Proof check

**Bingk** · July 14th 2010, 06:05 AM

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

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

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

**Also sprach Zarathustra** · July 14th 2010, 06:10 AM

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:

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

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!

**Prove It** · July 14th 2010, 06:13 AM

Not all laws of exponentiation extend to complex numbers.

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

**Bingk** · July 14th 2010, 06:25 AM

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

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