Proving f(x)>0 if x>0 with Function Proof

[jgens]

Homework Statement

Prove that if f is the function which is not always zero, that satisfies f(x+y)=f(x)+f(y) and f(xy)=f(x)f(y), we have that f(x)>0 if x>0

Homework Equations

So far I've managed to prove that f(x)=x if x \in \mathbb{Q} and that f must be odd.

The Attempt at a Solution

Suppose not, then if x > 0 and irrational we have that f(-x) > 0 > f(x). Since any rational number b > 0 can be expressed as the sum of two irrational numbers - x + (b-x) for instance - we have that b = x + y > 0 where x,y are irrational. This implies that,

b = f(b) = f(x+y) = f(x) + f(y) > 0

Clearly, both x,y cannot be negative since this would imply that x+y < 0 a contradiction. We also have x,y cannot both be positive since this would imply that f(x)+f(y)<0 another contradiction.

I'm not positive that any of this is correct (probably isn't) and I would appreciate any corrections along with suggestions on how to complete the proof. Thanks!

 

[tiny-tim]

Hi jgens!

Hints: what is f(√x)?

If f(y) = 0, what is f(x/y)?

 

[rasmhop]

I can't really follow your logic in 3). You first assume we have one positive irrational number x which is a counterexample to the problem statement. You then choose an arbitrary positive rational number b, and consider the irrational number y=b-x (about which you know very little). You say that x, y can't both be negative which is correct (since b is positive and also since by definition x > 0 so that can't be negative), but I don't see why we couldn't have y < 0 < x. x+y could still be positive if the absolute value of x is greater than that of y, and while f(x) < 0 you don't necessarily know that f(y) < 0 (remember you only assumed that there was a contradiction at one point x, not at all irrational points).

I think you're over complicating the problem slightly. If x > 0 we can write it as x = a^2 for some a > 0. We then have:
f(x) = f(aa) = f(a)^2
Also note that if a \not = 0 then a has a multiplicative inverse so:
1 = f(1) = f(a)f(1/a)
Try to see if you can show f(x) > 0 from this. You don't need a complete formula for f(x).