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Prove that <math> \forall n\in{\mathbb N}, n^5-n </math> is a multiple of n. | Prove that <math> \forall n\in{\mathbb N}, n^5-n </math> is a multiple of n. | ||
| − | Base case: n=0... 0^5=0 as we want | + | Base case: n=0... <math> 0^5=0 </math> as we want |
Inductive step: Assume that 5 divides <math> n^5-n </math> and show that 5 divides <math> (n+1)^5-(n+1) </math> | Inductive step: Assume that 5 divides <math> n^5-n </math> and show that 5 divides <math> (n+1)^5-(n+1) </math> | ||
Revision as of 15:39, 7 September 2008
Example: Prove that $ \forall n\in{\mathbb N}, n^5-n $ is a multiple of n.
Base case: n=0... $ 0^5=0 $ as we want
Inductive step: Assume that 5 divides $ n^5-n $ and show that 5 divides $ (n+1)^5-(n+1) $
