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To prove that two real numbers are equal, prove that neither is greater than the other

Quick description

Two real numbers x and y are equal if it is not the case that x<y and it is not the case that y<x. This is often the best way to prove that x=y 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 r is \pi r^2, then one way to do it is to inscribe a regular 2n-gon. One then cuts the 2n-gon into ``slices'' by joining its centre to the 2n 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 2n-gon. Now let s<r. Then if n is large enough, the altitude of each slice will be greater than s and the perimeter of the 2n-gon will be greater than \pi s. These assertions are not trivial but are not too hard to prove geometrically. It follows that for every s you can find inside the circle a polygon of area greater than \pi s^2. Therefore, the area of the circle cannot be less than \pi r^2. A similar argument shows that it cannot be greater than \pi r^2, and therefore it is equal to \pi r^2.

The point here is that we did not do some calculation that ended up with \pi r^2. 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 x\geq 0 and x<\epsilon for every positive \epsilon then x=0. With the help of the Archimedean principle we can also state a useful variant of this: if x\geq 0 and x<1/n for every positive integer n, then x=0.

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 z is a complex number and C is a simple closed contour that contains z in its interior, then f(z)=\frac 1{2\pi i}\int_C\frac{f(w)dw}{w-z}.

Now let z and z' be two points and let C be a simple closed contour that contains both z and z' in its interior. If we now calculate f(z)-f(z') using Cauchy's integral formula, we obtain \frac 1{2\pi i}\int_Cf(w)dw(\frac 1{w-z}-\frac 1{w-z'}, which equals \frac 1{2\pi i}\int_Cf(w)dw\frac{z-z'}{(w-z)(w-z')}. If f is bounded above in modulus by M, and C is a circle of radius R with R at least twice the maximum of |z| and |z'|, then this comes out to be at most \frac 1{2\pi}2\pi RM\frac{|z-z'|}{(R/2)^2}, which tends to 0 as R tends to infinity. Therefore, |f(z)-f(z')| 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 f is a continuous function, that a<b, and that f(a)<z<f(b). To prove that there exists c between a and b such that f(c)=z, one defines c to be the supremum of the set f(x)<z\}. Then instead of proving directly that f(c)=z, one obtains a contradiction from the assumption that f(c)>z and another contradiction from the assumption that f(c)<z.