{
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  "body": "\n## Example\n\nDefine $f \\colon \\mathbb{R} \\to \\mathbb{R}$ as follows.\nFor irrational $x$, $f(x) = 0$. For rational $x = p/q$ in lowest terms with $q \u003e 0$, $f(x) = 1/q$.\nThen $f$ is [[continuous|0EEE2CZB2A]] at each irrational number and discontinuous at each rational number.\n\n## Remarks\n\nLet $a$ be irrational and let $\\varepsilon \u003e 0$. Only finitely many $p/q$ near $a$ have $1/q \\ge \\varepsilon$. So $f(x) \u003c \\varepsilon$ for all $x$ close enough to $a$.\nAt a rational $a$, irrational points near $a$ have value $0$, but $f(a) \u003e 0$.\nThe function is periodic with period $1$.\nNo function on $\\mathbb{R}$ is continuous exactly at the rationals.\n\n## Explanations\n\n### L1 human\n\nThe value $1/q$ is small when the denominator $q$ is large.\nNear an irrational point, all rationals have large denominators, so the values there are small.\n",
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  "name": "Thomae's function",
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