Relations & Functions: Equivalence Relations & Equivalence Classes

For any $(a, b) \in \mathbb{N} \times \mathbb{N}$, since addition in $\mathbb{N}$ is commutative: $$a + b = b + a \implies (a, b) R (a, b)$$ Hence $R$ is reflexive.

Let $(a, b) R (c, d) \implies a + d = b + c$. This implies $b + c = a + d \implies c + b = d + a$. Hence $(c, d) R (a, b)$, proving $R$ is symmetric.

Let $(a, b) R (c, d)$ and $(c, d) R (e, f)$. Then: $$a + d = b + c \quad \text{and} \quad c + f = d + e$$ Adding the two equations: $$(a + d) + (c + f) = (b + c) + (d + e)$$ $$a + f + (c + d) = b + e + (c + d) \implies a + f = b + e$$ Hence $(a, b) R (e, f)$, proving $R$ is transitive. Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation.

$$[(2, 5)] = \{ (x, y) \in \mathbb{N} \times \mathbb{N} : (x, y) R (2, 5) \}$$ $$(x, y) R (2, 5) \iff x + 5 = y + 2 \iff y = x + 3$$ Since $x \in \mathbb{N} = \{1, 2, 3, \dots\}$: $$[(2, 5)] = \{ (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), \dots \}$$