Proving the Identity Function: Composed Functions

andmcg
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Suppose that f composed with g equals g composed with f for all functions g. Show that f is the identity function.

I really just don't know where to start.
 
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That was probably a bad tip. I'll try to think of something better.

OK, here we go...

If f isn't the identity, there's an x such that f(x)≠x. Now choose a g that contradicts that.
 
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Thread 'Finding the nth roots of a complex number'

There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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