I have these two terms:
$$b+a\frac{cx+dy}{ax+by}$$ $$a-b\frac{cx+dy}{ax+by}$$
The requirements are: $x,y\in\mathbb{Z};a,b,\frac{cx+dy}{ax+by}\in\mathbb{R}\backslash\mathbb{Q};c,d\in\mathbb{R}$. So basically, x & y are whole numbers, a & b are irrational numbers, and c & d are real numbers. And also the whole term $\frac{cx+dy}{ax+by}$ is definitely also irrational.
Based on those conditions I want to try to prove that one of the two terms mentioned above is definitely irrational no matter which values the variables take on. If this is not true I would like to see a counterexample or any other kind of proof falsifying the claim.
I have already tried to algebraically manipulate the term to find a more suitable shape but have not succeeded.
$$a-b\frac{cx+dy}{ax+by}=\frac{\frac{a^2}bx+ay-cx-dy}{\frac a b x+y}$$
I have also tried to go about it by considering the general rules for adding and multiplying rational and irrational numbers, like rational * irrational = irrational and such. But this also did not seem to get me anywhere.
The claim is false. The choices $a = \frac13 (2 + \sqrt2)$, $b = \frac13 (1 - 2 \sqrt2)$, $c = \sqrt2 - \frac23$, $d = 0$, $x = 1$, and $y = 1$, for which $\dfrac{cx+dy}{ax+by}=\sqrt2$, yield $b+a\dfrac{cx+dy}{ax+by} = 1$ and $a-b\dfrac{cx+dy}{ax+by} = 2$.
Brainstorming: the specific form $z=\dfrac{cx+dy}{ax+by}$ is a red herring—since $c$ and $d$ can be arbitrary real numbers, $z$ can be an arbitrary irrational number regardless of the values of $a$ and $b$. So we're essentially asking whether one of $b+az$ and $a-bz$ must be irrational if $a$, $b$, and $z$ are all irrational. This seems unlikely, though, since we have three degrees of freedom and only two counterexample-constraints to meet.
For a specific example, let's try to solve the equations $b+az=1$ and $a-bz=2$ for $a$ and $b$: we obtain $a = \dfrac{z+2}{z^2+1}$ and $b = \dfrac{-2z+1}{z^2+1}$. Choosing $z=\sqrt2$ (a random simple irrational number) makes $a=\frac13(2+\sqrt2)$ and $b=\frac13(1-2\sqrt2)$ also irrational.
Finally, there are lots of ways to make $\dfrac{cx+dy}{ax+by}=\sqrt2$: for example, we can just take $x=y=1$ and $d=0$ for simplicity, and solving yields $c=\sqrt2-\frac23$.