Page 3 - mathsvol1ch1to3ans

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(v) (B − A) ∩ C = (B ∩ C) − A = B ∩ (C − A).
Solution:
Let a ∈ (B − A) ∩ C ⇔ a ∈ B, a /∈ A, a ∈ C

Since a ∈ B, a ∈ C ⇔ a ∈ B ∩ C
and a ∈ B, a ∈ C, a /∈ A ⇔ a ∈ (B ∩ C) − A
We have (B − A) ∩ C = (B ∩ C) − A

Let a ∈ (B − A) ∩ C ⇔ a ∈ B, a /∈ A, a ∈ C
Since a ∈ C, a /∈ A ⇔ a ∈ (C − A)

and a ∈ B, a ∈ (C − A) ⇔ a ∈ B ∩ (C − A)
Not For Sale - Veeraragavan C S veeraa1729@gmail.com
We have (B − A) ∩ C = B ∩ (C − A).

(vi) (B − A) ∪ C = (B ∪ C) − (A − C) by taking suitable A, B, C.
Solution: For this problem let us consider the sets given below.
Let A = {a, b, c, d}, B = {c, d, e, f}, C = {a, b, e, f}
B − A = {e, f}
(B − A) ∪ C = {a, b, e, f}
B ∪ C = {a, b, c, d, e, f}
A − C = {c, d}
(B ∪ C) − (A − C) = {a, b, e, f}
(5) Justify the trueness of the statement:
“An element of a set can never be a subset of itself.”
Solution: By the definition of a power set of a set contains the all the subsets of it. It turns out
that each element in the power set is a set. Since each set is a subset of itself, we can easily
show that every element of the power set is a subset of itself. For example, if A = {1, 2} and
B = {1, {1, 2}, 3, 4}, then A ∈ B.
(6) If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, find n(A ∩ B).
Solution:
n(P(A)) = 1024 ⇒ n(A) = 10
n(P(B)) = 32 ⇒ n(B) = 5
n(A ∩ B) = n(A) + n(B) − n(A ∪ B)
= 10 + 5 − 15 = 0
n(A ∩ B) = 0
Both sets are disjoint to each other
(7) If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A∆B)).
Solution:
A∆B = (A ∪ B) − (A ∩ B)
n(A∆B) = n(A ∪ B) − n(A ∩ B)
= 10 − 3
n(A∆B) = 7
n(P(A∆B)) = 2 7
= 128
(8) For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the
elements of A.
Solution: n(A × A) = 16. n(A) = 4. Let the elements of A be {a, b, c, d}. Clearly the two
elements given can be represented as (a, b) and (c, d). Hence the elements of A are {0, 1, 2, 3}.
(9) Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B,
find A and B, where x, y, z are distinct elements.
Solution: A = {x, y, z} and B = {1, 2}