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  "body": "\n## Proof\n\n1. By the definition of the degree of an [[algebraic number|0MB4TGMXG6]], there is a polynomial of degree $d$ with rational coefficients that vanishes at $\\alpha$. Multiplying by a common denominator of its coefficients, we get such a polynomial $f$ with integer coefficients.\n2. Claim: $f$ has no rational root. Suppose $f(r) = 0$ with $r \\in \\Q$. By [[division by the monic polynomial|0MES3KS0ZC]] $X - r$ over $\\Q$, $f = (X - r) g + c$ with $c \\in \\Q$ and $g \\in \\Q[X]$ of degree $d - 1$. Evaluating at $r$ gives $c = 0$.\n3. Since $d \\ge 2$, $\\alpha$ is not rational, so $\\alpha \\ne r$. Then $0 = f(\\alpha) = (\\alpha - r) g(\\alpha)$ gives $g(\\alpha) = 0$. So $\\alpha$ is a root of a nonzero rational polynomial of degree $d - 1 \u003c d$, which contradicts the minimality of $d$. This proves the claim.\n4. The [[derivative|0MH1WK89PP]] $f'$ is a polynomial, so it is continuous. By the [[extreme value theorem|0WS1A8E2P9]], $\\abs{f'}$ has a maximum $M \\ge 0$ on $[\\alpha - 1, \\alpha + 1]$.\n5. Put $c = \\dfrac{1}{2 \\max(1, M)}$. Then $0 \u003c c \\le \\frac{1}{2}$. Let $p$ and $q$ be integers with $q \u003e 0$.\n6. Case $\\abs{\\alpha - p/q} \\ge 1$. Then $\\abs{\\alpha - p/q} \\ge 1 \u003e \\frac{1}{2} \\ge c \\ge c/q^d$.\n7. Case $\\abs{\\alpha - p/q} \u003c 1$. Write $f = \\sum_{j=0}^{d} b_j X^j$ with integers $b_j$. Then $q^d f(p/q) = \\sum_{j=0}^{d} b_j p^j q^{d-j}$ is an integer.\n8. By step 2, $f(p/q) \\ne 0$. So $\\abs{q^d f(p/q)} \\ge 1$, that is, $\\abs{f(p/q)} \\ge 1/q^d$.\n9. Also $p/q \\ne \\alpha$, since $\\alpha$ is irrational. By the [[mean value theorem|0BARQY52N5]], $f(p/q) - f(\\alpha) = f'(t)(p/q - \\alpha)$ for some $t$ between $\\alpha$ and $p/q$. Then $t \\in [\\alpha - 1, \\alpha + 1]$.\n10. Since $f(\\alpha) = 0$, steps 4, 8 and 9 give $\\frac{1}{q^d} \\le \\abs{f(p/q)} \\le M \\abs{\\alpha - p/q}$.\n11. In particular $M \u003e 0$, and $\\abs{\\alpha - p/q} \\ge \\frac{1}{M q^d} \u003e \\frac{c}{q^d}$, because $c \u003c 1/M$.\n12. In both cases $\\abs{\\alpha - p/q} \u003e c/q^d$. This proves the first part.\n13. Now let $x$ be a [[Liouville number|0J1R5NP36S]]. Suppose that $x$ is algebraic, of degree $d$. Since a nonzero constant has no root, $d \\ge 1$.\n14. Suppose $d = 1$. Then $u x + v = 0$ for some integers $u \\ne 0$ and $v$, as in step 1. So $x = a/b$ with integers $a$, $b$ and $b \\ge 1$. Choose $n$ with $2^{n-1} \u003e b$.\n15. The definition gives integers $p$, $q$ with $q \\ge 2$ and $0 \u003c \\abs{a/b - p/q} \u003c 1/q^n$. The number $\\abs{aq - bp}$ is a positive integer, so $\\abs{a/b - p/q} = \\abs{aq - bp}/(bq) \\ge 1/(bq)$.\n16. So $1/(bq) \u003c 1/q^n$, that is, $q^{n-1} \u003c b$. But $q^{n-1} \\ge 2^{n-1} \u003e b$, a contradiction. So $d \\ge 2$.\n17. Let $c \u003e 0$ be the constant of the first part for $x$. By the [[Archimedean property|0A3GWZ86AY]] and $2^k \\ge k$, there is an integer $n \u003e d$ with $2^{n-d} \\ge 1/c$.\n18. The definition gives integers $p$, $q$ with $q \\ge 2$ and $\\abs{x - p/q} \u003c 1/q^n$.\n19. Since $q^{n-d} \\ge 2^{n-d} \\ge 1/c$, we get $\\frac{1}{q^n} = \\frac{1}{q^d q^{n-d}} \\le \\frac{c}{q^d}$. So $\\abs{x - p/q} \u003c c/q^d$, against the first part.\n20. So $x$ is not algebraic, that is, $x$ is [[transcendental|079EGSD32Y]].\n\n## Idea\n\nIf $p/q$ is not a root of the integer polynomial $f$, then $f(p/q)$ is a fraction with denominator $q^d$, so it is at least $1/q^d$ in size.\nBut $f(\\alpha) = 0$, so the mean value theorem makes $f(p/q)$ small when $p/q$ is close to $\\alpha$.\nA Liouville number has approximations that are too good for any fixed $d$, so it cannot be algebraic.\n",
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