Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 1 - Section 1.7 - Introduction to Proofs - Exercises - Page 91: 9

Answer

See proof.

Work Step by Step

Suppose by contradiction that the sum of an irrational number and a rational number is rational. That is, $\dfrac{p}{q}+r=\dfrac{s}{t}$ where $p,q,s,t$ are integers and $r$ is irrational. It follows that $r=\dfrac{s}{t}-\dfrac{p}{q}$. Finding a common denominator we get that $r=\dfrac{sq-pt}{tq}$ where $sq-pt$ and $tq$ are integers. Thus, $r$ must be a rational number but this contradicts the fact that $r$ is irrational. Therefore, the sum of a rational number and an irrational number is irrational.
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