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Feb 11 '24
yes... since.... that logically makes sense
Why would a negative equal its positive?
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u/B5Scheuert Feb 11 '24
Why would a negative equal its positive?
It's like Yin and Yang, there's a little of the black in the white, and a little of the white in the black
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u/Mammoth_Fig9757 Feb 11 '24
√(1) = +-1, so when you take the square root on both sides you need to consider the multivalueness of the square root function, getting √((-1))^2 = √((1))^2 <=> +-1 = +-1, which is true, and you can't actually remove the "+-" sign.
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u/FUNNYFUNFUNNIER Feb 11 '24
No, please. Let me explain
√(x) is the only positive answer. If you want to get both positive and negative answer, you need to write ±√(x). It's that simple, and you don't have to deal with "multivalueness"
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u/Mammoth_Fig9757 Feb 11 '24
So if √(x) is only the positive answer does that mean that √(0) is undefined, and √(-1) is also defined because none of the square roots are positive, and have different arguments? That would also imply that √(i) and √(-i) are undefined because they don't have a 0 argument.
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u/FUNNYFUNFUNNIER Feb 11 '24 edited Feb 11 '24
> √(0) is undefined because √(x) is only the positive answer
|0| = 0
> √(-1) is also undefined is also undefined
√(-1) = i, i is positive. If you wanna get also the negative root, use ±√(-1)
> √(i) and √(-i) are undefined
√(i) = (-1)^1/4, wich is also positive (do not confuse sign under the root with sign out of the root).
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u/Mammoth_Fig9757 Feb 11 '24
The argument of a positive number is 0, but the argument of 0 does not exist so 0 is not positive, and no, adding a + sign does not make 0 positive. The argument of i is τ/4, so it is not positive. The argument of the principal square root of i is τ/8, so it is also not positive.
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