Bijection Problem?

Question

Prove that the function defined by f(x) = (x-1/2) / [ x * (x-1) ] defines a bijection between (0,1) and R.

(R defining the set of real numbers)

Answer 1

Since f is given by the ratio of the functions x --> x -1/2 and x --> x * (x -1), both continuous on (0,1) it follows f is continuous. Hence, f((0,1)) is an interval.

In addition, we see that, if x --> 0+, then the numerator goes to -1/2 and the denominator goes to 0-, so that f(x) --> oo. And when x --> 1-, the numerartor goes to 1/2 and the denominator goes to 0-, so that f(x) --> -oo. Since f is continuous, it follows f((0,1)) = R, the range of f is the whole R.

It remains to show that f is one-to-one. To see this, observe that

f'(x) = [x(x-1) - (x -1/2)(2x -1)]/[x(x-1)]^2 = [x^2 - x - (2x^2 - x - x+1/2)]/ [x(x-1)]^2 = (-x^2 -x -1/2)/[x(x-1)]^2 = -(x^2 + x +1/2)/[x(x-1)]^2

For x>=0, the numerator is clearly decreasing and takes -1/2 < 0 for x = 0. So, this numerator is negative in (0,1). And since the denominator is always positive, it follows f' is negative in (0,1), which implies f is strictly decreasing. So, f is one-to-one, that is a bijection between (0,1) and R.

This exercise provides an interesting conclusion: R and (0,1) have the same cardinality.

ADDENDUM:

If you haven't studied derivatives yet, another way to prove f is a bijection is to solve the equation

y = (x-1/2) / [ x * (x-1)] Then,

y x^2 - y x = x -1/2 ==> y x^2 -(1+ y) x + 1/2 = 0 By Bhaskara, this gives 1 or 2 real solutions (I didn't do the algebra, but the previous conclusion allows me to say this), x1 and x2. If x1 and x2 are distinct, I bet one, and only one, will be in (0,1). And if the discriminant is 0, then x1 = x2 is in (0,1). So, each real y is the image of f under one, and only one, x in (0,1).

Answer 2

To show that it is a bijection, we want to show that (i) it is 1-1 (injective), and (ii) that it is onto (surjective).

First, though, I just want to point out that the function is defined and differentiable over the whole interval (0,1).

(i) It suffices to show that f'(x) < 0 on (0,1). This will mean that f is always strictly decreasing on the interval, and hence that f is 1-1. (A strictly decreasing function is obviously 1-1, since if f(b) < f(a) for b>a, then clearly f(b)≠f(a) for b>a.)
I probably should leave the derivative-taking as an exercise... In fact, I will. :-)

(ii) To show that f is surjective, it suffices to show (thanks to the Intermediate Value Theorem) that
lim (x→0+) f(x) = +∞, and
lim (x→1-) f(x) = -∞.
I'll leave this to the reader, too; you can do a sign chart, or whatever.


This completes the proof, modulo a few exercises for the reader.