Functional Equation (20 points)

f(0 + 0) = f(0) + f(0) => f(0) = 2f(0) => f(0) = 0 .............(1)
f(a - a) = f(a) + f(-a) => f(a) = -f(a)....................................(2)

It can be easily proven with M.I. that f(ab) = a f(b) if a is a positive integer, by (1) and (2), it is true for all integers.

For p,q are integers.
f(1) = f(q * 1/q) = q f(1/q)
f(p/q) = p f(1/q) = p/q f(1).

Finally, if f(x) is continuous, since all reals has a rational sequence as it's limit, we can let

x = lim n->infinity xn.
f(x) = f(lim n->infinity xn) = lim n->infinityf(xn) = lim n->infinityxnf(1) =xf(1)

Let c = f(1), we have f(x) = cx, done.