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Recently my friend showed me this proof. I know this is wrong but I don't know accurate reasons. Can anyone help?

His proof:

$6 = \sqrt{\frac{-9}{-4}} = \sqrt{-9} \times \sqrt{-4} = 3i × 2i = 6×(i^2) = -6$

Robert Soupe
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Pratik Garai
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  • I guess something is wrong in the 2nd step... but what? – Pratik Garai Dec 29 '17 at 11:31
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    $\sqrt{ab} =\sqrt{a} \sqrt{b}$ holds only for non negative real numbers $a,b$ – Bijesh K.S Dec 29 '17 at 11:31
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    Every non-zero complex number has two square roots. All this boils down to is saying "$6^2=(-6)^2$ so $6=-6$". – Angina Seng Dec 29 '17 at 11:31
  • @Pratik.garai As said in this Math SE post, there are two answers for $\sqrt{-9}$ ($+3i$ and $-3i$), and two answers for $\sqrt{-4}$ ($+2i$ and $-2i$). When you multiply them, you get the four possibilities of $-6$, $6$, $6$ and $-6$. However, only the first and last possibilities make sense, and your friend is just selecting the one that doesn't make sense. – Toby Mak Dec 29 '17 at 11:41
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    When we square root a number, it is just convention that the principal square root is the positive one. The negative one still exists. – Toby Mak Dec 29 '17 at 11:41

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