>>219
Well, so far, we have that:
1. If f(1) = 0, then f is the zero function.
2. If f(1) is not 0, then f is the identity function on the rationals.
If f has to be continuous, then (2) implies that f is the identity on the reals as well, but without continuity I can't connect the rationals to non-rationals so easily.
Part 7: if f(1) is not zero, then f is id on all real nth roots of all rationals.
Let n be a power and r a non-negative rational.
Let b be the unique real nth root of r, b = root(r,n)
r
= f(r)
= f(b*b* ... (n times) ... *b)
= f(b)*f(b)*f(b)* ... (n times) ... *f(b)
Since f(b) multiplied by itself n times is equal to r, we have that f(b) is an nth root of r. Because real roots are unique, this implies that f(b) = b. XD