IF a=[1,2,3,4, and B=[1,2,3,5,6]
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Given, A={1,2,3,4,5},B={2,4,6,8} and C={3,4,5,6}
For the LHS:
Union of two sets will have the elements of both sets.
So, B∪C={2,3,4,5,6,8}
A−(B∪C) will have elements of A which are not in (B∪C)
So, A−(B∪C)={1} ..... (1)
For the RHS:
A−B will have elements of A which are not in B.
So, A−B={1,3,5}
A−C will have elements of A which are not in C.
So, A−C={1,2}
Intersection of two sets has the common elements of both the sets.
⇒(A−B)∩(A−C)={1} ..... (2)
From (1) and (2), we have
A−(B∪C)=(A−B)∩(A−C)
Hence, the given expression is true.
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