$f(x) = \frac {ax+b}{cx+d}$ , bijection $f: \Bbb R \to\Bbb R$?

Question

For which $a,b,c,d \in \Bbb R$ does $f(x) = \frac {ax+b}{cx+d}$ define a bijection $f: \Bbb R \to \Bbb R$?

I'm guessing I need a system of equations and I know that $cx + d \ne 0$.

Thanks!

Answer 1

I assume you mean a bijection $\,f:\mathbb{R} \to \mathbb{R}$, since I don’t think the notation you’ve used makes sense.

For injectivity, we want to show that $\,f(x_1) = f(x_2)$ only if $x_1 = x_2$. So we want to find conditions on $a,b,c,d$ such that $$\frac{ax_1 + b}{cx_1 + d} = \frac{ax_2 + b}{cx_2 + d}$$ is true. Let's rearrange them to try to isolate $a,b,c,d$.

First we multiply up and expand the denominators: $$ \begin{align*} (ax_1 + b)(cx_2 + d\,) &= ac x_1 x_2 + bc x_2 + ad x_1 + bd \\ (ax_2 + b)(cx_1 + d\,) &= ac x_1 x_2 + bcx_1 + ad x_2 + bd \end{align*} $$ And since these are equal, we must have $$ac x_1 x_2 + bc x_2 + ad x_1 + bd = ac x_1 x_2 + bcx_1 + ad x_2 + bd$$ Cancelling off terms that occur on both sides, we have $$bc x_2 + ad x_1 = bcx_1 + ad x_2$$ If we pull all the terms on to one side, then we see that we get a rather nice factorisation appearing: $$(ad - bc)(x_1 - x_2) = 0 \tag{$*$}$$ Now, we want this to be true only if $x_1 = x_2$.

If $ad - bc = 0$, then this is true for any choice of $x_1, x_2$, and in particular, we have $\,f(x_1) = f(x_2)$ for $x_1 \neq x_2$. So $ad - bc = 0$ breaks injectivity.

But then we see that if $ad - bc \neq 0$, then equation $(*)$ is only true if $x_1 = x_2$. This means that our condition for injectivity is $$\boldsymbol{ad - bc \neq 0}$$ For surjectivity, we need to find an $x$ such that $\,f(x) = y$ for any $y \in \mathbb{R}$. So we write out the equation: $$\frac{ax+b}{cx+d} = y$$ Since we know that $cx + d \neq 0$, we can multiply up on both sides: $$ax + b = (cx+d\,)\, y$$ Now rearranging to get all the terms in $x$ on one side, we have: $$(a - cy)\,x = dy - b$$ If $a - cy \neq 0$, then we can divide through to find an $x$ that gives rise to $y$; $$x = \frac{dy - b}{a - cy}$$ We didn't use any conditions on $a,b,c,d$, so our injectivity condition stands. But we do need to cover the case $a - cy = 0$. If this is allowed, then it breaks bijectivity, because we need $x \to \infty$.

If $a - cy = 0$, then we have $y = a/c$. The way to get around this is to say that $c = 0$, so that this isn’t a valid choice for $y$.

Thus our condition for subjectivity is $$\boldsymbol{c = 0}$$ Combining these two conditions, we have that a condition for bijectivity is $$\boldsymbol{c = 0}, \; \boldsymbol{ad \neq 0}$$ Note that this also gives us $cx + d \neq 0$, for if $cx + d = 0$, then $d = 0$, which breaks injectivity.

Thus maps of this form are bijective if they have the form $$f(x) = \frac{ax + b}{d}, \qquad a, b, d \in \mathbb{R}, \; ad \neq 0$$

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Addendum: while I was writing this answer, I knew I’d seen maps of this form before, but I couldn’t remember where. Turns out I was thinking of Möbius maps, which extend maps of the form $f(x) = (ax+b)/(cx+d)$ to the “extended complex plane”.

If you’re familiar with complex numbers, then this means taking $\mathbb C$ and adding a point $\infty$. This gets us around the $c = 0$ restriction, because we can define $f(\infty) = a/c$.

You might notice that the condition $ad - bc \neq 0$ looks suspiciously like the determinant of the $2\times 2$ matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ In the world of Möbius maps, the map $f(x) = (ax+b)/(cx+d)$ gets associated with that matrix, and we require it be non-singular/have non-zero determinant.

If you haven’t come across these before, then don’t worry: it’s not important to understand this question, but I thought I would add it because it was bugging me that I couldn’t remember. :)

Answer 2

At least one among $c$ and $d$ must be different from $0$; but, if $c\ne0$, the map is not defined on the whole of $\mathbb{R}$. Therefore $c=0$ and $d\ne0$.

Then $f(x)=\frac{ax+b}{d}$; if $a=0$ it is constant, so not bijective. Therefore $a\ne0$. But in this case the equation $$ \frac{ax+b}{d}=y $$ has a unique solution for all $y\in\mathbb{R}$, namely $$ x=\frac{dy-b}{a}. $$ Thus the condition is $ad\ne0$ and $c=0$.

Answer 3

Now surjectivity, we need $\forall y\in\Bbb R:\exists x\in\Bbb R:\frac{ax+b}{cx+d}=y$

$\Rightarrow ax+b=y(cx+d)\Rightarrow x(a-cy)=(dy-b)\Rightarrow x=\frac{dy-b}{a-cy}\in \Bbb R$

But then we need $a-cy\ne 0,\,\forall y\in\Bbb R$, which holds iff $c=0$.

So we have the map $\frac{ax+b}{d}=kx+l$, for $k,l$ constants.

Now we look at injectivity:

$f(x_1)=f(x_2)\Rightarrow kx_1+l=kx_2+l\Rightarrow x_1=x_2$

So the map is injective. (with no constraints on $k,l$)

So for bijectivity we must have $c=0$. as well as $a,d\ne 0$

Answer 4

First note that since the equation $cx+d = 0$ always has a solution $d$ for every $x \in \mathbb R$ if $c \neq 0$, we must have $c = 0$ and $d \neq 0$.

You can think of the function $f$ as acting like a matrix in the following manner: $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ 1 \end{pmatrix} = \begin{pmatrix} ax+b \\ cx + d \end{pmatrix} , $$ where we identify $x \in \mathbb R$ with any scalar multiple of the vector $(x,1) \in \mathbb R^2$. Check that composing $f$ with another such function $g$ corresponds to multiplying the matrices associated to $f$ and $g$.

Then your question becomes:

When is a 2 by 2 matrix invertible?

Precisely when the determinant $ad-bc$ is nonzero. Since we have already $c = 0$, this means that neither $a$ nor $d$ may be zero.