Let a relation on $\mathbb{Z}$, check if that is a equivalence relation on $\mathbb{Z}$ and write the equivalence class of $0$

Question

I have this exercise:

$$\mathcal{R} = \left \{ (a,b) \in \mathbb{Z} \times \mathbb{Z} \quad \mbox{ such that } \quad 15 \mid 4b + 11a\right \}$$

i.e. $a \mathcal{R} b \iff 15 \mid 4b + 11a$.
i) Check if $\mathcal{R}$ is a equivalence relation on $\mathbb{Z}$;
ii) Write the equivalence class of $0$.

development:

point i)
$\mathcal{R}$ is reflexive in fact,
$\forall a \in \mathbb{Z} \times \mathbb{Z}, \,a \mathcal{R} a, $ since $15 \mid 4a+11a \iff 15 \mid 15a$

$\mathcal{R}$ is symmetric in fact,
$a \mathcal{R} b \Rightarrow 15 \mid 4b+11a$
and also
$\forall a,b \in \mathbb{Z} \times \mathbb{Z}$ we have $15 \mid 15b+15a$
therefore
$15 \mid 15b+15a-4b-11a=4a+11b$
so $b \mathcal{R} a$

$\mathcal{R}$ is transitivve, in fact,
assuming $a \mathcal{R}b \mbox{ and } b\mathcal{R}c$ we have:
$a \mathcal{R} b \Rightarrow 15 \mid 4b+11a $
$b \mathcal{R} c \Rightarrow 15 \mid 4c+11b $
therefore $a \mathcal{R} c$

point ii)
The class of zero is:
$a \mathcal{R} 0 \iff 15 \mid 11a$ i.e. the multiples of $15$.
END

this is a solved exercise but there are some point I don't understand.
- why we have that $15 \mid 15b + 15a$, and also why $15 \mid 4a+11b$ ?
- why the class of zero is defined in that way?

Please, can you help me? Thanks!

Answer 1

$15|15b+15a$ because $15b+15a=15(a+b)$ and $a$ and $b$ are both integers and thus, so is their sum. For $15|4a+11b$ since we know $15|15a+15b$ then and since we have $a\mathscr Rb$ implies $15|4b+11A$ implies $(-a)\mathscr R(-b)$ implies $15|-4b-11a$ Then $15|(15a+15b)+(-4a-11b)$ becuase $15$ divides each thus is divides the sum. which gives us $15|4a+11b$

And the equivalence class is all the numbers that would be related to $0$. Then $a\mathscr R 0$ if $15|11a+4(0)$ which gives $15|11a$