Let \(\sim \) be the relation on the integers whereby \(a \sim b\) if and only if \(a - b = 3k\) for some integer \(k\) (i.e., \(a-b\) is divisible by 3).
Proof. We will prove that \(\sim \) is an equivalence relation by showing that it is reflexive, symmetric, and transitive.
To show that \(\sim \) is reflexive, we need to show that \(a \sim a\) for all integers \(a\). So let \(a\) be an integer. Then \(a - a = 0 = 3 \times 0\) so \(a \sim a\). Thus \(\sim \) is reflexive.
To show that \(\sim \) is symmetric, suppose \(a\) and \(b\) are integers and \(a \sim b\), i.e., \(a - b = 3k\) for some integer \(k\). Then \(b - a = -3k = 3(-k)\). Since \(k\) is an integer, so is \(-k\), and thus \(b - a\) is divisible by 3 and so \(b \sim a\). We have now shown that \(\sim \) is symmetric.
To show that \(\sim \) is transitive, let \(a\), \(b\), and \(c\) be integers such that \(a \sim b\) and \(b \sim c\). Thus \(a - b = 3k\) for some integer \(k\) and \(b - c = 3n\) for some integer \(n\). Now
Since integers are closed under addition, we have shown that \(a - c\) is divisible by 3, and so \(a \sim c\). Therefore \(\sim \) is transitive.
We have now shown that \(\sim \) is reflexive, symmetric, and transitive, and therefore that \(\sim \) as defined above is an equivalence relation. □