Irrationality from a Limit Lemma

Lemma

Let \(\delta\in\mathbb{R}^{+}\). Assume there is a sequence of distinct positive rational numbers \(r_{n}/s_{n}\) (written in reduced form) such that

\(\displaystyle s_{n}\left|\delta-\frac{r_{n}}{s_{n}}\right|\rightarrow 0\),

as \(n\rightarrow\infty\). Then \(\delta\) is irrational.

Proof

Assume \(\delta=p/q\) in reduced form. There is at most one index \(n=n_{0}\), for which \(\delta=r_{n}/s_{n}\). Then, for \(n\gt n_{0}\), it follows that

\(\displaystyle s_{n}\left|\delta-\frac{r_{n}}{s_{n}}\right|=\frac{|ps_{n}-qr_{n}|}{q}\ge\frac{1}{q}\).

This contradicts the assumption on \(s_{n}\).

Theorem

\(\sqrt{2}\) is irrational.

Proof

Let \(\delta=\sqrt{2}\). Construct two sequences of positive integers by

\( \begin{align} r_{n+1} &= r_{n}+2s_{n}, \\ s_{n+1} &= r_{n}+s_{n}, \end{align} \)

starting with \(r_{0}=s_{0}=1\). By induction, \(r_{n}\ge s_{n}\) and

\(2s_{n}^2 = r_{n}^{2}+(-1)^{n}\).

This yields

\(\displaystyle \frac{1}{s_{n}^{2}}=\left|2-\frac{r_{n}^2}{s_{n}^2}\right| = \left|\sqrt{2}-\frac{r_{n}}{s_{n}}\right|\cdot \left|\sqrt{2}+\frac{r_{n}}{s_{n}}\right|\).

Therefore,

\(\displaystyle s_{n}\left|\sqrt{2}-\frac{r_{n}}{s_{n}}\right|\le s_{n}\left|2-\frac{r_{n}^2}{s_{n}^2}\right|\le \frac{1}{s_{n}}\rightarrow 0.\)

By Lemma, \(\sqrt{2}\notin\mathbb{Q}\).

References

  1. V. H. Moll, Numbers and Functions: From a Classical-Experimental Mathematician's Point of View, AMS, 2012, p. 45

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