# 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

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

• The issue is whether $f(a, b)$ is defined everywhere (and in particular takes values in $\Bbb Q$). Note that $f(a, b)$ is not defined if the denominator in the definition, $a + b - 3ab$ is zero; if that never occurs, then $f(a, b)$ is a (defined) ratio of two integers for all $a, b$. – Travis May 3 '15 at 14:18
• Okay, and it turns out that the denominator never becomes zero, because, solving the equation of the denominator for zero, we have fractions, and a and b must be integers. – Benedict Voltaire May 3 '15 at 14:22

## 1 Answer

In order for $f$ to be a function, it must satisfy two things:

1. It's defined on every point $(a,b)\in \Bbb Z^*\times \Bbb Z^*$
2. For every point $(a,b)\in \Bbb Z^*\times \Bbb Z^*$, the value of $f(a,b)$ is not ambiguous, meaning with this that $((a,b),q)\in (\Bbb Z^*\times \Bbb Z^*)\times \Bbb Q$ lies in the relation for at most one $q$.

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.

• But how do we know that condition is 2 is satisfied rather than eye-balling it or depending upon intuition? – Benedict Voltaire May 3 '15 at 14:23
• It rests on the well definition of basic operations as sum, multiplication and division: If $(a,b)=(c,d)$, then $ab=cd$, $a+b=c+d$, $a+b-3ab=c+d-3cd$ and $\frac{a+b}{a+b-3ab}=\frac{c+d}{c+d-3cd}$ so if you have not any issue in assuming those operations are well defined, $f$ will be well defined. – Daniel May 3 '15 at 14:27
• Okay, I see that. Just to make sure that I understand why $a+b-3ab=0$ will never be true when $a,b \mathbb{Z^*}$, is because it turns out that if we solve the equation for $b$, say, we have the quotient $b=\frac{a}{3a-1}$. But how do we know that there does not exist a value for $a$ such that the fraction may be reduced in such a way as to make $b \in \mathbb{Z}$ ? – Benedict Voltaire May 3 '15 at 14:35
• @BenedictVoltaire I think you're concerned about 2 because you're confusing this with functions of the type $\Bbb Q\to \text{[some set]}, a/b\mapsto \text{[some expression with$a,b$]}$. With these you have to prove that the returned value doesn't depend on the representatives $a,b$. If the function is from $\Bbb Z,\Bbb Z^*$, or $\Bbb Z^*\times\Bbb Z^*$ there is no ambiguity. – GPerez May 3 '15 at 14:36
• @BenedictVoltaire $a$ and $3a-1$ are relatively primes since $-1\cdot (3a-1)+3\cdot (a)=1$ so $3a-1$ can't divide $a$ unless $|3a-1|=|a|=1$. – Daniel May 3 '15 at 14:39