If I remember correctly i = sqrt(-1) is just the definition. If you change the exponent, it's no longer defined as i.

but
((-1)^2)^(1/2) = 1^(1/2) = 1

Anyway, nice doodles. <3

Aureolen said:
If I remember correctly i = sqrt(-1) is just the definition. If you change the exponent, it's no longer defined as i.

((-1)^(1/2))^2 = i^2 = -1

but
((-1)^2)^(1/2) = 1^(1/2) = 1

Anyway, nice doodles. <3

Close. i has 4 possible exponents defined for it, 1 through 4. i^1 = sqrt(-1) as you said, but as math goes, i^2 = -1, i^3 = -i (-1 * sqrt(-1), or -1 * i), and i^4 = 1 (because (-1)^2)

Vaporius said:
Close. i has 4 possible exponents defined for it, 1 through 4.

Also, to further nerd because I realized I missed a thing; i can have more than 4 powers, but it's basically "power mod 4", so i^7 = i^3 (7 mod 4 = 3), i^4 technically = 1 because x^0=1 (except where x=0), and i^714 = i^2

Aureolen said:
If I remember correctly i = sqrt(-1) is just the definition. If you change the exponent, it's no longer defined as i.

((-1)^(1/2))^2 = i^2 = -1

but
((-1)^2)^(1/2) = 1^(1/2) = 1

Anyway, nice doodles. <3

What the hell

I just noticed a 3D OWO there.

is it bad that I want this sylveon to step on me

Rainbow_Sheep0666 said:
What the hell

The math on the post

I also want to nerd out since I'm a math major. Not the place I thought I'd put my degree to use but oh well...

Square root function is discontinuous on the complex plane and we usually use the principal branch, i.e. there's a discontinuation in the negative real axis. Because of this reason some familiar properties aren't true generally, e.g. (xy)^(1/2) != x^(1/2)*y^(1/2) and (x^y)^(1/2) != (x^(1/2))^y.

The error in the calculation happens when the artist separates (-1)^(2/4) into ((-1)^2)^(1/4).

Don't tell anyone you learned this from a porn site.

foxxy~ said:
I also want to nerd out since I'm a math major. Not the place I thought I'd put my degree to use but oh well...

Square root function is discontinuous on the complex plane and we usually use the principal branch, i.e. there's a discontinuation in the negative real axis. Because of this reason some familiar properties aren't true generally, e.g. (xy)^(1/2) != x^(1/2)*y^(1/2) and (x^y)^(1/2) != (x^(1/2))^y.

The error in the calculation happens when the artist separates (-1)^(2/4) into ((-1)^2)^(1/4).

Don't tell anyone you learned this from a porn site.

Thank you! this is a very good explanation, I won't tell anyone I learned it here

I take pride that I learned this on a porn site, I think it reflects upon who I am.

Ok, but is no one gonna mention the fact that the sylveon is getting oral and anal at the same time by the same eevee?

Princess bride but with eeveelutions instead, I'd watch it

foxxy~ said:
I also want to nerd out since I'm a math major. Not the place I thought I'd put my degree to use but oh well...

Square root function is discontinuous on the complex plane and we usually use the principal branch, i.e. there's a discontinuation in the negative real axis. Because of this reason some familiar properties aren't true generally, e.g. (xy)^(1/2) != x^(1/2)*y^(1/2) and (x^y)^(1/2) != (x^(1/2))^y.

The error in the calculation happens when the artist separates (-1)^(2/4) into ((-1)^2)^(1/4).

Don't tell anyone you learned this from a porn site.

I know I'm a few years late to reply to this but I also love this stuff. Great response.

I just wanted to add on for anyone else who is trying to understand that while 1^(1/4) is equal to 1, 1 actually has 4 potential 4th roots in the complex plain, those being 1,i, -1, and -i, all of which equal 1 when put to the 4th power, so in that sense you can claim any of those numbers are equal to 1 using similar logic:

let a = any of {1,i,-1,-i}

a^4 = 1
(a^4)^(1/4) = 1^(1/4)
a = 1^(1/4)
a = 1

The key detail is really that fractional (non-integer) exponents can represent multiple values in the negative/complex numbers.