1.21 Proof Rudin Principles of Mathematical Analysis - Uniqueness of the n-roots in R

help with rudin 1.21 PoMA

  • help! rudin 1.21

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  • #1

Main Question or Discussion Point

Hi!

I need some help here, please.

In Principles of Mathematical Analysis, Rudin prove the uniqueness of the n-root on the real numbers. For doing so, he prove that both $y^n < 0$ and that $y^n > 0$ lead to a contradiction.

In the first part of the proof, he chooses a value $h$ such that
1. $0<h<1$
2 $h<\frac{x-y^n}{n(y+1)^(n-1)}$

My question is: how does he know such a value $h$ exists?

I know this value is positive, but I am not sure how to prove is less than 1.

To add some context:
x is a real positive number
n is a integer positive number
y is a real positive number

If someone can help me with this, I will be very grateful!

Thanks
 

Answers and Replies

  • #2


Even if it is not less than 1 , you still can choose h less than this fraction and in the same time 0<h<1.
 
  • #3


Don't get it. How can it be not less than 1, and at the same time, less than one?

I think the more likely answers are that:

1) No need to prove the existence of h
2) There is something I'm missing about y=sup E, where E is the set all positive real numbers t, such that t^n<x

By the way, perhaps I am not writing correctly the equations so that they show clearly in latex style.
 
  • #4


I have Rudin at my hand now, so no source of confusion. My understanding is the following:
Given any F>0 (which is the fraction you wrote above), it is possible to choose h such that :
0 < h < 1 < F (in case 1<F)
or
0 < h < F <1 (in case F<1)

In other words all what you need is to make sure that the fraction is greater than zero which is actually the case here.
 
  • #5


Thanks. I am trying to get it now. But the fraction I wrote above is h, NOT x.
 
  • #6


The fraction is NEITHER x NOR h. I edited my post and renamed it F as I realized calling it x is a bad idea.

F= (x-yn)/n(y+1)n-1
and F>0
 
  • #7


Thank you.
 

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