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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?

Thank you for your time.

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:

         | \
         |  \
         |   \
         |    \
         |     \
         |      \
         |       \
         |        \
         --------- 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 

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
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Associated Topics:
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