### 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 .

## Comments

## Post new comment

(Note: commenting is not possible on this snapshot.)