Thread: 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

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 exists.

Indeed it does, consider the function and apply the indetermediate value theorem. Note, this function is one-to-one because .

Originally Posted by ThePerfectHacker

Nothing to do with number theory.

You are asking whether exists.

Indeed it does, consider the function and apply the indetermediate value theorem. Note, this function is one-to-one because .

Actually it is asking that you show that only one such exists. I did this in the set theory review in my Topology book, but I can't remember how I did it.

-Dan

Originally Posted by topsquark

Actually it is asking that you show that only one such 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.

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

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
Note that .
So if k is even then n-k+1 is even so that is a positive term. Moreover, if k is odd then is also a positive term. Thus the only way is that , in other words, b is a unique root.