Shadow Posted December 7, 2008 Share Posted December 7, 2008 Hey all, Where am I making a mistake? [math]-i=(-1)\sqrt{-1}=\sqrt{(-1)(-1)(-1)}=\sqrt{-1}=i[/math] Cheers, Gabe Link to comment Share on other sites More sharing options...

DJBruce Posted December 7, 2008 Share Posted December 7, 2008 I think your mistake is that: [math](-1)(\sqrt{-1})\neq\sqrt{(-1)(-1)(-1)}[/math] Link to comment Share on other sites More sharing options...

D H Posted December 7, 2008 Share Posted December 7, 2008 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. Link to comment Share on other sites More sharing options...

Petanquell Posted December 7, 2008 Share Posted December 7, 2008 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... Link to comment Share on other sites More sharing options...

Shadow Posted December 7, 2008 Author Share Posted December 7, 2008 If [math]\sqrt{-1}[/math] is illegal, how can [math]i[/math] be defined? aspon myslim... I told you, no Czech; only English Link to comment Share on other sites More sharing options...

D H Posted December 7, 2008 Share Posted December 7, 2008 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! Link to comment Share on other sites More sharing options...

Shadow Posted December 7, 2008 Author Share Posted December 7, 2008 Ah, I see now. My bad. Thanks for explaining guys ;-) Cheers, Gabe Link to comment Share on other sites More sharing options...

MolecularEnergy Posted December 20, 2008 Share Posted December 20, 2008 (edited) 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 Link to comment Share on other sites More sharing options...

Klaynos Posted December 21, 2008 Share Posted December 21, 2008 [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. Link to comment Share on other sites More sharing options...

MolecularEnergy Posted December 21, 2008 Share Posted December 21, 2008 Why... i thought the algebra i used was fine? Obviously not though, if you are adament it is wrong. Link to comment Share on other sites More sharing options...

Bignose Posted December 21, 2008 Share Posted December 21, 2008 But the algebra wasn't fine. You made a mistake. Check it again. Klaynos showed you the step you made the mistake on. Link to comment Share on other sites More sharing options...

MolecularEnergy Posted December 21, 2008 Share Posted December 21, 2008 I know, i see it now. I am sorry. Link to comment Share on other sites More sharing options...

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