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 4.2 - Page 169: 18

Answer

true

Work Step by Step

The original statement "If r and s are any two rational numbers, then (r+s)/2 is rational" is indeed true. Proof: 1. Let r and s be any two rational numbers. 2. By definition, a rational number can be expressed as the ratio of two integers, where the denominator is not zero. 3. Let r = a/b and s = c/d, where a, b, c, and d are integers, and b ≠ 0 and d ≠ 0. 4. The sum of two rational numbers is also a rational number, so r + s = (a/b) + (c/d) = (ad + bc)/(bd), which is a ratio of two integers with $bd\ne0$. 5. Dividing the sum by 2, we have (r + s)/2 = [(ad + bc)/(bd)]/2 = (ad + bc)/(2bd). 6. Since ad + bc and 2bd are integers (as the sum and product of integers), (r + s)/2 is expressed as the ratio of two integers with a non-zero denominator. 7. Therefore, (r + s)/2 is a rational number. So, the original statement holds true.
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