A relation R is defined on Z by x R y if 11x - 5y is even.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Show the equivalence classes of the following, if able please do it step by step in detail, show why you take the steps and the motivation behind them, I am absolutely clueless on how to approach equivalence classes, I understand equivalence relations.

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A relation R is defined on Z by x R y if 11x - 5y is even. Then R is an equivalence
relation.
Proof First, we show that R is reflexive. Let a € Z. Then 11a - 5a = 6a = 2(3a). Since 3a is
an integer, 11a - 5a is even. Thus, a R a and R is reflexive.
Next we show that R is symmetric. Assume that a R b, where a, b = Z. Thus, 11a – 5b
is even. Therefore, 11a - 5b = 2k, where k = Z. Observe that
11b - 5a = (11a − 5b) + (−16a+ 16b)
= 2k - 16a + 16b = 2(k − 8a+ 8b).
-
Since k 8a+ 8b is an integer, 11b - 5a is even. Hence, b R a and R is symmetric.
Finally, we show that R is transitive. Assume that a R b and b R c. Hence, 11a - 5b
and 11b - 5c are even. Therefore, 11a - 5b = 2k and 11b - 5c = 2l, where k, l € Z.
Adding these equations, we obtain (11a - 5b) + (11b — 5c) = 2k + 2l. Solving for
11a - 5c, we have
11a-5 5c = 2k + 2l
-
6b = 2(k+l - 3b).
Since k + l - 3b is an integer, 11a - 5c is even. Hence, a R c and R is transitive. There-
fore, R is an equivalence relation.
Transcribed Image Text:Result A relation R is defined on Z by x R y if 11x - 5y is even. Then R is an equivalence relation. Proof First, we show that R is reflexive. Let a € Z. Then 11a - 5a = 6a = 2(3a). Since 3a is an integer, 11a - 5a is even. Thus, a R a and R is reflexive. Next we show that R is symmetric. Assume that a R b, where a, b = Z. Thus, 11a – 5b is even. Therefore, 11a - 5b = 2k, where k = Z. Observe that 11b - 5a = (11a − 5b) + (−16a+ 16b) = 2k - 16a + 16b = 2(k − 8a+ 8b). - Since k 8a+ 8b is an integer, 11b - 5a is even. Hence, b R a and R is symmetric. Finally, we show that R is transitive. Assume that a R b and b R c. Hence, 11a - 5b and 11b - 5c are even. Therefore, 11a - 5b = 2k and 11b - 5c = 2l, where k, l € Z. Adding these equations, we obtain (11a - 5b) + (11b — 5c) = 2k + 2l. Solving for 11a - 5c, we have 11a-5 5c = 2k + 2l - 6b = 2(k+l - 3b). Since k + l - 3b is an integer, 11a - 5c is even. Hence, a R c and R is transitive. There- fore, R is an equivalence relation.
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