Student converted $\sqrt{x^2}$ and ended up with just $x$ instead of $|x|$

Question

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 $\sqrt[n]{a^m}=a^{m/n}$.

Answer 1

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).

Answer 2

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.

Answer 3

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.

Answer 4

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".