I was wondering how I could prove
aRb if and only if 5 | (a + 4b) , on the set of all integers
I'm used to proving for sets of numbers so I have no idea how to start out for this... Equivalence relation I know means reflexive, symmetric and transitive, but how can I prove it for that relation?
And also how can I show the equivalence classes of R (the above relation)?
$\def\r{\mathrel{\rm R}}$Remember that in proof problems, correct logic and clear setting out are both vital.
Reflexive. For any integer $a$ we have $5\mid 5a$ by definition of divisibility. This can be written as $$5\mid (a+4a)\ ,$$ and therefore $a\r a$. Thus $\r$ is reflexive.
Symmetric. Let $a,b$ be integers and suppose that $a\r b$. By definition, this means that $$5\mid(a+4b)\ .$$ But clearly $5\mid5(a+b)$, and so by well-known properties of divisibility, $$5\mid(5(a+b)-(a+4b))\ ,$$ that is, $$5\mid(b+4a)\ .$$ Thus $b\r a$. We have proved that if $a\r b$ then $b\r a$; so $\r$ is symmetric.
Transitive. The layout of the proof will be similar to the proof that $\r$ is symmetric (though of course the content will be different). Please try it for yourself.
Equivalence classes. For example, the equivalence class of $1$ is by definition the set of all $a$ such that $a\r1$; that is, the set of all $a$ such that $5\mid(a+4)$. This divisibility statement can be written $a+4=5k$, or $a=5k-4$. So the equivalence class of $1$ is $$[1]=\{5k-4:k\in{\Bbb Z}\}=\{\ldots,-9,-4,1,6,11,16,\ldots\}\ .$$ If you work out the rest for yourself I think you will soon realise that this relation is in fact just congruence modulo $5$, written in a slightly disguised way.
Good luck!
