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If anyone has a better method please feel free to comment it below.

My proof:

This is a direct proof. Assume $x \equiv_k y$ is true. From $x \equiv_k y$ we know that:

x = y + kq, where q is q ∈ Z $\space\space\space\space\space\space$ (1)

Now if we take (1) and solve for y we get:

y = x - kq $\space\space\space\space\space\space$ (2)

Equation (2) tells us that $y \equiv_k x$ is possbile provided that the value for $kq$ used in (1) is negated. $\blacksquare$

Also is my proof right?

jame_smith
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    $k,|,(x-y)\iff k,|,(y-x)$ – lulu Jan 04 '20 at 23:14
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    Presumably, your proof would need to use $$y = x-kq = x+k(-q)$$ and note that $-q \in \mathbb{Z}$ from which the conclusion of $y \equiv_k x$ follows. This is more clear and formal than "is possible provided that the value for $kq$ used in (1) is negated" – Brian Moehring Jan 04 '20 at 23:18
  • Use $\in$ for $\in$. – Shaun Jan 04 '20 at 23:47

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