-i = i [Disproven]

[Shadow]

Hey all,

 

Where am I making a mistake?

 

[math]-i=(-1)\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i[/math]

 

Cheers,

 

Gabe

[DJBruce]

I think your mistake is that:

 

[math](-1)(\sqrt{-1})\neq\sqrt{(-1)(-1)(-1)}[/math]

[D H]

You are attempting to apply the relation [math]a^ma^n=a^{m+n}[/math] to a situation where this relation does not apply: complex numbers. You are also abusing the square root symbol, which strictly applies only to non-negative reals.

[Petanquell]

Hey all,

 

Where am I making a mistake?

 

[math]-i=(-1)\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i[/math]

 

Cheers,

 

Gabe

 

[math] (-1)(\sqrt{-1})\neq\sqrt{(-1)(-1)(-1)} [/math] because [math] \sqrt{a^2}=\left|a \right| [/math]

 

aspon myslim...

[Shadow]

If [math]\sqrt{-1}[/math] is illegal, how can [math]i[/math] be defined?

 

aspon myslim...

 

I told you, no Czech; only English

[D H]

The error in the original post ultimately lies with the implicit use of [math]\sqrt{(-1)^2}=-1[/math]. There is no need for imaginary numbers here. For example, using [math](-1)^4 = 1 = ((-1)^2)^2[/math],

 

[math]\aligned

1 &= \sqrt{1} \\

&= \sqrt{(-1)^4} \\

&= \sqrt{(-1)^2}\sqrt{(-1)^2} \\

&= \sqrt{(-1)^2}\sqrt{1} \\

&= (-1)\times 1 \\

&= -1

\endaligned[/math]

 

The square root function takes a branch cut at x=0. Proofs that -x=x involve an abuse of this branch cut. Don't do that!

[Shadow]

Ah, I see now. My bad. Thanks for explaining guys ;-)

 

Cheers,

 

Gabe

[MolecularEnergy]

You can proove a negative very easily, by saying,

 

[math]x=zy^2[/math]

 

Add x to both sides;

 

[math]2x=zy^2+x[/math]

 

Divide by 2 on both sides

 

[math]x= \frac{1}{2} (zy^2+x)[/math]

 

Manipulating through algebra gives:

 

[math]\frac{1}{2} x-x= \frac{1}{2} (zy^2)[/math]

 

Solve the equation;

 

[math]\frac{1}{2} x= -\frac{1}{2} zy^2[/math]

 

Replace [math]\frac{1}{2}[/math] x with [math]\frac{1}{2}[/math] zy^2 then

 

[math]\frac{1}{2} x= -\frac{1}{2} x[/math]

 

Which is my own derivation.

Edited December 20, 2008 by MolecularEnergy

[Klaynos]

[math]

\frac{1}{2} x-x= \frac{1}{2} (zy^2)

[/math]

 

Here lies your mistake.

 

It should be:

 

 

[math]

x= \frac{1}{2} (zy^2) + \frac{1}{2} x

[/math]

 

 

[math]

x - \frac{1}{2} x= \frac{1}{2} (zy^2)

[/math]

 

Which works out fine.

[MolecularEnergy]

Why... i thought the algebra i used was fine?

 

Obviously not though, if you are adament it is wrong.

[Bignose]

But the algebra wasn't fine. You made a mistake. Check it again. Klaynos showed you the step you made the mistake on.

[MolecularEnergy]

I know, i see it now. I am sorry.