- #1

unintuit

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## Homework Statement

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

^{n}=x

## Homework Equations

0<y

_{1}<y

_{2}⇒ y

_{1}

^{n}<y

_{2}

^{n}

E is the set consisting of all positive real numbers t s.t. t

^{n}<x

t=[x/(x+1)]⇒ 0≤t<1. Therefore t

^{n}≤t<x. Thus t∈E and E is non-empty.

t>1+x ⇒ t

^{n}≥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 b

^{n}-a

^{n}=(b-a)(b

^{n-1}+ab

^{n-2}+...+a

^{n-2}b+a

^{n-1}) yields the inequality b

^{n}-a

^{n}<(b-a)nb

^{n-1}when 0<a<b.

## The Attempt at a Solution

Assume y

^{n}<x. Choose h so that 0<h<1 and

h<(x-y

^{n})/(n(y+1)

^{n-1}).

Put a=y, b=y+h. Then (y+h)

^{n}-y

^{n}<hn(y+h)

^{n-1}<hn(y+1)

^{n-1}<x-y

^{n}⇒ (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 y

^{n}>x.

*Put k=(y

^{n}-x)/(ny

^{n-1}) Then 0<k<y. If t≥y-k, we may conclude

y

^{n}-t

^{n}≤y

^{n}-(y-k)

^{n}<kny

^{n-1}=y

^{n}-x.

Thus t

^{n}>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 y

^{n}=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.