PROPERTIES OF UNION AND INTERSECTION - Unit 17 - Solved Exercise 17.3 - Mathematics For Class X Science

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Unit 17: SETS AND FUNCTIONS
Solved Exercise 17.3

1. Verify the commutative property of union and intersection for the following sets.

1. A = {a,b,c,d,e} and B = {a,e,i,o,u}
2. P = {x | x ∈ Z ⋀ -3 < x< 3} and Q = {y | y ∈ E⁺ ⋀ y ≤ 4}

(i) A = {a,b,c,d,e} and B = {a,e,i,o,u}
SOLUTION:
VERIFICATION OF COMMUTATIVE PROPERTY OF UNION

∵ AUB = BUA

Proof:
Taking L.H.S
AUB = {a,b,c,d,e} U {a,e,i,o,u}
AUB = {a,b,c,d,e,i,o,u}

Taking R.H.S
BUA = {a,e,i,o,u} U {a,b,c,d,e}
BUA = {a,b,c,d,e,i,o,u}
∵ AUB = BUA
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF COMMUTATIVE PROPERTY OF INTERSECTION

∵ A∩B = B∩A

Proof:
Taking L.H.S
A∩B = {a,b,c,d,e} ∩ {a,e,i,o,u}
A∩B = {a,e}

Taking R.H.S
B∩A = {a,e,i,o,u} ∩ {a,b,c,d,e}
B∩A = {a,e}
∵ A∩B = B∩A
∴ L.H.S = R.H.S
Hence Proved

(ii) P = {x | x ∈ Z ⋀ -3 < x< 3} and Q = {y | y ∈ E⁺ ⋀ y ≤ 4}
SOLUTION:
P = {x | x ∈ Z ⋀ -3 < x< 3} = {-2, -1, 0, 1, 2}
Q = {y | y ∈ E⁺ ⋀ y ≤ 4} = {2, 4}

VERIFICATION OF COMMUTATIVE PROPERTY OF UNION

∵ PUQ = QUP

Proof:
Taking L.H.S
PUQ = {-2,-1,0,1,2} U {2,4}
PUQ = {-2, -1, 0, 1, 2, 4}

Taking R.H.S
QUP = {2,4} U {-2 -1,0,1,2}
QUP = {-2, -1, 0, 1, 2, 4}
∵ PUQ = QUP
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF COMMUTATIVE PROPERTY OF INTERSECTION

∵ P∩Q = Q∩P

Proof:
Taking L.H.S
P∩Q = {-2,-1,0,1,2} ∩ {2,4}
P∩Q = {2}

Taking R.H.S
Q∩P = {2,4} ∩ {-2,-1,0,1,2}
Q∩P = {2}
∵ P∩Q = Q∩P
∴ L.H.S = R.H.S
Hence Proved

2. Verify the associative property of union and intersection for the following sets.

1. A = {1,2,4,5,10,20}, B = {5,10,15,20} and C = {1,2,5,10}
2. A = N, B = P and C = Z

SOLUTION:
(i) A = {1,2,4,5,10,20}, B = {5,10,15,20} and C = {1,2,5,10}
VERIFICATION OF ASSOCIATIVE PROPERTY OF UNION

∵ AU(BUC) = (AUB)UC

Proof:
Taking L.H.S
AU(BUC) = {1,2,4,5,10,20} U [{5,10,15,20} U {1,2,5,10}]
AU(BUC) = {1,2,4,5,10,20} U {1,2,5,10,15,20}
AU(BUC) = {1,2,4,5,10,15,20}

Taking R.H.S
(AUB)UC = [{1,2,4,5,10,20} U {5,10,15,20}] U {1,2,5,10}
(AUB)UC = {1,2,4,5,10,15,20} U {1,2,5,10}
(AUB)UC = {1,2,4,5,10,15,20}
∵ AU(BUC) = (AUB)UC
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF ASSOCIATIVE PROPERTY OF INTERSECTION

∵ A∩(B∩C) = (A∩B)∩C

Proof:
Taking L.H.S
A∩(B∩C) = {1,2,4,5,10,20} ∩ [{5,10,15,20} ∩ {1,2,5,10}]
A∩(B∩C) = {1,2,4,5,10,20} ∩ {5,10}
A∩(B∩C) = {5,10}

Taking R.H.S
(A∩B)∩C = [{1,2,4,5,10,20} ∩ {5,10,15,20}] ∩ {1,2,5,10}
(A∩B)∩C = {5,10,20} ∩ {1,2,5,10}
(A∩B)∩C = {5,10}
∵ A∩(B∩C) = (A∩B)∩C
∴ L.H.S = R.H.S
Hence Proved

(ii) A = N, B = P and C = Z
SOLUTION:
A = N = {1, 2, 3, 4, 5, ......}
B = P = {2, 3, 5, 7, 11, .....}
C = Z = {0, ±1, ±2, ±3, ±4, .....}

VERIFICATION OF ASSOCIATIVE PROPERTY OF UNION

∵ AU(BUC) = (AUB)UC

Proof:
Taking L.H.S
AU(BUC) = {1, 2, 3, 4, 5, ......} U [{2, 3, 5, 7, 11, .....} U {0, ±1, ±2, ±3, ±4, .....}]
AU(BUC) = {1,2,4,5,10,20} U {0, ±1, ±2, ±3, ±4, .....}
AU(BUC) = {0, ±1, ±2, ±3, ±4, .....} = Z

Taking R.H.S
(AUB)UC = [{1, 2, 3, 4, 5, ......} U [{2, 3, 5, 7, 11, .....}] U {0, ±1, ±2, ±3, ±4, .....}
(AUB)UC = {1, 2, 3, 4, 5, ......} U {0, ±1, ±2, ±3, ±4, .....}
(AUB)UC = {0, ±1, ±2, ±3, ±4, .....} = Z
∵ AU(BUC) = (AUB)UC
∴ L.H.S = R.H.S
Hence Proved

VERIFICATION OF ASSOCIATIVE PROPERTY OF INTERSECTION

∵ A∩(B∩C) = (A∩B)∩C

Proof:
Taking L.H.S
A∩(B∩C) = {1, 2, 3, 4, 5, ......} ∩ [{2, 3, 5, 7, 11, .....} ∩ {0, ±1, ±2, ±3, ±4, .....}]
A∩(B∩C) = {1, 2, 3, 4, 5, ......} ∩ {2, 3, 5, 7, 11, .....}
A∩(B∩C) = {2, 3, 5, 7, 11, .....} = P

Taking R.H.S
(A∩B)∩C = [{1, 2, 3, 4, 5, ......} ∩ {2, 3, 5, 7, 11, .....}] ∩ {0, ±1, ±2, ±3, ±4, .....}
(A∩B)∩C = {2, 3, 5, 7, 11, .....} ∩ {0, ±1, ±2, ±3, ±4, .....}
(A∩B)∩C = {2, 3, 5, 7, 11, .....} = P
∵ A∩(B∩C) = (A∩B)∩C
∴ L.H.S = R.H.S
Hence Proved

3. Verify
(a) Distributive property of union over intersection.
(b) Distributive property of intersection over union for the following sets.

1. A = {1,2,3,...,10}, B = {2,3,5,7} and C = {1,3,5,7,9}
2. A = N, B = P and C = W

(i) A = {1,2,3,...,10}, B = {2,3,5,7} and C = {1,3,5,7,9}
SOLUTION:
(a) VERIFICATION OF DISTRIBUTIVE PROPERTY OF UNION OVER INTERSECTION:

AU(B∩C) = (AUB) ∩ (AUC)

Proof:
Taking L.H.S
AU(B∩C) = {1,2,3,...,10} U [{2,3,5,7} ∩ {1,3,5,7,9}]
AU(B∩C) = {1,2,3,...,10} U {3,5,7}
AU(B∩C) = {1,2,3,...,10}

Taking R.H.S
(AUB) ∩ (AUC) = [{1,2,3,...,10} U {2,3,5,7}] ∩ [{1,2,3,...,10} U {1,3,5,7,9}]
(AUB) ∩ (AUC) = {1,2,3,...,10} ∩ {1,2,3,...,10}
(AUB) ∩ (AUC) = {1,2,3,...,10} ∵ AU(B∩C) = (AUB) ∩ (AUC)
∴ L.H.S = R.H.S
Hence Proved

(b) VERIFICATION OF DISTRIBUTIVE PROPERTY OF INTERSECTION OVER UNION:

A∩(BUC) = (A∩B) ∩ (A∩C)

Proof:
Taking L.H.S
A∩(BUC) = {1,2,3,...,10} ∩ [{2,3,5,7} U {1,3,5,7,9}]
A∩(BUC) = {1,2,3,...,10} ∩ {1,2,3,5,7,9}
A∩(BUC) = {1,2,3,5,7,9}

Taking R.H.S
(A∩B) U (A∩C) = [{1,2,3,...,10} ∩ {2,3,5,7}] U [{1,2,3,...,10} ∩ {1,3,5,7,9}]
(A∩B) U (A∩C) = {2,3,5,7} U {1,3,5,7,9}
(A∩B) U (A∩C) = {1,2,3,5,7,9}
∵ A∩(BUC) = (A∩B) U (A∩C)
∴ L.H.S = R.H.S
Hence Proved

(ii) A = N, B = P and C = W
SOLUTION:
A = N = {1,2,3,4,5,.......}
B = p = {2,3,5,7,11,......}
C = W = {0,1,2,3,4,.......}

(a) VERIFICATION OF DISTRIBUTIVE PROPERTY OF UNION OVER INTERSECTION:

AU(B∩C) = (AUB) ∩ (AUC)

Proof:
Taking L.H.S
AU(B∩C) = {1,2,3,4,5,.......} U [{2,3,5,7,11,......} ∩ {0,1,2,3,4,.......}]
AU(B∩C) = {1,2,3,4,5,.......} U {2,3,5,7,11,......}
AU(B∩C) = {1,2,3,4,5,.......} = N

Taking R.H.S
(AUB) ∩ (AUC) = [{1,2,3,4,5,.......} U {2,3,5,7,11,......}] ∩ [{1,2,3,4,5,.......} U {0,1,2,3,4,.......}]
(AUB) ∩ (AUC) = {1,2,3,4,5,.......} ∩ {0,1,2,3,4,.......}
(AUB) ∩ (AUC) = {1,2,3,4,5,.......} = N
∵ AU(B∩C) = (AUB) ∩ (AUC)
∴ L.H.S = R.H.S
Hence Proved

(b) VERIFICATION OF DISTRIBUTIVE PROPERTY OF INTERSECTION OVER UNION:

A∩(BUC) = (A∩B) U (A∩C)

Proof:
Taking L.H.S
A∩(BUC) = {1,2,3,4,5,.......} ∩ [{2,3,5,7,11,......} U {0,1,2,3,4,.......}]
A∩(BUC) = {1,2,3,4,5,.......} ∩ {0,1,2,3,4,.......}
A∩(BUC) = {1,2,3,4,5,.......}

Taking R.H.S
(A∩B) U (A∩C) = [{1,2,3,4,5,.......} ∩ {2,3,5,7,11,......}] U [{1,2,3,4,5,.......} ∩ {0,1,2,3,4,.......}]
(A∩B) U (A∩C) = {2,3,5,7,11,......} U {1,2,3,4,5,.......}
(A∩B) U (A∩C) = {1,2,3,4,5,.......} = N
∵ A∩(BUC) = (A∩B) U (A∩C)
∴ L.H.S = R.H.S
Hence Proved

4. Verify De Morgan's laws if U = {1,2,3...,12}, A = {1,2,3,4,6,12} and B = {2,4,6,8}.
SOLUTION:
According To De Morgan's Law

• (AUB)' = A'∩B'
• (A∩B)' = A'UB'

VERIFICATION OF DE MORGAN"S LAW:
(i) (AUB)' = A'∩B'
Proof:
Taking L.H.S
(AUB)' = U - (AUB)
(AUB)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [{1,2,3,4,6,12} U {2,4,6,8}]
(AUB)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [1,2,3,4,6,8,12}
(AUB)' = {5,7,9,10,11}

Taking R.H.S
A'∩B' = (U - A) ∩ (U - B)
A'∩B' = [{1,2,3,4,5,6,7,8,9,10,11,12} - {1,2,3,4,6,12}] ∩ [{1,2,3,4,5,6,7,8,9,10,11,12} - {2,4,6,8}]
A'∩B' = {5,7,8,9,10,11} ∩ {1,3,5,7,9,10,11}
A'∩B' = {5,7,9,10,11}
∵ (AUB)' = A'∩B'
∴ L.H.S = R.H.S
Hence Proved

(ii) (A∩B)' = A'UB'
Proof:
Taking L.H.S
(A∩B)' = U - (A∩B)
(A∩B)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [{1,2,3,4,6,12} ∩ {2,4,6,8}]
(A∩B)' = {1,2,3,4,5,6,7,8,9,10,11,12} - [2,4,6}
(A∩B)' = {1,3,5,7,8,9,10,11,12}

Taking R.H.S
A'UB' = (U - A) U (U - B)
A'UB' = [{1,2,3,4,5,6,7,8,9,10,11,12} - {1,2,3,4,6,12}] U [{1,2,3,4,5,6,7,8,9,10,11,12} - {2,4,6,8}] A'UB' = {5,7,8,9,10,11} U {1,3,5,7,9,10,11,12}
A'UB' = {1,3,5,7,8,9,10,11,12}
∵ (A∩B)' = A'UB'
∴ L.H.S = R.H.S
Hence Proved

5. If, A and B are subset of U then prove the following using properties.

1. AU(A∩B) = A∩(AUB)
2. AUB = AU(A'∩B)
3. B = (A∩B)U(A'∩B)
4. B = AU(A'∩B), if A⊆B

SOLUTION:
Let U = {1,2,3,4,5,6,7,8}
A = {1,2,3,4}
B = {2,4,6,8}

(i) AU(A∩B) = A∩(AUB)
Proof:
Taking L.H.S
AU(A∩B) = {1,2,3,4} U [{1,2,3,4} ∩ {2,4,6,8}
AU(A∩B) = {1,2,3,4} U {2,4}]
AU(A∩B) = {1,2,3,4}

Taking R.H.S
A∩(AUB) = {1,2,3,4} ∩ [{1,2,3,4} U {2,4,6,8}
A∩(AUB) = {1,2,3,4} ∩ {1,2,3,4,6,8}
A∩(AUB) = {1,2,3,4}
∵ AU(A∩B) = A∩(AUB)
∴ L.H.S = R.H.S
Hence Proved

(ii) AUB = AU(A'∩B)
Proof:
Taking L.H.S
AUB = {1,2,3,4} U {2,4,6,8}
AUB = {1,2,3,4,6,8}

Taking R.H.S
AU(A'∩B) = AU [(U - A) ∩ B] AU(A'∩B) = {1,2,3,4} U [({1,2,3,4,5,6,7,8} - {1,2,3,4}) ∩ {2,4,6,8}]
AU(A'∩B) = {1,2,3,4} U [{5,6,7,8} ∩ {2,4,6,8}]
AU(A'∩B) = {1,2,3,4} U {6,8}
AU(A'∩B) = {1,2,3,4,6,8}
∵ AUB = AU(A'∩B)
∴ L.H.S = R.H.S
Hence Proved

(iii) B = (A∩B)U(A'∩B)
Proof:
Taking L.H.S
B = {2,4,6,8}

Taking R.H.S
(A∩B)U(A'∩B) = [(A∩B)] U [(U - A) ∩ B]
(A∩B)U(A'∩B) = [{1,2,3,4} ∩ {2,4,6,8}] U [({1,2,3,4,5,6,7,8} - {1,2,3,4}) ∩ {2,4,6,8}]
(A∩B)U(A'∩B) = {2,4} U [{5,6,7,8} ∩ {2,4,6,8}
(A∩B)U(A'∩B) = {2,4} U {6,8}
(A∩B)U(A'∩B) = {2,4,6,8}
∵ B = (A∩B)U(A'∩B)
∴ L.H.S = R.H.S
Hence Proved

(iv) B = AU(A'∩B), if A⊆B
SOLUTION:
Let U = {1,2,3,4,5,6,7,8}
A = {2,4}
B = {2,4,6,8}
Proof:
Taking L.H.S
B = {2,4,6,8}

Taking R.H.S
AU(A'∩B) = A U [(U - A) ∩ B]
AU(A'∩B) = {2,4} U [({1,2,3,4,5,6,7,8} - {2,4}) ∩ {2,4,6,8}]
AU(A'∩B) = {2,4} U [{1,3,5,6,7,8} ∩ {2,4,6,8}]
AU(A'∩B) = {2,4} U {6,8}
AU(A'∩B) = {2,4,6,8}
∵ B = AU(A'∩B), if A⊆B
∴ L.H.S = R.H.S
Hence Proved