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I started a debate in a youtube video (something you should never do...) because of this question. For me, the answer has to be $-4$, simplifying powers for example. But I don't know if I'm having serious definition failures. Another possibility would be to swap the powers of order and do the root first, taking the complex value and squaring it. However for the vast majority of people who commented, the result was $+4$.

Sorry for the triviality and thanks for being the best mathematics forum in the world.

N. F. Taussig
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gabriel sobrino
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    Equivalent to $\sqrt{16}$ – PC1 Jun 20 '23 at 14:04
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    $\sqrt{n^2} = |n|$ – Peter Phipps Jun 20 '23 at 14:04
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    In a nutshell, you have to consider the difference between two different (but strictly linked) questions: (i) the "function" square root (in order to be a function we need a single value) and (ii) the roots of the equation $x^2=a$. – Mauro ALLEGRANZA Jun 20 '23 at 14:08
  • But $\sqrt{x^2} = (x^2)^{1/2}=x,$ not equal to $|x|$, right? – gabriel sobrino Jun 20 '23 at 14:13
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    $\sqrt{x^2} = \vert x \vert$ for all $x\in\mathbb{R}$ as said above multiple times. – stange Jun 20 '23 at 14:24
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    @gabrielsobrino: It is true that $\sqrt{x^2}=(x^2)^{1/2}$. It is not true, in general, that $(a^b)^c=a^{bc}$; that last statement is only guaranteed to hold when $a$ is positive. And indeed, in this case, $(x^2)^{1/2}\neq x$ when $x$ is negative. – Joe Jun 20 '23 at 14:30
  • For nonnegative inputs, this $\sqrt{\cdots}$ thing (the radical sign) gives the principal square root, which is nonnegative. And as $(-4)^2 = 4^2 = 16$, if you take the principal square root by putting $(-4)^2$ and $4^2$ under $\sqrt{\cdots}$, the result is the same nonnegative number whose square is $16$. – peterwhy Jun 20 '23 at 14:46
  • You might be interested in this discussion on a similar topic from the MathEducators SE. https://matheducators.stackexchange.com/questions/25677/can-it-be-defended-that-sqrt-4-is-both-2-and-2-and-likewise-for-general – johnnyb Jun 20 '23 at 16:12

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I don't feel that the linked answers completely solves you problem, so let me spell it out a little bit.

You write in your comments that

$\begin{equation} \sqrt{x^2}=(x^2)^{\frac{1}{2}}=x \end{equation}$

This is not correct. Sure, for a real number $x$ and integers $a$ and $b$, you have $(x^a)^b=x^{ab}$, but this is not true for non-integer exponents.

What might have confused you, is that the above rule is true when $x$ is a positive real number, and a lot of texts might ommit this crucial assumption when doing symbolic manipulations as the one suggested in the wrong equation above.

As some comments suggest, the correct equation is

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

You can maybe convince yourself why this is true by considering what happens when $x$ is either negative or positive. Else let me know, and I'll happily elaborate.

Edit: Comprehensive proof that $\sqrt{x^2}=|x|$.

Let $x$ be any real number. If $x\ge 0$, we have by definition of the squareroot function that

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

The first equality is true, since $x$ is the unique non-negative number satisfying the equation $y^2=x^2$ (where we solve for $y$). The second equality is true since the absolute value of a non-negative number is the number itself.

Let's now assume that $x< 0$. By this assumption, we have that $|x|=-x$. From this, we get

$x^2=1\cdot x^2=(-1)^2x^2=(-x)^2=|x|^2$

From this we get that

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

Since $|x|$ is a positive number, we conclude from the previous that

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

Richard Jensen
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  • Thank you very much for all the answers and the clarity in them. I understand that to put exponents together they must be integers. But, would it be wrong to swap them? I mean, would it be ok to do this for a real $x$?

    $$\sqrt{x^2} = (\sqrt{x})^2$$.

    I understand that you can't, because if you can, you would have to solve by complexes... I would still like to know why if you are so kind. Thanks for sharing the knowledge, and I leave the trivialities.

     – gabriel sobrino Jun 20 '23 at 14:40
  • That is also wrong. You have to remember that when you write $\sqrt{x}$, unless explicitly stated, it should be the function taking a positive real number to its positive square root. From this, you can see that what you write doesn't make sense, since if $x$ is negative, $\sqrt{x}$ is not defined. – Richard Jensen Jun 20 '23 at 14:46
  • But considering the complex plane of course it is well defined, without having to take a positive or negative value convention... And looking at it this way the result would be $(2i)^2=-4$, but I guess it's wrong.

    Thanks!!!

     – gabriel sobrino Jun 20 '23 at 14:52
  • It is actually also problematic when extending to the complex plane. Yes, every complex number has a square root, but defining a square root function is tricky. If you are interested, this wikipedia section explains it pretty well, but it's complicated stuff: https://en.wikipedia.org/wiki/Square_root#Square_roots_of_negative_and_complex_numbers – Richard Jensen Jun 20 '23 at 16:02
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There is a flaw in your argument,, Because if your argument is true then 4 =√16=√((-4)^2)= -4 NOW how can 4=-4 ,, there is a flaw. So you have to choose either √16 = 4 or -4, So let's do another experiment If √(16) = -4 Then let's calculate √(16)×√(16) = -4×-4=16, but it can also be written as √(16×16) =√(-16)^2= -16 How come 16=-16 ,, Look I can give you tonnes of examples like this ,your logic is correct but if we were to use your rule then whole mathematics will collapse just like this . Any rule you make should not only be logical but also consistent with previous rules.

So in conclusion your logic is correct but it just doesn't fit in right with other rules of mathematics..

Spidee
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For the notation $\sqrt{}$ to make any sense, it must be a well-defined function - that is, for any input $x$, it must give one, and only one result for $\sqrt{x}$.

I think we can all agree that $\sqrt{16}=+4$. We have to pick either $+4$ or $-4$, because remember, we need one and only one answer, so let's arbitrarily pick $+4$ (following normal convention).

However, I think we can all recognize the fact that $(-4)^2=16$. Therefore, $\sqrt{(-4)^2}=\sqrt{16}=+4$. It cannot be true that $\sqrt{(-4)^2}=-4$, because if that was the case, it would contradict our previously established fact that $\sqrt{16}=+4$, since $(-4)^2=16$.

In summary, the claim that $\sqrt{(-4)^2}=-4$ while simultaneously $\sqrt{16}=4$ violates the rules of logic.

K.defaoite
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    It is not true that for it "to make any sense" it has to be a well-defined function. Plenty of things make perfect sense which result in multiple possible values. – johnnyb Jun 20 '23 at 15:30
  • Not true, the square root is exactly defined to be the positive possibility by convention , hence we have $\sqrt{x^2}=|x|$ for every real $x$ , there is no ambiguity. – Peter Jun 20 '23 at 17:11
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    unreal hatin... – K.defaoite Jun 20 '23 at 17:25
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    @Peter: What is not true about this answer? It states that $\sqrt{16}=4$ and that the $\sqrt x$ must refer to one and only one number. – Joe Jun 20 '23 at 20:07