Samacheer Kalvi Class 9 Maths Solutions Chapter 1 Set Language Exercise 1.5

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.5

 

Question 1. Using the adjacent Venn diagram, find the following sets:
(i) A - B
(ii) B - C
(iii) A'∪B'
(iv) A'∩B'
(v) (B∪C)'
(vi) A – (B∪C)
(vii) A – (B∩C)

Answer: From the given diagram, we can identify the elements of each set:
Universal set \( U = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8\} \)
Set \( A = \{-2, -1, 3, 4, 6\} \)
Set \( B = \{-2, -1, 5, 7, 8\} \)
Set \( C = \{-3, -2, 0, 3, 8\} \)

First, let's find the complements and intersections needed for the calculations:
\( A' = U – A = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8\} – \{-2, -1, 3, 4, 6\} \)
\( = \{-3, 0, 1, 2, 5, 7, 8\} \)

\( B' = U – B = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8\} – \{-2, -1, 5, 7, 8\} \)
\( = \{-3, 0, 1, 2, 3, 4, 6\} \)

\( B \cup C = \{-2, -1, 5, 7, 8\} \cup \{-3, -2, 0, 3, 8\} \)
\( = \{-3, -2, -1, 0, 3, 5, 7, 8\} \)

\( B \cap C = \{-2, -1, 5, 7, 8\} \cap \{-3, -2, 0, 3, 8\} \)
\( = \{-2, 8\} \)

Now we can find the required sets:
(i) \( A – B = \{-2, -1, 3, 4, 6\} – \{-2, -1, 5, 7, 8\} \)
\( = \{3, 4, 6\} \)

(ii) \( B – C = \{-2, -1, 5, 7, 8\} – \{-3, -2, 0, 3, 8\} \)
\( = \{-1, 5, 7\} \)

(iii) \( A' \cup B' = \{-3, 0, 1, 2, 5, 7, 8\} \cup \{-3, 0, 1, 2, 3, 4, 6\} \)
\( = \{-3, 0, 1, 2, 3, 4, 5, 6, 7, 8\} \)

(iv) \( A' \cap B' = \{-3, 0, 1, 2, 5, 7, 8\} \cap \{-3, 0, 1, 2, 3, 4, 6\} \)
\( = \{-3, 0, 1, 2\} \)

(v) \( (B \cup C)' = U – (B \cup C) = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8\} – \{-3, -2, -1, 0, 3, 5, 7, 8\} \)
\( = \{1, 2, 4, 6\} \)

(vi) \( A – (B \cup C) = \{-2, -1, 3, 4, 6\} – \{-3, -2, -1, 0, 3, 5, 7, 8\} \)
\( = \{4, 6\} \)

(vii) \( A – (B \cap C) = \{-2, -1, 3, 4, 6\} – \{-2, 8\} \)
\( = \{-1, 3, 4, 6\} \)
In simple words: We find what elements are in each set from the Venn diagram. Then, we use these elements to do different set operations like finding elements unique to one set (difference), finding elements in either set (union), finding elements common to both sets (intersection), or finding elements not in a set (complement).

🎯 Exam Tip: When using Venn diagrams, carefully list all elements in the Universal set and each individual set before starting calculations. This prevents errors in complements and intersections.

 

Question 2. If K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h} then find the following:
(i) K∪(L∩M)
(ii) K∩(L∪M)
(iii) (K∪L)∩(K∪M)
(iv) (K∩L)∪(K∩M)
and verify distributive laws.

Answer: Given the sets:
\( K = \{a, b, d, e, f\} \)
\( L = \{b, c, d, g\} \)
\( M = \{a, b, c, d, h\} \)

(i) Find \( K \cup (L \cap M) \):
First, find \( L \cap M \):
\( L \cap M = \{b, c, d, g\} \cap \{a, b, c, d, h\} \)
\( = \{b, c, d\} \)
Now, find \( K \cup (L \cap M) \):
\( K \cup (L \cap M) = \{a, b, d, e, f\} \cup \{b, c, d\} \)
\( = \{a, b, c, d, e, f\} \)

(ii) Find \( K \cap (L \cup M) \):
First, find \( L \cup M \):
\( L \cup M = \{b, c, d, g\} \cup \{a, b, c, d, h\} \)
\( = \{a, b, c, d, g, h\} \)
Now, find \( K \cap (L \cup M) \):
\( K \cap (L \cup M) = \{a, b, d, e, f\} \cap \{a, b, c, d, g, h\} \)
\( = \{a, b, d\} \)

(iii) Find \( (K \cup L) \cap (K \cup M) \):
First, find \( K \cup L \):
\( K \cup L = \{a, b, d, e, f\} \cup \{b, c, d, g\} \)
\( = \{a, b, c, d, e, f, g\} \)
Next, find \( K \cup M \):
\( K \cup M = \{a, b, d, e, f\} \cup \{a, b, c, d, h\} \)
\( = \{a, b, c, d, e, f, h\} \)
Now, find \( (K \cup L) \cap (K \cup M) \):
\( (K \cup L) \cap (K \cup M) = \{a, b, c, d, e, f, g\} \cap \{a, b, c, d, e, f, h\} \)
\( = \{a, b, c, d, e, f\} \)

(iv) Find \( (K \cap L) \cup (K \cap M) \):
First, find \( K \cap L \):
\( K \cap L = \{a, b, d, e, f\} \cap \{b, c, d, g\} \)
\( = \{b, d\} \)
Next, find \( K \cap M \):
\( K \cap M = \{a, b, d, e, f\} \cap \{a, b, c, d, h\} \)
\( = \{a, b, d\} \)
Now, find \( (K \cap L) \cup (K \cap M) \):
\( (K \cap L) \cup (K \cap M) = \{b, d\} \cup \{a, b, d\} \)
\( = \{a, b, d\} \)

Verification of distributive laws:
From (i) and (iii), we see that \( K \cup (L \cap M) = \{a, b, c, d, e, f\} \) and \( (K \cup L) \cap (K \cup M) = \{a, b, c, d, e, f\} \).
Since both results are the same, \( K \cup (L \cap M) = (K \cup L) \cap (K \cup M) \). This verifies the distributive law of union over intersection.

From (ii) and (iv), we see that \( K \cap (L \cup M) = \{a, b, d\} \) and \( (K \cap L) \cup (K \cap M) = \{a, b, d\} \).
Since both results are the same, \( K \cap (L \cup M) = (K \cap L) \cup (K \cap M) \). This verifies the distributive law of intersection over union.
In simple words: We are given three sets of letters. We first do the operations inside the brackets, like finding common letters (intersection) or all letters together (union). Then we do the outer operation. We showed that the 'distributive laws' work here, meaning that combining sets in one way gives the same answer as combining them in another, similar way. This is a fundamental rule in set theory, much like distributing multiplication over addition in numbers.

🎯 Exam Tip: Always perform operations inside parentheses first. Double-check your elements in unions and intersections, especially for common elements, to avoid mistakes.

 

Question 3. For A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C = {-4, -1, 0, 2, 3, 4} verify A∪(B∩C) = (A∪B) ∩ (A∪C).

Answer: First, let's write the given sets in roster form:
Set \( A \): integers between -2 (not including -2) and 4 (including 4).
\( A = \{-1, 0, 1, 2, 3, 4\} \)
Set \( B \): whole numbers less than or equal to 5.
\( B = \{0, 1, 2, 3, 4, 5\} \)
Set \( C \) is already given as: \( C = \{-4, -1, 0, 2, 3, 4\} \)

Now, let's verify \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \).

**Left Hand Side (LHS): \( A \cup (B \cap C) \)**
First, find \( B \cap C \):
\( B \cap C = \{0, 1, 2, 3, 4, 5\} \cap \{-4, -1, 0, 2, 3, 4\} \)
\( = \{0, 2, 3, 4\} \)
Next, find \( A \cup (B \cap C) \):
\( A \cup (B \cap C) = \{-1, 0, 1, 2, 3, 4\} \cup \{0, 2, 3, 4\} \)
\( = \{-1, 0, 1, 2, 3, 4\} \) ........(1)

**Right Hand Side (RHS): \( (A \cup B) \cap (A \cup C) \)**
First, find \( A \cup B \):
\( A \cup B = \{-1, 0, 1, 2, 3, 4\} \cup \{0, 1, 2, 3, 4, 5\} \)
\( = \{-1, 0, 1, 2, 3, 4, 5\} \)
Next, find \( A \cup C \):
\( A \cup C = \{-1, 0, 1, 2, 3, 4\} \cup \{-4, -1, 0, 2, 3, 4\} \)
\( = \{-4, -1, 0, 1, 2, 3, 4\} \)
Now, find \( (A \cup B) \cap (A \cup C) \):
\( (A \cup B) \cap (A \cup C) = \{-1, 0, 1, 2, 3, 4, 5\} \cap \{-4, -1, 0, 1, 2, 3, 4\} \)
\( = \{-1, 0, 1, 2, 3, 4\} \) ........(2)

From (1) and (2), we see that \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \). This shows the distributive property of union over intersection holds true for these sets. The elements must match exactly on both sides for verification.
In simple words: We are given three sets with different numbers. We need to check if a special rule, called the distributive law, works for these sets. We do the math on the left side of the rule and then on the right side. If the final list of numbers is the same for both sides, then the rule is proven true for these sets.

🎯 Exam Tip: When given sets in set-builder form, the first and most crucial step is to convert them accurately into roster form. Errors here will propagate through all subsequent calculations.

 

Question 3. Verify A∪(B∩C) = (A∪B) ∩ (A∪C) using Venn diagrams.

Answer:
To verify \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \) using Venn diagrams, we compare the shaded regions of both sides.

(i) Diagram for \( A \): (Shade circle A entirely)

(ii) Diagram for \( B \cap C \): (Shade the intersection of circles B and C)
(iii) Diagram for \( A \cup (B \cap C) \): (Shade circle A and the intersection of B and C)
(iv) Diagram for \( A \cup B \): (Shade the union of circles A and B)
(v) Diagram for \( A \cup C \): (Shade the union of circles A and C)
(vi) Diagram for \( (A \cup B) \cap (A \cup C) \): (Shade the intersection of the shaded regions from (iv) and (v))

By comparing diagram (iii) for \( A \cup (B \cap C) \) and diagram (vi) for \( (A \cup B) \cap (A \cup C) \), we can see that the shaded regions are identical. This graphically verifies the distributive law of union over intersection.

🎯 Exam Tip: When drawing Venn diagrams for verification, make sure to shade each intermediate step clearly. The final shaded region for both sides of the equation must be exactly the same for a successful verification.

 

Question 5. If A = {b, c, e, g, h}, B = {a, c, d, g, f}, and C = {a, d, e, g, h}, then show that A – (B∩C) = (A – B) ∪ (A – C).

Answer: Given the sets:
\( A = \{b, c, e, g, h\} \)
\( B = \{a, c, d, g, f\} \)
\( C = \{a, d, e, g, h\} \)

We need to show that \( A – (B \cap C) = (A – B) \cup (A – C) \).

**Left Hand Side (LHS): \( A – (B \cap C) \)**
First, find \( B \cap C \):
\( B \cap C = \{a, c, d, g, f\} \cap \{a, d, e, g, h\} \)
\( = \{a, d, g\} \)
Now, find \( A – (B \cap C) \):
\( A – (B \cap C) = \{b, c, e, g, h\} – \{a, d, g\} \)
\( = \{b, c, e, h\} \) ........(1)

**Right Hand Side (RHS): \( (A – B) \cup (A – C) \)**
First, find \( A – B \):
\( A – B = \{b, c, e, g, h\} – \{a, c, d, g, f\} \)
\( = \{b, e, h\} \)
Next, find \( A – C \):
\( A – C = \{b, c, e, g, h\} – \{a, d, e, g, h\} \)
\( = \{b, c\} \)
Now, find \( (A – B) \cup (A – C) \):
\( (A – B) \cup (A – C) = \{b, e, h\} \cup \{b, c\} \)
\( = \{b, c, e, h\} \) ........(2)

From (1) and (2), we can clearly see that \( A – (B \cap C) = (A – B) \cup (A – C) \). This result holds true, verifying the given set identity.
In simple words: We have three groups of letters. We calculate the elements on the left side of the given equation and then on the right side. We want to show that the final list of letters from both calculations is the same. This proves a property similar to De Morgan's laws for set differences.

🎯 Exam Tip: Remember that set difference \( A-B \) means elements in A that are NOT in B. Be careful not to include elements common to both A and B when calculating \( A-B \).

 

Question 6. If A= {x : x = 6n, n∈W and n < 6}, B = {x : x = 2n, n∈N and 2 < n ≤ 9} and C = {x : x = 3n, n∈N and 4 ≤ n < 10}, then show that A – (B∩C) = (A – B) ∪ (A – C)

Answer: First, let's write the given sets in roster form:
For Set \( A \): \( x = 6n \), where \( n \) is a whole number (W) and \( n < 6 \).
So, \( n = \{0, 1, 2, 3, 4, 5\} \).
\( A = \{6 \times 0, 6 \times 1, 6 \times 2, 6 \times 3, 6 \times 4, 6 \times 5\} \)
\( A = \{0, 6, 12, 18, 24, 30\} \)

For Set \( B \): \( x = 2n \), where \( n \) is a natural number (N) and \( 2 < n \leq 9 \).
So, \( n = \{3, 4, 5, 6, 7, 8, 9\} \).
\( B = \{2 \times 3, 2 \times 4, 2 \times 5, 2 \times 6, 2 \times 7, 2 \times 8, 2 \times 9\} \)
\( B = \{6, 8, 10, 12, 14, 16, 18\} \)

For Set \( C \): \( x = 3n \), where \( n \) is a natural number (N) and \( 4 \leq n < 10 \).
So, \( n = \{4, 5, 6, 7, 8, 9\} \).
\( C = \{3 \times 4, 3 \times 5, 3 \times 6, 3 \times 7, 3 \times 8, 3 \times 9\} \)
\( C = \{12, 15, 18, 21, 24, 27\} \)

Now, we will show that \( A – (B \cap C) = (A – B) \cup (A – C) \).

**Left Hand Side (LHS): \( A – (B \cap C) \)**
First, find \( B \cap C \):
\( B \cap C = \{6, 8, 10, 12, 14, 16, 18\} \cap \{12, 15, 18, 21, 24, 27\} \)
\( = \{12, 18\} \)
Next, find \( A – (B \cap C) \):
\( A – (B \cap C) = \{0, 6, 12, 18, 24, 30\} – \{12, 18\} \)
\( = \{0, 6, 24, 30\} \) ........(1)

**Right Hand Side (RHS): \( (A – B) \cup (A – C) \)**
First, find \( A – B \):
\( A – B = \{0, 6, 12, 18, 24, 30\} – \{6, 8, 10, 12, 14, 16, 18\} \)
\( = \{0, 24, 30\} \)
Next, find \( A – C \):
\( A – C = \{0, 6, 12, 18, 24, 30\} – \{12, 15, 18, 21, 24, 27\} \)
\( = \{0, 6, 30\} \)
Now, find \( (A – B) \cup (A – C) \):
\( (A – B) \cup (A – C) = \{0, 24, 30\} \cup \{0, 6, 30\} \)
\( = \{0, 6, 24, 30\} \) ........(2)

From (1) and (2), we see that \( A – (B \cap C) = (A – B) \cup (A – C) \). This shows the identity holds true for these sets. This property is known as De Morgan's Law for set difference.
In simple words: We start with three sets defined by rules. We first list all the numbers in each set. Then, we apply set operations step-by-step on both sides of the equal sign. If the final set of numbers is the same for both sides, then the equation is true, like showing that a math rule works.

🎯 Exam Tip: Carefully convert set-builder notation to roster form, paying close attention to the conditions (e.g., natural numbers vs. whole numbers, strict vs. inclusive inequalities). A single mistake in listing elements will lead to an incorrect verification.

 

Question 7. If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that A – (B∪C) = (A – B) ∩ (A – С).

Answer: Given the sets:
\( A = \{-2, 0, 1, 3, 5\} \)
\( B = \{-1, 0, 2, 5, 6\} \)
\( C = \{-1, 2, 5, 6, 7\} \)

We need to show that \( A – (B \cup C) = (A – B) \cap (A – C) \).

**Left Hand Side (LHS): \( A – (B \cup C) \)**
First, find \( B \cup C \):
\( B \cup C = \{-1, 0, 2, 5, 6\} \cup \{-1, 2, 5, 6, 7\} \)
\( = \{-1, 0, 2, 5, 6, 7\} \)
Now, find \( A – (B \cup C) \):
\( A – (B \cup C) = \{-2, 0, 1, 3, 5\} – \{-1, 0, 2, 5, 6, 7\} \)
\( = \{-2, 1, 3\} \) ........(1)

**Right Hand Side (RHS): \( (A – B) \cap (A – C) \)**
First, find \( A – B \):
\( A – B = \{-2, 0, 1, 3, 5\} – \{-1, 0, 2, 5, 6\} \)
\( = \{-2, 1, 3\} \)
Next, find \( A – C \):
\( A – C = \{-2, 0, 1, 3, 5\} – \{-1, 2, 5, 6, 7\} \)
\( = \{-2, 0, 1, 3\} \)
Now, find \( (A – B) \cap (A – C) \):
\( (A – B) \cap (A – C) = \{-2, 1, 3\} \cap \{-2, 0, 1, 3\} \)
\( = \{-2, 1, 3\} \) ........(2)

From (1) and (2), we can see that \( A – (B \cup C) = (A – B) \cap (A – C) \). This identity, often considered a variation of De Morgan's Laws, is successfully verified for the given sets. This principle is fundamental in understanding how set operations relate to each other.
In simple words: We are given three sets of numbers. We need to check if taking elements from A that are not in the combined group of B and C gives the same result as finding elements only in A (not B) and then finding elements only in A (not C), and then taking what is common to these two new groups. We do the calculations for both sides and compare the answers.

🎯 Exam Tip: Pay close attention to the order of operations and the specific definitions of union, intersection, and difference. A common mistake is to confuse union and intersection when applying De Morgan's Laws or related identities.

 

Question 8. IF A = {y : y = \( \frac{a + 1}{2} \), a ∈ W and a ≤ 5}, B = {y : y = \( \frac{2n – 1}{2} \), n ∈ W and n < 5} and C = {-1, \( -\frac{1}{2} \), 1, \( \frac{3}{2} \), 2} then show that A – (B∪C) = (A – B) ∩ (A – C).

Answer: First, let's write the sets A and B in roster form based on their definitions:
For Set A: \( y = \frac{a + 1}{2} \), where \( a \) is a whole number (W) and \( a \leq 5 \).
So, \( a = \{0, 1, 2, 3, 4, 5\} \).
When \( a = 0, y = \frac{0 + 1}{2} = \frac{1}{2} \)
When \( a = 1, y = \frac{1 + 1}{2} = \frac{2}{2} = 1 \)
When \( a = 2, y = \frac{2 + 1}{2} = \frac{3}{2} \)
When \( a = 3, y = \frac{3 + 1}{2} = \frac{4}{2} = 2 \)
When \( a = 4, y = \frac{4 + 1}{2} = \frac{5}{2} \)
When \( a = 5, y = \frac{5 + 1}{2} = \frac{6}{2} = 3 \)
So, \( A = \{ \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3 \} \)

For Set B: \( y = \frac{2n – 1}{2} \), where \( n \) is a whole number (W) and \( n < 5 \).
So, \( n = \{0, 1, 2, 3, 4\} \).
When \( n = 0, y = \frac{2(0) – 1}{2} = -\frac{1}{2} \)
When \( n = 1, y = \frac{2(1) – 1}{2} = \frac{1}{2} \)
When \( n = 2, y = \frac{2(2) – 1}{2} = \frac{3}{2} \)
When \( n = 3, y = \frac{2(3) – 1}{2} = \frac{5}{2} \)
When \( n = 4, y = \frac{2(4) – 1}{2} = \frac{7}{2} \)
So, \( B = \{ -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2} \} \)

Set C is given as: \( C = \{-1, -\frac{1}{2}, 1, \frac{3}{2}, 2\} \)

Now we need to show that \( A – (B \cup C) = (A – B) \cap (A – C) \).

**Left Hand Side (LHS): \( A – (B \cup C) \)**
First, find \( B \cup C \):
\( B \cup C = \{ -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2} \} \cup \{ -1, -\frac{1}{2}, 1, \frac{3}{2}, 2 \} \)
\( = \{ -1, -\frac{1}{2}, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, \frac{7}{2} \} \)
Now, find \( A – (B \cup C) \):
\( A – (B \cup C) = \{ \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3 \} – \{ -1, -\frac{1}{2}, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, \frac{7}{2} \} \)
\( = \{3\} \) ........(1)

**Right Hand Side (RHS): \( (A – B) \cap (A – C) \)**
First, find \( A – B \):
\( A – B = \{ \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3 \} – \{ -\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2} \} \)
\( = \{1, 2, 3\} \)
Next, find \( A – C \):
\( A – C = \{ \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3 \} – \{ -1, -\frac{1}{2}, 1, \frac{3}{2}, 2 \} \)
\( = \{ \frac{1}{2}, \frac{5}{2}, 3 \} \)
Now, find \( (A – B) \cap (A – C) \):
\( (A – B) \cap (A – C) = \{1, 2, 3\} \cap \{ \frac{1}{2}, \frac{5}{2}, 3 \} \)
\( = \{3\} \) ........(2)

From (1) and (2), we see that \( A – (B \cup C) = (A – B) \cap (A – C) \). This identity is verified for the given fractional sets. It is important to be careful with fractional values to ensure accurate set representation and calculations.
In simple words: We start with three sets that contain fractional numbers. First, we list all the exact numbers in each set using the given rules. Then, we perform the set operations on the left side of the equation, followed by the operations on the right side. If the final set of numbers on both sides is the same, then the equation is proven correct for these specific sets.

🎯 Exam Tip: When dealing with sets defined by formulas (set-builder notation), calculate each element carefully by substituting the allowed values for the variable (e.g., 'a' or 'n'). Fractional values require precise comparison during set operations.

 

Question 9. Verify A- (B∩C) = (A – B) ∪ (A – C) using Venn diagrams.

Answer: To verify \( A – (B \cap C) = (A – B) \cup (A – C) \) using Venn diagrams, we draw each part and compare the final shaded regions.

(i) Diagram for \( A – (B \cap C) \): (Shade area A, excluding the common part of B and C)

(ii) Diagram for \( A – B \): (Shade area A that is outside of B)

(iii) Diagram for \( A – C \): (Shade area A that is outside of C)

(iv) Diagram for \( (A – B) \cup (A – C) \): (Shade the union of the regions from (ii) and (iii))

By comparing the shaded region of diagram (i) representing \( A – (B \cap C) \) with diagram (iv) representing \( (A – B) \cup (A – C) \), we can see that both diagrams have identical shaded areas. This visually verifies the given set identity.
In simple words: We draw Venn diagrams for each side of the equation. First, we shade the part of set A that is not in the common area of B and C. Then, we shade the part of A that is not in B, and separately shade the part of A that is not in C, and combine these two shaded areas. If the final shaded patterns look exactly the same, the rule is proven by pictures.

🎯 Exam Tip: When shading for set differences like A-B, ensure you only shade the part of A that does not overlap with B. For unions, combine all shaded regions from the individual parts.

 

Question 10. If U = {4, 7, 8, 10, 11, 12, 15, 16}, A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan's Laws for complementation.
(i) (A∪B)' = A'∩B'
(ii) (A∩B)' = A'∪B'

Answer: Given the sets:
\( U = \{4, 7, 8, 10, 11, 12, 15, 16\} \)
\( A = \{7, 8, 11, 12\} \)
\( B = \{4, 8, 12, 15\} \)

(i) Verify \( (A \cup B)' = A' \cap B' \):
**Left Hand Side (LHS): \( (A \cup B)' \)**
First, find \( A \cup B \):
\( A \cup B = \{7, 8, 11, 12\} \cup \{4, 8, 12, 15\} \)
\( = \{4, 7, 8, 11, 12, 15\} \)
Next, find \( (A \cup B)' \):
\( (A \cup B)' = U – (A \cup B) = \{4, 7, 8, 10, 11, 12, 15, 16\} – \{4, 7, 8, 11, 12, 15\} \)
\( = \{10, 16\} \) ........(1)

**Right Hand Side (RHS): \( A' \cap B' \)**
First, find \( A' \):
\( A' = U – A = \{4, 7, 8, 10, 11, 12, 15, 16\} – \{7, 8, 11, 12\} \)
\( = \{4, 10, 15, 16\} \)
Next, find \( B' \):
\( B' = U – B = \{4, 7, 8, 10, 11, 12, 15, 16\} – \{4, 8, 12, 15\} \)
\( = \{7, 10, 11, 16\} \)
Now, find \( A' \cap B' \):
\( A' \cap B' = \{4, 10, 15, 16\} \cap \{7, 10, 11, 16\} \)
\( = \{10, 16\} \) ........(2)

From (1) and (2), we see that \( (A \cup B)' = A' \cap B' \). This verifies the first De Morgan's Law.

(ii) Verify \( (A \cap B)' = A' \cup B' \):
**Left Hand Side (LHS): \( (A \cap B)' \)**
First, find \( A \cap B \):
\( A \cap B = \{7, 8, 11, 12\} \cap \{4, 8, 12, 15\} \)
\( = \{8, 12\} \)
Next, find \( (A \cap B)' \):
\( (A \cap B)' = U – (A \cap B) = \{4, 7, 8, 10, 11, 12, 15, 16\} – \{8, 12\} \)
\( = \{4, 7, 10, 11, 15, 16\} \) ........(1)

**Right Hand Side (RHS): \( A' \cup B' \)**
From part (i), we already found:
\( A' = \{4, 10, 15, 16\} \)
\( B' = \{7, 10, 11, 16\} \)
Now, find \( A' \cup B' \):
\( A' \cup B' = \{4, 10, 15, 16\} \cup \{7, 10, 11, 16\} \)
\( = \{4, 7, 10, 11, 15, 16\} \) ........(2)

From (1) and (2), we see that \( (A \cap B)' = A' \cup B' \). This verifies the second De Morgan's Law. These laws are very important in set theory and logic, helping to simplify complex expressions.
In simple words: We are given a main group of numbers (Universal Set) and two smaller groups (A and B). De Morgan's Laws tell us two specific rules about how 'not in A or B' relates to 'not in A and not in B', and how 'not in A and B' relates to 'not in A or not in B'. We test both rules by doing the calculations on each side of the equal sign. Since both sides give the same final group of numbers, the laws are proven true for these sets.

🎯 Exam Tip: De Morgan's Laws are fundamental. Remember the pattern: the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements. Always list the Universal Set clearly before computing complements.

 

Question 11. Verify (A∩B)' = A'∪B' using Venn diagrams.

Answer: To verify \( (A \cap B)' = A' \cup B' \) using Venn diagrams, we compare the shaded regions of both sides of the equation.

(i) Diagram for \( A \cap B \): (Shade the intersection area of circles A and B)

(ii) Diagram for \( (A \cap B)' \): (Shade everything outside the intersection of A and B)

(iii) Diagram for \( A' \): (Shade everything outside circle A)

(iv) Diagram for \( B' \): (Shade everything outside circle B)

(v) Diagram for \( A' \cup B' \): (Shade the union of the regions from (iii) and (iv))

By comparing the shaded region of diagram (ii) representing \( (A \cap B)' \) with diagram (v) representing \( A' \cup B' \), we can see that both diagrams have identical shaded areas. This graphically verifies the second De Morgan's Law. This law is very useful for simplifying complex set operations by changing how complements, unions, and intersections are applied.
In simple words: We draw pictures to show that 'not (A and B)' is the same as '(not A) or (not B)'. First, we shade the area that is NOT where A and B overlap. Then, we separately shade everything outside A, and everything outside B, and combine those two shaded areas. If the final shaded patterns are identical, the rule is proven using Venn diagrams.

🎯 Exam Tip: When shading complements, remember to shade the entire universal set except for the specified set. For the union of complements, ensure all areas outside both A and B are shaded, including the area outside their union.