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  "body": "\n## Statement\n\nLet $f \\colon \\R \\to \\R$ be an [[additive function|KF719SZD5T]], that is, $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\R$.\nAssume that there is no constant $c$ with $f(x) = cx$ for all $x \\in \\R$.\nThen the [[graph|0JBJYT69ZV]] $\\set{(x, f(x)) : x \\in \\R}$ is a [[dense subset|0BSJSX7ME2]] of the plane $\\R^2$.\n\n## Remarks\n\nAdditivity gives $f(qx) = q f(x)$ for all rational $q$ and real $x$.\nSince $f$ is not linear, there are nonzero $x_1, x_2$ with $f(x_1)/x_1 \\ne f(x_2)/x_2$.\nThen the vectors $v_1 = (x_1, f(x_1))$ and $v_2 = (x_2, f(x_2))$ are linearly independent in $\\R^2$.\nThe graph contains every point $q_1 v_1 + q_2 v_2$ with $q_1, q_2 \\in \\Q$, because $f(q_1 x_1 + q_2 x_2) = q_1 f(x_1) + q_2 f(x_2)$.\nThese points are dense in $\\R^2$, since $v_1, v_2$ form a basis of $\\R^2$ and $\\Q$ is dense in $\\R$.\nA typical instance is the [[coordinate function of a Hamel basis|FFTNS66XG6]].\n\n## Explanations\n\n### L1 human\n\nA line through the origin is a thin set in the plane.\nA non-linear additive function is the opposite: its graph comes arbitrarily close to every point of the plane.\nSo such a function is as far from a drawable curve as possible.\n\n### L2 human\n\nThis proposition explains why any mild regularity makes an additive function linear.\nA dense graph meets every small rectangle, so the function is unbounded above and below on every interval.\nHence boundedness on one interval, as in [[Darboux's theorem|2F30ZHN1RT]], already rules out the non-linear case.\nThe proof only uses that $f$ is $\\Q$-linear, which every additive function on $\\R$ is.\n",
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