1 = -1 ??? Watch

Evil Phoenix
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Ok, follow this reasoning step by step... seems to make sense if it wasnt that it breaks just about every axiom of mathematics

1 = √1 = √(-1)(-1) = (√(-1))(√(-1)) = (i) (i) = i² = -1

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Eru Iluvatar
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i think the √(-1)(-1) is invalid, as you can't route negative numbers, but i may be totally wrong
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manps
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(Original post by Eru)
i think the √(-1)(-1) is invalid, as you can't route negative numbers, but i may be totally wrong
You can root negative numbers (post alevel anyway)
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kitsune
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the imaginary number!!! (complex numbers)
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Gaz031
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(Original post by nicolo)
Ok, follow this reasoning step by step... seems to make sense if it wasnt that it breaks just about every axiom of mathematics

1 = √1 = √(-1)(-1) = (√(-1))(√(-1)) = (i) (i) = i² = -1

comments? questions?
The error is that \sqrt{\alpha \beta}=\sqrt{\alpha}\sqrt{\beta} is only true if \alpha and \beta are real.
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kitsune
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ohhh so cool
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Evil Phoenix
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Oooooooooh

haha i knew there had to be something wrong with it :p:

well thanx :cool:
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Neapolitan
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(Original post by Gaz031)
The error is that \sqrt{\alpha \beta}=\sqrt{\alpha}\sqrt{\beta} is only true if \alpha and \beta are real.
\alpha and \beta are real. I think you mean positive.
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Gaz031
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(Original post by Neapolitan)
\alpha and \beta are real. I think you mean positive.
Yes, thank you.
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JohnSPals
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(Original post by Gaz031)
The error is that \sqrt{\alpha \beta}=\sqrt{\alpha}\sqrt{\beta} is only true if \alpha and \beta are real.
I thought it was that √1 = ±1, too.
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JohnSPals
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(Original post by Gaz031)
The error is that \sqrt{\alpha \beta}=\sqrt{\alpha}\sqrt{\beta} is only true if \alpha and \beta are real.
I thought it was that √1 = ±1, too.
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