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 $y=\pm \sqrt{k}$.
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Did you define $\sqrt[n]{x}$ to be the positive $n$th root of $x > 0$ when $n$ is even? – N. F. Taussig Dec 19 '22 at 21:58
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For even powers $x^{2n}$, we can rewrite them as $(x^n)^2$ which must be positive, so $x^{2n}=(-x)^{2n}$, so in order to be unique their inverse $\sqrt[n]{x}$ is defined over only positives, thus $\sqrt[n]{x^n}=|x|$. This isn't true for odd powers as $x^{2n+1}=xx^{2n}$ which must have the same sign as $x$. – Numeral Dec 19 '22 at 22:02
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Over at Simplifying $\sqrt{3 - \sqrt{8}}$, I wrote: In the complex world, principal root has no universal definition and $\sqrt[3] {-1}$ could mean either $e^{i \frac\pi3}$ (smallest nonnegative argument) or $-1$ (real), so it is common to allow surd symbols only inputs from $[0,\infty)$. If you adopt this convention (in which case principal root just means nonnegative root), then $$\sqrt[n]{a^n}\equiv|a|\quad\quad(n\in\mathbb N).$$ – ryang Feb 22 '23 at 04:30
1 Answers
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
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Your last sentence is my biggest challenge as a math teacher. Helping the stronger students while not leaving the other weaker students in the dust. My M.O. is that everything in math is provable from the set of axioms, but we use many results in algebra that are only provable via calculus. e.g. each $y>0$ has a unique $n$-th root. – Chris Christopherson Dec 19 '22 at 22:16
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2@ChrisChristopherson I honor you for facing that challenge. That said, I think it reasonable to work with the real numbers without constructing them (even in most calculus courses the existence of $n$th roots is assumed, not proved). But maybe explaining that, say, $x^0 = 1$ for positive $x$ is a definition that we make because it makes sense in context is worthwhile. Your call of course - you are in the thick of the struggle. – Ethan Bolker Dec 19 '22 at 22:39