Prove $\sqrt[n]{x^n} =x$ when $n$ is odd and $\sqrt[n]{x^n}=|x|$ when $n$ is even.

Question

I was teaching my algebra 2 class and the question of why $$((-5)^4)^\frac{1}{4} \neq (-5)^\frac{4}{4}=-5$$ but it rather is $5$ came about. I chose to explain this a couple of ways: first, we can think of this as an order of operations problem. i.e. we first multiply $-5$ to itself $4$ times, ridding the negative and then take the fourth root, leaving us with positive 5; second, I referenced this result $$\sqrt[n]{x^n} = |x|$$ when $n$ is even. My "proof" would be very flimsy though as I would kind of explain the same thing i just did. Finally, I said that in general the rule $(a^p)^q=a^{pq}$ only holds for $a>0,$ so we could not apply it in this situation. Is there an elementary proof of $\sqrt[n]{x^n} = |x|$ for even $n$, that I can give to a high-school class? Or is there perhaps a better way to explain why $$\sqrt[\text{even}]{(\text{neg})^\text{even}}=\text{pos}?$$ We have proven the following: If $y^2=k$ then $y=\pm \sqrt{k}$.

Answer 1

For even $n$ you know $x^n$ is nonnegative for every real $x$.

The convention is that $\sqrt[n]{x}$ always refers to the positive root when $n$ is even. That's really what the last sentence in the question means. It leads to the $|x|$ for $\sqrt[n]{x^n}$ when $x < 0$.

It may be a little harder to convince your students that the rule $a^pa^q = a^{pq}$ may fail when $a < 0$. One way to do that might be to point out that since they know that $(-1)^{1/2}$ does not exist, it makes no sense to talk about $((-1)^{1/2})^2$.

Explaining where the "rules" for exponents come from (rather than "memorize these rules") may help your better students but confuse the weaker ones. See Proofs-request: Proofs that five exponention rules hold for positive real bases and rational exponents,using pre-calculus math