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Prove that the continuous function $f:mathbb{R} to mathbb{R}$ satisfying $fleft(xright)-fleft(yright) inmathbb{Q} iff fleft(x-yright) in mathbb{Q}$ is of the form $
fleft(xright)=ax+b.$

My Attempt.

I tried considering the function $$gleft(xright)=frac{fleft(xright)-fleft(0right)}{fleft(1right)-fleft(0right)}$$ which also satisfies the property $gleft(aright)-gleft(bright) inmathbb{Q} iff gleft(a-bright) in mathbb{Q}$,

Now I am trying to prove that this function g is identity function and then I can prove that $fleft(xright)=left(fleft(1right)-fleft(0right)right)x+fleft(0right).$ And I am done.

Also This function has to be identity function because $gleft(0right)=0$ and $gleft(1right)=1$.

I tried assuming that the function g is such that $gleft(aright)neq a$ for some $ain mathbb{R}$. Then by continuity $gleft(xright)neq x$ for some $delta>0$ neighborhood of $a$.
But I cannot move further.

Also
Using a previously known result, I was able to prove that f must be monotonic. However I do not want to use any other result which is not known and not trivial.

If a function $f $ is continous in $left[a,bright]$ and $fleft(aright)=fleft(bright)$ then for any $epsilon >0$ there exists $m,n in left[a,bright] $such that $fleft(mright)=fleft(nright)$ and $m-n=epsilon$.

In this case choose $epsilon in mathbb{R}-mathbb{Q} $ and get $m,n in left[a,bright] $ such that $fleft(mright)-fleft(nright)=0 in mathbb{Q}$ but $m-n inmathbb{R}-mathbb{Q}$.

For any $y$, we have either $f(x+y) - f(x) in mathbb Q$ for all $x$, or $f(x+y) - f(x) notin mathbb Q$ for all $x$, depending on whether or not $f(y) in mathbb Q$. But $f(x+y)-f(x)$ is continuous, so by the Intermediate Value Theorem we conclude $f(x+y) - f(x)$ is constant. Thus
$$ f(x+y) - f(x) = f(y) - f(0) $$
This says $f(x) - f(0)$ is an additive function. And it's not hard to prove that continuous additive functions are linear.