Prove that $f(x)>f(y)$ if $x>y$ I am working on Spivaks calculus book and have the following question:
We have a function with properties (Chapter 3, Ex. 17, (d)):
$$f(x+y)=f(x)+f(y)$$
$$f(xy)=f(x)f(y)$$
and we are trying to prove that $f(x)=x$ for $x$ a real number.
In this step (d) we want to prove that $f(x)>f(y)$ if $x>y$ and here I am struggling.
My attempt was to do the following:
$x>y$ is the same as $x-y>0$.
Then
$$f(x-y) = f(x)+f(-y) = f(x)+f(-1)f(y)=f(x)+f(y)$$
because from (a) we know that
$$f(1)=1=f((-1)(-1))=f(-1)^2$$
therefore
$$f(-1)=1$$
From (c) we know that $f(x)>0 $ if $x>0$ so
$$f(x-y)=f(x)+f(y)>0$$
but i want $f(x)-f(y)>0$.
The solution manual says instantly $f(x-y)=f(x)-f(y)$, but why is that? And what is wrong in my reasoning?
 A: Note that $f(x+y)=f(x)+f(y)$ implies that:

*

*$f(0)=0$ because $f(0)=f(0+0)=f(0)+f(0)$

*$f(-x)=-f(x)$ because $0=f(0)=f(x-x)=f(x)+f(-x)$

*$f(x-y)=f(x)-f(y)$ because $f(x-y)=f(x)+f(-y)=f(x)-f(y)$
These are classical properties of group homomorphisms.

There's something wrong with your reasoning though. $f(-1)^2=1$ implies that $f(-1)=\pm 1$ not $f(-1)=1$. And indeed, in reality it is $-1$, because ultimately $f(x)=x$ is supposed to be a solution, so $f(-1)=-1$. So either $f(x)=x$ is not a solution (spoiler alert: it actually is) or you've made a mistake.
Here's a different idea. First of all if $f(x_0)=0$ for some $x_0\neq 0$ then $f(x)=0$ for all $x$. That's because
$$f(x)=f(x_0x_0^{-1}x)=f(x_0)f(x_0^{-1}x)=0\cdot f(x_0^{-1}x)=0$$
Note that constant $0$ is a valid solution.
Now assume that $f$ is not constant $0$.
For any $x> 0$ we have
$$f(x)=f(\sqrt{x}\sqrt{x})=f(\sqrt{x})\cdot f(\sqrt{x})=f(\sqrt{x})^2\geq 0$$
But as we've said $f(x)$ cannot be equal $0$ because $f$ is not constant $0$. And so $f(x)>0$.
Therefore if $x> y$ then $x-y> 0$ and so $f(x-y)> 0$ meaning $f(x)-f(y)> 0$ which finally leads to $f(x)> f(y)$.
