Let f:R→R satisfy

f(a)≠0 for some a in R

f(xf(y))=yf(x) for all x,y in R
Prove that f(x)=f(x) for all x in R.
(In reply to
re(2): A final input by Bractals)
I realized I made a mistake by assuming f(x^r)=f(x)^r for all real numbers r at some point, although I have proven it only for rational r. Maybe there are more than two solutions after all... I still think they'd have to be discontinuous everywhere.

Posted by JLo
on 20061230 06:58:23 