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Base case: n=0...    0^5=0  as we want
 
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)
+
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 15:36, 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) $

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