Nikita: the latest news

Nikita expands his internet presence by attacking the Russian functional programming community: http://ru-declarative.livejournal.com/108182.html

Cute!

>>177

So you want a function f: R -> R such that:

for all a,b in R, f(a + b) = f(a) + f(b)
for all a,c in R, f(ac) = f(a)f(c)

and then 0 and 1 are fixed?

Is that it? I'm not that great in ring theory.

>>180
Yeah, and by 0 and 1 being fixed I hope you mean that f(0)=0 and f(1)=1.

I guess 1 can't be fixed if the zero function is a solution, and zero being fixed is a consequence of satisfying f(a + b) = f(a) + f(b), what with:

f(a + 0) = f(a) + f(0)
f(a) = f(a) + f(0)
f(a) - f(a) = f(a) - f(a) + f(0)
0 = f(0)

so I'm good. One proof coming right up! Would you like fries with it?