In what circumstances can $\dfrac{aA+b}{cA+d}$ be rational?

Question

I am working on the chapter one practice problems in Hardy and cannot seem to figure it out. My attempt has actually left me with a result contrary to what the question is looking for.

The Question

In what circumstances can $\dfrac{aA+b}{cA+d}$ be rational, where $a,b,c,d$ are rational and $A$ is irrational.

My Attempt

Any rational number can be written as the ratio of two relatively prime integers. Replacing each of the rational numbers with such ratios, we have:

$\dfrac{p}{q}=\dfrac{ \dfrac{y}{z} A+ \dfrac{w}{x}} { \dfrac{u}{v} A + \dfrac{s}{t}}$

Rearranging this expression and solving for A yields:

$A=\dfrac{qz}{pt} \dfrac{wt-sx}{uz-yv}$

However, since both the numerator and denominator are algebraic operations on integers, than A must be a rational number; being the ratio of two integers.

Answer 1

Your steps are fine. But since $A=\dfrac{qz}{pt} \dfrac{wt-sx}{uz-yv}$ is an irrational then, denominator must be $0$ and then $A$ to be defined, numerator must be $0$ too which gives $wt=sx$ and $uz=yv$ which implies $a:c::b:d$

Answer 2

assuming your calculations are correct, that means that you have a contradiction (and therefore $p$ and $q$ do not exist) unless $p(uz−yv)=0$ (as $t$ already can not be $0$)

Dividing by $0$ is very bad :)

My answer using this logic :

$$\frac{p}{q} = \frac{aA+b}{cA+d} \Longrightarrow p(cA+d)=q(aA+b) \Longrightarrow (pc-qa)A = qb-pd$$

therefore, as $A$ is irrationnal, $(pc-qa)$ must be $0$, and therefore $qb-pd=0$ too.

This leads (baring zero cases) to $\frac{p}{q}=\frac{a}{c}=\frac{b}{d}$, and therefore $(a,c)$ and $(b,d)$ must be proportionnal.

julien's comment gives the reciprocate.

Answer 3

$\begin{array}{c}\rm \color{#C00}{A\not\in\Bbb Q},\quad\\ \rm\color{#C00}{ a,b,c,d,e\in\Bbb Q}\,\end{array}\bigg\rbrace\ $ $\rm \dfrac{aA\!+\!b}{cA\!+\!d}\, =\, e\:\Rightarrow\: aA\!+\!b\, =\, ecA\!+\!ed\:\Rightarrow\:(a\!-\!ec)A\, =\,ed\!-\!b\:\color{#C00}{\Rightarrow}\ \begin{eqnarray} a \,&=&\,\rm ec\\ \rm b \,&=&\,\rm ed\end{eqnarray}$

Remark $\ $ The proof doesn't depend on $\Bbb Q,\,$ it works for any field.