Discrete Mathematics with Applications 4th Edition

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

Chapter 8 - Relations - Exercise Set 8.2 - Page 458: 18

Answer

Reflexive Symmetric Transitive

Work Step by Step

Define a relation Q on R as follows: For all real numbers x and y, x Q y ⇔ x − y are rational. Reflexive: for all x ∈ Z xAx true since x-x = 0 (note that 0 is rational = 0/1) Symmetric: for all x,y ∈ Z if xQy then yQx xQy <=> x-y = $\frac{a}{b}$ b ≠ 0 -x+y = $\frac{-a}{b}$ (muliplie by -1) y - x = $\frac{-a}{b}$ (commutive , b ≠ 0 hence rational) yQx by definition of Q Transitive: for all x,y, and z ∈ Z if (xQy and yQz) then xQz xQy <=> x-y = $\frac{a}{b}$ , b ≠ 0 yQz <=> y-z = $\frac{c}{d}$ , d ≠ 0 add (1) and (2) x-z = $\frac{ad+cb}{bd}$ , bd ≠ 0 (x-z is rational) xQz (By definition of Q) [that's what we needed to show]
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