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$\sqrt {-3}$ multiplied by $\sqrt{-3}$ is $-3$. But this can also be written as $\sqrt {-3} \cdot \sqrt{-3} = \sqrt {(-3).(-3)}= \sqrt{9} =3$
So my question is why is this not possible?

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    Your problem is that “$\sqrt{}$” is not a well-defined function outside of the positive reals. In fact, it’s not even well definable. If you define $\sqrt{-3}$ to be that number in the upper half-plane whose square is $-3$, then the rule $\sqrt{ab}=\sqrt a\sqrt b$ fails. – 2017-01-02
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    http://math.stackexchange.com/questions/1096917/can-someone-prove-why-sqrtab-sqrta-sqrtb-is-only-valid-when-a-and-b-ar – 2017-01-02
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    Try to read http://math.stackexchange.com/questions/534940/why-sqrt-1-cdot-sqrt-1-1-rather-than-sqrt-1-cdot-sqrt-1-1-pr?rq=1 – 2017-01-02
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    Asked zillion times here. – 2017-01-02

3 Answers 3

3

Because $\sqrt{ab}=\sqrt{a} \sqrt{b}$ if and only if $a,b\geq 0$

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    If one of a or b is $\ge 0$ then it is also applicable@J.Salieri – 2017-01-02
2

The reason is the notation $\sqrt{\phantom{0}}$ can be used with two different meanings:

  • by abuse of language, $\sqrt{-3}$ denotes any complex number with square equal to $-3$. There are two of them.
  • the ‘normal’ $\sqrt{9}$ denotes, of the two (real) numbers with square equal to $9$, the positive one.
0

While teaching complex numbers, my teacher mentioned it as a theorem:

If a and b are positive real numbers, then $\sqrt{-a}\times\sqrt{-b}=-ab$.