# Intro to Proofs: Properties of Relations

1. May 14, 2012

Hello, I would like to check my arguments for this problem.

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

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

2. Relevant equations

Supposing a relation $R$ on a set $A$.

Reflexivity: Relation $R$ is reflexive if $\forall x \in A, xRx$.

Symmetry: Relation $R$ is symmetric if $\forall x,y \in A, xRy \Rightarrow yRx$.

Transitivity: Relation $R$ is transitive if $\forall x,y,z \in A, \left((xRy) \land (yRz)\right) \Rightarrow xRz$.

3. The attempt at a solution

Reflexivity:

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

Symmetry:

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

Transitivity:

If $xRy$ and $yRz$ are both integers, then adding them yields another integer.

2. May 14, 2012

### vela

Staff Emeritus
It depends on how nit-picky you want to get, but it's probably better to say $y-x \in \mathbb{Z}$, rather than $-(x-y) \in \mathbb{Z}$, implies yRx.

Right idea, but you should write out explicitly how xRy and yRz imply x-z is in Z.

Last edited: May 14, 2012
3. May 14, 2012

### LCKurtz

You have the idea but you need to rewrite the transitive one. xRy is not an integer and neither is yRz, and you can't add them. Write more carefully what you mean.

4. May 14, 2012