/sci/ - Science & Math » Thread #12300105

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Anonymous Mon 02 Nov 20:12:02 2020 No.12300105 View Reply Original Report

Quoted By: >>12300118 >>12300547 >>12302297 >>12302509 >>12303775

So why doesn't this work?
$ i^2 = -1 = (-i)^2 $
So
$ \sqrt{-1} = \sqrt{i^2} = i $
and
$ \sqrt{-1} = \sqrt{(-i)^2} = -i $
Hence
$ i = -i $

Anonymous

Anonymous Mon 02 Nov 2020 20:16:09 No.12300118 Report

Quoted By: >>12300143

>>12300105
$\sqrt{x^2} \neq x$ in general

Anonymous

Anonymous Mon 02 Nov 2020 20:23:42 No.12300143 Report

Quoted By: >>12300407

>>12300118
So when I was told (a^b)^c = a^(bc) I was lied to?

Anonymous

Anonymous Mon 02 Nov 2020 22:08:09 No.12300407 Report

Quoted By:

>>12300143
it's true only for a >= 0, otherwise you have to be careful

Anonymous

Anonymous Mon 02 Nov 2020 23:01:11 No.12300547 Report

Quoted By:

>>12300105
>i=?i
there's nothing wrong with that. Where do you think the complex conjugate comes from?

Anonymous

Anonymous Tue 03 Nov 2020 15:36:40 No.12302297 Report

Quoted By:

>>12300105
When a - 3 = 5, adding 3 to both sides of the equation gets you an equivalent equation "because" adding three is a bijective function. Squaring is not a bijective function, and neither is taking a square root.

Anonymous

Anonymous Tue 03 Nov 2020 16:56:37 No.12302509 Report

Quoted By:

>>12300105
$\sqrt{-1}=\sqrt{(-i)^2}=\sqrt{(-1)^2i^2}=\sqrt{i^2}=i$

Anonymous

Anonymous Tue 03 Nov 2020 23:21:30 No.12303775 Report

Quoted By: >>12303784

>>12300105
you need the +/- cancel square and square root
$sqrt(-1)=sqrt(i^2)=\pm i$
actually it should be absolute value but im just gonna stick with +/- so no absolute value of complex numbers
$sqrt(-1)=sqrt((-i)^2)=\pm (-i)=\mp i$

Anonymous

Anonymous Tue 03 Nov 2020 23:25:55 No.12303784 Report

Quoted By:

>>12303775
i messed up sqrt

Then you get two possible solutions for i and you remove the negative solution and get +i which is how you defined sqrt(-1)

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