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}=n^2 \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. | 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 \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. | ||
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+ | Colball |
Revision as of 12:43, 15 June 2019
Let us assume that is rational where is a nonperfect square positive integer. Then it can be written as . But no perfect square times a nonperfect square positive integer is a perfect square. Therefore is irrational.
Thank you,
Colball