# Rudin 1.21 Problem understanding proof of unique positive root to the

1. Nov 25, 2011

### EdMel

Firstly, the set E is defined:
"Let E be the set of all positive real numbers t such that tn<x."

Later on the proof goes:
"Assume yn>x. Put k=yn-x / nyn-1. Then 0 < k <y. If ty - k, we conclude that
yn-tn ≤ yn-(y-k)n < knyn-1 = yn-x.
Thus tn>x, and t is not a member of E. It follows that y - k is an upper bound of E."

Why does it follow? Is it because the possibility that t = y - k combined with the fact that tn>x mean that y-k always has to be an upper bound of E? Or is there some other reasoning?

2. Nov 27, 2011

### EdMel

Re: Rudin 1.21 Problem understanding proof of unique positive root to the nth power.

Hi All,

I feel I understand this now.

There were two things I did not have right in my head.

1) To prove the assumption yn > x is false all you need to do is find one instance when it is false. Having t ≥ y - k provides one instance. I was not recognizing this.

I was getting confused about what would happen if you set t = y - k, or t > y - k, or t < y - k; it seems that using the either of the first two would suffice in finding one instance to disprove the assumption, but you can treat them together as is done in the proof. Using t < y - k doesn't get you anywhere.

2) It is obvious that y - k is an upper bound of E if you picture E on the number line and y-k somewhere to the right of it, and the set of t ≥ y - k pointing from here out to the right.

Peace out.

3. Nov 9, 2013

My question is why didn't Rudin simply define t as $$t=y-k$$ instead of $$t\ge y-k$$?