How do I prove transitivity of $R = \{(x, y) : x - y \in \mathbb{Z}\}$?

Question

For the relation $R = \{(x, y) : (x - y) \in \mathbb{Z}\}$ over $\mathbb{Q}$, how do I prove transitivity in this case?

For any number $n \in \mathbb{Q}$, it's going to contain a fractional part (any rational is just an integer plus a fractional part). If $(a, b) \in R$, then both $a$ and $b$ share a fractional part $\delta_1$, and if $(b, c) \in R$, then $b$ and $c$ share a fractional part $\delta_2$. But since both statements include $b$, then $\delta_1 = \delta_2$.

More declaratively, for $\{a, b, c\} \in \mathbb{Q}$ there is an $\{m, n, p\} \in \mathbb{Z}$ and fractional parts $\{\delta_1, \delta_2, \delta_3\} \in \mathbb{Q}/\mathbb{Z}$ such that $m+\delta_1 = a, n+\delta_2 = b, p+\delta_3 = c$:

$$(a, b) \in R \implies (m+\delta_1, n+\delta_2) \in R \implies \delta_1 = \delta_2 \\ (b, c) \in R \implies (n+\delta_2, p+\delta_3) \in R \implies \delta_2 = \delta_3 \\ \therefore \delta_1 = \delta_3 \implies (m + \delta_1, p+\delta_3) \in R \implies (a, c) \in R$$

Thus the relation is transitive.

Is there a better way to get this idea across of using fractional parts of rational numbers?

Answer 1

Suppose $(a,b), (b,c)\in R $. Then $a-c=(a-b)+(b-c)\in\mathbb Z $, because $\mathbb Z $ is closed under addition. So $(a,c)\in R $.

Answer 2

In fact, $R$ is actually an equivalence relation, because $\mathbb{Z}$ is a subgroup of $\mathbb{Q}$.

More generally, given any group $G$ and any subset $S$ of $G$, one can define a relation $R$ on $G$ where $x R y \iff x*y^{-1} \in S$. Then $R$ will be:

• reflexive if and only if $e_{G} \in {S}$
• symmetric if and only if $S$ is closed under inversion
• transitive if and only if $S$ is a subsemigroup of $G$
• a preorder if and only if $S$ is a submonoid of $G$
• a partial order if and only if $S$ is a submonoid of $G$ that does not contain the inverse of any of its non-identity elements
• an equivalence relation if and only if $S$ is a subgroup of $G$
• an equivalence relation respecting multiplication if and only if $S$ is a normal subgroup of $G$