Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 10: 27


When the signs are the same, $\frac{1}{b}>\frac{1}{a}$ For the other case, $\frac{1}{a}>\frac{1}{b}$

Work Step by Step

$b^{-1}=\frac{1}{b}$, $a^{-1}=\frac{1}{a}$ In this case, a and b serve as the denominators. Consider the case when $a>b$ and both are positive. As the denominator increases, the overall value of the fraction decreases. Therefore, $\frac{1}{a}$ is less than $\frac{1}{b}$. When $a>b$ and both are negative, dividing by a more negative number, b, would result in a number that is closer to 0 as supposed to a less negative number. Therefore, $\frac{1}{a}$ is less than $\frac{1}{b}$. If a > 0 and b < 0, $\frac{1}{a}$ is positive and $\frac{1}{b}$ is negative. Therefore, $\frac{1}{a}>\frac{1}{b}$
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