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!

  • 2
    $\begingroup$ I'd suggest you write out what $ a+c\sim b+c$ means in terms of the definition of $\sim$ $\endgroup$ – J. W. Tanner May 3 at 13:50
  • $\begingroup$ @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} $? $\endgroup$ – Olly Reynolds May 3 at 13:56
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    $\begingroup$ $\iff a-b\in\mathbb Z \iff a\sim b\;$; that's it! $\endgroup$ – J. W. Tanner May 3 at 14:00
  • $\begingroup$ ohhhhh, thank you so much, discrete math always gets me and I never see these simple tricks :( $\endgroup$ – Olly Reynolds May 3 at 14:01

$a\sim b\iff a-b\in\mathbb Z \iff (a+c)-(b+c) \in \mathbb Z \iff a+c\sim b+c$

  • $\begingroup$ because $(a+c)-(b+c)=a-b,$ as J.G. noted $\endgroup$ – 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 $(a+c)-(b+c)=a-b$.


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