# Math Help - show the root exists

1. ## show the root exists

Greetings, I hope someone can help on this problem

If n is an odd positive integer and a < 0, prove that there is exactly one negative b such that b^n = a

Thanks

2. Originally Posted by Recklessid
Greetings, I hope someone can help on this problem

If n is an odd positive integer and a < 0, prove that there is exactly one negative b such that b^n = a

Thanks
Nothing to do with number theory.

You are asking whether $\sqrt[n]{a}$ exists.

Indeed it does, consider the function $f(x) = x^n$ and apply the indetermediate value theorem. Note, this function is one-to-one because $f'(x) = nx^{n-1} >0$.

3. Originally Posted by ThePerfectHacker
Nothing to do with number theory.

You are asking whether $\sqrt[n]{a}$ exists.

Indeed it does, consider the function $f(x) = x^n$ and apply the indetermediate value theorem. Note, this function is one-to-one because $f'(x) = nx^{n-1} >0$.
Actually it is asking that you show that only one such $\sqrt[n]{a}$ exists. I did this in the set theory review in my Topology book, but I can't remember how I did it.

-Dan

4. Originally Posted by topsquark
Actually it is asking that you show that only one such $\sqrt[n]{a}$ exists. I did this in the set theory review in my Topology book, but I can't remember how I did it.

-Dan
But that is easy to show! Just show this function is strictly monotone. That will imply it is one-to-one.

5. Originally Posted by ThePerfectHacker
But that is easy to show! Just show this function is strictly monotone. That will imply it is one-to-one.
Actually now that I've had time to think about it the proof was part of a many step problem which started with the order properties of the real line and went into things like "can you always insert a rational number between two other rational numbers" and such. I was dealing with functions, but not their graphs.

It wasn't that hard to do, I just can't remember the method.

-Dan

6. The TPH solution is correct and complete!
But let’s say we need to do it without calculus.
We know that n is odd. Suppose that the are two values b & c such that $b^n = a = c^n$
Note that $\left( {b^n - c^n } \right) = \left( {b - c} \right)\left( {\sum\limits_{k = 0}^{n - 1} {b^{n - k + 1} c^k } } \right)$.
So if k is even then n-k+1 is even so that ${b^{n - k + 1} c^k }$ is a positive term. Moreover, if k is odd then ${b^{n - k + 1} c^k }$ is also a positive term. Thus the only way $\left( {b^n - c^n } \right) = 0$ is that $b=c$, in other words, b is a unique root.