Page 88 - ICSE Math 7

For example, let E = {3, 4, 5, 6, 7, 8} and F = {2, 4, 6, 8, 10, 12, 14}, then E – F = {3, 5, 7} and
F – E = {2, 10, 12, 14}.

As, E – F ≠ F – E, the difference of two sets is not commutative.

Example 12: Let A = {9, 11, 13, 17, 21} and B = {5, 7, 8, 9, 11, 15, 17}. Find the following sets.
(a) A – B (b) B – A
Solution: Common elements of A and B are 9, 11, 17.

(a) A – B = {13, 21} (Removing 9, 11, 17 from set A and writing the remaining
elements of set A)

(b) B – A = {5, 7, 8, 15} (Removing 9, 11, 17 from set B and writing the remaining
elements of set B)

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Example 13: If A = {x | x = 2n + 1, n ∈ N, n ≤ 4}, B = {x | x = n + 1, n ∈ N, n ≤ 5} and
C = {x | x is a factor of 20}, verify that:
(a) A – (B ∪ C) = (A – B) ∩ (A – C) (b) A – (B ∩ C) = (A – B) ∪ (A – C)
Solution: The sets in roster form are given by:
A = {3, 5, 7, 9}, B = {2, 5, 10, 17, 26}, C = {1, 2, 4, 5, 10, 20}

(a) B ∪ C = {2, 5, 10, 17, 26, 1, 4, 20}, A – B = {3, 7, 9}, A – C = {3, 7, 9}
A – (B ∪ C) = {3, 5, 7, 9} – {2, 5, 10, 17, 26, 1, 4, 20} = {3, 7, 9}
( A – B) ∩ (A – C) = {3, 7, 9} ∩ {3, 7, 9} = {3, 7, 9}

\ A – (B ∪ C) = (A – B) ∩ (A – C)
(b) B ∩ C = {2, 5, 10}, A – B = {3, 7, 9}, A – C = {3, 7, 9}
A – (B ∩ C) = {3, 5, 7, 9} – {2, 5, 10} = {3, 7, 9}
( A – B) ∪ (A – C) = {3, 7, 9} ∪ {3, 7, 9} = {3, 7, 9}
\ A – (B ∩ C) = (A – B) ∪ (A – C)

Cardinal Properties of Sets
For any non-empty sets A and B,
• n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
• If A and B are disjoint sets, then A ∩ B = f. Therefore, n(A ∩ B) = 0.

Hence, n(A ∪ B) = n(A) + n(B)
• n(A – B) = n(A) – n(A ∩ B)
• n(B – A) = n(B) – n(A ∩ B)

Example 14: Let P and Q be two sets such that P ∪ Q has 60 elements, P has 28 elements and Q
has 42 elements. How many elements are there in P ∩ Q?
Solution: n(P ∪ Q) = 60, n(P) = 28, n(Q) = 42
n(P ∪ Q) = n(P) + n(Q) – n(P ∩ Q)

⇒ 60 = 28 + 42 – n(P ∩ Q)
⇒ 60 = 70 – n (P ∩ Q)
\ n(P ∩ Q) = 70 – 60 = 10
Hence, the number of elements in P ∩ Q is 10.

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