# sols4 - MATH 135 Solutions to Assignment 4 W5 This...

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MATH 135 Solutions to Assignment 4 W5 This assignment is due at 8:30 a.m. on Wednesday, February 3, in the drop box outside the Tutorial Centre, MC 4067. Remember to show all of your computational work. 1. [ 5 marks ] Find a test for divisibility by 8. Answer : We notice that 10 2(mod 8), 10 2 4(mod 8), and 10 3 0(mod 8). This means that 10 n 0(mod 8) for any n 3, so given an integer a r a r - 1 . . . a 1 a 0 , we have 10 r a r + . . . + 10 3 a 3 + 10 2 a 2 + 10 a 1 + a 0 0 + . . . + 0 + 10 2 a 2 + 10 a 1 + a 0 (mod 8) 10 2 a 2 + 10 a 1 + a 0 (mod 8) Therefore, our integer will be divisible by 8 if and only if the number one gets by considering only its last three digits (10 2 a 2 + 10 a 1 + a 0 ) is divisible by 8. 2. [ 9 marks total: 3 marks for each of (a), (b), and (c) ] For each of the following relations, determine whether it is reﬂexive, symmetric, and transitive and explain why in each case. (a) Let S be the integers Z and let aRb if and only if | a - b | ≥ 5. Answer : R is symmetric because | a - b | = | b - a | for all integers a and b . R is not reﬂexive since | 0 - 0 | 6≥ 5, and it is not transitive since 0 R 5 and 5 R ( - 1), but 0 is not related to - 1 (or any other counterexamples). (b) Let S be the real numbers R and let xRy if and only if | x | + | y | = 1. Answer : R is symmetric because | x | + | y | = | y | + | x | for any reals x and y . However, it is not reﬂexive because 1 is not related to 1 (or any other counterexample), and it is not transitive because 0 R 1 and 1 R 0, but 0 is not related to 0 (or any other counterexample). (c) Let

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## This note was uploaded on 02/21/2010 for the course MATH 135 taught by Professor Andrewchilds during the Spring '08 term at Waterloo.

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sols4 - MATH 135 Solutions to Assignment 4 W5 This...

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