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 $(-x)^n=y$


  • $\begingroup$ 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? $\endgroup$ – WoolierThanThou Aug 14 at 15:42
  • $\begingroup$ You mean real-valued solutions? $\endgroup$ – Wuestenfux Aug 14 at 15:44
  • $\begingroup$ 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 $\endgroup$ – Carlos Bacca Aug 14 at 15:51
  • $\begingroup$ Yes real valued $\endgroup$ – Carlos Bacca Aug 14 at 15:52

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.

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