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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==== Proof that 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}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \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}</math> is irrational.

Latest revision as of 14:17, 19 June 2019

Proof that any nonperfect square positive integer is irrational

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