Rudin PMA Theorem 1.21 Existence of nth roots of positive reals

[unintuit]

Homework Statement

For every real x>0 and every n>0 there is one and only one positive real y s.t. yⁿ=x

Homework Equations

0<y₁<y₂ ⇒ y₁ⁿ<y₂ⁿ
E is the set consisting of all positive real numbers t s.t. tⁿ<x
t=[x/(x+1)]⇒ 0≤t<1. Therefore tⁿ≤t<x. Thus t∈E and E is non-empty.
t>1+x ⇒ tⁿ≥t>x, s.t. t∉E, and 1+x is an upper bound of E.
By the existence of an ordered field ℝ which has the least-upper-bound property we see y=SupE.

The identity bⁿ-aⁿ=(b-a)(b^n-1+ab^n-2+...+a^n-2b+a^n-1) yields the inequality bⁿ-aⁿ<(b-a)nb^n-1 when 0<a<b.

The Attempt at a Solution

Assume yⁿ<x. Choose h so that 0<h<1 and
h<(x-yⁿ)/(n(y+1)^n-1).

Put a=y, b=y+h. Then (y+h)ⁿ-yⁿ<hn(y+h)^n-1<hn(y+1)^n-1<x-yⁿ ⇒ (y+h)ⁿ<x and y+h∈E but y+h>y in contradiction to y=SupE.*Now herein lies my problem understanding this proof.*

Assume yⁿ>x.

*Put k=(yⁿ-x)/(ny^n-1) Then 0<k<y. If t≥y-k, we may conclude
yⁿ-tⁿ≤yⁿ-(y-k)ⁿ<kny^n-1=yⁿ-x.
Thus tⁿ>x, and t∉E.
It follows that y-k is an upper bound of E. However, y-k<y in contradiction to y=SupE.
Hence yⁿ=x.

I have tried to go through this proof repeatedly in order to understand why Rudin has chosen an equality and the particular 'k' for this case.
If anyone could help elucidate this problem, i.e. the seemingly random selection of the k-equality, I would greatly appreciate the light shed upon this dark spot.

Thank you for your effort.

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?