# Proving $x^n=y$ has 2 solutions if n even natural and $y>0$

Hi how is my proof for the question with $$n$$ an even natural number and $$y>0$$ $$x^n=y$$ has two solutions.

Assume n is some even natural number and $$y>0$$. The equation $$x^n=y$$ has one $$x>0$$ which satisfies it by the existence of roots.

Also $$x^n=(-x)^n$$ for all $$x$$. Therefore

$$x^n=y$$ if and only if $$(-x)^n=y$$

Thanks

• The last part is fine. What exactly do you mean by existence of roots? x^2+1 has no root in $\mathbb{R}$, so what about the equation in question are you using? – WoolierThanThou Aug 14 at 15:42
• You mean real-valued solutions? – Wuestenfux Aug 14 at 15:44
• Thanks I meant to say that I'm assuming the existence of a unique x>0 which satisfies x^n=y and I can do this because of an assumption about nth roots – Carlos Bacca Aug 14 at 15:51
• Yes real valued – Carlos Bacca Aug 14 at 15:52

## 1 Answer

Your proof needs a little touching-up in some places:

• You say there is one $$x>0$$ such that $$x^n=y$$. You also need to say this $$x$$ is unique, so that the desired conclusion can be reached.
• After showing $$x^n=y\iff(-x)^n=y$$, you need to state that since there is a unique positive $$x$$ satisfying the original equation, this "reflection identity" means there is also a unique negative $$x$$ satisfying it.
• The case of $$x=0$$ should also be dealt with, but that is easy. Combining positive, negative and zero $$x$$, you may conclude the desired result.