Prove: For every natural number $m$,
$(Z_m , +_m)$ is an algebraic system that is associative and commutative with identity element $[0]$. Every element has an inverse.
$(Z_m , ._m)$ is an algebraic system is associative and commutative. If $m>1$, the system has identity element $[1]$.
I'm new to this topic so my understanding is pretty low plus I don't really understand congruent classes all that well.
We define a relation $\sim$ on $\mathbb{Z}$ as follows: $a \sim b$ if an only if $m|(a-b).$ I can be checked that $\sim$ is actually an equivalence relation on $\mathbb{Z}.$ Let's denote the equivalence class of $a$ by $[a].$ Let $\mathbb{Z}_m$ is the collection of all equivalence classes. For $[a], [b] \in \mathbb{Z},$ define $[a] +_m [b] := [a+b]$ where $a+b$ is the usual addition in $\mathbb{Z}.$ First we need to show that it is well defined, i.e. if $[a] = [a'], [b] = [b']$ then $[a]+_m[b] = [a']+_m[b'].$ In other words, we need to show that $[a+b] = [a' + b'].$ But this happen precisely when $m|(a+b -a' -b').$ Now $[a] = [a'] \Rightarrow m|(a-a'), [b] = [b'] \Rightarrow m|(b - b') \Rightarrow m|(a+b -a'-b').$ This show that the operation $+_m$ defined on $\mathbb{Z}_m$ is well defined.
To prove that it is associative, we need to show that $([a] +_m [b]) +_m [c] = [a] +_m ([b] +_m [c]).$ Note that $([a] +_m [b]) +_m [c] = [a+b] +_m [c] = [(a+b) + c]$ and $[a] +_m [(b] +_m [c]) = [a] +_m [b+c] = [a+(b+c)].$ Since $a, b, c \in \mathbb{Z},$ we have $(a+b)+c = a+ (b+c).$ Hence we have the associativity. Similarly, we will have the commutativity.
$[0]$ is an identity element, because $[a]+_m[0] = [a +0] = [a] = [0+a] = [0] +_m [a].$
In a similar way, you can prove for $(\mathbb{Z}_m, \cdot_m).$ Here the operation is defined by $[a] \cdot_m[b] = [ab],$ where $ab$ stands for the standard multiplication in $\mathbb{Z}.$ I hope this will help.