The function is this: $f(\frac{a}{b},\frac{c}{d})=\frac{a+c}{b+d}$ where
$0\lt \frac{a}{b} \lt 1$
$0\lt \frac{c}{d}\lt 1$
$a,b,c,d$ are all integers
$a/b$ and $c/d$ are in lowest terms
Are there real numbers, for no matter what integers you plug in for $a,b,c$ and $d$, $f(\frac{a}{b},\frac{c}{d})$ will never equal to? If we classify all impossible numbers in one group, group $A$, does $A$ have finite or $\infty$ terms? Or even none at all? Thanks a bundle.
It is easy to show that the range of $f$ is $\mathbb{Q} \cap (0,1)$. Clearly $f$ is positive, and $$\frac{a}{b} < 1, \frac{c}{d} < 1 \;\; \Rightarrow \;\; a < b, c < d \;\; \Rightarrow \;\; a + c < b + d \;\; \Rightarrow \;\; \frac{a+c}{b+d} < 1.$$ Now take $y \in \mathbb{Q} \cap (0,1)$. Then $y = \frac{p}{q}$ for some $p,q \in \mathbb{N}$ with $\gcd(p,q) = 1$. We have to show two rationals the function maps to $y$. But we can do this easily by letting $a = c = p$ and $b = d = q$: $$f\left(\frac{a}{b},\frac{c}{d}\right) = \frac{a+c}{b+d} = \frac{2p}{2q} = \frac{p}{q}.$$
By construction, $f$ is not irrational, leaving us with the range mentioned above. To answer the original question, the set of all real numbers that $f$ never attains is $\mathbb R \setminus(\mathbb{Q} \cap (0,1)),$ which has uncountably many elements.