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Proof that the square root of any nonperfect square positive integer is irrational

Revision as of 12:35, 15 June 2019 by Colball (talk | contribs) (Created page with "Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be expressed as <math>\frac{p}{q}</math>. Thus <...")
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Let us assume that $\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be expressed as $\frac{p}{q}$. Thus $\frac{p^2}{q^2}=n$. That means that $(q^2)n=p^2$. But no perfect square times an integer is a perfect square. Therefore $\sqrt{n}$ is irrational.

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