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The following "proof" is given:

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

Could someone please explain to me where the logic has broken Down?

Evie123
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  • $\sqrt a\sqrt b=\sqrt{ab}$ doesn't hold in the complex. But $\sqrt a\sqrt b=\pm\sqrt{ab}$ does. –  Mar 12 '17 at 15:55

1 Answers1

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This may be closed soon due to the fact that it has been asked so many times, but note that:

$$\sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \iff a,b \geq 0$$

Thus you don't have equality in the statement:

$$\sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1}$$

Faraad Armwood
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