How do I show that the relation $R : \mathbb{Q} \times \mathbb{Q} → \mathbb{Q}$ defined by $R((a/b, c/d)) = ((a+c) / (b+d))$ is not a function?

Question

I don't understand how to establish the relation $$R ((a/b, c/d)) = ((a+c) / (b+d))$$

Answer 1

An easy approach is just to say, let $a=1,c=1,b=1,d=-1$. Then $R(\frac{1}{1},\frac{1}{-1})=\frac{1+1}{1-1}$ which is undefined.

However, you'll learn more through observing that $R$ is not even well defined, by which I mean that you can give the same argument to $R$ twice, and get different results. Since you have no canonical form for the argument, you could input $\frac{1}{2}$ as $\frac{2}{4}$. However, $R(\frac{1}{2},\frac{1}{1})=\frac{1+1}{2+1}=\frac{2}{3}$. This is not the same as $R(\frac{2}{4},\frac{1}{1})=\frac{2+1}{4+1}=\frac{3}{5}$ even though $(\frac{1}{2},\frac{1}{1})=(\frac{2}{4},\frac{1}{1})$, so $R$ is not well defined.

Answer 2

Note that $(\frac ab,\frac cd)\in \Bbb Q\times \Bbb Q$ is the same point as $(\frac ab, \frac{2c}{2d})$. As such, if $R$ were a well-defined function, it should give the same result for them both. But it doesn't.

Try, for instance, with $a=b=d=1$ and $c=2$, which gives the value $\frac32$ in the first case, and $\frac 53$ in the other.

Answer 3

It is not a function, because it is ill-defined as a map from $\mathbb{Q}\times\mathbb{Q}$ to $\mathbb{Q}$.

For instance, $R(1/2,0/2) = 1/4$, while $R(2/4,0/2) = 1/3$ --- but $(1/2,0/2) = (2/4,0/2)$ in $\mathbb{Q}\times\mathbb{Q}$, so it is ill-defined where this element is supposed to map to.

Answer 4

$R(a/b, c/d) \ne R((na)/(nb), c/d)$ even though $a/b = (na)/(nb)$.

Is this what you want?

Answer 5

If something is a function then it cannot has two (or more) different values at one point.

Now suppose $a=\frac{1}{2}, ~b=\frac{1}{1}$ and you obtain $R(\frac{1}{2}, \frac{1}{1}) = \frac{2}{3}$, but then take arguments $a=\frac{2}{4}$ $b=\frac{1}{1}$ (that is definitely the same) and the value of this "function" becomes $R(\frac{2}{4},\frac{1}{1}) = \frac{3}{5}$.