>>184
That comment makes me think I have the wrong idea of what we are talking about here, but here is my proof of what I gathered from
>>180-182
Suppose that f is a function on the reals, satisfying the following:
P1: for all a,b in R, f(a + b) = f(a) + f(b)
P2: for all a,c in R, f(ac) = f(a)f(c)
Part 1: f(0) = 0.
let a,b = 0 from P1.
f(0 + 0) = f(0) + f(0)
f(0) = f(0) + f(0)
f(0) - f(0) = (f(0) + f(0)) - f(0)
0 = f(0) + 0
0 = f(0) XD
Part 2: f(1) = 1, or f(1) = 0
f(1) might be zero. In the case that f(1) is not zero, we have the following:
let a,c = 1 from P2.
f(1*1) = f(1)f(1)
f(1) = f(1)f(1)
f(1)(1/f(1)) = (f(1)f(1))(1/f(1))
1 = f(1)1
1 = f(1) XD
Part 3: f(1/x) = 1/f(x)
let x be a real, with x not 0
1 = f(1) = f(x(1/x)) = f(x)f(1/x)
because f(x)f(1/x) = 1, f(x) is the multiplicative inverse of f(1/x).
Hence, 1/f(x) = f(1/x) XD
Part 4: f(-x) = -f(x)
0 = f(0) = f(x + -x) = f(x) + f(-x)
Because 0 = f(x) + f(-x), f(x) is the additive inverse of f(-x).
Hence, f(-x) = -f(x). XD
Part 5: f is the zero function or the identity.
to be continued.
Sorry, I can't do any more right now, not that any of what I've written so far is hard or original.
The fries might be cold.