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
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.