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Please have a look :

Problem

The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$, where $a$,$b$,$c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$, $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$. Find the unique number that is not in the range of $f$.

The solution can be found here

It states, without proof, that if we have the functional equality:

$$\frac{px+q}{rx+s}=x$$ then $r=q=0.$

At the first solution, line $3$ why does it have to be $q = r = 0$? [ I understand that the opposite is true, i.e. , if $q = r = 0 $ then the fraction reduces to $x$ when $p = s$ ]

Community
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adnan2nd
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3 Answers3

3

Basically, a polynomial with infinitely many roots must be a zero polynomial.

If $\frac{px+q}{rx+s}=x$ for infinitey many $x$ then:

$$px+q=rx^2+sx$$ for infinitely many $x$, and thus:

$$rx^2+(s-p)x-q=0$$

for infinitely many $x.$ So this must be a zero polynomial, which means that $r=0,s-p=0,$ and $q=0.$

Thomas Andrews
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  • So, as $x$ is used to define the function in question and nothing is said about it's domain, we have to assume that $x$ can be any real number, as a result $\frac{px + q} {rx + s} = x$ is 'a polynomial with infinitely many roots'? – adnan2nd Jun 06 '18 at 04:31
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    Well, $\frac{px+q}{rx+s}=x$ is not a polynomial, but $rx^2+(s-p)x-q$ is a polynomial – Thomas Andrews Jun 06 '18 at 14:28
  • I should have said " ... as a result $\frac{px+q} {rx+s}=x$ is 'a polynomial equation with infinitely many roots" – adnan2nd Jun 06 '18 at 16:48
  • Well, even that's not accurate. What is accurate is that it is equivalent to a polynomial which has infinitely many zeros. (An equation has solutions, not zeros. A polynomial has zeros.) – Thomas Andrews Jun 07 '18 at 15:13
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One way to come to this conclusion is to put different values of $x$. For example, putting $x=\pm 1$ in $$\frac{px + q}{rx + s}=x\tag{1}$$ results in $q+p=r+s$ and $q-p=r-s$ which imply $q=r$, and $p=s$. Similarly, plugging $x=0$ in $(1)$ results in $q=0$. As such, $r=q=0$.

Math Lover
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Note that homographies are uniquely defined via a proportionality coefficient.

First remark that if $(ad-bc)=0$ or $k=\frac ac=\frac bd$ then $f(x)=\dfrac{ax+b}{cx+d}=\dfrac{kcx+kd}{cx+d}=k$

In the case the homography is not degenerated to a constant then

$\dfrac{ax+b}{cx+d}=\dfrac{Ax+B}{Cx+D}\iff (aC-Ac)x^2+(aD+bC-Ad-Bc)x+bD-Bd = 0$

You get two conditions $\begin{cases}aC=Ac\iff A=\lambda a,\ C=\lambda c\\ bD=Bd\iff B=\mu b,\ D=\mu d\end{cases}$

The central term then factorizes to $(\lambda-\mu)(ad-bc)x=0\iff \lambda=\mu$ since we are not in the degenerated case.

Finally you get that two homographies are equal only when their coefficients are all proportional with the same proportionality constant.

In our simple case $\dfrac{px+q}{rx+s}=x=\dfrac{1x+0}{0x+1}$ you get $q=0\lambda, r=0\lambda, p=1\lambda, s=1\lambda$

Or more simply $q=r=0$ and $p=s$.

zwim
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  • That first sentence is a whopper, given that you have not defined "homography" and "proportionality coefficient." – Thomas Andrews Jun 05 '18 at 19:52
  • Feel free to write a better sentence, I admit it is not best choice of wording. Speaking of Mobius transforms is also maybe too much here. – zwim Jun 05 '18 at 20:58