Given the relation $R=\{(a,b)\in\mathbb{N}\times\mathbb{N}\ |\ 3b\leq a<3b+3\}$
I have to prove whether or not this is a function.
To be a function, the relation must have 2 properties:
Every element $a$ in the domain $\mathbb{N}$ must have at least one partner $b$ in the codomain $\mathbb{N}$: $\forall a(a\in\mathbb{N}\rightarrow\exists b(b\in\mathbb{N}\ \land (a,b)\in R))$
Every element $a$ in the domain $\mathbb{N}$ must have at most one partner $b$ in the codomain $\mathbb{N}$: $\forall a,b_1,b_2((a\in\mathbb{N}\land b_1\in\mathbb{N}\land b_2\in\mathbb{N}\land (a,b_1)\in R\land (a,b_2)\in R)\rightarrow b_1=b_2)$
Now, the first part is easy: Fix an $a\in\mathbb{N}$ and let $b=\lfloor \frac{a}{3}\rfloor$. Now we have covered the whole domain, such that every $a\in\mathbb{N}$ has at least a partner $b\in\mathbb{N}$.
The second part is where I'm stuck. My proof starts like this:
Fix some $a,b_1,b_2\in\mathbb{N}$.
Further, let $(a,b_1)\in R$ and $(a,b_2)\in R$.
Now I somehow need to prove that $b_1=b_2$.
I also have an idea:
$3b\leq a< 3b+3\iff \frac{a}{3}-1<b\leq\frac{a}{3}$
Since the domain and the codomain are both $\mathbb{N}$, the next natural number after $\frac{a}{3}-1$ must be $\lceil\frac{a}{3}\rceil$ if $\frac{a}{3}-1\geq 0$
and $0$ if $\frac{a}{3}-1< 0$.
The previous natural number before $\frac{a}{3}$ is $\lfloor\frac{a}{3}\rfloor$... and so on. Now I don't know how to go on.
Am I on the right track with this one?
How do I prove that every $a$ has at most one partner $b$?
You may suppose for a contradiction that (wolog) $b_1 < b_2$. Then we have $$3b_1 \leq a < 3b_1+3; \quad 3b_2 \leq a < 3b_2+3$$ But then $$a < 3b_1+3 \leq 3b_2 \leq a$$