# Question about a Symmetric relation

I have the following relation R ⊆ N × N and is defined as R = {(x, y); 3 | (x − y)}

A symmetric relation is defined as follows ∀x, y ∈ A : (xRy ⇒ yRx)

What i did was, i defined R = {(x, y); 3 | (x − y)} as x - y = 3k and y - x = -3k to get 3 | (y - x) in order to prove the implication xRy ⇒ yRx

I know this relation is symmetric i just dont understand the reason behind it, so any help is greatly appreciated.

Your proof is right. That is the the reason behind it. If $$x,y \in \mathbb{N}$$ and $$xRy$$, then there exists $$k \in \mathbb{Z}$$ such that $$x-y = 3k$$ so $$y-x = -3k$$. Hence $$yRx$$.
• @newplayer $y-x=-(x-y)=-(3k)=-3k$ – Grešnik Jun 18 at 9:29
• Multiply the equation $3k = x-y$ by $-1$. Then you get $y-x$ on the right hand side and $3 \cdot (-k)$ on the left hand side. The thing is that you need to change $k$. For example $3$ is a divisor of $9-6$ since $3 \cdot 1 = 9-6 = 3$. But $3$ is also a divisor of $6-9$! Here you can see that since $3 \cdot (-1) = 6-9 = -3$. – ThorWittich Jun 18 at 9:31
What you did was basically correct. Let $$(x,y) \in R$$, i.e. we have that $$3$$ is a divisor of $$x - y$$, such that we can write $$3k = x-y$$ for a suitable $$k$$. Since $$x - y = -(y - x)$$ and $$3 \mid x-y$$ we also know that $$3$$ is a divisor of $$y-x$$, since $$3 \cdot (-k) = y - x$$.