# Irrationality of $\sqrt{2}$ via a Difference Equation

$\sqrt{2}$ is irrational.

The general solution of the difference equation $x_{n+2}=-2x_{n+1}+x_{n},$ where $n=0,1,2,\ldots,$ is expressed as $\displaystyle x_{n}=ar_{1}^{n}+br_{2}^{n},$ where $a,b$ are arbitrary coefficients, and $r_{1,2}$ are the roots of the characteristic equation $r^{2}+2r-1=0,$ say, $r_{1}=\sqrt{2}-1$ and $r_{2}=-(\sqrt{2}+1).$

Assume that $\sqrt{2}$ is rational, so that $\displaystyle r_{1}=\frac{p}{q},$ with $p,q$ integers and $q\ne 0.$ By taking $a=q$ and $b=0,$ we have $x_{0}=q\in\mathbb{Z}$ and $x_{1}=p\in\mathbb{Z}$ so that by induction $x_{n+2}=-2x_{n+1}+x_{n}\in\mathbb{Z},$ for all $n=0,1,2,\ldots$

On the other hand, since $0\lt r_{1}\lt 1,$ and $q\ne 0,$ $\displaystyle x_{n}=qr_{1}^{n}\ne 0,$ for all $n,$ and $\displaystyle\lim_{n\rightarrow\infty}x_{n}=\lim_{n\rightarrow\infty}qr_{1}^{n}=0.$ A contradiction.

### References

- José Ángel Cid Araújo,
__A Difference Equation Leading to the Irrationality of $\sqrt{2},$__*Amer. Math. Monthly*, Vol. 121, No. 5 (2014), p. 443.

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