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Show that for any two given integers, $a$ and $b$, either $[a] = [b]$ or $[a]\cap [b] = \emptyset$.

Every integer can be written as such $a=mq+r$ for some integer $q$ Now we also have that $[a]=\{a+mq\}$.

So basically its just saying that when an integer $a$ and $b$ are divided by $m$ they have the same remainder $r$. Clearly if the remainder is the same for $a$ and $b$ then $[a]=[b]$.

However if the remainder is different when divided by an integer $m$ then they share no common elements for the classes so the intersection will be empty.

I'm having a little trouble cleaning up this proof I understand that if two integers have the same remainder then the equivalence class of the set of all integers with the same remainder will be identical. Also if they have a different remainder then the equivalence classes will be different also.

Can't use the fact its an equivalence relation in this proof

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    Well, if you've proven that the relation given by $[a] = [b]$ is an equivalence relation, then you know in general that equivalence classes are disjoint, by transitivity. – 2017-02-21
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    Oh I forgot to mention I can't use that technique that was a seperate exercise – 2017-02-21
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    If $[a]\cap [b]=\emptyset$ then there is nothing more to prove. So, suppose that $[a]\cap [b]\neq\emptyset$. Let $x\in[a]\cap [b]$. Then we get $a\equiv x \pmod m$ and $x\equiv b\pmod m$ of which (by transitivity) we get $a\equiv b\pmod m$ and hence $[a]=[b]$. – 2017-02-21

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So suppose $a$ and $b$ are given. Then one of two things can happen upon division by $m$:

1) $a$ and $b$ leave the same remainder. Then, if $x \in [a]$, then $x = a + mq$ for some $q$. Since $b$ and $a$ leave the same remainder, $a - b= km$ for some $k$, so $$x = a + mq = (a-b) + b+ mq = b + mk + mq = b+ (q+k)m$$

This shows that $x \in [b]$. I leave you to show that $y \in [b] \implies y \in [a]$ (It is exactly the same as the previous proof). Hence, $[a]=[b]$.

2) $a$ and $b$ don't leave the same remainder. Now, let $x \in [a]$. Then, $x = a + mq$, where $q \in \mathbb Z$. Now, since $b$ and $a$ don't leave the same remainder, we can write $a - b = km + r$, where $0 < r < m$. (It can't be equal to zero!). Now, we write: $$ x = a + mq = (a-b) + b+ mq = b + mk + mq + r = b+ (q+k)m + r $$

Now, if , by contradiction, $x \in [b]$, then $x = b + mp$ for some $p$, but then: $$ r = x-b-(q+k) m = (p-q-k)m $$

Which means that $r$ is a multiple of $m$, but this can't happen because we assumed that $0 < r < m$. This contradiction shows that $x \notin [b]$. Consequently, $[a] \cap [b] = \emptyset$.

Please get back if there are any questions!