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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.

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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 $yRx$.

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  • $\begingroup$ What confuses me the most here is this part y - x = -3k why negative 3? What would happen if it was y - x = 3k instead? $\endgroup$ – newplayer Jun 18 at 9:26
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    $\begingroup$ @newplayer $y-x=-(x-y)=-(3k)=-3k$ $\endgroup$ – Grešnik Jun 18 at 9:29
  • $\begingroup$ 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 $3 \cdot (-1) = 6-9 = -3$. $\endgroup$ – ThorWittich Jun 18 at 9:31
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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 $3 \cdot (-k) = y - x$.

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