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1. The problem statement, all variables and given/known data

Consider the relation [itex] R = \{(x,y) \in \mathbb{R} \times \mathbb{R}: x-y \in \mathbb{Z}\}[/itex] on [itex]\mathbb{R} [/itex]. Prove that this relation is symmetric, reflexive, and transitive.

2. Relevant equations

Supposing a relation [itex]R[/itex] on a set [itex]A[/itex].

Reflexivity: Relation [itex] R [/itex] isreflexiveif [itex] \forall x \in A, xRx [/itex].

Symmetry: Relation [itex] R [/itex] issymmetricif [itex] \forall x,y \in A, xRy \Rightarrow yRx [/itex].

Transitivity: Relation [itex] R [/itex] istransitiveif [itex] \forall x,y,z \in A, \left((xRy) \land (yRz)\right) \Rightarrow xRz [/itex].

3. The attempt at a solution

Reflexivity:

We choose any [itex] x \in \mathbb{R} [/itex] and discover that [itex] x - x = 0 [/itex], which is in [itex] \mathbb{Z} [/itex]. Therefore, we have [itex] xRx [/itex], showing that [itex] R [/itex] is reflexive on [itex] \mathbb{R} [/itex].

Symmetry:

We can argue directly by assuming [itex] xRy [/itex]. This relation means we have [itex] (x - y) \in \mathbb{Z} [/itex]. It follows that [itex] -(x - y) [/itex], which is [itex] yRx [/itex], is also in [itex] \mathbb{Z} [/itex].

Transitivity:

If [itex] xRy [/itex] and [itex] yRz [/itex] are both integers, then adding them yields another integer.

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# Homework Help: Intro to Proofs: Properties of Relations

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