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  "body": "\n## Proof\n\nWe argue by contradiction, as in the proof for $\\sqrt{2}$.\nLet $p$ be a [[prime number|0SN0MAEKYX]]. Since $p \u003e 0$, the [[square root|0VB8BFRABH]] $\\sqrt{p}$ is a real number with $(\\sqrt{p})^2 = p$.\n\n1. Suppose that $\\sqrt{p}$ is not [[irrational|080T2TXVQV]], that is, $\\sqrt{p}$ is rational.\n2. By the [[existence of a fraction in lowest terms|0R64A2CVN4]], $\\sqrt{p} = a / b$ with integers $a$ and $b$, $b \u003e 0$ and $\\gcd(a, b) = 1$.\n3. Squaring gives $a^2 / b^2 = p$. Multiplying by $b^2$ gives $a^2 = p b^2$.\n4. So $p$ [[divides|096PW7Q014]] $a \\cdot a$. By [[Euclid's lemma|051VYKA1EX]], $p$ divides $a$. Write $a = p c$ with $c$ an integer.\n5. Then $p^2 c^2 = p b^2$. Since $p \\neq 0$, dividing by $p$ gives $b^2 = p c^2$.\n6. So $p$ divides $b \\cdot b$, and Euclid's lemma gives $p \\mid b$.\n7. Thus $p$ is a common divisor of $a$ and $b$. By the definition of the [[greatest common divisor|02WF878BH1]], $\\gcd(a, b) \\ge p \\ge 2$.\n8. This contradicts $\\gcd(a, b) = 1$. So $\\sqrt{p}$ is irrational.\n\n## Idea\n\nIf $\\sqrt{p} = a / b$ in lowest terms, then $a^2 = p b^2$.\nBecause $p$ is prime, $p$ must divide $a$, and then $p$ must divide $b$ as well.\nThis contradicts the choice of lowest terms.\n",
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