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Difference between revisions of "Proof that the square root of any nonperfect square positive integer is irrational"
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Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be written as <math>\frac{p}{q} \Rightarrow \frac{p^2}{q^2}=n</math>. That means that <math>(q^2)n=p^2</math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.
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Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be written as <math>\frac{p}{q}=n^2 \Rightarrow \frac{p^2}{q^2}=n</math> \Rightarrow (q^2)n=p^2<math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore </math>\sqrt{n}$ is irrational.

Revision as of 12:41, 15 June 2019

Let us assume that $\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\frac{p}{q}=n^2 \Rightarrow \frac{p^2}{q^2}=n$ \Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore$\sqrt{n}$ is irrational.

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