# proof of -1 = +1

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• Jan 24th 2009, 01:52 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
yes, proof by counter-example is obvious here. i was hoping for something else. because this seems to imply a larger problem, namely, the distribution of powers over a product. if it doesn't work for the 1/2 power, that would mean it probably won't work for all or some other powers, right? a proof i would like to see is when this move is illegal, or is it always illegal? is $(ab)^3 = a^3b^3$ false for $a,b \in \mathbb{C}$? etc

The equation $(ab)^n = a^nb^n$ is true for $a,b\in \mathbb{C}, ~ n\in \mathbb{Z}$.
Furthermore, $(a^b)^n = a^{bn}$.
However, $(a^b)^c = a^{bc}$ --> fails!
• Jan 25th 2009, 08:14 AM
lebanon
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 PM
Jhevon
Quote:

Originally Posted by lebanon
there is an error because
example x^2=(-)*(-)=+
or=(+)*(+)=+
but you can say that redical -1 * -1 =1
then -1 is undifined.

??

see post #16, it tells you the rule you are using does not work in general for complex numbers.
• Feb 2nd 2009, 09:04 PM
Yendor
Let $f:\mathbb{C}\to \mathbb{C}$ be defined as $f(x)=x^\frac{1}{2}$. Then $f$ is not a well defined map, because there exists $x\in\mathbb{C}$ such that $f(x)$ maps to two different elements in $\mathbb{C}$

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

And we know that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ .
example: $\sqrt{4}\sqrt{9} = \sqrt{4(9)} = \sqrt{36}$

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

So $6=-6$. gg.

Where is the problem with this?

Quote:

Originally Posted by mnova
[I hope I did the LaTex correctly.]

We know that $\textit{i}^2 = \sqrt{-1}^2$ and the square root and square operations are inverses so they cancel and $\textit{i}^2 = -1$ .

And we know that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ .
example: $2\textit{i} = \sqrt{4}\sqrt{-1} = \sqrt{4(-1)} = \sqrt{-4}$

So $\textit{i}^2 = \textit{ii} = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{+1} = +1$ .

So -1 = +1
QED

Where's the problem with this?

• Feb 2nd 2009, 09:13 PM
ThePerfectHacker
Quote:

Originally Posted by Yendor
Let $f:\mathbb{C}\to \mathbb{C}$ be defined as $f(x)=x^\frac{1}{2}$. Then $f$ is not a well defined map, because there exists $x\in\mathbb{C}$ such that $f(x)$ maps to two different elements in $\mathbb{C}$

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: $\sqrt{z} = \sqrt{|z|}e^{i\arg(z)},z\not = 0$. Once you settle with this definition the square root function is well-defined.
• May 4th 2009, 06:49 AM
another 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 AM
Chop Suey
$\infty - \infty$ is an indeterminate expression.
• May 7th 2009, 12:40 PM
Musab
Quote:

Originally Posted by Chop Suey
The property $\sqrt{ab} = \sqrt{a} \sqrt{b}$ is defined for $a, b \geq 0$. And hence, as TPH mentioned, your step here:
$
\textit{i}^2 = \textit{ii} = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{+1} = +1
$

is erroneous.

Yes, that's right and the actual solution is:

${i}^2 = \textit{ii} = \sqrt{-1}\sqrt{-1} = \textit{(-1)}^ {0.5+0.5}= -1$
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