Question

Prove by ' Induction' that the number of all subsets of set A which contains n distinct elements is 2^n ..........
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Answer 1

Answer:

Here is a proof by induction on n. This proof assumes that we have defined 2n by recursion as 20=1 and 2n+1=2n⋅2.

This is true for n=0 because ∅ has exactly one subset, namely ∅ itself.

Now assume that the claim is true for sets with n many elements. Given a set Y with n+1 many elements, we can write Y=X∪{p} where X is a set with n many elements and p∉X. There are 2n many subsets A⊂X, and each subset A⊂X gives rise to two subsets of Y, namely A∪{p} and A itself. Moreover, every subset of Y arises in this manner. Therefore the number of subsets of Y is equal to 2n⋅2, which in turn is equal to 2n+1.

If a set contains 'n' elements, then the number of subsets of the set is 22. Number of Proper Subsets of the Set: If a set contains 'n' elements, then the number of proper subsets of the set is 2n - 1. In general, number of proper subsets of a given set = 2m - 1, where m is the number of elements.