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$\sqrt{x}=\sqrt{1\cdot x}=\sqrt{(-1)^2\cdot x} = \sqrt{(-1)^2} \cdot \sqrt{x} = (-1) \cdot \sqrt{x}=-\sqrt{x}$

The idea popped into my head while I was evaluating an integral. I have a feeling that I made some obvious mistake because the "proof" is so simple, but I don't see any flaw. Of course, there must be a flaw somewhere. What is it?

Eric
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    You seem to be using the false statement $\sqrt{(-1)^2}=-1$. – Git Gud May 29 '14 at 15:34
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    @AsafKaragila I don't think this should be marked as a duplicate of your given question; this one seems to show confusion about the fact that $\sqrt{a^2}=|a|$ rather than the fact that the identity $\sqrt{ab}=\sqrt{a}\sqrt{b}$ does not always hold whenever $a,b\in\mathbb C$. – user26486 May 29 '14 at 16:28
  • @mathh: There are duplicates regarding that issue as well. But this "idea" is based on the notion that $\sqrt{(-1)^2}=\sqrt{-1}^2$, which is what happens in the suggested duplicate. – Asaf Karagila May 29 '14 at 16:36
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    @AsafKaragila It is my opinion that we should help the question askers with the exact problems they have. I highly doubt the OP thought of it as $\sqrt{(-1)^2}=\sqrt{-1}\sqrt{-1}$ rather than $\sqrt{a^2}=a$. The question you've pointed to wouldn't show the OP that $\sqrt{a^2}=|a|$ and hence he would stay baffled about it. – user26486 May 29 '14 at 16:51
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    I agree with @mathh, this question is based on a different conceptual error than the one linked to. See my comment on kang zhou's answer below. – mweiss May 29 '14 at 17:14

3 Answers3

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Most fake proofs involving square roots rest, at some point, on the false identity $$\sqrt{a^2} = a$$ This identity seems natural and true, which is why it fools us. The correct identity is $$\sqrt{a^2} = |a|$$

mweiss
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$$\sqrt{(-1)^2} = \sqrt{(-1)(-1)} = \sqrt 1 = 1 \neq -1$$

Perhaps you are making the mistake of thinking $(\sqrt{(-1)^2} = (-1)^{2/2} = (-1)^{1} = -1$, but this too is mistaken because $(-1)^{1/2}$ does not exist in the real numbers, so $\Big((-1)^{1/2}\Big)^2$ does not exist, whereas $\Big((-1)^{2}\Big)^{1/2} = 1$ does indeed exist.

amWhy
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$\sqrt{(-1)^2 \cdot x}=\sqrt{(-1)^2}\cdot \sqrt{x}$and $\sqrt{(-1)^2}=-1$ are wrong.

Kang Zhou
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    The first one is correct. – Git Gud May 29 '14 at 16:01
  • Agree with @GitGud, which is precisely why I think this question is not a duplicate of the one people are linking to; that question incorrectly uses $\sqrt{ab}=\sqrt{a}\sqrt{b}$ while this question correctly uses it. – mweiss May 29 '14 at 17:13
  • $\sqrt{(-1)^2 \cdot (-5)}=\sqrt{(-1)^2} \cdot \sqrt{-5}$ is right?!! – Kang Zhou May 30 '14 at 05:15
  • I remember that $\sqrt{ab}=\sqrt{a} \sqrt{b}$ is right only for $a,b \geq 0$. – Kang Zhou May 30 '14 at 05:31
  • The first equality is correct assuming that both radicands are real and positive; note that nowhere in the OP are square roots of negative quantities involved. – mweiss May 30 '14 at 15:55