I've been having a bit of a a think about this question but still haven't come up with a particuarly good answer yet.
So what's wrong with the following proof that x^n=1 for all non negative intergers n, and all non zero real x. Let P(n) be the proposition that x^n=1.
The base case x^0=1 (so that's fine)
Inductive hypothesis: suppose n greater than equal to 0, and that P(m) true for all m less than or equal to n. (i.e we're assuming everything up to P(n) is true. Then x^n=x^(n-1)=1 based on our assumption.
So p(n+1) is true. So by complete induction P(n) is true.
Thanks in advance
Just been looking at the proof a bit more, think I may have got somewhere.
We've assumed that P(m) is true for all m less than or equal to n, and n is greater than or equal to 0.
From this we got x^n=x^(n-1)=1.
If we choose n=0 then we're told to assume x^n=1. But n-1=-1, which isn't greater than or equal to 0. As such we can't assume that x^n-1=1.
Hence x^(n+1)= (x^n*x^n)/x^(n-1)=1.1/1=1 only when n is greater than or equal to 1. So the problem is getting between the base case and the n=1 step (one doesn't lead to the other).
This seems right as if you know x^1=1 it follows that x^2=x*x=1, x^3=x*x*x=1 and so on. But just knowing x^0=1 doesn't enable you to say x^1=1.
Does this sound along the right lines?