Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Note that if $(a,b)$ and $(c,d)$ rationally are proportional, you get a a rational number no matter what $A$ is is. You potentially divided by $0$... – 1015 Mar 5 '13 at 16:45
up vote 2 down vote accepted

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 $a:c::b:d$

share|cite|improve this answer

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.

share|cite|improve this answer

$\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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.