# Function Proof . . .

1. Aug 9, 2009

### jgens

1. The problem statement, all variables and given/known data

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 $x>0$

2. Relevant equations

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

3. 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!

2. Aug 9, 2009

### tiny-tim

Hi jgens!

Hints: what is f(√x)?

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

3. Aug 9, 2009

### 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).