Why does $\sqrt{x^2}$ seem to equal $x$ and not $|x|$ when you multiply the exponents?

Question

I understand that $\sqrt{x^2} = |x|$ because the principal square root is positive.

But since $\sqrt x = x^{\frac{1}{2}}$ shouldn't $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?

Also, doesn't $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn't $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?

How do I reconcile all this? What rules am I not aware of?

Edit: Since someone voted to close my question, I should probably explain the difference between my question and Proving square root of a square is the same as absolute value, regardless of how much I think the difference should be obvious to anyone who reads the questions. Cole Johnson was asking if there's any way to prove that $\sqrt{x^2} = |x|$. I am not asking that; I already accept the equation as fact. I'm asking how to resolve some apparent contradictions that arise when considering square roots of squares, and how I should approach these types of problems. (Cameron, please read.)

The issue is that $x^{\frac{1}{2}}$ and $\sqrt{x}$ are not the same thing. $x^{\frac{1}{2}}$ is the set of all $y$ such that $y^2 = x$. $\sqrt{x}$ is the set of all $y$ such that $y^2 = x$ and $y \ge 0$. $(x^{\frac{1}{2}})^2$ should really be interpreted as: square all of the values $y$ such that $y^2 = x$, but this is the same thing as ${y^2: y^2 = x}$, which is just $x$. — Cameron Williams, Dec 11 '15 at 01:45
Because the so-called "rule" $a^x a^y = a^{x+y}$ is only for positive numbers $a$. — GEdgar, Dec 11 '15 at 01:58
@Cameron: You're the first source I've ever seen say that $x^\frac 1 2$ is not $\sqrt x$. That makes sense, though. So $x^\frac 1 2$ is both square roots of $x$ rather than the principal square root? Doesn't that still create a problem though? It seems to me that the implication is that $(x^2)^{\frac{1}{2}} = \pm x$ instead of $x$. — Kyle Delaney, Dec 11 '15 at 02:11
@GEdgar: I didn't use that rule. I used the rule that $(a^x)^y=a^xy$. Although I don't see why the rule you stated should only work for positive values of $a$. — Kyle Delaney, Dec 11 '15 at 02:13
@CameronWilliams In my experience, $\sqrt x=x^{\frac12}$. I guess you could support this because of complex-analysis background. But it is generally true that the square root function is defined only for positive numbers (generally), but some of us are a bit past that. — Simply Beautiful Art, Dec 11 '15 at 02:16
@Simple Art: Yes, although 5xum seems to be contesting that statement. — Kyle Delaney, Dec 11 '15 at 02:20
"Whenever one learns a new mathematical operation, it is imperative also to learn the limitations under which the operation may be performed. Lack of this additional knowledge can lead to the employment of the new operation in a blindly formal manner in situations where the operation is not properly applicable, perhaps resulting in absurd and paradoxical conclusions." -- Howard Eves, Great Moments in Mathematics. — Daniel R. Collins, Dec 19 '15 at 18:26
Possible duplicate of Proving square root of a square is the same as absolute value — Cameron Williams, Oct 06 '16 at 19:37

score 11 · Answer 1 · answered Dec 11 '15 at 01:41

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The rule $(x^a)^b = x^{ab}$ is only true for positive values of $x$. With negative values, you need to be much more careful.

For example, $\sqrt x \sqrt x = \sqrt{x\cdot x}$ is only true for positive values of $x$, because for negative values, the left side is not defined.

answered Dec 11 '15 at 01:41

5xum

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I thought the square roots of negative values are imaginary. Are imaginary numbers not defined? – Kyle Delaney Dec 11 '15 at 01:54
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@KyleDelaney $i$ is defined as "The solution to $z^2=-1$" which is not the same as "The square root of $-1$". The $\sqrt{}$ fucntion is not well defined on the whole real line (you have to specify the branch cut you use!) – 5xum Dec 11 '15 at 01:55
@KyleDelaney They are not defined since the square root of a complex number is not a complex number: it is, in some sense, two complex numbers. For example, the square rootS of $i$ ARE defined as the complex numberS $z$ such that $z^{2}=i$. You have $z=e^{i\frac{\pi}{4}}$ and $z=e^{-i\frac{3\pi}{4}}$ – MoebiusCorzer Dec 11 '15 at 01:59
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@MoebiusCorzer But positive numbers also have two square roots. The symbol $\sqrt{}$ means principal square root. – Kyle Delaney Dec 11 '15 at 02:18
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@KyleDelaney I mean, if you talk about $\sqrt{\cdot}$ as a function, any positive real number has one and only one square root. Actually, with what you're saying, the complex number $1$ would have $4$ complex square roots. We usually talk about the principal square root, unless I'm mistaken. – MoebiusCorzer Dec 11 '15 at 02:20
So you make no distinction between square root and principal square root? – Kyle Delaney Dec 11 '15 at 02:27
Like @KyleDelaney, I disagree with some of the comments here. I think it would be better to simply clarify what context this discussion is in. Say, for negative values, $\sqrt{x}$ is not defined in real numbers; or else point to a definition that $\sqrt{-x} = \sqrt{x}i$ only for positive real x. – Daniel R. Collins Dec 11 '15 at 04:12
Also, what are the four complex square roots of $1$? – Kyle Delaney Dec 12 '15 at 03:53
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@MoebiusCorzer: Every non-zero complex number has exactly two square roots in the complex numbers (whereas zero has exactly one square root, namely itself). The statement that $1$ has four square roots in the complex numbers is incorrect. – Zev Chonoles Dec 12 '15 at 06:06
@ZevChonoles You're completely right. I was mistaken and I don't understand how I've come up with such a comment. I'm sorry for that. Yet, $i^(1/2)$, for example, is not defined as a complex number. Does there exist a so-called principal square root of a complex number? – MoebiusCorzer Dec 12 '15 at 09:48
Moreover, I think the confusion of the OP comes from the fact he seems to not know the definition of $\sqrt{}$. One possible definition is $\sqrt:\mathbb{R}^{+}\to\mathbb{R}^{+}:x\mapsto \sqrt{x}=x^{1/2}$. Since $x\mapsto x^{2}$ is also a function, we should see $\sqrt{x^{2}}$ as the composite of these two functions. They are clearly not inverse functions since they are not defined on the same domain/codomain. Since the image of some positive $y$ by $\sqrt{}$ is a positive $\sqrt{y}$, it is quite clear that $\sqrt{x^{2}}=|x|$. – MoebiusCorzer Dec 12 '15 at 10:09
@MoebiusCorzer You should delete the comments that can create confusion for future readers. – Jul 01 '16 at 18:13

Daniel R. Collins · Accepted Answer · 2015-12-12T06:02:22.480

A common source of confusion or "paradoxes" comes from not paying close attention to the (perhaps rarely exercised) restrictions or boundary conditions. These restrictions are necessary to ensure that paradoxes like you're considering do not arise (i.e., otherwise the definitions would fail to be well-defined). For example, here's a proper definition of rational exponents from Michael Sullivan's College Algebra:

Note that we only consider real numbers here. Now, to answer your questions:

But since $\sqrt{x} = x^{\frac{1}{2}}$ shouldn't $\sqrt{x^2} = (x^2)^{\frac{1}{2}} = x^{\frac{2}{2}} = x$ because of the exponents multiplying together?

The first assertion is not generally true; $\sqrt{x} = x^{\frac{1}{2}}$ only provided that $\sqrt{x}$ exists (that is, not for negative $x$). In your chained equality, the second equality is false, because the exponent in $x^{\frac{2}{2}}$ contains common factors (i.e., is not in lowest terms). These statements would, however, be true if $x$ was restricted to positive real numbers only.

Also, doesn't $(\sqrt{x})^2$ preserve the sign of $x$? But shouldn't $(\sqrt{x})^2 = (\sqrt{x})(\sqrt{x}) = \sqrt{x^2}$?

All of these equalities are false for negative $x$, because in that case the expression $\sqrt{x}$ does not exist in real numbers (i.e., it's undefined). Likewise, if you look carefully at the rule for multiplying radicals, then you'll see the same restriction against square roots of negative numbers.

Edit: Added text from Precalculus: a right triangle approach by Ratti & McWaters. Hopefully this clarifies the rule for products of square roots (namely that only positive radicands can be generally combined or separated). Also, note the warning from the section on complex numbers that doing so in that case is illegitimate.

The proper definition in the photo says $a$ need only be a real number, implying that it can be negative. It doesn't say $\sqrt a$ needs to be a real number. So using that definition, the paradox seems to apply.
Supposing that $\sqrt x = x^\frac 1 2$ only provided that $\sqrt x$ exists, what does $x^\frac 1 2$ equal if $x$ is negative?

Why should the equalities be false for negative $x$ because $\sqrt x$ is imaginary? As soon as the two imaginary numbers multiply together they become a real number. — Kyle Delaney, Dec 12 '15 at 04:33
No, this whole definition is only in the context of real numbers; the clause "provided that $\sqrt[n]{a}$ exists" is in fact saying that it must be a real number (otherwise it would be an empty statement). And again, multiplying radicals is only ever defined if at least one radicand is nonnegative. Seriously, you need to read the definitions carefully. — Daniel R. Collins, Dec 12 '15 at 05:25
Edited to add more clarifying text from the Ratti & McWaters Precalculus text. — Daniel R. Collins, Dec 12 '15 at 06:03

score 0 · Answer 3 · answered Dec 11 '15 at 01:53

First of all, $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only when $a$ and $b$ are non-negative. $\sqrt{a}$ and $a^{\frac{1}{2}}$ are not the same function, as Cameron Williams points out in the comments above. Secondly, we can break your square of square root up. For positive values, $\sqrt{x^2} = x$, as one would expect. However, since $\sqrt{(-x)^2} = \sqrt{x^2}$, we notice that we get a positive answer from a negative value, getting $x$ from $-x$. We can represent this as flipping the function over the $x$-axis, or in other-words throwing a negative sign out front (For example, $y=-(x^2+5x)$ flips $y=x^2+5x$ across the $x$-axis). Thus, we get $$\sqrt{\gamma^2}=\begin{cases}\gamma & \gamma\geq 0\\-\gamma & \gamma<0,\end{cases}$$ And this is the definition of the absolute value function

Simply Beautiful Art · Answer 4 · 2015-12-11T02:29:56.440

Consider the following:$$1^{\frac1n}$$Look up roots of unity.

The nth root of a number produces n different answers, creating a regular n-sided shape with a "radius" of 1 on the complex plane.

But what does this mean? Well, you could consider this...

The nth root of a positive number in general.

This removes all of the other possible answers, leaving us with only one answer. Among the answers removed was the original answer, but because real-life math usually can ignore this, we're all good.

Now signs, as you may imagine, have been messed up. We, as a general society, have decided that it be positive. This has made math much easier for younger pupils still learning math. And as we take more roots, we pretty much lose signs. We end up with $i$'s, plusses, and minuses. It is not that you have messed up anywhere, it was just a simple decision made to make the square root function positive.

In many maths, you will find that we will say the n-th root of a number does indeed produce n amount of answers, but for lower level math, we stick with the positive answer, or the result of no positive answer.

I mean, think about it, what does $\sqrt1=-1,1$ do for us in real life? Not much, as it turns out. When we need both the positive and negative answers, it is known that that is what we are finding. But it doesn't have much application to real life.

So the rule is that square roots are positive, and just keep a spot in the back of your mind remembering that this is not the only answer.

No, that's not what I'm asking. I'm well aware that the principal square root is not the inverse of squaring. My question implies my understanding of this when I say that the principal square root is positive. — Kyle Delaney, Dec 11 '15 at 02:15
Oops, you wrote a completely different answer and now my comment makes no sense. — Kyle Delaney, Dec 12 '15 at 04:39

Why does $\sqrt{x^2}$ seem to equal $x$ and not $|x|$ when you multiply the exponents?

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