I know that $ad\neq bc $ is sufficient for $z$ irrational because if $ad = bc$ then $\frac{ax+b}{cx+d} = \frac{ax+b}{cx+d} \frac{cb}{ad} = \frac{cax+cb}{cax+da}\frac{b}{d}$ because $cb=da$ nominator and denominator are the same. Hence $\frac{cax+cb}{cax+da}\frac{b}{d} = \frac{b}{d}$ (Contradiction).
But I don't know to prove that $ad \neq bc$ is also necessary for $z$ being irrational.
$\endgroup$ 62 Answers
$\begingroup$Assume by contradiction that
$$\frac{ax+b}{cx+d}=\frac pq,$$ or
$$(qa-pc)x+qb-pd=0,$$ and $x$ is rational !
Note that this doesn't work when $qa-pc=qb-pd=0$, which is equivalent to $ad= bc$.
$\endgroup$ 3 $\begingroup$$\frac {ax+b}{cx + d} =z$
$ax + b = z(cx + d) $
$x(a-cz) = zd - b$. ($cz$ is irrational and $a$ is rational so $a \ne cz$ and $a -cz \ne 0$.)
$x =\frac {zd- b}{a-cz}$.
If $z$ is rational then so is $x$. But $x$ isn't so $z$ can't be either.
$\endgroup$
