Prove that the relation $f$: ($\mathbb{Z^*} \times \mathbb{Z^*}$) $\rightarrow$ $\mathbb{Q}$ by $f(a,b)$ = $\frac{a+b}{a+b-3ab}$ is a function
I believe that f is a function and I am attempting to prove it. I think my way of proving it is too easy to be true, but I would assume that the preimages $(a,b) = (c,d)$, and then deduce whether or not the images $f((a,b)) = f((c,d))$ are equal, which it turns out that this is the case.
But suppose that the way I proved it above is correct, and I extend this proof method to other functions that are simplistic in their own right. How would this proof strategy reveal particular preimage(s) that could be mapped to more than one image? In other words, do I actually have to sit down and begin guessing particular values for a and b until I reach a point where either f is not everywhere defined or not well-defined? Is there a more efficient way?
So my questions are twofold:
1) is f a function and is there a way to prove that it is, and
2) If we have a hunch that a relation is a function and we attempt to prove that it is a function, will the sequence of arguments reveal certain preimages that are mapped to more than one image?
Thanks
In order for $f$ to be a function, it must satisfy two things:
The problem here is not concerned with $2$, but $1$, you must prove that this relation it's defined on every point $(a,b)\in \Bbb Z^*\times \Bbb Z^*$, How could it not happen? Well, since the value of $f(a,b)$ is defined as a quotient, it won't be defined if the denominator is zero, i.e., if
$$a+b-3ab=0$$
However, this can't happen since $a,b\in \Bbb Z$ (why?), so the value of $f$ is defined everywhere and therefore $f$ is a function.