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  "body": "\n## Proof\n\nWe argue by contradiction, with a prime divisor of the denominator.\nLet $x$ be real, $n \\ge 1$ an integer and $m$ an [[integer|0C4JS9RF5M]] with $x^n = m$. Suppose that $x$ is not an integer.\n\n1. Suppose that $x$ is not [[irrational|080T2TXVQV]], that is, $x$ is rational.\n2. By the [[existence of a fraction in lowest terms|0R64A2CVN4]], $x = a / b$ with integers $a$ and $b$, $b \u003e 0$ and $\\gcd(a, b) = 1$.\n3. If $b = 1$, then $x = a$ is an integer, which is false. So $b \\ge 2$.\n4. By the [[existence of a prime divisor|0SPBEDVNW1]], there is a [[prime number|0SN0MAEKYX]] $p$ with $p \\mid b$.\n5. From $x^n = m$ we get $a^n / b^n = m$, so $a^n = m b^n$.\n6. Since $n \\ge 1$, $b$ [[divides|096PW7Q014]] $b^n$, and so $p$ divides $m b^n = a^n$.\n7. Claim: if $p \\mid a^k$ for an integer $k \\ge 1$, then $p \\mid a$. We use [[induction|0ASJVEFA9J]] on $k$. For $k = 1$ there is nothing to prove.\n8. If $p \\mid a^{k+1} = a^k \\cdot a$, then [[Euclid's lemma|051VYKA1EX]] gives $p \\mid a^k$ or $p \\mid a$. In the first case, the hypothesis of the induction gives $p \\mid a$.\n9. By steps 6 to 8, $p \\mid a$. Together with step 4, $p$ is a common divisor of $a$ and $b$.\n10. By the definition of the [[greatest common divisor|02WF878BH1]], $\\gcd(a, b) \\ge p \\ge 2$. This contradicts $\\gcd(a, b) = 1$.\n11. So $x$ is not rational. Since $x$ is real, $x$ is irrational.\n\n## Idea\n\nWrite the rational root $a / b$ in lowest terms with $b \u003e 1$.\nAny prime that divides $b$ divides $a^n = m b^n$, and so divides $a$.\nThis contradicts lowest terms, so the denominator is $1$ and the root is an integer.\n",
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