Which step is incorrect?

Question

I came up with a Complex "Proof" idly at work today that seemed to be rather simple, but proved counterintuitive on paper. Year 12 Student looking into pure mathematics, so all feedback is cool and interesting.

$-i = -1 \times {(-1)}^{1/2}$

$-i = i^2 \times {(-1)}^{1/2}$

$-i = {(i^4)}^{1/2} \times {(-1)}^{1/2}$

Since: $a^n \times b^n = {(ab)}^n$

$-i = (i^4 \times -1)^{1/2}$

And: $i^4 = 1$

$-i = {(-1)}^{1/2}$

$-i = i$

$\therefore -1 = 1$

Awesome.

I expect to use stackexchange religiously in the upcoming years, so I'll debut with this dumb loophole and hopefully come back again with some serious dilemmas.

Answer 1

You need to be very careful with $\sqrt{a}$ or $a^{\frac{1}{2}}$ even with the real numbers and even more so with the complex numbers.

With the real numbers, positive numbers have two square roots but it is easy to pick one as special (the positive one) and define $\sqrt{a}$ or $a^{\frac{1}{2}}$ to mean the positive square root. With some care, you can get sensible results though mistakes are still possible. With this convention, $\sqrt{ab} = \sqrt{a} \sqrt{b}$. Of course, you have to remember the negative numbers have no square root.

With the complex numbers, it gets harder. All numbers, except $0$, have two square roots and it is not so easy to say which $\sqrt{a}$ represents. You can set arbitrary rules to pick one but, even if you do, you cannot be sure that $\sqrt{ab} = \sqrt{a} \sqrt{b}$, it could be that $\sqrt{ab} = - \sqrt{a} \sqrt{b}$.

Answer 2

I think your 3rd statement is wrong.

See, in complex numbers we don't multiply two surds as we do normally with a real numbers. Like, for example $\sqrt{-1}×\sqrt{-1}=-1$ not $1$. We don't use the identity $a^n×b^n=(ab)^n$.

We instead multiply conserving the minus sign or you can say conserving the iota like,

$\sqrt{-1}×\sqrt{-1}=\sqrt{(-1)^2}=-1$.

So in your third statement your third statement it instead of using the identity. You should use the following method

$-i=(i^4)^{1/2}×(-1)^{1/2}=\sqrt{(-1)^2}×(-1)^{1/2}$. Here the square and the under root should cancel out which is basically what you started with.

So as long as you don't use the identity your proofs should be fine. Though the identity works fine with real numbers it can give rise to to false proofs with Complex numbers.