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Worked Examples · Example 13

Q.Show that the relation RR in the set ZZ of integers given by R={(a,b):3 divides a−b}R = \{(a, b) : 3 \text{ divides } a - b\} is an equivalence relation.

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"33 divides a−ba-b" is congruence modulo 33; proving all three properties on Z\mathbb{Z} shows it is an equivalence relation.

R={(a,b)∈Z×Z:3∣(a−b)}⟺a≡b(mod3)R=\{(a,b)\in\mathbb{Z}\times\mathbb{Z}: 3\mid(a-b)\}\quad\Longleftrightarrow\quad a\equiv b\pmod 3

"3∣n3\mid n" means n=3kn=3k for some integer kk. RR is an equivalence relation if reflexive, symmetric, and transitive.

  1. Check Reflexivity. For any a∈Za\in\mathbb{Z}:

a−a=0=3×0a-a=0=3\times0

Since 00 is (trivially) a multiple of 33, 3∣(a−a)3\mid(a-a), so (a,a)∈R(a,a)\in R for every a∈Za\in\mathbb{Z}. Hence RR is reflexive.

  1. Check Symmetry. Suppose (a,b)∈R(a,b)\in R, i.e. 3∣(a−b)3\mid(a-b), so a−b=3ka-b=3k for some integer kk. Then

b−a=−(a−b)=−3k=3(−k)b-a=-(a-b)=-3k=3(-k)

Since −k-k is also an integer, 3∣(b−a)3\mid(b-a), so (b,a)∈R(b,a)\in R. Hence RR is symmetric.

  1. Check Transitivity. Suppose (a,b)∈R(a,b)\in R and (b,c)∈R(b,c)\in R, i.e. a−b=3ka-b=3k and b−c=3mb-c=3m for integers k,mk,m. Adding:

a−c=(a−b)+(b−c)=3k+3m=3(k+m)a-c=(a-b)+(b-c)=3k+3m=3(k+m)

Since k+mk+m is an integer, 3∣(a−c)3\mid(a-c), so (a,c)∈R(a,c)\in R. Hence RR is transitive. …

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