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Cauchy's functional equation asks for functions $f:\mathbb{R}\to \mathbb{R}$ satisfyingThe identity must hold for every pair of real numbers, and a solution is an additive function. It says that combining two inputs before applying $f$ gives the same result as combining their outputs.^[ref]

For example, $f(x)=3x$ satisfies the identity because $3(x+y)=3x+3y$. At $x=2/3$ and $y=5/6$ the two sides giveFor positive widths, a rectangle of height three makes this visible: cutting its width into $x$ and $y$ partitions its area into $3x$ and $3y$. The same identity also holds for negative inputs, where $3x$ is a signed output rather than an ordinary area.

Every function $f(x)=cx$ with a fixed real number $c$ solves Cauchy's functional equation. The surprising question is whether these are all the solutions. Rational homogeneity of an additive function forces $f(q)=qf(1)$ for every rational number $q$, but reaching an irrational number requires an additional argument. Continuity at just one point or being a bounded function on an interval of positive length supplies that argument; without such a condition, discontinuous additive functions exist.^[ref]

This makes Cauchy's functional equation a useful test of the boundary between algebra and analysis: a rule about addition fixes all rational multiples, while a tiny amount of continuity fixes the entire real line. For example, the continuous exponential functional equation reduces to an additive function after taking a logarithm, so this result explains why its nonzero solutions are exponential functions.