Quick description
Two real numbers and
are equal if it is not the case that
and it is not the case that
. This is often the best way to prove that
when it is not obvious.
Example 1 (and general discussion)
This basic idea, phrased very differently, goes back at least as far as the ancient Greeks. If you want to prove that the area of a circle of radius is
, then one way to do it is to inscribe a regular
-gon. One then cuts the
-gon into ``slices'' by joining its centre to the
vertices, and then rearranges the slices into a parallelogram by alternating upwards-pointing slices and downwards-pointing slices. Each slice is an isosceles triangle, and the parallelogram has height the altitude of these isosceles triangles and base half the perimeter of the
-gon.
Now let
. Then if
is large enough, the altitude of each slice will be greater than
and the perimeter of the
-gon will be greater than
. These assertions are not trivial but are not too hard to prove geometrically. It follows that for every
you can find inside the circle a polygon of area greater than
. Therefore, the area of the circle cannot be less than
. A similar argument shows that it cannot be greater than
, and therefore it is equal to
.
The point here is that we did not do some calculation that ended up with . Rather, we proved that the area could not be greater than this and could not be less than this.
A similar idea is often used to prove that a non-negative number is zero: if and
for every positive
then
. With the help of the Archimedean principle we can also state a useful variant of this: if
and
for every positive integer
, then
.
Example 2
A nice example of this is the standard proof of Liouville's theorem in complex analysis, which states that a bounded analytic function on the whole complex plane is constant. To prove this, one uses Cauchy's integral formula, which states that if is a complex number and
is a simple closed contour that contains
in its interior, then
.
Now let and
be two points and let
be a simple closed contour that contains both
and
in its interior. If we now calculate
using Cauchy's integral formula, we obtain
, which equals
. If
is bounded above in modulus by
, and
is a circle of radius
with
at least twice the maximum of
and
, then this comes out to be at most
, which tends to 0 as
tends to infinity. Therefore,
is less than any positive real number, which forces it to be zero.
Example 3
One of the standard proofs of the intermediate value theorem uses the law of trichotomy in a fairly explicit way. Suppose that is a continuous function, that
, and that
. To prove that there exists
between
and
such that
, one defines
to be the supremum of the set
. Then instead of proving directly that
, one obtains a contradiction from the assumption that
and another contradiction from the assumption that
.