Let's define $x=\sqrt{ab}$, where $a=-1$ and $b=-1$.

Does $x=1$, as $-1\cdot-1=1\implies\sqrt{-1\cdot-1}=\sqrt{1}=1$?

Or maybe $x=-1$, as $\sqrt{a^2}=a\implies\sqrt{-1\cdot-1}=\sqrt{(-1)^2}=-1$?

I assume that at least one of the following is true:

  1. The former solution is true, and the fallacy is in assuming $\sqrt{a^2}=a$ for a negative base $a$. This doesn't seem to hold, given that this is equivalent to saying that $a^{2^{\frac12}}\not=a^{2\cdot\frac12}=a$ for a negative $a$, where the value of $a$ seemingly has no bearing on the arithmetic performed with the exponents.
  2. The former solution is true, and the fallacy is in assuming the positive root is meant in the latter solution; in fact, $\sqrt{-1\cdot-1}=-\sqrt{(-1)^2}=--1=1$. But this implies that, depending on how the radicand is factored, one is forced to take the positive root for one factoring and the negative root for another factoring to get the same answer, which doesn't seem right.
  3. The latter solution is true, and the fallacy is in assuming the positive root is meant in the former solution; in fact, $\sqrt{-1\cdot-1}=-\sqrt1=-1$. But, in addition to the problem with the previous solution, this additionally means that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ holds even for negative $a,b$.

Which of my reasonings is incorrect? Or are all of them correct, and there's an additional solution that I'm not seeing?

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    $\begingroup$ $\sqrt{x^2} = |x|$ $\endgroup$ – krirkrirk Jan 3 '19 at 20:57
  • $\begingroup$ Every complex number, apart from zero, has two square roots. $\endgroup$ – Lord Shark the Unknown Jan 3 '19 at 21:01
  • $\begingroup$ You can do the same thing with any square: $\sqrt{25}$ = $\sqrt{-5 \cdot -5}$. for example. But we take the positive branch and define $\sqrt{25}$ = 5. $\endgroup$ – Joel Pereira Jan 3 '19 at 21:03
  • $\begingroup$ @JoelPereira I know, I thought I'd take the most basic case, though. $\endgroup$ – DonielF Jan 3 '19 at 21:04
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    $\begingroup$ I don't understand the downvotes on this. It's perfectly reasonable to ask where the fallacious step is in a well-known fallacy, and it's clear enough what's being ssked. Also the questioner's own thoughts are included in quite a lot of detail $\endgroup$ – timtfj Jan 4 '19 at 1:38

For $a \in \mathbb{R}^+$, $\sqrt a $ is defined as the positive real $b $ such that $b^2=a $. Thus $\sqrt{x^2} = |x|$ because by definition a square root is positive. Hence in your case $\sqrt {(-1)^2} =1$.

Regarding your other points :

  • As you noticed $\sqrt {ab} = \sqrt a \sqrt b $ only applies for $a,b $ positive so it cant be used here

  • $\sqrt{a^2}=(a^2)^\frac 12$ is true, but $(a^x)^y = a^{xy} $ for all $x,y \in \Bbb R $ is only true for $a>0$. This is because in general $a^b = e^{b\ln a} $ which only makes sense for $a>0$. With your example, you can see why this will not work : if $(a^x)^y = a^{xy} $ then $((-1)^2)^\frac 12 = (-1)^1 \iff 1=-1$

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    $\begingroup$ Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.” $\endgroup$ – DonielF Jan 4 '19 at 1:44

The following statement is not true:

$$\sqrt{x^2} = x$$

The square root function returns the non-negative square root, or the principal square root, so it’s actually

$$\sqrt{x^2} = \vert x\vert = \begin{cases} x; \quad x\geq 0 \\ -x; \quad x < 0 \end{cases}$$

So, in your case, $x = -1 < 0$, so $\sqrt{(-1)^2} = \vert -1\vert = 1$.

  • $\begingroup$ Why should $\sqrt{x}=|x|$ when $x^{2^{\frac12}}=x^{2\cdot\frac12}=x\not=|x|$? $\endgroup$ – DonielF Jan 3 '19 at 21:02
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    $\begingroup$ $\sqrt{x^2} = \vert x\vert$ because the square root function gives the non-negative value by convention. It’s just how it works. $\endgroup$ – KM101 Jan 3 '19 at 21:03
  • $\begingroup$ Then how can $\sqrt{x}=x^\frac12$? $\endgroup$ – DonielF Jan 3 '19 at 21:05
  • $\begingroup$ Why shouldn’t that be the case? $\endgroup$ – KM101 Jan 3 '19 at 21:07
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    $\begingroup$ @DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $\ln (a) $ does. $\endgroup$ – krirkrirk Jan 3 '19 at 21:13

Based on the definition of the square root which is positive, it should be 1. See here to know more about that.

  • $\begingroup$ So what’s wrong with $\sqrt{-1\cdot-1}=\sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems? $\endgroup$ – DonielF Jan 3 '19 at 20:59
  • $\begingroup$ @DonielF it's not correct based on the definition of square root. just this. $\endgroup$ – OmG Jan 3 '19 at 20:59
  • $\begingroup$ @DonielF $\sqrt{x^2} = x$ is false. $\endgroup$ – KM101 Jan 3 '19 at 21:00
  • $\begingroup$ @DonielF: base on what do you think that $\sqrt{(-1)^2}=-1?$ $\endgroup$ – user587192 Jan 3 '19 at 22:54
  • $\begingroup$ @user587192 Read the immediately preceding line. Since $\sqrt{a^2}=a$, shouldn’t $\sqrt{(-1)^2}=-1$? $\endgroup$ – DonielF Jan 4 '19 at 1:42

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