# Proof that the square root of a negative number is real.

So I stumbled upon this weird result when experimenting with fractional exponents. Suppose you have some negative, real number, for example -8. We know $$\sqrt{-8}$$ is not a real number.But $$\sqrt{-8} = {(-8)}^{\frac1{2}}$$ and $$\frac1{2} = \frac2{4}$$ so $$\sqrt{-8} = {(-8)}^\frac2{4} = ((-8)^2)^{1/4} = 64^{1/4}$$ which is obviously real!
I can't seem to find out what the error of the proof is, any thoughts are appreciated!

• it's not the case that $(-8)^{2/4} = ((-8)^2)^{1/4}$. these exponentiation rules do not necessarily hold for complex numbers, mostly because raising something to the $\frac{1}{4}$ power is not well defined. For example, you can't say $1^{1/2} = 1$, since also $(-1)^2 = 1$. – mathworker21 Nov 23 '18 at 12:43

You're using the power rule $$a^{bc} = (a^b)^c$$, but this rule does not hold in general unless $$a$$ is a positive real (and $$b, c$$ are both real too), or $$b, c$$ are both integers.

The trouble is that extending exponentiation to complex numbers most naturally produces a multiple-valued function -- and if you use such a definition, you can generally choose values for each of the exponentiations that make $$(a^b)^c$$ come out the same as $$a^{bc}$$ -- but multiple-valued functions don't really work with equational reasoning, so something still has to give. The most you could conclude is that there is some multiple-valued function whose value for some input can be both $$\sqrt8$$ and $$i\sqrt 8$$, but that does not make those two outputs equal.

If you want a single-valued exponentiation operation for complex numbers, it will necessarily contain more-or-less arbitrary discontinuities and branch choices that makes $$a^{bc}= (a^b)^c$$ fail for some values.

A square root is specifically a quantity whose square is equal to the argument of the square root. While it is true that $$x^2=-8$$ implies $$x^4=+64$$, the converse does not hold. While the latter equation has real solutions, those real solutions are extraneous to the defining equation for any square root of $$-8$$.

• I agree with this point, but this does not answer my question completely. I am using square roots and n-roots only as exponents and not as solutions to equations. – Magnus E-F Nov 23 '18 at 12:58

You tell it yourself, $$\sqrt{-8}$$ is not a real number. So you have to wonder what is the legitimacy of raising that beast to a power: what's the meaning of

$$\left(\sqrt{-8}\right)^2 ?$$

Mathematicians have found answers to this question, but it turns out that

$$\left(\sqrt{-8}\right)^{2}=\sqrt{(-8)^2}$$ cannot hold !