Often we are required to prove that a function *f*,
is one-to-one, or an injection. A proof of
this depends on how the function is given to us and what properties the
function has.
For functions that are
given by some formula there is a standard way. We use the contrapositive
of the definition of one-to-one, namely that if *f*(*u*) = *f*(*v*) then *u* = *v*.
For example, suppose that is defined
by where **N** denotes the positive integers. We start by
assuming there are integers *u* and *v* for which *f*(*u*) = *f*(*v*). Thus,
and so . Then, (*u*-*v*)(*u*+*v*) = 0. Since the product
is zero then *u*-*v* = 0 or *u*+*v* = 0. The first condition tells us that *u*=*v*.
The second condition says that *v* = -*u*. However, since *u* > 0 this would
imply *v* < 0 which is impossible since *v* is positive. Hence, *u* =*v*
as required and *f* is one-to-one.
There are other methods used to prove a function is one-to-one. In calculus
for example, if *f* is differentiable then it is sufficent to show that
the derivative is positive everywhere or negative everywhere. In linear
algebra, if *f* is a linear transformation it is sufficient to show that
the kernal of *f* contains only the zero vector. If *f* is a function
with finite domain it is sufficient to look through the list of images of each
domain element and check that no image occurs twice in the list.

Wed Aug 21 23:10:39 PDT 1996