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I'm reading Precalculus from James Stewart. In the book, the author says that $\sqrt{x^2}$ is not equal to $\big(\sqrt{x}\big)^2$. I was performing a couple of proofs and I ended up here:

If you have, for example:

$\sqrt{5^2} ≟ \big(\sqrt{5}\big)^2 \implies (\sqrt{5})² = \sqrt{5}\cdot\sqrt{5} = \sqrt{5}\cdot5 = \sqrt{25} = \big(\sqrt{5}\big)^2$

So, for me, $\sqrt{5^2}= \big(\sqrt{5}\big)^2$.

Can you proof that I'm wrong?

leomestizo
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    You can't prove a positive proposition by an example. – Kenny Lau Sep 09 '17 at 14:52
  • @KennyLau Proposition: there exists a Besicovitch set of measure 0.

    I would be very interested if you proved this without an example

     – mathworker21 Sep 09 '17 at 14:57
  • @mathworker21 you obviously know what I meant. – Kenny Lau Sep 09 '17 at 14:58
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    @KennyLau OP may have accidentally called it a "proof" but he was looking for a counter-example to Stewart's claim. He wasn't trying to prove anything. – Andrew Tawfeek Sep 09 '17 at 15:00
  • @AndrewTawfeek OP wanted us to prove that OP is wrong, which presumes that OP has done any proof at all (his example of $\sqrt{5^2}=(\sqrt5)^2$ obviously cannot be wrong) – Kenny Lau Sep 09 '17 at 15:01

4 Answers4

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It is true for all positive and zero $x$, but if $x \lt 0, \sqrt x$ is not defined in the reals, so $(\sqrt x)^2$ is also not defined. On the other hand $\sqrt {x^2}$ is defined and equals $|x|$ for all real $x$

Ross Millikan
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  • I understand your point, but what if you have (√-1)² = √-1.√-1 = √(-1).(-1) = √1² = 1. I'm a bit confused right now – leomestizo Sep 09 '17 at 15:09
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    @leomestizo Your example is the reason that the full statement of the rule you are trying to use is: $\sqrt{ab} = \sqrt{a}\sqrt{b}$ provided that not both $a$ and $b$ are negative. – Jim H Sep 09 '17 at 15:18
  • If you work in the complex numbers you have to be careful about the branch of the square root function you use. In the reals we agree that the square root sign is always the positive square root. In the complex plane there is no way to consistently choose the square root to make all the usual real laws of exponents work. Are you using real or complex numbers? I assumed reals because your example is real. – Ross Millikan Sep 09 '17 at 15:38
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$\sqrt{(-1)^2} = \sqrt1 = 1$ whereas $(\sqrt{-1})^2 = i^2 = -1$.

Kenny Lau
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$\sqrt{a}$ is just a symbol for the following problem: solve for $x$ the equation: $x^2=a$. For positive $a$, this equation has 2 solutions with the same absolute value and opposite sign. Usually the positive solution is indicated by $\sqrt[+]{a}$. So in the end: $$\sqrt{a}=\pm\sqrt[+]{a},~\forall a\ge0$$

Back to your example: $$(\sqrt{5})^2=(\pm\sqrt[+]{5})^2=(\sqrt[+]{5})^2=5$$ while $$\sqrt{5^2}=\pm\sqrt[+]{5^2}=\pm\sqrt[+]{25}=\pm5$$

trying
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  • $x^2 = -1 \implies x = \pm i$. $a$ need not be positive. Also I have never seen $\sqrt[+]{x}$ stand for positive solution, maybe it is just me. –  Sep 09 '17 at 15:15
  • @A---B I think that the OP don't know complex numbers, that's why I limited arguments to positive reals only. – trying Sep 09 '17 at 15:18
  • Fine with me.$~~$ –  Sep 09 '17 at 15:22
  • @trying Usually by $\sqrt{u}$ we denote the non-negative number whose square is $u$. Also, for any real number $x,$ $\sqrt{x^2}=|x|.$ So, by following standard math notation, $\sqrt{5^2}=5>0,$ not $-5.$ – KS_ Sep 09 '17 at 15:36
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You are not wrong about $\sqrt{(5)^2} = \left( \sqrt{5}\right) ^2$.

But if you think that this proves that $\sqrt{(-5)^2}$ is equal to $\left( \sqrt{-5}\right) ^2$, then you are mistaken.

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

While $\left( \sqrt{-5}\right) ^2 = \left( \sqrt{-5}\right)\left( \sqrt{-5}\right) = -5$ (or it does not exist if our domain of discourse if the real numbers.)

Deniz Tuna Yalçın
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Jim H
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