# Proof of a theorem (y^n=x)

1. Mar 18, 2012

### jwqwerty

the theorem goes likes this:
For every x>0 and every integer n>0 there is one and only one real y>0 such that y^n=x

The book starts the proof by stating E as set consisting of all positive real numbers t such that t<x^n. Then it states that:

If t= x/(1+x) then 0<t<1. Hence t^n<t<x. Thus t exists in E and E is not empty
If t>1+x then t^n>t>x so that t does not exist in E. Thus 1+x is an upper bound of E.

My questions is this:
Why does it divide into two cases, t= x/(1+x) and t>1+x? And instead, why can't we divide into t> x/(1+x) and t=1+x? Don't they still have the same meaning as the previous one in the way that 0<t<1 and t>1?

Last edited: Mar 18, 2012
2. Mar 18, 2012

### mathman

The original statement is wrong. If n is even there are two values of y, one positive and the other negative.

3. Mar 18, 2012

### jwqwerty

oh sorry i changed the statement from all real y to all real y>0