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>. | ||
− | Thus <math>\frac{p^2}{q^2}=n</math>. That means that <math>(q^2)n=p^2</math>. But no perfect square times | + | 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 is rational where is a nonperfect square positive integer. Then it can be expressed as . Thus . That means that . But no perfect square times a nonperfect square positive integer is a perfect square. Therefore is irrational.