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### One-to-one Function

```
Date: 02/05/98 at 13:19:32
From: Melissa Daley
Subject: College analysis

If A = (0,1) and B = (a,b) show that there is a one-to-one function
from A onto B.

I don't even know where to begin with this question. Could you direct
me to the right path?

```

```
Date: 02/05/98 at 15:59:31
From: Doctor Sam
Subject: Re: College analysis

Hi Melissa,

The problem looks really strange when stated in this notation. But if
you think of almost any simple function from algebra, that might show
you the way.

Consider the function f(x) = 2x. If you restrict the domain to values
of x in A, that is, 0 < x < 1, then the range of values output by f
is 0 < y < 2.

So f is an example of a function that is one-to-one (each input
corresponds to exactly one output and vice-versa. (Or each x
corresponds to exactly one y and vice-versa).

Likewise, the function y = 2^x takes numbers in A and produces numbers
between 2^0 and 2^1 (that is, between 1 and 2).

So that's an algebraic start to the problem. But I like a more visual
or geometric approach. You want to make a correspondence between the
interval (0,1) and the interval (a, b), so try drawing them. (I can't
draw here so I hope my description will be clear enough.

For (a, b) mark any interval on the x-axis.

For (0, 1) mark the segment on the horizontal line y = 1 from x = 0
to x = 1.

You should see two parallel segments, one on the line y = 1 and one on
the x-axis. Join the endpoints. This should form a trapezoid.

Now extend the sides of the trapezoid until they intersect. Call this
point of intersection P. It is the vertex of a triangle whose base is
(a,b). Here is my attempt at a drawing:

P
| \
|  \
|   \
|    \
|     \
|      \
|       \
|        \
--------- y = 1
|          \
|           \
|            \
--------------
a             b

Now draw any line through P to a point inside the base of the
triangle. This line intersects the small segment on y = 1 and the base
on the x-axis. The two points of intersection correspond to each
other. EVERY point on (0,1) will correspond to some point on (a,b) and
EVERY point on (a,b) will correspond to some point on (0, 1).

This is a nice visual to show that the correspondence exists. Now you
have to find an algebraic equation whose domain is (0,1) and whose
range is (a,b). I suggest that you try for the simplest of functions:
a linear function.

How to begin? Since (0, 1) is supposed to be mapped onto (a,b) we
want our function to take x = 0 and produce y = a and x = 1 and
produce y = b. That is, f(0) = a and f(1) = b. But if we are looking
for a *linear* function this is all the information we need because
two points determine a line. In this case the points are (0,a) and
(1,b).

The slope of the line is (b-a)/1 or b-a. Further, the point (0,a) is
the y-intercept of the line. So the function f(x) = (b-a)x + a will
do it.  f (0) = a and f (1) = b. Since lines are continuous, any
number between 0 and 1 will be mapped into a point between a and b
and vice-versa.

You can actually prove this algebraically. Assume that 0 < c < 1, so
c is a number between 0 and 1. We want to show that f (c) is between a
and b. But f (c) = (b-a)c + a. Since b-a > 0, f (c)>a and since c < 1,
f (c) < b - a + a = b.

To go in the other direction, assume that a < p < b. That is, p is a
number between a and b. Is there a number between 0 and 1 that is
mapped into p by our line?  To see that it is just substitute p for
f(x) and solve:

p = (b - a)x + a   so   x = (p - a)/(b - a)

Since p > a and b > a this x > 0.

Since p < b, p-a < b-a so the fraction (p-a)/(b-a) is a proper
fraction. That is, x < 1. So the number (p-a)/(b-a) is between 0 and 1
and is mapped by our line into p. This completes the proof.

I hope that helps!

-Dr. Sam  The Math Forum
Check out our Web site  http://mathforum.org/dr.math/
```
Associated Topics:
High School Functions

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