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 1.8 - Proof Methods and Strategy - Supplementary Exercises - Page 112: 18

Answer

1) The truth value for $\vee_{i=1}^{100}\left(p_{i} \wedge p_{i+1}\right)$ is false. The reason is: since there is one even integer and one odd integer among $i$ and $i+1$, the truth value of $p_{i} \wedge p_{i+1}$ is always false, and so this is an or-statement of all false statements, which is false. 2) The truth value for $\bigwedge_{i=1}^{100}\left(p_{i} \vee p_{i+1}\right)$ is true. The reason is : since there is one even integer and one odd integer among $i$ and $i+1$, the truth value of $p_{i} \vee p_{i+1}$ is always true, and so this is an and-statement of all the true statements, which is true.

Work Step by Step

1) The truth value for $\vee_{i=1}^{100}\left(p_{i} \wedge p_{i+1}\right)$ is false. The reason is: since there is one even integer and one odd integer among $i$ and $i+1$, the truth value of $p_{i} \wedge p_{i+1}$ is always false, and so this is an or-statement of all false statements, which is false. 2) The truth value for $\bigwedge_{i=1}^{100}\left(p_{i} \vee p_{i+1}\right)$ is true. The reason is : since there is one even integer and one odd integer among $i$ and $i+1$, the truth value of $p_{i} \vee p_{i+1}$ is always true, and so this is an and-statement of all the true statements, which is true
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