Show that conguence $\bmod m\,$ is an equivalence relation 
Let $m \in \mathbb{N}$. Show that $\,a \text{ R } b \Leftrightarrow a
\equiv b \text{ }(\text{mod } m)\,$ is an equivalence realtion.

We need to show 3 things:
(i) reflexive: $\forall a \in \mathbb{Z}: a \equiv a \text{ mod } m$
(ii) symmetric: $\forall a,b \in \mathbb{Z}: a \equiv b \text{ mod } m \text{ } \Rightarrow b \equiv a \text{ mod } m$
(iii) transitive: $\forall a,b,c \in \mathbb{Z}: a \equiv b \text{ mod } m \text{ } \wedge b \equiv c \text{ mod } m \Rightarrow a \equiv c \text{ mod } m$
So I tried it like this:
(i) $m \text{ | } 0 \Rightarrow m \text{ | } a-a \Rightarrow a \equiv a \text{ mod } m$
(ii) $a \equiv b \text{ mod } m \Rightarrow a \text{ mod } m = b \text{ mod } m \Rightarrow b\text{ mod } m = a \text{ mod } m \Rightarrow b \equiv a \text{ mod } m$
(iii) $a \equiv b \text{ mod } m \text{ } \wedge \text{ }b \equiv c \text{ mod } m \Rightarrow m \text{ | } (a-b) \text{ } \wedge \text{ } m \text{ | } (b-c) \Rightarrow m \text{ | } (a-b) + (b-c) \Rightarrow m \text{ | } a-c \Rightarrow a \equiv c \text{ mod } m$
Because every condition is fulfilled, we are done (?)
Please help me and tell me if it would be correct like this? Maybe there is better way to prove it?
 A: Your proof looks OK, but I would correct it by:


*

*Explaining which definition of $a\equiv b\pmod m$ you are using. I believe you are using the definition $a\equiv b\pmod m \iff m\mid (a-b)$.

*Sticking to that definition through all 3 steps of the proof. Currently, in your proof of $ii$, you are using the expression $a\bmod m$ which is not defined (I presume you mean "the remainder", and that's perfectly valid, but for clarity, I would take the same approach as in $i$ and $iii$: you know that $a\equiv b\pmod m$, meaning that $m\mid(a-b)$, and since $a-b=-(b-a) $, you know that $m\mid(b-a)$, so $b\equiv a\pmod m$.



Alternatively, you could use the less standard, but equivalent, definition $$a\equiv b\pmod m\iff a\bmod m = b\bmod m$$
where $a\bmod m$ is the remainder of $a$ when divided by $m$. In that case, you could change proofs of $i$ and $iii$ and leave the proof of $ii$ as it is.
A: We have $\ a\,R\,b {\overset{\rm\ def}{\color{}\iff}} a-b \in  S,\,$ for $\,S = m\Bbb Z = $ all multiples of $\,m.\,$ Suppose more generally that $\,S\subset \Bbb Z\,$ is arbitrary set of integers containing $\,0.\,$ Checking the equivalence relation properties

*

*reflexive: $\quad\ \ \, a\, R\, a \iff 0\in S,\ $ which is true by hypothesis


*symmetric: $\,\ (a\, R\, b\,\Rightarrow\, b\, R\, a)\iff (a\!-\!b\in S\,\Rightarrow\, b\!-a\!\in  S)$


*transitive: $\ \ \ (a\, R\, b,\ b\,R\,c\,\Rightarrow\, a\, R\, c)\iff (a\!-\!b,\,b\!-c\in S\,\Rightarrow\,a\!-\!c\in S)$
If $\,S\,$ is closed under negation: $\,a-b\in S\,\Rightarrow\, b-a\in S,\,$ so $\,R\,$ is symmetric. If $\,S\,$ is closed under addition then $\,R\,$ is transitive: $\,a-b,\,b-c\in S$ $\,\Rightarrow$ $\, a-b\, +\, b-c = a-c\in S$.
The converse holds too, i.e. if $\,S\,$ satisfies these implications then it is closed under negation and addition, since $\,b=0\,$ in "symmetric" yields $\,a\in S\,\Rightarrow\, -a\in S,\,$ i.e. $\,S\,$ is closed under negation. Further $\,b=0,\, c=-d\,$ in the "transitive" implication shows that $\,S\,$ is closed under addition.
Thus $\,R\,$ is an equivalence relation precisely because $\,S = m\Bbb Z\,$ is closed under negation and addition (or, equivalently, $\,S\,$ is closed under subtraction). If you know group theory then you will recognize this as the subgroup test, i.e. $\,R\,$ is an equivalence relation $\,\iff S\,$ is a subgroup of the additive group of integers. The innate structure is clarified algebraically when one studies quotient groups and quotients rings (here $\,\Bbb Z/m\Bbb Z \cong $ integers mod $m).$
Remark $\ $ In your proof you are also using "mod" as a binary operator (in addition to the ternary equivalence relation of modular congruence). Be careful not to confuse the two. See this answer and this one for more on this distinction.
If you proceed as you do in $(ii)$ using $\, x\equiv y\pmod m\!\iff\! (x\bmod m) = (y\bmod m)$ then the proof is a special case of the fact that relations of the form $\ x\sim y {\overset{\rm\ def}{\iff}} f(x) = f(y)\ $ are always equivalence relations - with obvious proof. In your case $\,f(x) := x\bmod m.\,$ Again this is clarified when one studies quotient structures.
