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

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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 $f(c)<z$ .