Given a function f:R->R such that f(xy) =f(x)f(y) and f(x+y) = f(x) + f(y) for all real x and y, give, with proof, and explicit expression for f(x).

OK, I think I can do this. Note that '>' means 'greater than or equal to' because I can't get the character to display properly. Assume f(x) is nonzero for some x.

Let c>b. Then f(c) = f(c-b) + f(b) > f(b) since f(c-b)>0

=> f is increasing

Now given any real number x, we can find sequences of rationals t(n) above and u(n) below which tend to x. f(t(n)) > f(x) > f(u(n)) => t(n) > f(x) > u(n) for all n