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I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$

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Maybe someone can create sample general reference question for $$ \sqrt{x^2} = x $$ fallacy. Then we can just mark similar questions as dupes. – default locale Aug 21 '13 at 8:40
@newcomer you are confusing different roots of unity. – oldrinb Aug 21 '13 at 16:28
I think people are being overly complicated in bringing up multivaluedness of $\sqrt{}$. It does play a role, but by interpreting everythingn in the normal way on the complex plane, mrf's solution is the simplest, and doesn't require excuses about nonprincipal square roots. – rschwieb Aug 23 '13 at 16:49
@defaultlocale I don't see this as assuming $\sqrt{x^2} = x$. The right-most equality is achieved by $\sqrt{\left(-1\right)^6} = \sqrt{1} = 1$, which is clearly valid. The suspect part is $\left(-1\right)^{6/2} = \left(\left(-1\right)^6\right)^{1/2} = \sqrt{\left(-1\right)^6}$ – Tim Goodman Mar 20 '14 at 6:16

15 Answers 15

The "rule" $(a^b)^c = a^{bc}$ doesn't necessarily hold when $a < 0$.

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In my opinion this is the only answer that exactly says where the problem is. – Git Gud Aug 20 '13 at 18:48
I would add that it holds when b and c are integers but not necessarily when one (or both of them) aren't. – ypercubeᵀᴹ Aug 20 '13 at 23:28
And I guess you mean that the main issue is not the (1)^3 = (-1)^(6/2) equality but the next one with the root. But some readers may be confused that you mean the second equality and not the third. – ypercubeᵀᴹ Aug 20 '13 at 23:30
@ypercube The second equality only uses that $6/2=3$. I don't see how anyone could think that's what I mean. – mrf Aug 21 '13 at 15:12
$(a^b)^c = a^{bc}$ doesn't hold when $a^b$ or $a^c$ have cuts in the complex plane. If we defined $sqrt(1) := -1$ the paradox presented in this question wouldn't hold. – jimifiki Aug 22 '13 at 6:56

$x^2=y$ does not imply $x=\sqrt y$. In fact both $+\sqrt y$ and $-\sqrt y$ both satisfy $x^2=y$.

So $(-1)^6=1$, does not imply $(-1)^3=\sqrt1=1$

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Which equality are you objecting to in the above question? I don't see the question using $x^2 = y$ implies $x = \sqrt{y}$ at all. Rather it uses $x^{a/2} = \sqrt{x^a}$. – Tim Goodman Mar 20 '14 at 6:25
Hmm, had a good chat with default locale about this, I guess it depends on what justification the OP wants to claim for $(-1)^{6/2} = \sqrt{(-1)^6}$. – Tim Goodman Mar 20 '14 at 8:00
This is the best answer IMO. The fact is that $\sqrt{1}$ can equal $1$ or $-1$. – Kidburla Mar 7 at 22:10

There is a simpler version of this fallacy: $-1 = (-1)^{2/2} = \sqrt{(-1)^2} = \sqrt{1} = 1$. The mistake comes from the fact that the function $f(x)=x^2$ is not invertible so you cannot conclude that for any real number $x$ it is the case that $x = \sqrt{x^2}$.

There is a version of the same mistake that uses the fact that $log$ is not invertible on $\mathbb{C}$ to prove that all numbers equal 1:

$x = e^{\ln(x)} = e^{\ln(x) * (2\pi i) / (2\pi i)} = (e^{2\pi i})^{\ln(x)/2\pi i} = (\cos(2\pi)+i \sin(2\pi))^{\ln(x) / 2\pi i} = 1^{\ln(x) / 2\pi i} = 1$

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I think the first part of your answer is correct. But for the second part, I think the problem is not that $\log$ is not invertible, but is still as the first answer says that $(a^b)^c=a^{bc}$ is not true for all $a$ in general. – L. Xu Aug 22 '13 at 19:58
@L.Xu I disagree. The problem in OP's set of equalities is that $x=\sqrt{x^2}$ is true when $x^2$ is invertible, e.g., if we assume that $x$ is a positive real number. (To be precise, we have $x=\sqrt{x^2}$ if $x^2$ is invertible because we are working with strictly negative numbers; in that case, its inverse is $-\sqrt{\cdot}$. The problem in the fake proof that all numbers equal $1$ is that $x=e^{\ln(x)}$ is true when $ln(x)$ is invertible, e.g., if we are working in $\mathbb{R}$ but not if it is not, e..g, if we are working in $\mathbb{C}$. – exk Aug 23 '13 at 0:59
One way to see that this is about invertability and not about the equality $(a^b)^{c}$ per se is to see that the same mistake can creep in if you use non-invertable function other than $x^n$. Consider the following fallacy: $x=sin^{-1}(sin(x))=sin^{-1}(sin(2\pi x)=2\pi x$. Hence $\forall x, x=2\pi x$ which implies $\forall x, x=0$. – exk Aug 23 '13 at 1:07
Dear @exk : The invertibility is tied together with the failure of $(a^b)^c=(a^c)^b$. If you interpret in complex numbers, $-1=(-1^{1/2})^2=i^2$, but $((-1)^2)^{1/2}=1$. In other words the order you do the exponents matters when you have multiple valued roots. I think you are both right :) – rschwieb Aug 23 '13 at 16:44

What is wrong here is assuming that $\sqrt{x^2} = x$ when the fact is $\sqrt{x^2} = |x|$. Let $x=−1$ and use $\sqrt{x^2} = |x|$ in the problem above, you should arrive at a valid equation.

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I'm surprised no one wrote that earlier @subzero...nice addition – Eleven-Eleven Aug 21 '13 at 11:58
Thanks @ChristopherErnst – Subzero-273K Aug 21 '13 at 12:02
Hear hear (they did say it, just not in so many word, which really is surprising). – Jonathan Y. Aug 21 '13 at 17:30
What you say is correct, but how does this help clear up the OP's problem? – us2012 Aug 21 '13 at 20:56
@us2012 let $x=-1$ and use that $\sqrt{x^2}=|x|$ in the problem above, you should arrive at a correct equation. – Subzero-273K Aug 22 '13 at 10:08

As other say the square root may be two valued. However you use it as a function, so it's single valued. It really depends on your definition of used functions. I think that $(-1)^{6/2} = -1$ but $\sqrt{(-1)^6} = ((-1)^6)^{1/2} = 1$. So @mrf is right that $(a^b)^c ≠ a^{bc}$ in general and the third equality of your equation is the one that doesn't hold.

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Important point mentioning there's no problem with taking the square root. – Git Gud Aug 20 '13 at 18:50

The problem here is that the square root function, $\sqrt{-},(-)^\frac{1}{2}$, is not a single-valued function.

As PVAL says, it is a two-valued function, meaning you have two consistently choose which square root you're talking about. That's why you often will have problems when you have chains of equalities as above.

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The function $x \mapsto \sqrt{x}$ is single valued for all $x \ge 0$. By definition $\sqrt{2}$ is the positive solution to the equation $y^2 = 2$. That is why we write $\pm \sqrt{2}$. It is the function $x \mapsto x^{1/2}$ which is multi-valued. – Fly by Night Aug 20 '13 at 18:58
@FlybyNight I'm not sure many people make this distinction between $\sqrt{x}$ and $x^{1/2}$. The reasoning in the OP breaks down even when we think only about single valued functions, here. – rschwieb Aug 23 '13 at 16:56
@rschwieb But they should... For example $3^{1/2} = \pm\sqrt{3}$ and not just $\sqrt{3}$. Similarly $1^{1/4} = \pm 1, \pm\operatorname{i}$ and not just $\sqrt[4]{1} = 1$. According to Wikipedia "Every non-negative real number a has a unique non-negative square root, called the principal square root, which is denoted by $\sqrt{a}$, where $\sqrt{}$ is called the radical sign or radix." – Fly by Night Aug 23 '13 at 17:51
Dear @FlybyNight : Maybe, but I'm reluctant to moralize about it. This is more likely to confuse readers at a basic level rather than help. Maybe a little less so now, since we've discussed it directly :) – rschwieb Aug 23 '13 at 18:11
Dear @FlybyNight I do not believe that multifunctions are standard high school fare (at least in the US) nor do I think they should be as long as students continue to struggle with single-valued functions (I'll moralize on that :) ). You needn't explain details of your argument: I'm not faulting it anywhere! I just think it's not as accessable a reason. – rschwieb Aug 23 '13 at 18:29

Taking square roots is in a sense a two valued function, because every non-zero complex number $z$ has two distinct complex numbers $w_1, w_2$ for which $w_1^2=w_2^2=z$.

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$\sqrt{x^2} = +x$ or $-x$

The fault with the "proof" is the false assumption that you can choose the positive root and still have everything hold. In fact, you need to select the correct root based on context or accept two possible answers.

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You say, $(-1)^{6/2}= [(-1)^6]^{1/2}$

I'd rather see it as, $(-1)^{6/2}=[(-1)^{1/2}]^6$

which is equal to $i^6$ or $(-i)^6 = -1$

(Just for fun : $DMAS$ rule for powers? :P)

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The rule $a^{b/c}=\sqrt[c]{a^b}$ is always true when $a > 0$. If $a$ is negative, it may not be true.

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What you have done is correct but the third and so the fourth equality is not valid in mathematics. $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$

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Here in your problem: $$\sqrt{(-1)^6}=\sqrt{(-1)^3.(-1)^3}=\sqrt{-1.-1} $$

Here since both a and b are negative

Therefore your third equality was wrong and so the fourth one.

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The correct use of ${\sqrt{}}$ in this context would be

$$ −1 = (−1)^3 = (−1)^{6 / 2} = -\sqrt{(-1)^6} = -\sqrt{1} = -1$$

and this is simply a consequence of the inverse of $x^2$ being $-\sqrt{x}$ not $+\sqrt{x}$ when $x < 0$.

$$ -1 = \sqrt{(-1)^2} = \sqrt{1} = 1$$

is an equally invalid chain of equalities becuase the square root function is not injective.

If we start at one value, and apply a function whose inverse is not injective, we can easily (by choice) end up at a different value; ie if $f^-1(x)$ is not injective, then

$$f^-1 \circ f (x) \in S$$

where S contains values other than $x$. For mathematical rigor we have to specify in which domain we are working, so that we don't simply "choose" the inverse value. If from the start we have said $x < 0$, then the inverse of $x^2$ is $-\sqrt{x^2}$ not $+\sqrt{x ^2 }$ and the fallacy would be avoided.

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The problem is in step: $$(-1)^3 = (-1)^{6/2} $$ You are squaring a number and taking root. As many people pointed out: $$ (x^2)^{1/2} $$ can be +x or -x.

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Dear vinash: I hate to say it, but I find this a little misleading. While it's possible to interpret $x^{1/2}$ in a multivalued way, it isn't really the core of the problem. I'm afraid beginners are going to get the wrong idea about the square root function, which in most basic math classes is defined to be single valued... – rschwieb Aug 23 '13 at 16:52
No, in that step, he just rewrites the $3$ as $6/2$, which is absolutely unproblematic. It's the next step that's the problem, namely equating $(-1)^{6/2}$ with $((-1)^6)^{1/2}$. – celtschk Apr 12 '14 at 19:32

Think about operator associativity in $$\sqrt{(-1)^6}$$. This is constructed in following way

-1 --> -1^6 --> (-1^6)^0.5

let us solve this

(-1^6)^0.5 = 1^0.5 = 1

for more

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Hi this is not a fallacy but this problem overlooks some facts i.e. there will be two square roots to every number like. $$ \sqrt{4} = +2 \text{ as well as} -2 $$ similarly $$ \sqrt{(-1)^6} = +1 \text{ or } -1 $$

So I think writing $$\sqrt{(-1)^6} = 1$$in this proof is wrong

And also people who are saying that $$(a^b)^c != a^{b*c}$$ for a <0 or something like that please calculate this in your calculator so that it proves you worng $$10^{2.5*log10(-1)} $$

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$\sqrt{4}=2$ as well as $-2$ No, that is a common misconception as well. In the context of the real numbers, $\sqrt{a}$ always denotes the nonnegative solution to $x^2-a=0$. You're right that for $a>0$ there are always two roots to $x^2-a$, but the positive one is singled out for $\sqrt{a}$. – rschwieb Aug 23 '13 at 12:40
This might actually support my point. :) Why is only +ve root considered?? – tejas Aug 23 '13 at 16:18
Because that is how the square root function is defined. I see above some people are being (imo overly generous) in considering $\sqrt$ as a multivalued function. Mrf's answer does not require one to mess with multivalued functions. – rschwieb Aug 23 '13 at 16:40
Before anyone gets the wrong idea, I know you can consider it as a multifunction, but I don't see why this discussion has to depart from basic highschool algebra like that. – rschwieb Aug 23 '13 at 16:53
I'm not denying you your freedom to think about multifunctions, I'm just saying that mainstream math education uses the principal roots to talk about these as single-valued functions, and so it's a little reckless and confusing to go against the mainstream convention. You are going to confuse lower level students :/ – rschwieb Aug 23 '13 at 17:07

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