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  "body": "\n## Problem\n\nFind all [[functions|0EPAT51DXQ]] $f \\colon \\R \\to \\R$ such that\n$$f(x + y) = f(x) + f(y) \\quad \\text{for all } x, y \\in \\R.$$\n\n## Remarks\n\nThe problem is solved.\nEvery solution satisfies $f(0) = 0$, $f(-x) = -f(x)$ and $f(qx) = q f(x)$ for every rational $q$, by induction on sums and then division.\nSo $f(q) = cq$ for all rational $q$, where $c = f(1)$.\nIf $f$ is also [[continuous|0X3WGX95FE]], then by density of $\\Q$ in $\\R$ the solution is $f(x) = cx$ for all real $x$.\nThe same conclusion holds under each of these weaker conditions on a solution $f$:\n\n1. $f$ is continuous at one point;\n2. $f$ is [[monotone|0A7KV2WB7A]] on some interval;\n3. $f$ is bounded above on some interval;\n4. $f$ is measurable for [[Lebesgue measure|2Z2QVWZ262]].\n\nWithout such a condition there are other solutions.\nTake a [[basis|0KCQ9Y3QKA]] of $\\R$ as a vector space over $\\Q$, which exists by the [[axiom of choice|0HTGB13FE2]].\nEvery $\\Q$-linear map $\\R \\to \\R$ solves the equation, and choosing its values on the basis freely gives solutions that are not of the form $cx$.\nThe graph of such a non-linear solution is dense in the plane $\\R^2$.\n\n## Explanations\n\n### L1 human\n\nThe equation asks for functions that turn sums into sums.\nThe lines through the origin, $f(x) = cx$, are the obvious answers, and they are the only reasonable ones.\nStrange solutions exist only when one allows wild functions, which need the axiom of choice to construct and cannot be drawn.\n",
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