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I found the following proof online and am confused as to where the error occurs.

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

My guess is that the error occurs here: $\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}$, but I'm not sure how to show that.

Is my guess correct? How might I prove this?

Thank you!

  • You are absolutely correct that they error is exactly there. Think back. When were you told and why do we believe $\sqrt {ab}=\sqrt {a}\sqrt {b} $? Why should that be true? It's sometimes true. It's true if $a\ge 0$ and $b\ge 0$ but otherwise it isn't true. – fleablood Aug 24 '16 at 07:17
  • Oh, it sounds like Ivm asking those questions rhetorically. I'm not. They are very subtle and difficult questions. Ponder them and think about them. Part of the problem is that we really abuse what $\sqrt {a} $ means. What does it mean? how many values does it have? Is it unambiguous in meaning? – fleablood Aug 24 '16 at 07:23

3 Answers3

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Indeed what you are proving is that in the complex numbers you don't have (in general) $$\sqrt{xy}=\sqrt{x}\sqrt{y}$$ Because you find a counterexample.

EQJ
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0

Well,

$$\sqrt{ab} = \sqrt a \sqrt b$$

is only true when $a$ and $b$ are non-negative.

MathMajor
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For complex numbers $a$ and $b$, it is not in general true that $(ab)^{1/2} = a^{1/2}b^{1/2}$, so when they write $$ \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)}, $$ this is incorrect. The issue is that $a$ and $b$ must each by nonnegative.

Alex Ortiz
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