Proving that √3 is irrational using the well-ordering property

If \(\sqrt{3}\) is rational, then there are two integers \(a\) and \(b\) such that:

$$\begin{align} \sqrt{3} &= \frac{a}{b} \\ b\sqrt{3} &= a \end{align}$$

This means there exists integers \(b\) such that \(b\sqrt{3}\) is an integer. Let \(S\) be a set that contains these integers:

$$ S = \{ b\sqrt{3} \quad | \quad b \text{ and } b\sqrt{3} \text{ are positive integers}\}$$

According to the well-ordering property, \(S\) contains a smallest element \(s\). Let \(s=t\sqrt{3}\). This means:

$$\begin{gather} s\sqrt{3} = 3t \\ (t \in \mathbb{N}) \wedge (s \in \mathbb{N}) \implies 3t - s \in \mathbb{Z} \end{gather}$$

We can rewrite this as:

$$\begin{gather} 3t - s = \\ s\sqrt{3} - t\sqrt{3} = \\ (s-t)\sqrt{3} \in \mathbb{Z} \end{gather}$$

Since \([s = t\sqrt{3}]\) and \([\sqrt{3} \gt 1]\), then \(s \gt t\). Which means \([s-t]\) is a positive integer. This would also make \([(s-t)\sqrt{3}]\) would also be a positive integer. By definition of \(S\), \([(s-t)\sqrt{3}]\) is also in \(S\).

Since \(1.9^2 = 3.61\), then \(1.9 \gt \sqrt{3}\).This means:

$$\begin{gather} \sqrt{3} \lt 2 \\ t\sqrt{t} \lt 2t \\ s \lt 2t \\ s-t \lt t \\ (s-t)\sqrt{3} \lt t\sqrt{3} \\ (s-t)\sqrt{3} \lt s \end{gather}$$

Since \([(s-t)\sqrt{3}]\) exists in \(S\) and \([(s-t)\sqrt{3} \lt s]\), then we have a contradiction.

Proving That √3 Is Irrational Using The Well-Ordering Property