What is wrong with this proof of $i = 0$?

Question

This is a proof I made a year ago and at that time, I didn't see any problems with it. Could anyone point out what is wrong here?

Consider the following expresion: $(-1)^{(4n+3)/2}$, where $n \in \Bbb Z_+$. We have that: \begin{align} (-1)^{(4n+3)/2} &= [(-1)^{4n+3}]^{1/2} = (-1)^{1/2} = i \\ (-1)^{(4n+3)/2} &= [(-1)^{1/2}]^{4n+3} = i^{4n+3} = -i \end{align} That means $i=-i$ $\implies$ $i=0$.

Why do you think $(-1)^{1/2} = i$? This expression has two values, $\pm i$. — MPW, Mar 22 '19 at 17:16
Similar (if not duplicate): https://math.stackexchange.com/questions/472227/math-fallacy-problem-1-13-16-2-sqrt-16-1. — Martin R, Mar 22 '19 at 17:18

score 1 · Accepted Answer · answered Mar 22 '19 at 17:25

1

A 'square root' is not a function in complex numbers - there are two distinct values of $\sqrt{-1},$ and if you take the other one, which is $-i$, the result would be $-i=-i.$

answered Mar 22 '19 at 17:25

CiaPan

13,884

What is wrong with this proof of $i = 0$?

1 Answers1