Can someone help me with this question please

Question

A relation $M$ is defined on the set $\mathbb{Z}$ by “$(a,b)\in M$,if $a – b$ is divisible by $5$” for $a, b\in\mathbb{Z}$.

i. Examine if $M$ is an equivalence relation on $\mathbb{Z}$.

Not sure if this is the correct way to answer the question, but i gave it a go.

Reflexive: $a - a$ divided by $5$ is $0$, does that mean it is divisible by $5$ since I got $0$?

Symmetric: $a - b$ divided by $5$ and $b - a$ divided by $5$? $0 - 1$ divided by $5 = -0.2$ $1 - 0$ divided by $5 = 1$? I'm guessing it's right since it includes a negative integer?

Transitivity: $a - b$ divided by $5$, and $b - c$ divided by $5$ then $a - c$ divided by $5$?

How do I work out the answer for this type of question? Im not sure if I am doing it right like the way I've answered it.

Answer 1

You need to use the definition: $x$ is divisible by $5$ if there exists a $n\in\mathbb{Z}$ such that $x=5n$.

Reflexivity: $a-a=0$, which is divisible by $5$, by taking $n=0$.

Symmetry: We can find $n\in\mathbb{Z}$ such that $a-b=5n$, so take $-n$ to see that $b-a=-5n$, so $b-a$ is divisible by $5$.

Transitivity: We can find $n,m\in\mathbb{Z}$ such that $a-b=5n$ and $b-c=5m$, so adding the equalities we obtain $a-c=(a-b)+(b-c)=5n+5m=5(n+m)=5k$, so we found $k=n+m\in\mathbb{Z}$ such that $a-c=5k$, so $a-c$ is divisible by $5$.

Therefore $M$ is an equivalence relation on $\mathbb{Z}$.

Remark: The usual approach is to try to prove it, and if it doesn't seem to work, use the part that impedes your proof to construct a counterexample (sometimes a counterexample is trivial and spotted directly). Unfortunately this exercise gives an equivalence relation, so I can't really give you an example of this.

Answer 2

Note that your relation is defined such that two integers $m$ and $n$ are related if $m-n$ is a multiple of 5.

Your relation is reflexive because for every integer $n$ we have ,$ n-n =0$ which is a multiple of 5.

It is symmetric because if $m-n$ is a multiple of 5 then $n-m$ is also a multiple of 5.

It is transitive because if $m-n$ and $n-k$ are multiples of 5, then their sum which is $ m-k$ is also a multiple of 5.

Therefore the relation is an equivalence relation.