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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> | ||
| + | |||
| + | P(1) = 1-1 = 0 (true,divisible by 5) | ||
| + | |||
| + | P(2) = 32-2 = 30 (true,divisible by 5) | ||
| + | |||
| + | let P(m) be true, P(m+1) = <math> (m+1)^5-m </math> | ||
| + | |||
| + | P(m+1) = <math> (m^5+1)+5*(other terms) </math>. | ||
| + | |||
| + | which is divisible by 5 since P(m) is true. | ||
| + | |||
| + | Hence 5| <math> n^5-n </math> | ||
Revision as of 13:48, 21 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) $
P(1) = 1-1 = 0 (true,divisible by 5)
P(2) = 32-2 = 30 (true,divisible by 5)
let P(m) be true, P(m+1) = $ (m+1)^5-m $
P(m+1) = $ (m^5+1)+5*(other terms) $.
which is divisible by 5 since P(m) is true.
Hence 5| $ n^5-n $
