What is wrong with my proof: $-1 = 1$?

Question

I have some theories about why this could by wrong but I still haven't something that convinces me. What is wrong with this proof:

$ -1 = i^2 = i.i = \sqrt{-1}.\sqrt{-1} = \sqrt{(-1).(-1)}= \sqrt1 = 1 $

This would imply that: $1 = -1$

Which is obviously false.

So my theory is that it's not a great idea to write $i = \sqrt{-1}$, but I'm not sure why...

I'm fairly sure this has been on this site before. At any rate, the rule that generally fails here is $$ \sqrt{ab} = \sqrt a \sqrt b $$ This is true when $a,b \geq 0$, but not in our generalization to the complex numbers. — Ben Grossmann, Jan 27 '15 at 16:43
@GitGud how are people still asking it? I found 4 duplicates and didn't know which one to flag it as. — dustin, Jan 27 '15 at 16:45
@barakmanos: usually the square root is defined to be a function and the positive branch is preferred. Hence $\sqrt1=1$. — , Jan 27 '15 at 17:08

score 6 · Answer 1 · answered Jan 27 '15 at 16:43

6

The step $\sqrt{-1}\times \sqrt{-1}=\sqrt{(-1)\times(-1)}$ is incorrect. Such a law is only valid for nonnegative arguments.

answered Jan 27 '15 at 16:43

Peter

score 5 · Accepted Answer · answered Jan 27 '15 at 16:44

5

The flaw is in assuming that the rule $\sqrt x\sqrt y=\sqrt{xy}$ holds with imaginary numbers. You just show us a counter-example.

answered Jan 27 '15 at 16:44

2 Answers2