Samacheer Kalvi Class 9 Maths Solutions Chapter 1 Set Language Exercise 1.4

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.4

Question 1. If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find
(i) (PUQ)UR
(ii) (PnQ)∩S
(m) (QnS)∩R
Answer:
Given sets:
\( P = \{1, 2, 5, 7, 9\} \)
\( Q = \{2, 3, 5, 9, 11\} \)
\( R = \{3, 4, 5, 7, 9\} \)
\( S = \{2, 3, 4, 5, 8\} \)

(i) \( (P \cup Q) \cup R \)
First, find \( P \cup Q \):
\( P \cup Q = \{1, 2, 5, 7, 9\} \cup \{2, 3, 5, 9, 11\} \)
\( P \cup Q = \{1, 2, 3, 5, 7, 9, 11\} \)
Now, find \( (P \cup Q) \cup R \):
\( (P \cup Q) \cup R = \{1, 2, 3, 5, 7, 9, 11\} \cup \{3, 4, 5, 7, 9\} \)
\( (P \cup Q) \cup R = \{1, 2, 3, 4, 5, 7, 9, 11\} \)

(ii) \( (P \cap Q) \cap S \)
First, find \( P \cap Q \):
\( P \cap Q = \{1, 2, 5, 7, 9\} \cap \{2, 3, 5, 9, 11\} \)
\( P \cap Q = \{2, 5, 9\} \)
Now, find \( (P \cap Q) \cap S \):
\( (P \cap Q) \cap S = \{2, 5, 9\} \cap \{2, 3, 4, 5, 8\} \)
\( (P \cap Q) \cap S = \{2, 5\} \)

(m) \( (Q \cap S) \cap R \)
First, find \( Q \cap S \):
\( Q \cap S = \{2, 3, 5, 9, 11\} \cap \{2, 3, 4, 5, 8\} \)
\( Q \cap S = \{2, 3, 5\} \)
Now, find \( (Q \cap S) \cap R \):
\( (Q \cap S) \cap R = \{2, 3, 5\} \cap \{3, 4, 5, 7, 9\} \)
\( (Q \cap S) \cap R = \{3, 5\} \)
In simple words: For union, we combine all elements without repeating. For intersection, we find elements that are common to both sets. We follow these steps for each part of the question.

🎯 Exam Tip: When dealing with multiple set operations, always solve the operations inside the parentheses first, just like in arithmetic, to avoid mistakes.

Question 2. Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Answer:
Let's define the sets based on the given description:
Set \( P \) contains all real numbers between 2 and 7. This means \( P = (2, 7) \).
Set \( Q \) contains all irrational numbers between 2 and 7. This means \( Q \) is a subset of \( P \), so \( Q \subset P \).

To test the commutative property for union:
\( P \cup Q = Q \cup P \)
When we combine all elements from \( P \) and \( Q \), since all numbers in \( Q \) are also in \( P \), their union will simply be \( P \).
So, \( P \cup Q = P \). Also, \( Q \cup P = P \).
Thus, \( P \cup Q = Q \cup P \). This shows that the union of sets is commutative.

To test the commutative property for intersection:
\( P \cap Q = Q \cap P \)
When we look for common elements between \( P \) and \( Q \), since all numbers in \( Q \) are already in \( P \), the common elements will be exactly the numbers in \( Q \).
So, \( P \cap Q = Q \). Also, \( Q \cap P = Q \).
Thus, \( P \cap Q = Q \cap P \). This shows that the intersection of sets is commutative.
In simple words: The commutative property means that the order of sets in an operation (like union or intersection) does not change the final result. For these specific sets, both union and intersection follow this rule.

🎯 Exam Tip: Remember that a set of irrational numbers is always a subset of real numbers. This relationship helps determine the union and intersection outcomes quickly.

Question 3. If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets.
Answer:
Given sets:
\( A = \{p, q, r, s\} \)
\( B = \{m, n, q, s, t\} \)
\( C = \{m, n, p, q, s\} \)

To verify the associative property of union, we need to show that \( A \cup (B \cup C) = (A \cup B) \cup C \).

First, let's calculate the Left Hand Side (LHS): \( A \cup (B \cup C) \)
Step 1: Find \( B \cup C \)
\( B \cup C = \{m, n, q, s, t\} \cup \{m, n, p, q, s\} \)
\( B \cup C = \{m, n, p, q, s, t\} \)
Step 2: Find \( A \cup (B \cup C) \)
\( A \cup (B \cup C) = \{p, q, r, s\} \cup \{m, n, p, q, s, t\} \)
\( A \cup (B \cup C) = \{m, n, p, q, r, s, t\} \) ........(1)

Next, let's calculate the Right Hand Side (RHS): \( (A \cup B) \cup C \)
Step 1: Find \( A \cup B \)
\( A \cup B = \{p, q, r, s\} \cup \{m, n, q, s, t\} \)
\( A \cup B = \{m, n, p, q, r, s, t\} \)
Step 2: Find \( (A \cup B) \cup C \)
\( (A \cup B) \cup C = \{m, n, p, q, r, s, t\} \cup \{m, n, p, q, s\} \)
\( (A \cup B) \cup C = \{m, n, p, q, r, s, t\} \) ........(2)

From (1) and (2), we can see that \( A \cup (B \cup C) = (A \cup B) \cup C \).
This verifies that the associative property of union holds true for the given sets. The associative property states that how you group the sets in a union operation does not change the final result.
In simple words: We checked if putting sets together in different orders gives the same final list of items. We first combined B and C, then added A. Then we combined A and B, then added C. Both ways gave the exact same list of items. This proves the associative rule for joining sets.

🎯 Exam Tip: When verifying properties, always clearly show the steps for both sides of the equation (LHS and RHS) and label them, then conclude by comparing the results.

Question 4. Verify the associative property of intersection of sets for A = {-11, √2, √5, 7}, B = {√3, √5, 6, 13} and C = {√2, √3, √5, 9}.
Answer:
Given sets:
\( A = \{-11, \sqrt{2}, \sqrt{5}, 7\} \)
\( B = \{\sqrt{3}, \sqrt{5}, 6, 13\} \)
\( C = \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\} \)

To verify the associative property of intersection, we need to show that \( A \cap (B \cap C) = (A \cap B) \cap C \).

First, let's calculate the Left Hand Side (LHS): \( A \cap (B \cap C) \)
Step 1: Find \( B \cap C \)
\( B \cap C = \{\sqrt{3}, \sqrt{5}, 6, 13\} \cap \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\} \)
\( B \cap C = \{\sqrt{3}, \sqrt{5}\} \)
Step 2: Find \( A \cap (B \cap C) \)
\( A \cap (B \cap C) = \{-11, \sqrt{2}, \sqrt{5}, 7\} \cap \{\sqrt{3}, \sqrt{5}\} \)
\( A \cap (B \cap C) = \{\sqrt{5}\} \) ........(1)

Next, let's calculate the Right Hand Side (RHS): \( (A \cap B) \cap C \)
Step 1: Find \( A \cap B \)
\( A \cap B = \{-11, \sqrt{2}, \sqrt{5}, 7\} \cap \{\sqrt{3}, \sqrt{5}, 6, 13\} \)
\( A \cap B = \{\sqrt{5}\} \)
Step 2: Find \( (A \cap B) \cap C \)
\( (A \cap B) \cap C = \{\sqrt{5}\} \cap \{\sqrt{2}, \sqrt{3}, \sqrt{5}, 9\} \)
\( (A \cap B) \cap C = \{\sqrt{5}\} \) ........(2)

From (1) and (2), we can see that \( A \cap (B \cap C) = (A \cap B) \cap C \).
This verifies that the associative property of intersection holds true for the given sets. The common elements remain the same regardless of how the sets are grouped for intersection.
In simple words: We checked if finding common items among sets in different orders gives the same final common item. We first found what B and C have in common, then found what that has in common with A. Then we found what A and B have in common, then found what that has in common with C. Both ways gave the single item \( \sqrt{5} \), proving the associative rule for finding common items.

🎯 Exam Tip: Be careful with elements like \( \sqrt{2} \) or \( -11 \); treat them as distinct elements in the set and ensure they are only included in an intersection if present in *all* sets being intersected.

Question 5. If A={ x : x = 2ⁿ, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets.
Answer:
Given sets:
Set A: \( A = \{x : x = 2^n, n \in W \text{ and } n < 4\} \)
Here, \( W \) represents whole numbers, so \( n \in \{0, 1, 2, 3\} \).
Let's list the elements of A:
For \( n=0, x = 2^0 = 1 \)
For \( n=1, x = 2^1 = 2 \)
For \( n=2, x = 2^2 = 4 \)
For \( n=3, x = 2^3 = 8 \)
So, \( A = \{1, 2, 4, 8\} \)

Set B: \( B = \{x : x = 2n, n \in N \text{ and } n \leq 4\} \)
Here, \( N \) represents natural numbers, so \( n \in \{1, 2, 3, 4\} \).
Let's list the elements of B:
For \( n=1, x = 2(1) = 2 \)
For \( n=2, x = 2(2) = 4 \)
For \( n=3, x = 2(3) = 6 \)
For \( n=4, x = 2(4) = 8 \)
So, \( B = \{2, 4, 6, 8\} \)

Set C: \( C = \{0, 1, 2, 5, 6\} \)

To verify the associative property of intersection, we need to show that \( A \cap (B \cap C) = (A \cap B) \cap C \).

First, let's calculate the Left Hand Side (LHS): \( A \cap (B \cap C) \)
Step 1: Find \( B \cap C \)
\( B \cap C = \{2, 4, 6, 8\} \cap \{0, 1, 2, 5, 6\} \)
\( B \cap C = \{2, 6\} \)
Step 2: Find \( A \cap (B \cap C) \)
\( A \cap (B \cap C) = \{1, 2, 4, 8\} \cap \{2, 6\} \)
\( A \cap (B \cap C) = \{2\} \) ........(1)

Next, let's calculate the Right Hand Side (RHS): \( (A \cap B) \cap C \)
Step 1: Find \( A \cap B \)
\( A \cap B = \{1, 2, 4, 8\} \cap \{2, 4, 6, 8\} \)
\( A \cap B = \{2, 4, 8\} \)
Step 2: Find \( (A \cap B) \cap C \)
\( (A \cap B) \cap C = \{2, 4, 8\} \cap \{0, 1, 2, 5, 6\} \)
\( (A \cap B) \cap C = \{2\} \) ........(2)

From (1) and (2), we get \( A \cap (B \cap C) = (A \cap B) \cap C \).
This verifies that the associative property of intersection holds for these sets. The order in which you find the common elements does not affect the final result.
In simple words: First, we listed the actual numbers for sets A and B based on their rules. Then, we checked the associative rule for finding common items. This means we found what numbers are common when grouping the sets in two different ways. Both ways gave us the same single number, {2}, which means the rule works.

🎯 Exam Tip: Always list the elements of sets defined by rules (set-builder notation) explicitly before performing operations to avoid errors. Pay close attention to the number system (Whole numbers W, Natural numbers N) and inequality signs.