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Can someone can tell me if my proof of the next propostion is correct?

Define the following relation: $$a\sim b \iff a-b=km, m\in \mathbb{Z}$$ Show $\sim$ is an equivalence relation

And so here's my attempt on showing it

Proof.

Let $a, b, c \in \sim$

i)Reflexivity of $\sim$: $$a\sim a \iff a-a = km \iff 0=km \iff k = 0 \in \mathbb{Z} \vee m=0 \in \mathbb{Z}$$ ii)Simmetry of $\sim$: $$a\sim b \iff a-b = km \iff b-a = (-k)m\iff a\sim b$$ iii)Transitivity of $\sim$: $$a\sim b \wedge b\sim c \iff a-b=km \wedge b-c=qm \iff a-b+b-c = km - qm $$ $$\iff a-c = km-qm \iff a-c = (k-q)m, (k-q)\in \mathbb{Z}$$

Alan Moreno de la Rosa
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    yah- this is correct. – voldemort Jan 29 '14 at 06:18
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    Please do not make your titles large. It takes up too much room on the main page. – Cameron Williams Jan 29 '14 at 06:18
  • The proof looks good to me! As for titles, note that (for my browser, at least) the content of the first few sentences of your post is already visible on the main page, and so I feel the title should be a very short summary, rather than a full restatement, of the problem. – Zach L. Jan 29 '14 at 06:21
  • I would just be careful using $\Longleftrightarrow$ when only $\Longrightarrow$ is true. But your proof is essentially correct. – breeden Jan 29 '14 at 06:27
  • @ZachL. Sorry about of the title stuff :/ and in which one only is true $\rightarrow$ on proving Transitivity? – Alan Moreno de la Rosa Jan 29 '14 at 06:31

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The essence of the matter is clearer as follows. Note $\ a\sim b\iff (a-b)/m\in\Bbb Z.$

i) $\ \ (a-a)/m \in \Bbb Z\ $ by $\ 0\in\Bbb Z$

ii) $\ \ (a-b)/m \in \Bbb Z \, \Rightarrow\, (b-a)/m \in \Bbb Z\ $ by $\ \Bbb Z\ $ closed under negation.

iii) $ \ (a-b)/m,\, (b-c)/m \in \Bbb Z\,\Rightarrow\, (a-c)/m \in \Bbb Z\ $ by $\ \Bbb Z\ $ closed under addition.

So congruence is an equivalence relation precisely because $\,\Bbb Z\,$ is an additive subgroup of $\,\Bbb Q.$

Bill Dubuque
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