- #1
jwqwerty
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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?
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?
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