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I asked a student I tutor what $\sqrt{x^2}$ was so I could show him why the solution is $|x|$ instead of just $x$. He ended up changing the problem to $(x^2)^{1/2}$ and then multiplied the exponents to get $x^{2/2}=x^1=x$. Now I'm well aware why the answer to the problem I gave him is $|x|$ and explained it, and the answer to $(x^2)^{1/2}$ is likewise $|x|$ but I wasn't able to give a real good explanation why he couldn't use the rule $\sqrt[n]{a^m}=a^{m/n}$.

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    $\begingroup$ because that rule only applies without qualification to non-negative $a$. $\endgroup$ – James S. Cook Feb 1 '16 at 1:02
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    $\begingroup$ my favorite: $1 = \sqrt{1 \cdot 1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = -1$. Oh noes. $\endgroup$ – James S. Cook Feb 1 '16 at 3:22
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    $\begingroup$ This whole business is a "gotcha"... I don't like to dwell on it. $\endgroup$ – paul garrett Feb 2 '16 at 3:17
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    $\begingroup$ Part of the issue is that most students learn formulas or rules without context. As the answers below indicate, the issue is that a manipulation $(a^m)^n = a^{mn}$ is not always valid, particularly when $a^m$ is undefined. But this requires paying attention to subtlety and carefully thinking through the logic of what you are doing -- almost the exact opposite mindset of most students who are trying to rush through the problem as quickly as possible to produce a "correct" answer (where "correct" is not tied to logical necessity but to approval from the person in charge of grading). $\endgroup$ – Michael Joyce Feb 2 '16 at 13:39
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    $\begingroup$ This shows up in some standard Calc I problems. E.g., finding $\lim_{x\to -\infty} \frac{\sqrt{x^2+7}}{2x+3}$. Like most things taught in algebra and precalculus, I imagine this was made a big deal of because of problems like these. $\endgroup$ – ncr Feb 6 '16 at 7:05
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This is a surprisingly tricky issue, and the exact answer "why" depends on the context (real or complex numbers?) and definitions in play (varies by textbook). I'm looking for a good general answer to effectively the same issue on this question here.

Restricting the discussion purely to real numbers, the most popular and straightforward response seems to be that the identity $(a^r)^s = a^{rs}$ requires some additional restriction on it, for example, that both $a^r$ and $a^s$ be defined in real numbers. Observing that restriction, the student's simplification is invalid for negative $x$, because then $x^\frac{1}{2}$ is undefined in real numbers. It's possible that may only be implied (not explicitly stated) in a given textbook that you might use.

For some exposition of the complex-valued case, consider the Wikipedia article here.

Edit: Note that as defined in complex numbers, $(z^m)^\frac{1}{n} = (z^\frac{1}{n})^m$ will be true if and only if $m$ and $n$ are relatively prime (assumes $n$, $z$ nonzero; Silverman, 1975).

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    $\begingroup$ Actually, $x^\frac{1}{2}$ is defined only on the non-negative real numbers, so the student's proof would only work for $x\geq 0$ if you used those restrictions. $\endgroup$ – Dan Christensen Feb 1 '16 at 3:56
  • $\begingroup$ @DanChristensen: I thought that's what I was trying to say, how is it different? $\endgroup$ – Daniel R. Collins Feb 1 '16 at 5:13
  • $\begingroup$ You said, "$x^\frac{1}{2}$ is undefined in real numbers." A typo? $\endgroup$ – Dan Christensen Feb 1 '16 at 5:47
  • $\begingroup$ @DanChristensen: "for negative $x$" is right before that... $\endgroup$ – Daniel R. Collins Feb 1 '16 at 7:35
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    $\begingroup$ ... But added "then" to try and further clarify. $\endgroup$ – Daniel R. Collins Feb 1 '16 at 15:36
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The identity $(x^m)^n=x^{mn}$ is true for real numbers $m,n$ and $x\ge 0$. It is not necessarily true otherwise.

An $n$th root of $b$ (where $n$ is a positive integer and $b$ is a real number) is a value of $x$ such that $x^n=b$. Note that if $b\ne 0$, then there are $n$ distinct values of $x$, some of which may be complex.

Although the expression $x^n=b$ implies many possible values for $x$, the notation $x=\sqrt[n]{b}$ implies only one value of $x$, the principal $n$th root of $b$. This is the unique positive real value of $x$ (if it exists), or the unique negative real value of $x$ otherwise (if it exists). (In my experience, the use of the notation $x=\sqrt[n]{b}$ to denote a unique value of $x$ is often done in primary and secondary education.)

For example, if $x^2=4$, then this means that either $x=2$ or $x=-2$. But $x=\sqrt{4}$ means that $x=2$ (and not $x=-2$). If $x^3=-8$, then there are three solutions (only one of which is real), but the notation $x=\sqrt[3]{-8}$ means $x=-2$ (and no other values).

The equation $x^2=b$ (where $b>0$) means that either $x=\sqrt{b}$ or $x=-\sqrt{b}$. If it is the former, then $\sqrt{x^2}=x$ but if it is the latter then $\sqrt{x^2}=-x$. So one cannot claim that, in general, $\sqrt{x^2}=x$. However, $\sqrt{x^2}=|x|$ holds for either case, so it is, in general, true.

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  • $\begingroup$ The case where none of the $n$ roots of $b$ are real numbers is interesting. For example, what is the principal square root of $-1$? $\endgroup$ – Joel Reyes Noche Feb 1 '16 at 1:27
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    $\begingroup$ So because the student had a fractional exponent for $n$ he cannot apply the identity, since it requires both $m$ and $n$ to be integers, correct? On a similar note, I've argued with many teachers why $\sqrt{49}$ is just 7, when they insist it is both 7 and -7. The teachers often don't seem to grasp that solving $x^2=49$ is different than simplifying $\sqrt{49}$ $\endgroup$ – Brendan Feb 1 '16 at 1:29
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    $\begingroup$ Yes, that's why it's important to understand the correct terminology (principal $n$th root of $b$) and the correct notation ($\sqrt[n]{b}$). $\endgroup$ – Joel Reyes Noche Feb 1 '16 at 1:35
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    $\begingroup$ I hope my latest edit fixes things. $\endgroup$ – Joel Reyes Noche Feb 1 '16 at 3:27
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The identity $(x^m)^n = x^{mn}$ is only true if $x>0$. In fact pretty much all identities involving exponents are only true for positive bases -- another example that often trips students up is $x^r \cdot y^r = (xy)^r$.

There are several ways to explain why these formulas are only true for positive bases, but at a fundamental level it has to do with the question of how we even define $x^r$ when $r$ is a real number. The conventional definition is that we define $$x^r = \exp \left(r \ln x \right) $$ (although in order to make this definition non-circular we have to have some way of defining what $\exp$ and $\ln$ are without invoking exponents.)

Based on the above definition it can be shown (for positive $x$) that $x^{r+s}=x^r x^s$. From this it can further be deduced that if the exponent $r$ is some integer $n$ then the definition above is consistent the more elementary notions based on "repeated multiplication." It further follows that $x^{1/n}$ is the unique positive $n^{th}$ root of $x$, and we can likewise prove that $(x^r)^s = x^{rs}$ for any real $r$ and $s$. But all of these provable properties hinge on how exactly we define $x^r$, and note that the definition given above only makes sense for $x>0$, because $\ln x$ is not defined (or, more precisely, is not single-valued) for negative values of $x$.

Update: Actually, the situation is more complicated than what I have written above; see https://math.stackexchange.com/a/2085269/124095, and scroll down to the second update, for more information regarding the question of in which contexts the various exponent rules apply.

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  • $\begingroup$ You can have (-8)^(1/3) = -2. $\endgroup$ – Dan Christensen Feb 1 '16 at 4:47
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    $\begingroup$ @mweiss I like the contributions you gave to Joel's answer below that resulted in his edit. I don't see anything wrong with this answer and it makes sense, but I would imagine the student being even more confused if I defined $a^x = \exp \left(r \ln a \right)$ since they have not yet learned about logarithms or what $\exp$ means, and will be confused whether $x^2$ is of the form $a^x$ $\endgroup$ – Brendan Feb 1 '16 at 14:11
  • $\begingroup$ @Brendan: Just so you know, the stance that mweiss takes here would be the natural perspective for someone working in the field of complex analysis. $\endgroup$ – Daniel R. Collins Feb 1 '16 at 17:26
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    $\begingroup$ @DanielR.Collins Thanks for the heads up. I do plan on pursuing a PhD. in Mathematics, and I have no disagreements with his answer. I understand the answer and it has helped me personally understand the student's error in an in-depth way that will make me a better instructor. I do thank mweiss for his very helpful input. If my comment sounded like I didn't appreciate his answer, I apologize. I was trying to say that if I were to use such an explanation while tutoring a student in Algebra I who is already struggling, it might end up confusing the student even more. $\endgroup$ – Brendan Feb 1 '16 at 17:34
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    $\begingroup$ @JamesS. Good point! I've edited throughout to make $x$ the base and used $r, s$ for real exponents. $\endgroup$ – mweiss Apr 3 '18 at 14:43
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Other answers have concentrated on the base, but the problem can also be ascribed to the exponents. $(x^m)^n=x^{mn}$ is true for any base, as long as m and n are integers. This follows simply from the definition of exponentiation as repeated multiplication. $x^m$ is m copies of x multiplied together. $(x^m)^n$ is n copies of (m copies of x multiplied together) multiplied together, and so is mn copies of x multiplied together. This works for any base, integer, rational, irrational, positive, negative, complex. In fact, it works in any ring.

The problem comes when we try to extend the concept of exponents to non-integers. If we want to define $y = x^{\frac{1}{n}}$, we can take $y^n$ and extend the rule to say $y^n = (x^{\frac{1}{n}})^n =x^{\frac{1}{n}n}=x$. With this extension, y is a number such that $y^n$ = x. This is a description of y, but it's not a definition of y, as it not does unique specify y. Thus, when we take $(x^2)^{\frac{1}{2}}$, we are finding a number that when squared is equal to $x^2$. And while it's tempting to look at the phrase "a number that when squared is equal to $x^2$" and say "Well, that's a rather complicated way of saying 'x' ", x isn't the only number that satisfies this description. Thus, we have to make qualifications such as "$x^{\frac{1}{n}}$ is the principal nth root of x", and its those qualifications that cause the rule to fail with negative numbers.

Thus, while we can save this rule by saying "x has to be a positive number", we can also save it by saying "m and have to be integers". Or combine these qualifications and say "This works for non-integers only when we have positive numbers".

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