Real Analysis Problem......proving x^n=p

[slb123]

I have to prove that it p is any real positive number and n is a positive integer, then there is a unique real number such that x^n=p

there are five parts to this problem and i have attatched them to this post.......PLEASE HELP any way you can!!

i know that this is an extention to the previous theorem: if p is any positive real number then there is a positive real number x such that x^2=p. i have attatched the proof.....

i know that i need to prove the first part by induction but i am having trouble showing it is true for n+1.

for part 2 i know that A is nonempty because it contains z. I need to find some M such that M is greater than or equal to x for all x in A. can i let x=z^n and M=p to show that A is bounded from above??

i am completely lost on parts 3 and 4.....please point me in the right direction

 

[daon]

You have to show the set of x's such that x^n<=p is bounded. Not the set of all x^n is bounded.

Notice you can assume that if x belongs to this set then x>1, so that x<x^n which is bounded by p.

 

[slb123]

thanks so much for you help......i can solve that part from there. do you have any words of wisdom for the rest of the proof?

 

[slb123]

You have to show the set of x's such that x^n<=p is bounded. Not the set of all x^n is bounded.

Notice you can assume that if x belongs to this set then x>1, so that x<x^n which is bounded by p.

so does this mean that x is a least upper bound? and p is your M??

 

[daon]

You may partition the set into the subset which is less than or equal to 1 and the subset which greater than 1. In the first case it is bounded by 1, so M=1 In the second case since x^n <= p, and x < x^n, we have M=p as our bound. So M=max{1,p} is an upper bound for the entire set.

It doesn't ask for the least upper bound. You want to show that it has one, that is all.

For 3, You are assuming \(\displaystyle \delta = min \{\frac{p-x^n}{c},1\} > 0\) and you want to show that

\(\displaystyle (x+\delta)^n \le p\)

By assumption on \(\displaystyle \delta\)

\(\displaystyle (x+\delta)^n \le (x+\frac{p-x^n}{c})^n\)

You'll notice c doesn't have an x^n term...

edit:

one more thing:

\(\displaystyle cx+p-x^n = \sum_{k=1}^{n-1} {n \choose k}x^k + p = c-1+p\)