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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 $f(b)=b$?

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    $\begingroup$ No. $f(x)=x+k$ where $k$ is some constant also can be the solution. $\endgroup$ – Satvik Mashkaria Apr 29 '14 at 3:53
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    $\begingroup$ No one has explained why the unique solution is $f(x)=x+k$, $k\in\mathbb R$. Well, substitute $b$ with $0$ and you have $f(a)=a+f(0)=a+k$. The $f(0)$ is just a plain constant. So if we do have real numbers $a,b$ satisfying $f(a)-f(b)=a-b$, then the function cannot be of any form other than $f(x)=x+k$ for some real $k$. We're still not sure whether the function of the form will always satisfy the equation. Substitute $f(a)$, $f(b)$ with $a+k$ and $b+k$ and you can easily see that it does satisfy $f(a)-f(b)=a-b$, hence the function is of the form $f(x)=x+k$ and of no other form. $\endgroup$ – user26486 Apr 29 '14 at 4:50
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    $\begingroup$ Your question is about some specific $a$ and $b$, right? I am led to this by the last line, which separately concludes $f(a)=a$ and $f(b)=b$ instead of summarizing $f(x)=x$. The answers and comments seem to assume the conditions are for all $a,b\in\mathbb{R}$. But there is no such function in that case, since you would be demanding $f(a)>0$ for all $a$. Would you clarify if your conditions are meant to hold for all $a$ and $b$, or just some specific $a$ and $b$? $\endgroup$ – alex.jordan Apr 29 '14 at 5:09
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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 $1$.

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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 $f:\mathbb{R}\rightarrow \mathbb{R}$.

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