RELATIONS Let R be a relation on Z defined according to the following rule: for
ID: 3186122 • Letter: R
Question
RELATIONS Let R be a relation on Z defined according to the following rule: for all m, nE Z, m R n if and only if 6 divides m+5in VI a. [6 points] Prove that R is an equivalence relation on Z. For each variable that appears in your proof, you must state what that variable represents. For each step in your proof, it must be made clear to the reader whether it is an assumption, something you are about to prove, or something that follows from a previous step, definition, or the relation's rule, etc.Explanation / Answer
1. Let a be an arbitrary integer in Z. Then a+5a = 6a which implies that a R a as 6a/6 = 6, an integer. Thus the relation R is reflexive.
2. Let a, b be 2 arbitrary integers in Z and let a R b i.e. let a+5b be divisible by 6 so that a+5b = 6k, where k is an integer. Then a = 6k-5b so that b+5a = b +5(6k-5b) = 30k -24b = 6(5k-4b). Now, since k and b are integers, hence 5k-4b is also an integer. This implies that b+5a is divisible by 6. Hence b R a so that the relation R is symmetric.
3. Let a, b, c be 3 arbitrary integers in Z and let a R b and b R c i.e. let a+5b be divisible by 6 and b+5c be divisible by 6. Then a+5b = 6m and b+5c = 6n, where m,n are integers. Then a = 6m-5b and 5c = 6n-b so that a+5c = 6m-5b+6n-b = 6(m+n) -6b = 6(m+n-b). Now,m since m,n, b are integers, hence m+n-b is also an integer. This implies that a+5c is divisible by 6 so that the relation R is transitive.
Finally,m since R is reflexive, symmetric and transitive, hence R is an equivalence relation.
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