Solution of the Additive Cauchy Equation

**Step 1: Determine $f(0)$.** Set $x=y=0$ in $f(x+y)=f(x)+f(y)$: $f(0)=f(0)+f(0)$, so $f(0)=0$.

Step 2: Extend to positive integers. For a positive integer $n$, set $x=(n-1)$ (or induct): $f(n\cdot 1) = f((n-1)\cdot 1 + 1) = f((n-1)\cdot 1) + f(1)$. By induction on $n$, $f(n) = n f(1)$ for every positive integer $n$ (base case $n=1$ trivial, inductive step just applied).

Step 3: Extend to negative integers. Set $y=-x$ in $f(x+y)=f(x)+f(y)$: $f(0) = f(x) + f(-x)$, and since $f(0)=0$, $f(-x) = -f(x)$. Combined with Step 2, $f(n) = n f(1)$ for every integer $n$ (positive, negative, or zero), writing $c = f(1)$.

Step 4: Extend to rationals. Let $q = p/r$ with $p\in\mathbb{Z}$, $r\in\mathbb{Z}_{>0}$. Since $r \cdot q = p$ (as integers under repeated addition, i.e. $q + q + \dots + q$ ($r$ times) $= p$), applying the integer case of Step 2/3 to the function evaluated $r$ times gives $f(rq) = r f(q)$ (by the same induction argument used for Step 2, now with $x=q$). But $rq = p$, so $f(p) = r f(q)$, i.e. $cp = r f(q)$, i.e. $f(q) = c \cdot \frac{p}{r} = cq$.

Step 5: Conclude the rational case. This shows $f(q) = cq$ for every $q \in \mathbb{Q}$, with $c = f(1)$ — the entire function on $\mathbb{Q}$ is determined by its value at a single point.

**Step 6: Extend to $\mathbb{R}$ under continuity.** Suppose now $f:\mathbb{R}\to\mathbb{R}$ is additive and continuous at even one point (continuity everywhere then follows from additivity: $f(x+h)-f(x) = f(h) \to 0$ as $h\to 0$ if continuous at $0$). For any real $x$, take a sequence of rationals $q_n \to x$. By Steps 1–5, $f(q_n) = c q_n$. Continuity gives $f(x) = \lim_n f(q_n) = \lim_n c q_n = cx$.

**Step 7: Extend to $\mathbb{R}$ under monotonicity or local boundedness (sketch).** If $f$ is monotonic, then for rationals $q_1 < x < q_2$ squeezing any real $x$, monotonicity forces $cq_1 \le f(x) \le cq_2$ (if $c>0$; reverse if $c<0$), and letting $q_1,q_2\to x$ pins $f(x)=cx$ by the same squeeze. If instead $f$ is bounded on some interval $I$, one shows $f$ is bounded near $0$ (using additivity to shift the interval), then $f(x/n)\to 0$ as $n\to\infty$ for fixed $x$ forces continuity at $0$, reducing to Step 6. In all three regularity cases (continuous, monotonic, bounded on an interval), the conclusion is the same: $f(x)=cx$ for all $x\in\mathbb{R}$.