The equation $\displaystyle (ab)^n = a^nb^n$ is true for $\displaystyle a,b\in \mathbb{C}, ~ n\in \mathbb{Z}$.

Furthermore, $\displaystyle (a^b)^n = a^{bn}$.

However, $\displaystyle (a^b)^c = a^{bc}$ --> fails!

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- Jan 24th 2009, 01:52 PMThePerfectHacker
- Jan 25th 2009, 08:14 AMlebanon
there is an error because

example x^2=(-)*(-)=+

or=(+)*(+)=+

but you can say that redical -1 * -1 =1

then -1 is undifined. - Jan 25th 2009, 01:20 PMJhevon
- Feb 2nd 2009, 09:04 PMYendor
Let $\displaystyle f:\mathbb{C}\to \mathbb{C}$ be defined as $\displaystyle f(x)=x^\frac{1}{2}$. Then $\displaystyle f$ is not a well defined map, because there exists $\displaystyle x\in\mathbb{C}$ such that $\displaystyle f(x)$ maps to two different elements in $\displaystyle \mathbb{C}$

This doesn't really apply to your proof, but in general be careful when you use the square root with complex numbers, as you'll often get multiple answers. In fact, when you use the square root with real numbers, there are multiple possible answers, but it's convention to ignore the negative parts. I could do the same proof without complex numbers.

We know that $\displaystyle (-6)^2=36$ and the square root and square operations are inverses so they cancel and $\displaystyle \sqrt{36}=\sqrt{(-6)^2}=-6$ .

And we know that $\displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab}$ .

example: $\displaystyle \sqrt{4}\sqrt{9} = \sqrt{4(9)} = \sqrt{36}$

So $\displaystyle 6 = 2(3) = \sqrt{4}\sqrt{9} = \sqrt{(4)(9)} = \sqrt{36} $ .

So $\displaystyle 6=-6$. gg.

Where is the problem with this?

- Feb 2nd 2009, 09:13 PMThePerfectHacker
Yes it is well-defined. There are two ways to define the square root, but the one used in post #1 is the principal square root: $\displaystyle \sqrt{z} = \sqrt{|z|}e^{i\arg(z)},z\not = 0$. Once you settle with this definition the square root function is well-defined.

- May 4th 2009, 06:49 AMadhyetaanother proof
another one...find out whats wrong...

http://latex.codecogs.com/gif.latex?...fty+1=%3E-1=+1

(Nerd)

perhaps its easier to prove -1=1 than to prove 1=1. lol. - May 4th 2009, 07:10 AMChop Suey
$\displaystyle \infty - \infty$ is an indeterminate expression.

- May 7th 2009, 12:40 PMMusab