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How do I show the equivalence classes of the following proof, I am completely clueless so explain in detail please, thank you in advance

How do I show the equivalence classes of the following proof, I am completely clueless so explain in detail please, thank you in advance

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How do I show the equivalence classes of the following proof, I am completely clueless so explain in detail please, thank you in advance.

 

Result Proof The relation R defined on Z by x R y if x + 3y is even is an equivalence relation. Before proving this result, let's be certain that we understand this relation. First, notice that 5 R 7 since 5 +3.7 = 26 is even. However, 8 R9 since 8+3.9 = 35 is not even. On the other hand, 4 R 4 because 4 + 3.4 = 16 is even. First, we show that R is reflexive. Let a € Z. Then a + 3a = 4a = 2(2a) is even since 2a € Z. Therefore, a R a and R is reflexive. Next, we show that R is symmetric. Assume that a R b. Thus, a + 3b is even. Hence, a + 3b = 2k for some integer k. So, a = 2k - 3b. Therefore, b+ 3a = b + 3(2k − 3b) = b + 6k - 9b = 6k - 8b = 2(3k - 4b). Since 3k - 4b is an integer, b + 3a is even. Therefore, b R a and R is symmetric. Finally, we show that R is transitive. Assume that a Rb and b R c. Hence, a + 3b and b + 3c are even; so a + 3b = 2k and b + 3c = 2l for some integers k and l. Adding these two equations, we obtain (a + 3b) + (b + 3c) = 2k +2l. So a + 4b + 3c = 2k + 2l and a + 3c = 2k + 2l − 4b = 2(k+ l − 2b). Since k + l − 2b is an integer, a + 3c is even. Hence, a R c and so R is transitive. Therefore, R is an equivalence relation. ■
Transcribed Image Text:Result Proof The relation R defined on Z by x R y if x + 3y is even is an equivalence relation. Before proving this result, let's be certain that we understand this relation. First, notice that 5 R 7 since 5 +3.7 = 26 is even. However, 8 R9 since 8+3.9 = 35 is not even. On the other hand, 4 R 4 because 4 + 3.4 = 16 is even. First, we show that R is reflexive. Let a € Z. Then a + 3a = 4a = 2(2a) is even since 2a € Z. Therefore, a R a and R is reflexive. Next, we show that R is symmetric. Assume that a R b. Thus, a + 3b is even. Hence, a + 3b = 2k for some integer k. So, a = 2k - 3b. Therefore, b+ 3a = b + 3(2k − 3b) = b + 6k - 9b = 6k - 8b = 2(3k - 4b). Since 3k - 4b is an integer, b + 3a is even. Therefore, b R a and R is symmetric. Finally, we show that R is transitive. Assume that a Rb and b R c. Hence, a + 3b and b + 3c are even; so a + 3b = 2k and b + 3c = 2l for some integers k and l. Adding these two equations, we obtain (a + 3b) + (b + 3c) = 2k +2l. So a + 4b + 3c = 2k + 2l and a + 3c = 2k + 2l − 4b = 2(k+ l − 2b). Since k + l − 2b is an integer, a + 3c is even. Hence, a R c and so R is transitive. Therefore, R is an equivalence relation. ■
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