# $i=-i$ false proof

I was playing around with complex numbers and came upon this false proof. I don't see the mistake here.

$$i^3=i^2i$$ $$i^3=-i$$ but $$i^3=i^{6/2}$$ $$i^3=√(i^6)$$ $$i^3=√((-1)^3)$$ $$i^3=√(-1)$$ $$i^3=i$$ so $$i=-i$$

• $(-i)^2=-1$ and $i^2=-1$, so there would be no contradiction if you picked $\sqrt{-1}=-i$ near the end. Long story short, you have to be careful with square roots (which are multi-valued functions). – angryavian Apr 5 '18 at 16:28
• Why bother with with complex numbers? $(-2)^3 = (-2)^{\frac 62} = \sqrt{(-2)^6} = \sqrt {64} = 8$ so $-8 = 8$. Same proof. Do you see the error now? – fleablood Apr 5 '18 at 16:32
• Ah yes, I see. Thank you very much – Austin Shiner Apr 5 '18 at 16:34

## 2 Answers

Don't bother with complex numbers and confusing concepts.

$-1 = (-1)^3 = (-1)^{\frac 62} = ((-1)^6)^{\frac 12} = 1^{\frac 12} = 1$.

You are assuming that $(x^a)^b = x^{ab} = (x^b)^a$ holds true even when $a,b$ are not nesc. integers and when $x$ is not nesc. positive.

When we say something like $\sqrt {x}$ is the number $y$ so that $y^2 = x$ there are always two such numbers. And the one we call "the" square root is positive one (if $x$ is positive real) or the complex number with an argument of less than $180^\circ$ if the result is positive. (These are called the "primary" roots)

Assuming defining $a^{\frac mn}$ as $(a^m)^{\frac 1n}$ will work if $a > 0; a \in \mathbb R$, or if $\gcd (m,n) =1$ and we arbitrarily stick to primary roots.

But if $a$ is negative or not real then $a^k = (a^m)^{\frac km}$ will add extraneous roots and not be true at all.

When you write $i^{6/2}=(i^6)^{1/2}$, that implies that you think that $(a^b)^c=a^{bc}$ is true for complex $a$ and rational $b,c$. It isn't, and you have just demonstrated why it can't be.

As a side note, please be very careful with using $\sqrt{\hphantom a\vphantom a}$ and fractional exponents with complex numbers as bases, and avoid it completely of possible. It only invites mistakes like this one.