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I've learnt that the square root of a number squared is equal to the absolute value of that number, but I haven't really understood why. I have looked through other questions on MSE but didn't really find a good answer.

As an example: for me there are two ways to arrive at a solution for $\sqrt{(-5)^2}$

First: $\sqrt{(-5)^2}= $ $\sqrt{25}=5$

Second: $\sqrt{(-5)^2}= $ $\ (-5)^{\frac{2}{2}}$$\ =(-5)^1=-5$

but according to the $\sqrt{x^2} = | x | $ rule 5 is the only solution. What is the flaw in my logic in getting -5 as a solution. I would really appreciate a comprehensive explanation that clears this up. Thanks in advance.


marked as duplicate by N. F. Taussig, Joel Reyes Noche, Hippalectryon, Johanna, quid Mar 1 '15 at 14:51

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  • 2
    $\begingroup$ It boils down to the fact that although the square function $ \text{sq} $ maps $ \mathbb{R} \to [0,\infty) $, the square-root function $ \sqrt{\cdot} $ maps $ [0,\infty) $ to $ [0,\infty) $, not back to $ \mathbb{R} $, assuming that you choose only non-negative square roots so that $ \sqrt{\cdot} $ does not become multi-valued. $\endgroup$ – Berrick Caleb Fillmore Mar 1 '15 at 11:31
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    $\begingroup$ If you regard $\sqrt{\,\,}$ as a function then it can only give one value. So even if $x^2=25$ may have two solutions, $x=\sqrt{25}$ can only have one. $\endgroup$ – Henry Mar 1 '15 at 11:32
  • $\begingroup$ Possible duplicate of: Significance of $\sqrt[n]{a^n}$. Also Proving square root of a square is the same as absolute value $\endgroup$ – Mufasa Mar 1 '15 at 11:36
  • $\begingroup$ You may also find this article useful: Absolute Value $\endgroup$ – Mufasa Mar 1 '15 at 11:40

We define $\sqrt k$ to be the positive square root of $k$. This means that taking square roots is a valid function on the positive reals - if we allowed it to have multiple values, then it wouldn't be a function.

This is different to the solutions of the equation $x^2=k$. Here we are looking for all real numbers whose square is $k$, so we must also consider the negative square root.

In you second case, the flaw in the argument is that you are implicitly taking the square root of a negative number - you are saying that $$\sqrt{(-5)^2}=\sqrt{-5}^2$$which gets messy, as taking the square root of a negative number, even if we allow complex numbers, requires a choice of root, and the rules of indices don't apply in the complex case in the same way they do in the real case.

  • $\begingroup$ k i got what u mean. thnx $\endgroup$ – vink007 Mar 1 '15 at 11:37

You can think this in this way.

Let $$f: \mathbb{R}\rightarrow\mathbb{R}^+$$ $$x\mapsto x^2$$ where $x\in\mathbb{R}$

and $$g:\mathbb{R}^+\rightarrow\mathbb{R}^+$$ $$y\mapsto \sqrt{y}$$ where $y\in\mathbb{R}^+$

So to solve this problem, we must follow this order of arithmetic just like your first illustration and your second should be $\sqrt{(-5)^2}=[(-5)^2]^{\frac{1}{2}}$


How do you know if $25$ is $5^2$ or $(-5)^2$ ? Numbers have no "memory".


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