...ignore that last entry.
Feb. 25th, 2006 04:50 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoyCharacteristic p implies (x+y)^p=x^p+y^p *when you have commutativity*. Upon trying an actual calculation with some noncommutative elements, my counterexample does not satisfy a^5=a. It simply does *not* work. Similarly for my generalization and interesting note.
...now I have no idea again.
FWIW, when I was actually trying to prove a^5=a implies commutativity, I was able to get 15xy=15yx and x²y=yx².
Time to go back to trying to prove it, to find a counterexample, or to just stop thinking about it?
In other news, the lounge clock is broken. There are other clocks but I'm just so used to looking at the main one in the lounge that I have to remember it's not actually always 7:45.
-Sniffnoy
...now I have no idea again.
FWIW, when I was actually trying to prove a^5=a implies commutativity, I was able to get 15xy=15yx and x²y=yx².
Time to go back to trying to prove it, to find a counterexample, or to just stop thinking about it?
In other news, the lounge clock is broken. There are other clocks but I'm just so used to looking at the main one in the lounge that I have to remember it's not actually always 7:45.
-Sniffnoy
