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 .
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.
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 .