If ξ = { 1, 2, 3, 4, … … 9 }, A = { 2, 4, 6, 8 } and B = { 1, 3, 5, 7, 9 }, then find
1. A′
2.B′
Answers
Answer:
From the question it is given that, ξ={1,2,3,4,5,6,7,8,9}
A={1,2,3,4,6,7,8}
B={4,6,8}
(i) A
′
=ξ−A={1,2,3,4,5,6,7,8,9}−{1,2,3,4,6,7,8}
A
′
={5,9}
(ii) B
′
=ξ−B={1,2,3,4,5,6,7,8,9}−{4,6,8}
B
′
={1,2,3,5,7,9}
(iii) A∪B={1,2,3,4,6,7,8}∪{4,6,8}
A∪B={1,2,3,4,6,7,8}
(iv) A∩B={1,2,3,4,6,7,8}∩{4,6,8}
A∩B={4,6,8}
(v) A−B={1,2,3,4,6,7,8}−{4,6,8}
A−B={1,2,3,7}
(vi) B−A={4,6,8}−{1,2,3,4,6,7,8}
B−A={}
(vii) (A∩B)
′
=ξ−(A∩B)
(A∩B)
′
={1,2,3,4,5,6,7,8,9}−{4,6,8}
(A∩B)
′
={1,2,3,5,7,9}
(viii) A
′
∪B
′
={5,9}∪{1,2,3,5,7,9}
A
′
∪B
′
={1,2,3,5,7,9}
Then, n(ξ)=9
n(A)=7
n(A
′
)=2
n(B
′
)=6
n(A∩B)=3
n((A∩B)
′
)=6
n(A
′
∪B
′
)=6
n(A−B)=4
n(B−A)=0
n(A∪B)=7
(a) (A∩B)
′
=A
′
∪B
′
(A∩B)
′
={1,2,3,5,7,9}
A
′
∪B
′
={1,2,3,5,7,9}
By comparing the results, (A∩B)
′
=A
′
∪B
′
(b) n(A)+n(A
′
)=n(ξ)
7+2=9
9=9
Therefore, by comparing the results, n(A)+n(A
′
)=n(ξ)
(c) n(A∩B)+n((A∩B)
′
)=n(ξ)
3+6=9
9=9
Therefore, by comparing the results, n(A∩B)+n((A∩B)
′
)=n(ξ)
(d) n(A−B)+n(B−A)+n(A∩B)=n(A∪B)
4+0+3=7
7=7
Therefore, by comparing the results, n(A−B)+n(B−A)+n(A∩B)=n(A∪B)
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