - n(C \cap P \cap H)) + (n(C \cap H) - n(C \cap P \cap H)) + n(C \cap P \cap H) ]$

$n(C \text{ only}) = n(C) - [n(C \cap P) - n(C \cap P \cap H) + n(C \cap H) - n(C \cap P \cap H) + n(C \cap P \cap H) ]$

$n(C \text{ only}) = n(C) - n(C \cap P) - n(C \cap H) + n(C \cap P \cap H)$

$n(C \text{ only}) = 200 - 60 - 30 + 10$

$n(C \text{ only}) = 200 - 90 + 10$

$n(C \text{ only}) = 120$

The number of students enrolled in only the computer science program is 120.

(iv) Number of students who are enrolled in only the physics program:

This is the number of students in P who are not in C or H.

$n(P \text{ only}) = n(P) - n(C \cap P) - n(P \cap H) + n(C \cap P \cap H)$

$n(P \text{ only}) = 150 - 60 - 40 + 10$

$n(P \text{ only}) = 150 - 100 + 10$

$n(P \text{ only}) = 60$

The number of students enrolled in only the physics program is 60.

(v) Number of students who are enrolled in none of these programs:

This is the number of students in the college (the universal set U) who are not in any of the three programs, i.e., $n((C \cup P \cup H)')$.

To find this, we need the total number of students in the college, $n(U)$. This information is not explicitly provided in the question text. If we assume the question implies a survey of 700 students (as in Question 2), then $n(U) = 700$. Let's proceed with this assumption.

Assuming the total number of students in the college (Universal Set U) is 700:

$n(\text{none}) = n(U) - n(C \cup P \cup H)$

$n(\text{none}) = 700 - 330$

$n(\text{none}) = 370$

Under the assumption that the total number of students in the college is 700, the number of students enrolled in none of these programs is 370.

We need to prove the distributive law of intersection over union: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

We will prove this identity using an element-wise argument.

Proof using Element-wise Argument:

To prove that two sets are equal, we must show that every element of the first set is an element of the second set, and vice versa.

Part 1: Prove that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$.

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

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 $x \in A$ and ($x \in B$ or $x \in C$).

Using the distributive property of logical conjunction over disjunction, $P \land (Q \lor R) \iff (P \land Q) \lor (P \land R)$, where P is the statement $x \in A$, Q is $x \in B$, and R is $x \in C$.

$(x \in A) \text{ and } (x \in B \text{ or } x \in C)$ is logically equivalent to $(x \in A \text{ and } x \in B) \text{ or } (x \in A \text{ and } x \in C)$.

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

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

So, we have $x \in (A \cap B)$ or $x \in (A \cap C)$.

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

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

Hence, $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$.

Part 2: Prove that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.

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

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

By the definition of intersection, $y \in (A \cap B)$ means $y \in A$ and $y \in B$.

By the definition of intersection, $y \in (A \cap C)$ means $y \in A$ and $y \in C$.

So, we have ($y \in A$ and $y \in B$) or ($y \in A$ and $y \in C$).

Using the distributive property of logical conjunction over disjunction in reverse: $(P \land Q) \lor (P \land R) \iff P \land (Q \lor R)$.

($y \in A$ and $y \in B$) or ($y \in A$ and $y \in C$) is logically equivalent to $y \in A$ and ($y \in B$ or $y \in C$).

By the definition of union, ($y \in B$ or $y \in C$) means $y \in (B \cup C)$.

So, we have $y \in A$ and $y \in (B \cup C)$.

By the definition of intersection, this means $y \in A \cap (B \cup C)$.

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

Hence, $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$.

Since $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$ and $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$, we conclude that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Venn Diagram Illustration:

Draw a rectangle representing the universal set U. Inside the rectangle, draw three overlapping circles representing sets A, B, and C.

Representing $A \cap (B \cup C)$ (LHS):

First, identify the region representing $B \cup C$. This is the entire area covered by circle B or circle C or both.

Then, find the intersection of set A with the region $(B \cup C)$. This is the area that is both inside circle A AND inside the combined region of B and C.

In the Venn diagram, shade the parts of circle A that overlap with circle B (representing $A \cap B$) and the parts of circle A that overlap with circle C (representing $A \cap C$). Note that the region where all three circles overlap ($A \cap B \cap C$) is part of both $A \cap B$ and $A \cap C$, and it is correctly included in the shaded region $A \cap (B \cup C)$.

Representing $(A \cap B) \cup (A \cap C)$ (RHS):

First, identify the region representing $A \cap B$. This is the overlapping area common to circle A and circle B.

Next, identify the region representing $A \cap C$. This is the overlapping area common to circle A and circle C.

Then, find the union of the regions $(A \cap B)$ and $(A \cap C)$. This is the total area covered by the region $A \cap B$ OR the region $A \cap C$ (or both).

In the Venn diagram, shade the region where A and B overlap. Then, shade the region where A and C overlap. The union $(A \cap B) \cup (A \cap C)$ is the total shaded area from both steps.

Comparison:

Upon comparing the shaded region for $A \cap (B \cup C)$ and the shaded region for $(A \cap B) \cup (A \cap C)$, we observe that both regions cover the exact same area in the Venn diagram: the parts of A that are also in B or C. This visually demonstrates the identity $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

The Identity Laws and Idempotent Laws are fundamental properties of set operations.

Identity Laws: These laws state that certain special sets ($\emptyset$ and U, the universal set) act like identity elements for union and intersection, respectively.

1. Identity Law for Union: $A \cup \emptyset = A$

Statement: The union of any set A with the empty set $\emptyset$ is the set A itself.

Proof:

To prove $A \cup \emptyset = A$, we show that $A \cup \emptyset \subseteq A$ and $A \subseteq A \cup \emptyset$.

Part 1: Prove $A \cup \emptyset \subseteq A$.

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

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

The empty set $\emptyset$ contains no elements. The statement "$x \in \emptyset$" is always false.

Therefore, "$x \in A$ or $x \in \emptyset$" simplifies to "$x \in A$ or False", which is equivalent to "$x \in A$".

So, if $x \in A \cup \emptyset$, then $x \in A$. This shows that every element in $A \cup \emptyset$ is also in A. Hence, $A \cup \emptyset \subseteq A$.

Part 2: Prove $A \subseteq A \cup \emptyset$.

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

By the definition of union, if $x \in A$, then $x \in A$ or $x \in \emptyset$ (because if "$x \in A$" is true, then "$x \in A$ or $x \in \emptyset$" is also true regardless of whether "$x \in \emptyset$" is true or false).

The statement "$x \in A$ or $x \in \emptyset$" means $x \in A \cup \emptyset$.

So, if $x \in A$, then $x \in A \cup \emptyset$. This shows that every element in A is also in $A \cup \emptyset$. Hence, $A \subseteq A \cup \emptyset$.

Since $A \cup \emptyset \subseteq A$ and $A \subseteq A \cup \emptyset$, we conclude that $A \cup \emptyset = A$.

2. Identity Law for Intersection: $A \cap U = A$

Statement: The intersection of any set A with the universal set U is the set A itself.

Proof:

To prove $A \cap U = A$, we show that $A \cap U \subseteq A$ and $A \subseteq A \cap U$.

Part 1: Prove $A \cap U \subseteq A$.

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

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

The statement "$x \in A$ and $x \in U$" implies "$x \in A$".

So, if $x \in A \cap U$, then $x \in A$. This shows that every element in $A \cap U$ is also in A. Hence, $A \cap U \subseteq A$.

Part 2: Prove $A \subseteq A \cap U$.

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

Since A is a set within the universal set U, any element of A must also be an element of U. By definition, U contains all possible elements under consideration.

So, if $x \in A$, then $x \in A$ and $x \in U$.

By the definition of intersection, "$x \in A$ and $x \in U$" means $x \in A \cap U$.

So, if $x \in A$, then $x \in A \cap U$. This shows that every element in A is also in $A \cap U$. Hence, $A \subseteq A \cap U$.

Since $A \cap U \subseteq A$ and $A \subseteq A \cap U$, we conclude that $A \cap U = A$.

Idempotent Laws: These laws state that combining a set with itself using union or intersection results in the same set.

3. Idempotent Law for Union: $A \cup A = A$

Statement: The union of any set A with itself is the set A.

Proof:

To prove $A \cup A = A$, we show that $A \cup A \subseteq A$ and $A \subseteq A \cup A$.

Part 1: Prove $A \cup A \subseteq A$.

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

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

The statement "$x \in A$ or $x \in A$" is equivalent to "$x \in A$".

So, if $x \in A \cup A$, then $x \in A$. This shows that every element in $A \cup A$ is also in A. Hence, $A \cup A \subseteq A$.

Part 2: Prove $A \subseteq A \cup A$.

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

By the definition of union, if $x \in A$, then $x \in A$ or $x \in A$ (because if "$x \in A$" is true, then "$x \in A$ or $x \in A$" is also true).

The statement "$x \in A$ or $x \in A$" means $x \in A \cup A$.

So, if $x \in A$, then $x \in A \cup A$. This shows that every element in A is also in $A \cup A$. Hence, $A \subseteq A \cup A$.

Since $A \cup A \subseteq A$ and $A \subseteq A \cup A$, we conclude that $A \cup A = A$.

4. Idempotent Law for Intersection: $A \cap A = A$

Statement: The intersection of any set A with itself is the set A.

Proof:

To prove $A \cap A = A$, we show that $A \cap A \subseteq A$ and $A \subseteq A \cap A$.

Part 1: Prove $A \cap A \subseteq A$.

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

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

The statement "$x \in A$ and $x \in A$" is equivalent to "$x \in A$".

So, if $x \in A \cap A$, then $x \in A$. This shows that every element in $A \cap A$ is also in A. Hence, $A \cap A \subseteq A$.

Part 2: Prove $A \subseteq A \cap A$.

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

If $x \in A$, then it is true that $x \in A$ and $x \in A$.

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

So, if $x \in A$, then $x \in A \cap A$. This shows that every element in A is also in $A \cap A$. Hence, $A \subseteq A \cap A$.

Since $A \cap A \subseteq A$ and $A \subseteq A \cap A$, we conclude that $A \cap A = A$.

Let U be the set of all students surveyed.

Let C be the set of students who play cricket.

Let F be the set of students who play football.

Let H be the set of students who play hockey.

Given:

$n(U) = 50$

$n(C) = 25$

$n(F) = 30$

$n(H) = 28$

$n(C \cap F) = 10$

$n(F \cap H) = 8$

$n(C \cap H) = 5$

$n(C \cap F \cap H) = 2$

We can find the number of students who play only a specific game by subtracting the numbers in the intersections involving that game from the total number playing that game, accounting for the students who play all three games.

(i) Number of students who play only cricket ($n(C \text{ only})$):

This is the number of students in C but not in F or H. We can calculate this as $n(C) - n(C \cap F \text{ only}) - n(C \cap H \text{ only}) - n(C \cap F \cap H)$.

First, find the numbers in the intersections of exactly two games:

$n(C \cap F \text{ only}) = n(C \cap F) - n(C \cap F \cap H) = 10 - 2 = 8$

$n(C \cap H \text{ only}) = n(C \cap H) - n(C \cap F \cap H) = 5 - 2 = 3$

$n(C \text{ only}) = n(C) - (n(C \cap F \text{ only}) + n(C \cap H \text{ only}) + n(C \cap F \cap H))$

$n(C \text{ only}) = 25 - (8 + 3 + 2)$

$n(C \text{ only}) = 25 - 13$

$n(C \text{ only}) = 12$

The number of students who play only cricket is 12.

(ii) Number of students who play only football ($n(F \text{ only})$):

This is the number of students in F but not in C or H.

First, find the number in the intersection of F and H only:

$n(F \cap H \text{ only}) = n(F \cap H) - n(C \cap F \cap H) = 8 - 2 = 6$

$n(F \text{ only}) = n(F) - (n(C \cap F \text{ only}) + n(F \cap H \text{ only}) + n(C \cap F \cap H))$

$n(F \text{ only}) = 30 - (8 + 6 + 2)$

$n(F \text{ only}) = 30 - 16$

$n(F \text{ only}) = 14$

The number of students who play only football is 14.

(iii) Number of students who play only hockey ($n(H \text{ only})$):

This is the number of students in H but not in C or F.

$n(H \text{ only}) = n(H) - (n(F \cap H \text{ only}) + n(C \cap H \text{ only}) + n(C \cap F \cap H))$

$n(H \text{ only}) = 28 - (6 + 3 + 2)$

$n(H \text{ only}) = 28 - 11$

$n(H \text{ only}) = 17$

The number of students who play only hockey is 17.

(iv) Number of students who play none of the games ($n((C \cup F \cup H)')$):

This is the total number of students minus the number of students who play at least one game. The number of students who play at least one game is $n(C \cup F \cup H)$.

Using the Principle of Inclusion-Exclusion:

$n(C \cup F \cup H) = n(C) + n(F) + n(H) - n(C \cap F) - n(F \cap H) - n(C \cap H) + n(C \cap F \cap H)$

$n(C \cup F \cup H) = 25 + 30 + 28 - 10 - 8 - 5 + 2$

$n(C \cup F \cup H) = 83 - 23 + 2$

$n(C \cup F \cup H) = 60 + 2 = 62$

The number of students who play at least one game is 62.

Now, we calculate the number of students who play none of the games:

$n(\text{none}) = n(U) - n(C \cup F \cup H)$

$n(\text{none}) = 50 - 62$

$n(\text{none}) = -12$

The result for the number of students who play none of the games is -12. Since the number of students cannot be negative, this indicates an inconsistency in the data provided in the question. The sum of students playing at least one game (62) is greater than the total number of students surveyed (50).

Based on the provided numbers for individual and intersection counts, 62 students play at least one sport. This contradicts the statement that only 50 students were surveyed.

We need to prove the set identity: $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.

This identity is a form of De Morgan's Law for set difference.

We will prove this identity using an element-wise argument.

To prove that two sets are equal, we must show that every element of the first set is an element of the second set, and conversely, every element of the second set is an element of the first set.

Part 1: Prove that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$.

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

By the definition of set difference, $x \in A \setminus (B \cap C)$ means that $x \in A$ and $x \notin (B \cap C)$.

By the definition of intersection, $x \notin (B \cap C)$ means that $x$ is not in the intersection of B and C. This means $x$ is not in B OR $x$ is not in C (or both).

Now, we use the distributive property of logical conjunction ("and") over logical disjunction ("or"): $P \land (Q \lor R) \iff (P \land Q) \lor (P \land R)$.

Let P be "$x \in A$", Q be "$x \notin B$", and R be "$x \notin C$".

By the definition of set difference, $(x \in A \text{ and } x \notin B)$ means $x \in A \setminus B$, and $(x \in A \text{ and } x \notin C)$ means $x \in A \setminus C$.

By the definition of union, $x \in (A \setminus B) \text{ or } x \in (A \setminus C)$ means $x \in (A \setminus B) \cup (A \setminus C)$.

Starting with $x \in A \setminus (B \cap C)$ and reaching $x \in (A \setminus B) \cup (A \setminus C)$ through a series of logical equivalences shows that every element in $A \setminus (B \cap C)$ is also in $(A \setminus B) \cup (A \setminus C)$.

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

Part 2: Prove that $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$.

Let $y$ be an arbitrary element such that $y \in (A \setminus B) \cup (A \setminus C)$.

By the definition of union, $y \in (A \setminus B) \cup (A \setminus C)$ means that $y \in (A \setminus B)$ or $y \in (A \setminus C)$.

By the definition of set difference, $y \in (A \setminus B)$ means $y \in A$ and $y \notin B$, and $y \in (A \setminus C)$ means $y \in A$ and $y \notin C$.

Now, we use the distributive property of logical conjunction over logical disjunction in reverse: $(P \land Q) \lor (P \land R) \iff P \land (Q \lor R)$.

Let P be "$y \in A$", Q be "$y \notin B$", and R be "$y \notin C$".

By De Morgan's Law for logic, $(y \notin B \text{ or } y \notin C)$ is equivalent to $y \notin (B \text{ and } C)$.

By the definition of intersection, "$y \in B \text{ and } y \in C$" is $y \in (B \cap C)$. So, "$y \notin B \text{ or } y \notin C$" is equivalent to "$y \notin (B \cap C)$".

By the definition of set difference, $y \in A \text{ and } y \notin (B \cap C)$ means $y \in A \setminus (B \cap C)$.

Starting with $y \in (A \setminus B) \cup (A \setminus C)$ and reaching $y \in A \setminus (B \cap C)$ through a series of logical equivalences shows that every element in $(A \setminus B) \cup (A \setminus C)$ is also in $A \setminus (B \cap C)$.

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

Since we have shown that $A \setminus (B \cap C) \subseteq (A \setminus B) \cup (A \setminus C)$ and $(A \setminus B) \cup (A \setminus C) \subseteq A \setminus (B \cap C)$, by the definition of set equality, we conclude that $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$.

To Prove: For any two finite sets A and B, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

Proof using Argument Based on Disjoint Sets:

Consider the sets A and B. The union $A \cup B$ can be expressed as the union of three pairwise disjoint sets:

1. The set of elements that are in A but not in B, denoted by $A \setminus B$.

2. The set of elements that are in B but not in A, denoted by $B \setminus A$.

3. The set of elements that are in both A and B, denoted by $A \cap B$.

Thus, we can write the union as:

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

The sets $(A \setminus B)$, $(B \setminus A)$, and $(A \cap B)$ are mutually disjoint. That is, the intersection of any two of these sets is the empty set ($\emptyset$).

For example, $(A \setminus B) \cap (A \cap B) = (A \cap B') \cap (A \cap B) = A \cap (B' \cap B) \cap A = A \cap \emptyset \cap A = \emptyset$. Similarly for other pairs.

For finite sets, the cardinality of the union of pairwise disjoint sets is the sum of their cardinalities.

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

Now, consider set A. Set A can be partitioned into two disjoint parts: elements in A but not in B ($A \setminus B$) and elements in A and B ($A \cap B$).

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

Since $(A \setminus B)$ and $(A \cap B)$ are disjoint, we have:

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

Rearranging this equation to solve for $n(A \setminus B)$:

Similarly, consider set B. Set B can be partitioned into two disjoint parts: elements in B but not in A ($B \setminus A$) and elements in A and B ($A \cap B$).

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

Since $(B \setminus A)$ and $(A \cap B)$ are disjoint, we have:

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

Rearranging this equation to solve for $n(B \setminus A)$:

Now, substitute the expressions for $n(A \setminus B)$ from (i) and $n(B \setminus A)$ from (ii) into the equation for $n(A \cup B)$:

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

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

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

Combine the terms involving $n(A \cap B)$:

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

This proves the formula using an argument based on partitioning the sets into disjoint regions.