How to proof /P(A)/=2ⁿ / / For modulus sign A: Given set P(A):Power of set A 2ⁿ: The number of possible elements. Hints: In the question it says that to prove it by induction method . /P(A)/=2ⁿ is possible when set A has 'n' number of elements.
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By induction on $n$. Let $P(n)$ be the sentence $P(n)=(|A|=n\Rightarrow |\mathcal{P}(A)|=2^n)$. Then $P(1)=(|A|=1\Rightarrow |\mathcal{P}(A)|=2)$, which is true, because if $|A|=1$ then $\mathcal{P}(A)=\{\varnothing, A\}$. Let now $|A|=n+1$. Let $x\in A$. Consider the set $A'=A-\{x\}$. Counting the subsets of $A$, we have all the subsets of $A'$ and all the subsets of $A'$ with $x$ added to them, hence in total $2^n+2^n=2^{n+1}$ subsets.
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