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title: 'The n-th root of m is either an integer or irrational',
clickbait: 'In other words, no power of a rational number that is not an integer is ever an integer.',
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text: 'There is an intuitive way to see that for any natural numbers $m$ and $n$, $\\sqrt[n]m$ will always either be an integer or an irrational number.\n\nSuppose that there was some $\\sqrt[n]m$ that was a rational number $\\frac{a}{b}$ that was not an integer. Suppose further that $\\frac ab$ is written as a [-reduced] fraction, such that the [-greatest_common_divisor] of $a$ and $b$ is $1$. Then, since $\\frac{a}{b}$ is not an integer, $b > 1$.\n\nSince $\\frac ab = \\sqrt[n]m$, we have conversely that $(\\frac ab)^n = m$, which is a natural number by our hypothesis. But let's take a closer look at $(\\frac ab)^n$. It evaluates to $\\frac{a^n}{b^n}$, which is still a reduced fraction. [todo: Proof of gcd(a^n, b^n) = 1.]\n\nBut since $b > 1$ before, we have that $b^n > 1$ as well, meaning that $(\\frac ab)^n$ cannot be an integer, contradicting the fact that it equals $m$, a natural number.',
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