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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2$\begingroup$ See math.stackexchange.com/questions/49169/… $\endgroup$– Gamma FunctionAug 20, 2013 at 19:02
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42$\begingroup$ Maybe someone can create sample general reference question for $$ \sqrt{x^2} = x $$ fallacy. Then we can just mark similar questions as dupes. $\endgroup$– default localeAug 21, 2013 at 8:40
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1$\begingroup$ @newcomer you are confusing different roots of unity. $\endgroup$– obatakuAug 21, 2013 at 16:28
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1$\begingroup$ 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. $\endgroup$– rschwiebAug 23, 2013 at 16:49
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3$\begingroup$ @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 $\left(-1\right)^{6/2} = \left(\left(-1\right)^6\right)^{1/2} = \sqrt{\left(-1\right)^6}$ $\endgroup$– Tim GoodmanMar 20, 2014 at 6:16
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15 Answers
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The "rule" $(a^b)^c = a^{bc}$ doesn't necessarily hold when $a < 0$.
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33$\begingroup$ In my opinion this is the only answer that exactly says where the problem is. $\endgroup$– Git GudAug 20, 2013 at 18:48
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10$\begingroup$ I would add that it holds when
bandcare integers but not necessarily when one (or both of them) aren't. $\endgroup$ Aug 20, 2013 at 23:28 -
1$\begingroup$ @ypercube The second equality only uses that $6/2=3$. I don't see how anyone could think that's what I mean. $\endgroup$– mrfAug 21, 2013 at 15:12
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1$\begingroup$ $(a^b)^c = a^{bc}$ 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. $\endgroup$– jimifikiAug 22, 2013 at 6:56
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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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$\begingroup$ 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. $\endgroup$– L. XuAug 22, 2013 at 19:58
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2$\begingroup$ @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 $\mathbb{C}$. $\endgroup$– exkAug 23, 2013 at 0:59
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1$\begingroup$ 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 $\forall x, x=0$. $\endgroup$– exkAug 23, 2013 at 1:07
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2$\begingroup$ 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 :) $\endgroup$– rschwiebAug 23, 2013 at 16:44
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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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1$\begingroup$ I'm surprised no one wrote that earlier @subzero...nice addition $\endgroup$ Aug 21, 2013 at 11:58
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$\begingroup$ Hear hear (they did say it, just not in so many word, which really is surprising). $\endgroup$ Aug 21, 2013 at 17:30
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4$\begingroup$ What you say is correct, but how does this help clear up the OP's problem? $\endgroup$– us2012Aug 21, 2013 at 20:56
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$\begingroup$ @us2012 let $x=-1$ and use that $\sqrt{x^2}=|x|$ in the problem above, you should arrive at a correct equation. $\endgroup$– BranAug 22, 2013 at 10:08
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$\begingroup$ Not clear to me how this relates to the original problem. Subzero, you write "let $x = -1$ and use that $\sqrt{x^{2}} = \vert x \vert$..." However, there IS NO $x$ int he original problem, so there is nothing that can be substituted. $\endgroup$– FlorianJul 11, 2017 at 9:53
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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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2$\begingroup$ Important point mentioning there's no problem with taking the square root. $\endgroup$– Git GudAug 20, 2013 at 18:50
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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.
answered Aug 20, 2013 at 18:44
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5$\begingroup$ 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. $\endgroup$ Aug 20, 2013 at 18:58
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$\begingroup$ @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. $\endgroup$– rschwiebAug 23, 2013 at 16:56
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$\begingroup$ @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." $\endgroup$ Aug 23, 2013 at 17:51
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$\begingroup$ 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 :) $\endgroup$– rschwiebAug 23, 2013 at 18:11
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1$\begingroup$ 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. $\endgroup$– rschwiebAug 23, 2013 at 18:29
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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)
answered Mar 25, 2014 at 19:32
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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.
answered Mar 9, 2014 at 2:45
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What you have done is correct but the third and so the fourth equality is not valid in mathematics. $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
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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2$\begingroup$ 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... $\endgroup$– rschwiebAug 23, 2013 at 16:52
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3$\begingroup$ 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 $((-1)^6)^{1/2}$. $\endgroup$– celtschkApr 12, 2014 at 19:32
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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 http://en.wikipedia.org/wiki/Operator_associativity
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Your error $$(-1)^{6/2} = \sqrt{(-1)^6}\tag✗$$ can be explained in either of two ways:
- in real analysis, for negative $a,$ the definition $$a^{\frac xy}:=\sqrt[y]{a^x}$$ conventionally requires $x$ and $y$ to be coprime such that $y$ is positive and odd;
- for nonzero complex $a,$ the law/theorem $$a^{xy}=(a^x)^y$$ requires $x$ and $y$ to be integers.
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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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3$\begingroup$ $\sqrt{4}=2$ 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 $\sqrt{a}$. $\endgroup$– rschwiebAug 23, 2013 at 12:40
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$\begingroup$ This might actually support my point. :) Why is only +ve root considered?? $\endgroup$– tejasAug 23, 2013 at 16:18
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$\begingroup$ 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. $\endgroup$– rschwiebAug 23, 2013 at 16:40
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$\begingroup$ 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. $\endgroup$– rschwiebAug 23, 2013 at 16:53
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4$\begingroup$ 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 :/ $\endgroup$– rschwiebAug 23, 2013 at 17:07