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  "body": "\n## Proof\n\nWe argue by contradiction.\nBy the definition of an [[irrational number|080T2TXVQV]], it suffices to show that no fraction $a / b$ of integers with $b \\neq 0$ has square $2$.\nThe [[square root|0VB8BFRABH]] $\\sqrt{2}$ is a real number with square $2$, so this shows that $\\sqrt{2}$ is irrational.\n\n1. Suppose that $a$ and $b$ are integers with $b \\neq 0$ and $(a / b)^2 = 2$.\n2. By the [[existence of a fraction in lowest terms|0R64A2CVN4]], there are integers $r$ and $s$ with $s \u003e 0$, $\\gcd(r, s) = 1$ and $r / s = a / b$.\n3. Then $(r / s)^2 = 2$. Multiplying by $s^2$ gives $r^2 = 2 s^2$.\n4. The number $2$ is a [[prime number|0SN0MAEKYX]]: a positive divisor of $2$ is at most $2$, so it is $1$ or $2$.\n5. By step 3, $2$ [[divides|096PW7Q014]] $r \\cdot r$. By [[Euclid's lemma|051VYKA1EX]], $2$ divides $r$. Write $r = 2 t$ with $t$ an integer.\n6. Then $4 t^2 = 2 s^2$, so $s^2 = 2 t^2$. Thus $2$ divides $s \\cdot s$, and Euclid's lemma gives $2 \\mid s$.\n7. So $2$ is a common divisor of $r$ and $s$. Then $\\gcd(r, s) \\ge 2$ by the definition of the [[greatest common divisor|02WF878BH1]].\n8. This contradicts $\\gcd(r, s) = 1$. So there are no such $a$ and $b$, and $\\sqrt{2}$ is irrational.\n\n## Idea\n\nWrite a supposed fraction for $\\sqrt{2}$ in lowest terms.\nThe equation $r^2 = 2 s^2$ forces $r$ to be even, and then $s$ to be even too.\nSo the fraction was not in lowest terms after all.\n",
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