Revision as of 10:46, 7 September 2008 by Jstraete (Talk)

Using Binomial Theorem, $ (a+b)^n=\binom{n}{0}a^n+ \binom n 1 a^{n-1} b+...+\binom{n}{n}b^n $.

We have $ \binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n $


Using Induction

Base case:
n=0: $ 2^0=1 $ Subsets with 0 elements: {∅}
n=1: $ 2^1=2 $ Subsets with 1 elements: {∅}, {1}

So we can assume a set S with n elements has $ 2^n $ subsets.

n+1: $ 2^(n+1) = 2^1 + 2^n = 2*2^n = 2^(n+1) $

-Jesse Straeter


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