Questions and Answers

Example 1: Identifying Sets

Question: Which of the following are sets? (i) The collection of all months of a year beginning with the letter J. (ii) The collection of ten most talented writers of India.

Solution: (i) This is a set. The collection is well-defined because we can definitively list its members: {January, June, July}. (ii) This is not a set. The criterion 'most talented' is subjective and not well-defined, as it can vary from person to person.

Example 2: Set-Builder to Roster Form

Question: Write the set $A = \{x : x \text{ is an integer and } -3 \le x < 7\}$ in roster form.

Solution: The integers that are greater than or equal to -3 and less than 7 are -3, -2, -1, 0, 1, 2, 3, 4, 5, and 6. Therefore, the set in roster form is: $$A = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$$

Example 3: Roster to Set-Builder Form

Question: Write the set $\{3, 6, 9, 12\}$ in set-builder form.

Solution: Each element in the set is a multiple of 3. Specifically, they are the first four multiples of 3. We can write this as: $$A = \{x : x = 3n, n \in \mathbb{N} \text{ and } 1 \le n \le 4\}$$

Example 4: Identifying the Null Set

Question: Which of the following is an example of the null set? (i) Set of odd natural numbers divisible by 2. (ii) Set of even prime numbers.

Solution: (i) An odd number cannot be divisible by 2. Therefore, this set has no elements and is the null set ($\phi$). (ii) The number 2 is an even prime number. The set is {2}, which is a singleton set, not the null set.

Example 5: Finite vs. Infinite Sets

Question: State whether the following set is finite or infinite: The set of lines which are parallel to the x-axis.

Solution: There are infinitely many lines that can be drawn parallel to the x-axis (e.g., y=1, y=2, y=2.5, etc.). Therefore, the set is infinite.

Example 6: Checking Set Equality

Question: Are the sets $A = \{2, 3\}$ and $B = \{x : x \text{ is a solution of } x^2 + 5x + 6 = 0\}$ equal?

Solution: First, find the elements of set B by solving the quadratic equation: $x^2 + 5x + 6 = 0 \implies (x+2)(x+3) = 0$. The solutions are $x=-2$ and $x=-3$. So, $B = \{-2, -3\}$. Set A is $\{2, 3\}$. Since A and B do not have the exact same elements, A is not equal to B (A ≠ B).

Example 7: Understanding Subsets

Question: Let $A = \{1, 2, \{3, 4\}, 5\}$. Which of the following statements are correct and which are incorrect? (i) $\{3, 4\} \subset A$ (ii) $\{3, 4\} \in A$

Solution: The elements of set A are 1, 2, the set {3, 4}, and 5. (i) Incorrect. For $\{3, 4\}$ to be a subset of A, its elements (3 and 4) must be elements of A. They are not. The element is the set itself. (ii) Correct. The object $\{3, 4\}$ is listed as a single element within set A.

Example 8: Writing Subsets

Question: Write down all the subsets of the set $\{1, 2, 3\}$.

Solution: The subsets are:

• Subsets with 0 elements: $\phi$
• Subsets with 1 element: $\{1\}, \{2\}, \{3\}$
• Subsets with 2 elements: $\{1, 2\}, \{1, 3\}, \{2, 3\}$
• Subsets with 3 elements: $\{1, 2, 3\}$ There are $2^3 = 8$ subsets in total.

Example 9: Intervals

Question: Write the set $\{x : x \in \mathbb{R}, -4 < x \le 6\}$ as an interval.

Solution: The set includes all real numbers greater than -4 and less than or equal to 6. The number -4 is excluded (open bracket) and 6 is included (closed bracket). The interval is $(-4, 6]$.

Example 10: Union of Sets

Question: If $A = \{x : x \text{ is a natural number and multiple of 3}\}$ and $B = \{x : x \text{ is a natural number less than 6}\}$, find $A \cup B$.

Solution: First, write the sets in roster form: $A = \{3, 6, 9, 12, ...\}$ $B = \{1, 2, 3, 4, 5\}$ The union includes all unique elements from both sets: $$A \cup B = \{1, 2, 3, 4, 5, 6, 9, 12, ...\}$$ This can also be written as $\{x : x=1,2,4,5 \text{ or x is a multiple of 3}\}$.

Example 11: Intersection of Sets

Question: If $A = \{3, 5, 7, 9, 11\}$ and $B = \{7, 9, 11, 13\}$, find $A \cap B$.

Solution: The intersection contains only the elements that are common to both sets A and B. The common elements are 7, 9, and 11. $$A \cap B = \{7, 9, 11\}$$

Example 12: Intersection with Union

Question: If $A = \{3, 5, 7, 9, 11\}$, $B = \{7, 9, 11, 13\}$, $C = \{11, 13, 15\}$, find $A \cap (B \cup C)$.

Solution:

1. First, find $B \cup C$: $B \cup C = \{7, 9, 11, 13, 15\}$.
2. Next, find the intersection of A with this new set: $A \cap \{7, 9, 11, 13, 15\}$. The common elements are 7, 9, and 11. $$A \cap (B \cup C) = \{7, 9, 11\}$$

Example 13: Identifying Disjoint Sets

Question: Are the sets $\{1, 2, 3, 4\}$ and $\{x : x \text{ is a natural number and } 4 \le x \le 6\}$ disjoint?

Solution: The first set is $A = \{1, 2, 3, 4\}$. The second set is $B = \{4, 5, 6\}$. Their intersection is $A \cap B = \{4\}$. Since the intersection is not the empty set, the sets are not disjoint.

Example 14: Difference of Sets

Question: If $A = \{3, 6, 9, 12, 15, 18, 21\}$ and $B = \{4, 8, 12, 16, 20\}$, find $A - B$ and $B - A$.

Solution: $A - B$ contains elements that are in A but not in B. Removing 12 from A gives: $A - B = \{3, 6, 9, 15, 18, 21\}$.

$B - A$ contains elements that are in B but not in A. Removing 12 from B gives: $B - A = \{4, 8, 16, 20\}$.

Example 15: Complement of a Set

Question: Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $A = \{1, 2, 3, 4\}$, $B = \{2, 4, 6, 8\}$. Find $(A \cup B)'$.

Solution:

1. First, find $A \cup B$: $A \cup B = \{1, 2, 3, 4, 6, 8\}$.
2. Next, find the complement of this set with respect to U. This means finding all elements in U that are not in $A \cup B$. $$(A \cup B)' = \{5, 7, 9\}$$

Example 16: Verifying De Morgan's Law

Question: Using the sets from the previous example, find $A'$ and $B'$ and verify that $(A \cup B)' = A' \cap B'$.

Solution: We have $U = \{1, ..., 9\}$, $A = \{1, 2, 3, 4\}$, $B = \{2, 4, 6, 8\}$. From the previous example, $(A \cup B)' = \{5, 7, 9\}$.

1. Find $A'$: $A' = U - A = \{5, 6, 7, 8, 9\}$.
2. Find $B'$: $B' = U - B = \{1, 3, 5, 7, 9\}$.
3. Find $A' \cap B'$: This is the set of elements common to $A'$ and $B'$. $$A' \cap B' = \{5, 7, 9\}$$ Since both calculations result in the same set, the law is verified.

Example 17: Complement in Words

Question: Taking the set of natural numbers as the universal set, write the complement of the set $\{x : x \text{ is a prime number}\}$.

Solution: The complement will be the set of all natural numbers that are not prime. This includes the number 1 and all composite numbers. $$A' = \{x : x \text{ is a composite number or } x = 1\}$$

Example 18: Set Equality from Operations

Question: Show that for any sets A and B, $A = (A \cap B) \cup (A - B)$.

Solution: This is best understood with a Venn diagram, but we can prove it using set properties. Let $x \in A$. There are two possibilities:

1. $x \in B$: If $x$ is in B, then $x \in A \cap B$. This means $x$ is in the right-hand side.
2. $x \notin B$: If $x$ is not in B, then $x \in A - B$. This also means $x$ is in the right-hand side. So, any element in A is in $(A \cap B) \cup (A - B)$.

Now let $y \in (A \cap B) \cup (A - B)$. This means $y \in (A \cap B)$ or $y \in (A - B)$.

• If $y \in (A \cap B)$, then $y \in A$.
• If $y \in (A - B)$, then $y \in A$. In both cases, $y \in A$. So, any element in the right-hand side is in A. Since both sets are subsets of each other, they are equal.

Example 19: Absorption Law

Question: Using properties of sets, show that $A \cup (A \cap B) = A$.

Solution: We know that for any set X, $X \subset A \cup X$. So, $A \subset A \cup (A \cap B)$. Now let's show that $A \cup (A \cap B) \subset A$. Let $x \in A \cup (A \cap B)$. This means $x \in A$ or $x \in (A \cap B)$.

• If $x \in A$, we are done.
• If $x \in (A \cap B)$, then $x \in A$ and $x \in B$. This also implies $x \in A$. In both cases, any element of $A \cup (A \cap B)$ is also an element of A. Therefore, $A \cup (A \cap B) \subset A$. Since both sets are subsets of each other, they are equal.

Example 20: Solving for Sets

Question: Let A, B, and C be sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$. Show that $B = C$.

Solution: To show that $B=C$, we need to prove that $B \subset C$ and $C \subset B$. Let $x$ be an arbitrary element of B ($x \in B$). Then $x \in A \cup B$. Since $A \cup B = A \cup C$, we have $x \in A \cup C$, which means $x \in A$ or $x \in C$. Case 1: If $x \in A$, then $x \in A \cap B$. Since $A \cap B = A \cap C$, we have $x \in A \cap C$, which implies $x \in C$. Case 2: If $x \in C$, then we are done. In both cases, we see that if $x \in B$, then $x \in C$. Thus, $B \subset C$. A symmetrical argument shows that $C \subset B$. Therefore, B = C.

Question: In a class of 100 students, 52 passed Mathematics, 47 passed Physics, and 40 passed Chemistry. Of these, 20 passed both Mathematics and Physics, 18 passed both Physics and Chemistry, 15 passed both Mathematics and Chemistry, and 5 passed all three. How many students passed at least one subject?

Solution: Use the principle of inclusion–exclusion for three sets M, P, C:

|M ∪ P ∪ C| = |M| + |P| + |C| − |M ∩ P| − |P ∩ C| − |M ∩ C| + |M ∩ P ∩ C|

= 52 + 47 + 40 − 20 − 18 − 15 + 5 = 91

So 91 students passed at least one subject.