/sci/ - Science & Math

[Anonymous No.12300105]

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

[Anonymous No.12300118]

>>12300105
[math]\sqrt{x^2} \neq x[/math] in general

[Anonymous No.12300143]

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

[Anonymous No.12300407]

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

[Anonymous No.12300547]

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

[Anonymous No.12302297]

>>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 No.12302509]

>>12300105
[eqn]\sqrt{-1}=\sqrt{(-i)^2}=\sqrt{(-1)^2i^2}=\sqrt{i^2}=i[/eqn]

[Anonymous No.12303775]

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

[Anonymous No.12303784]

>>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)