Functions as relations

Question

I´ve got just a short question. Is this relation a function?

$R = \{(a,b) \in N \times N | 3b \leq a < 3b+3\}$

In my opinion it is, because every a in N has got exactly one b in N. But I´m really not sure. Can an inequality be a function?

Thank you for your help =)

Answer 1

Yes, this relation is indeed a function. That inequality can be written in the following way: $$3b \leq a <3 (b+1) $$

And it is easy to see that each $a \in \mathbb{N}$ is always between a certain pair of consecutive multiples of $3$. So, for each $a $ there is exactly one $b $ that "matches" with it.

You could write it as: $$f: \mathbb{N} \rightarrow \mathbb{N}$$ $$b=f(a)=\lfloor \frac{a}{3} \rfloor $$

(Where $\lfloor x \rfloor $ is the floor function, that gives the biggest integer less or equal to $x $.)

By the way, you cannot write it in the form $a=g(b) $ because there is more than one $a$ for each $b $; for example, $f(1)=0$ and $f(2)=0$.

Note: I assumed that $0 \in \mathbb{N}$. If you don't use that convention, you have to exclude ${0,1,2} $ from the domain of the function.

Answer 2

A function can be seen as a relation in the sense that if $aRb$, then $f(a) = b$. As you know, a function has a special "property": one object does not correspond to different images.

Thus all you have to do is, for your relation $R$, check if for a given value of $a$ you can have different values of $b$.

Answer 3

No it is not a function. A function is of the form f(x), where f is a function and x is a parameter. What you presented is not a function. It would have to be presented differently and stated in terms of a function with parameters.