Is the relation on integers, defined by $(a,b)\in R\iff a=5q+b$, a function?

Question

Let $A=B=\mathbb N$. Relation $R$ is: $(a,b)\in R$ iff for some $q \in \mathbb Z$ we have $a=5q+b$

Given a relation, show that it's a function.

To Show:

1. $\forall a \in A \ \exists b \in B$ such that $(a,b)\in R$

2. $\forall a \in A $ there is a unique $b\in B$ such that $(a,b) \in R$.

Proof Outline: Let $a \in A$. Assume $(a,b)\in R$ and $(a,c) \in R$ where $b,c \in B$.

then some steps leading to.... $\therefore$ b=c. QED

What math should I do to complete the proof?

Answer 1

Suppose $a=1$ then each of $(1,6)$ and $(1,11)$ are in $R$ so it isn't a function.

Note we need to use negative $q$ values for these two pairs in the OP definition, which is OK according to phrasing of the OP's question. $1=5(-1)+6$ and $1=5(-2)+11$ give the two pairs, using respectively $q=-1,-2.$

Even if $q>0$ is required there would still be the pairs $(11,1)$ and $(11,6)$ with $q=2,1$ which would just as well show $R$ not a function. [It seems maybe that's what Arthur's now deleted answer was going for, just with the coordinates switched.]