# Thread: Proof check

1. ## Proof check

Not sure where to put this ... I saw this some time ago, but could never really figure out what/where the problem is:

$\displaystyle i=\sqrt{-1}$
$\displaystyle i^2=(\sqrt{-1})^2=-1$
$\displaystyle \sqrt{-1} \cdot \sqrt{-1}=-1$
$\displaystyle \sqrt{(-1) \cdot (-1)}=-1$
$\displaystyle \sqrt{1}=-1$
$\displaystyle 1=-1$

As far as I can tell, the process seems to be valid, but obviously 1 is not -1 ... so what exactly is wrong?

2. Originally Posted by Bingk
Not sure where to put this ... I saw this some time ago, but could never really figure out what/where the problem is:

$\displaystyle i=\sqrt{-1}$
$\displaystyle i^2=(\sqrt{-1})^2=-1$
$\displaystyle \sqrt{-1} \cdot \sqrt{-1}=-1$
$\displaystyle \sqrt{(-1) \cdot (-1)}=-1$
$\displaystyle \sqrt{1}=-1$
$\displaystyle 1=-1$

As far as I can tell, the process seems to be valid, but obviously 1 is not -1 ... so what exactly is wrong?
The mistake is in third line!

3. Not all laws of exponentiation extend to complex numbers.

E.g. $\displaystyle a^n\cdot b^n = (a\cdot b)^n$ is only true for REAL $\displaystyle a^n$ and $\displaystyle b^n$.

4. Ah!! Okay ... makes sense ... I feel sort of silly, hehehe. Thanks!