Let $c$ and $d$ be real numbers, not both zero, and let $f(x) = \frac{ax+b}{cx+d}$. Then $f$ is a function $S\to\mathbb R$ where $$ S = \{ x \in\mathbb R : cx + d \neq 0\}. $$ Under what conditions on $a$, $b$, $c$, and $d$ will $f$ be one-to-one?

Question on a recent proofs class examination. Studying set theory and onto/one-to-one functions. I was not able to figure out how to prove whether this function will be one to one and the conditions on $a$, $b$, $c$, and $d$.

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    $\begingroup$ Try computing the inverse. When does it work, when doesn't it? $\endgroup$ – Daniel Fischer Mar 14 '14 at 2:26

If $ac = 0$, then either $a = 0$ or $c = 0$ or $a = c = 0$. If $a = 0$, then if $b = 0, f(x) = 0$ and it is not one to one. So $b$ must be non-zero. if $c = 0$, then $f(x) = (ax + b)/d = ax/d + b/d$. Clearly $f$ is one to one if $a$ is non-zero. If $a = c = 0$, then $f(x) = 0$ and is not one to one.

If $ac$ is not zero, meaning both $a$ and $c$ are non-zero, then $$ f(x) = \frac{a(x + b/a)}{c(x + d/c)}. $$ Look at the part $(x + b/a)/(x + d/c)$ of $f(x)$. $f$ is one to one if this part (a function) is also one to one. Write it as: $1 + (b/a - d/c)/(x + d/c)$. From here it is quite clear that this part and hence $f(x)$ is one to one if $b/a - d/c$ is not zero or $bc - ad$ is not zero.

To sum this up, $f$ is one to one iff: $a = 0$ and $b, c$ are non-zero or $c = 0$ and $a$ is non-zero or $a, c$, and $bc - ad$ all non-zero.


One way to tell if a function is one-to-one is to see if it has an inverse. So let's assume that $f$ has an inverse -- call it $g$. The function $g$ has to satisfy the equation


Now we'll solve for $g(x)$. \begin{align*} x\cdot(cg(x)+d)&=ag(x)+b \\ cx\cdot g(x) + dx &= ag(x) + b \\ cx\cdot g(x) - ag(x) &= b-dx \\ g(x)\cdot(cx-a)&=(b-dx) \\ g(x) &= \frac{b-dx}{cx-a}. \end{align*}

If $f$ is one-to-one then it must have an inverse and the inverse must be given by the formula above. The only reason the formula for $g(x)$ might not work is if it were undefined at some point -- if we can plug in some $x\in S$ that makes us divide by zero. That is to say, $f$ will fail to be one-to-one precisely when $cx-a=0$ for some $x\in S$. In other words, $f$ is one-to-one precisely when $cx-a\neq0$ for all $x\in S$.

There are two cases.

  1. If $c=0$ then $cx-a=-a$, and it is nonzero if and only if $a\neq0$. So an inverse exists when $c=0$, $a\neq0$. What about $b$ and $d$? There are no additional conditions except that we can't have $d=0$, since in that case we would be dividing by zero in the definition for $f$ (and we would have $S=\varnothing$). Under these circumstances the function $f$ is a line with nonzero slope, which is clearly one-to-one.
  2. If $c\neq0$ then $cx-a=0$ for $x=a/c$. In this case $f$ is one-to-one if and only if $a/c\notin S$, if and only if $c(a/c) + d = 0$, if and only if $a=-d$. The variable $b$ may be chosen arbitrarily as long as one of $a$ and $b$ is nonzero (otherwise $f$ would be zero). Under these circumstances the function $f$ is a hyperboloid (a shifted version of the graph of $y=1/x$) which is lined up just right so as to be one-to-one.

In summary, there are two situations when $f$ is one-to-one:

  1. $a\neq 0$, $b\in\mathbb R$, $c=0$, $d\neq0$.
  2. $a\in\mathbb R$, $b\in\mathbb R$, $c\neq0$, $d=-a$, and one of $a$ or $b$ is nonzero.

Does that make sense?


If the function is injective it must preserve distinctness.

If it preserves distinctness then an inverse function exists, which uniquely maps the image of x back unto itself. $\forall x \in \mathbb{S}, x \mathop{\longmapsto}^{f^{-1}\circ f} x$

Assuming the function is injective, find this inverse function, $f^{-1}: f^{-1}(f(x))=x, \forall x \in \mathbb{S}$.

Let $y=f(x)$

$ y = \frac{ax+b}{cx+d} \implies y(cx+d)=(ax+b) \implies (yc-a)x=(b-yd) \implies x=\frac{b-yd}{yc-a}$

$\therefore f^{-1} : y \mapsto -\frac{dy-b}{cy-a}$

Testing preservation of distinctness.

$f^{-1}\circ f : x \mapsto -\frac{d\frac{ax+b}{cx+d}-b}{c\frac{ax+b}{cx+d}-a}$

$f^{-1}\circ f : x \mapsto -\frac{d(ax+b)-b(cx+d)}{c(ax+b)-a(cx+d)}$, since $(cx+d)\neq 0, \forall x \in \mathbb{S} $

$f^{-1}\circ f : x \mapsto -\frac{(ad-bc)x}{cb-ad}$

$\therefore f^{-1}\circ f : x \mapsto x \iff (cb-ad\neq 0) $


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