Monotone Additive Function is Linear

Theorem
Let $f:\R\to\R$ be a monotonic function which is additive, i.e., $f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ for all $x,y\in\R$.

Then, there exists $a\in\mathbb{R}$ such that $f\left(x\right)=ax$ for all $x\in\R$.

Proof
Let $a=f\left(1\right)$. Then $f\left(1\right)=a\times1$. Supposing, by induction, that $f\left(n\right)=an$ for some $n\in\N$, we have $f\left(n+1\right)=f\left(n\right)+f\left(1\right)$ which is $an+a=a\left(n+1\right)$. So:

$\forall n\in\mathbb{N},\,\,\,\,\,\, f\left(n\right)=an $

We also have: $f\left(1\right)=f\left(0+1\right)=f\left(0\right)+f\left(1\right)$, that is:

$f\left(0\right)=0$

Another milestone: for all $x\in\mathbb{R}$, we have $0=f\left(0\right)=f\left(x+\left(-x\right)\right)=f\left(x\right)+f\left(-x\right)$, and therefore the function $f$ is odd:

$\forall x\in\mathbb{R},\,\,\,\,\,\, f\left(-x\right)=-f\left(x\right)$

Then it follows that $f\left(-n\right)=-\left(an\right)=a\left(-n\right)$. Since we have already proved to the naurals, to zero, and to the negaive integers, we see that is holds for all integers:

$\forall p\in\mathbb{Z},\,\,\,\,\,\, f\left(p\right)=ap$

Now, we have $f\left(1x\right)=1f\left(x\right)\,\forall x\in\mathbb{R}$. If we suppose, by induction, that $f\left(nx\right)=nf\left(x\right)$, we have $f\left(\left(n+1\right)x\right) = f\left(nx+x\right) = f\left(nx\right)+f\left(x\right) = nf\left(x\right)+f\left(x\right) = \left(n+1\right)f\left(x\right)$. It also holds for $n=0$: $f\left(0x\right)=f\left(0\right)=0=0f\left(x\right)\,\forall x\in\mathbb{R}$. For the negative integers, we have: $f\left(\left(-n\right)x\right)=-f\left(nx\right)=-\left(nf\left(x\right)\right)=\left(-n\right)f\left(x\right)$. Therefore:

$\forall p\in\mathbb{Z},x\in\mathbb{R},\,\,\,\,\,\, f\left(px\right)=pf\left(x\right) $

Given $q\in\mathbb{Z},q\neq0$, we have $a=f\left(1\right)=f\left(\frac{q}{q}\right)=f\left(q\frac{1}{q}\right)=qf\left(\frac{1}{q}\right)$, i.e.:

$\forall q\in\mathbb{Z},q\neq0,\,\,\,\,\,\, f\left(\frac{1}{q}\right)=\frac{a}{q}$

Given $p,q\in\mathbb{Z},q\neq0, f\left(\frac{p}{q}\right)=f\left(p\frac{1}{q}\right)=pf\left(\frac{1}{q}\right)=p\frac{a}{q}=a\frac{p}{q}$, and then:

$\forall r\in\mathbb{Q},\,\,\,\,\,\, f\left(r\right)=ar $

Let $x\in\mathbb{R}-\mathbb{Q}$. Let $\left(r_{n}\right)$ be crescent, wih $r_{n}\in\mathbb{Q}$ for each $n\in\mathbb{N}$, and with $\lim r_{n}=x$. Likewise, let $\left(s_{n}\right)$ be decrescent, with each term a rational number, such that $\lim s_{n}=x$. It is always possible to construct sequences like these, for $\mathbb{Q}$ is dense in $\R$. Now, we will assume (without loss of generality) that $f$ is crescent. Then we have $f\left(r_{n}\right)<f\left(x\right)<f\left(s_{n}\right)$ for all $n$. But $f\left(r_{n}\right)=ar_{n}$, and $f\left(s_{n}\right)=as_{n}$, and then:

$ar_{n}<f\left(x\right)<as_{n}$

Since both sequences $ar_{n}$ and $as_{n}$ converge to $ax$, $f\left(x\right)$ must be $ax$. (because $f\left(x\right)$ must be $\geq\lim ar_{n}$, and $\leq\lim as_{n}$).