# Equivalence relation $a\sim b$ $\iff a-b\in\mathbb{Z}$

I have this equivalence relation defined on $$\mathbb{Q}$$ and $$a\sim b$$
$$\iff a-b\in\mathbb{Z}$$

I know this is an equivalence relation and have proven so already. But how can I prove that for rational numbers $$a,b,c$$ we have $$a\sim b$$ $$\iff a+c\sim b+c$$?

I was wondering how I could go about proving this? I was thinking to combine the relations since they are obviously related to each other so $$a+c\sim b+c \iff a-b\in\mathbb{Z}$$, then would I try and prove the 3 criteria? And if so how? Thank you!

• I'd suggest you write out what $a+c\sim b+c$ means in terms of the definition of $\sim$ – J. W. Tanner May 3 at 13:50
• @J.W.Tanner Can I just add c to both sides so $a+c\sim b+c \iff (a+c)-(b+c) \in \mathbb{Z}$? – Olly Reynolds May 3 at 13:56
• $\iff a-b\in\mathbb Z \iff a\sim b\;$; that's it! – J. W. Tanner May 3 at 14:00
• ohhhhh, thank you so much, discrete math always gets me and I never see these simple tricks :( – Olly Reynolds May 3 at 14:01

$$a\sim b\iff a-b\in\mathbb Z \iff (a+c)-(b+c) \in \mathbb Z \iff a+c\sim b+c$$
• because $(a+c)-(b+c)=a-b,$ as J.G. noted – J. W. Tanner May 3 at 14:08
For any $$c\in\Bbb Q$$, we can reason that $$a\sim b\iff\exists k\in\Bbb Z(a-b=k)\iff\exists k\in\Bbb Z((a+c)-(b+c)=k)\iff a+c\sim b+c$$ because $$(a+c)-(b+c)=a-b$$.