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Why doesnt √-4 equal 2 if using the principle x to the power of m/n equals the nth root of x to the power of m causes √-4 = ∜(-4)^2 = ∜16 = 2?

Orion
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  • Because $\sqrt{-4}$ is not well defined. It should refer to a certain number whose square is $-4$. Certainly it can not be $2$ because $2^2=4\not = -4$. Actually, the two square roots of $-4$ exist and are the complex numbers $2i$ and $-2i$. You may be interested by this related question https://math.stackexchange.com/questions/438/why-sqrt-1-times-1-neq-sqrt-12 – Suzet Jul 19 '18 at 04:59
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    because $2^2\neq -4$ – Thomas Jul 19 '18 at 05:02
  • $x^{m/n}=\sqrt [n]{x^n} $ for non negative $x $. – fleablood Jul 19 '18 at 05:45

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$\sqrt x$ is, by definition, the non-negative real number which when multiplied by itself becomes $x$.

So, $\sqrt{-4}$ would be a positive number which when multiplied by itself becomes $-4$. But there is no such number. Therefore $\sqrt{-4}$ is undefined.

If you look carefully in your math book, you will see that the manipulations you do are specifically defined only when whatever is under the square root sign is $\geq0$. If they haven't specified that in your book then it's a bad book.

To those who know about complex numbers, I personally do not like to mix square roots symbols and complex numbers, because they lead to all sorts of problems (the manipulation in the question post included), and do not help much. So to me, $\sqrt{-4}$ is still undefined in that context. Thus, for instance, $i$ is not really defined as $i=\sqrt{-1}$, but as $i^2=-1$.

Arthur
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