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  text: '$\\sqrt 2$, the unique [-positive] [-4bc] whose square is 2, is not a [-4zq].\n\n#Proof\n\nSuppose $\\sqrt 2$ is rational. Then $\\sqrt 2=\\frac{a}{b}$ for some integers $a$ and $b$; [-without_loss_of_generality] let $\\frac{a}{b}$ be in [-lowest_terms], i.e. $\\gcd(a,b)=1$. We have\n\n$$\\sqrt 2=\\frac{a}{b}$$\n\nFrom the definition of $\\sqrt 2$,\n\n$$2=\\frac{a^2}{b^2}$$\n$$2b^2=a^2$$\n\nSo $a^2$ is a multiple of $2$. Since $2$ is [4mf prime], $a$ must be a multiple of 2; let $a=2k$. Then\n\n$$2b^2=(2k)^2=4k^2$$\n$$b^2=2k^2$$\n\nSo $b^2$ is a multiple of $2$, and so is $b$. But then $2|\\gcd(a,b)$, which contradicts the assumption that $\\frac{a}{b}$ is in lowest terms! So there isn't any way to express $\\sqrt 2$ as a fraction in lowest terms, and thus there isn't a way to express $\\sqrt 2$ as a ratio of integers at all. That is, $\\sqrt 2$ is irrational.',
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