Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.7 - Page 213: 19


$\sqrt 5$ is irrational.

Work Step by Step

Assume that $\sqrt 5$ is rational. Then, $\sqrt 5 = \frac{a}{b}, 5 = \frac{a^2}{b^2}, 5b^2 = a^2$ for some $a,b$ where either $a$ or $b$ is not divisible by 5.. Then, for this to work, $a$ must be divisible by $5$ (otherwise the $5b^2 = a^2$ equation fails. Then, we get $5b^2 = 25(a')^2$ for some $a' = \frac{a}{5}$, $b^2 = 5(a')^2$. Now for this to work, $b$ must be divisible by $5$. However, from the beginning, both $a$ and $b$ can't be both divisible by 5, so this is a contradiction and $\sqrt 5 $ must be irrational.
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