I thought you couldn't distribute exponents to added terms?
Watch this: [math](x + y)^p = x^p + y^p[/math]
guess that's correct, if x and y are orthogonal vectors and we're talking about taking the inner product.
you can if the field has characteristic [math]p[\math]
>If I leave away important information people will think it's not a tautology, even though it is when I assume [math] x,y \in \mathbb{Z} ~,~ p ~\text{is prime}[/math] and it's actually all mod p.
>LeL brainlets
2*1
Let [math] x=1, y=3, p=2 [/math]
>Use non-standard notation
>Make false statement that is only true in the particular ring I was thinking about
Haha tricked you brainlets can't read my mind
k
>taking the pth power is non standard notation
b r a i n l e t
if you didn't see and use this identity a lot in your algebra sequence, you shouldn't post in math threads
If p = 0 it would be 1 = 2.