Properties of $f(a)-f(b)=a-b$

Question

Let:

$f(a)-f(b)=a-b$

and $f(a)>f(b)$ and $f(a),f(b)>0$ and $a>b$

Based on the above facts is it sufficient enough to say that $f(a)=a$ and $f(b)=b$?

Answer 1

Divide throughout by $a-b$ to get

$$\frac{f(a) - f(b)}{a-b} = 1$$

Notice how this resembles the gradient of a straight line graph.

In other words, $f$ is a straight line with gradient $1$.

Answer 2

Counterexample: Let $f(x)=x+d$ for $d\neq 0$ real number. Then for $a>b$, $f(a)-f(b)=a+d-b-d=a-b>0$. So by choosing any $a,b$ such that $a>b$ you have all the properties you mentioned. Here we assume $f:\mathbb{R}\rightarrow \mathbb{R}$.