| Line 10: | Line 10: | ||
n=1: <math>2^1=2</math> Subsets with 1 elements: {∅}, {1} | n=1: <math>2^1=2</math> Subsets with 1 elements: {∅}, {1} | ||
| − | So we can assume a set ''S'' with ''n'' elements has | + | So we can assume a set ''S'' with ''n'' elements has <math>2^n</math> subsets. |
n+1: <math>2^(n+1) = 2^1 + 2^n = 2*2^n = 2^(n+1)</math> | n+1: <math>2^(n+1) = 2^1 + 2^n = 2*2^n = 2^(n+1)</math> | ||
Revision as of 09:46, 7 September 2008
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
