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Difference between revisions of "Proof that the square root of any nonperfect square positive integer is irrational"

(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 <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>.
 
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>.
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Thus <math>\frac{p^2}{q^2}=n</math>. That means that <math>(q^2)n=p^2</math>. But no perfect square times an integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.
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Thus <math>\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.

Revision as of 12:37, 15 June 2019

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 a nonperfect square positive integer is a perfect square. Therefore $\sqrt{n}$ is irrational.

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