Square Root of 2 is Irrational, Proof 29

Cesare Palmisani,
22 December, 2017

The following is based on M. Jacobson and H. Williams' book.

We define two sequences of positive integers $(d_n)$ and $(s_n ).$ "The law of formation of these numbers is explained by Theon of Smyrna, and is as follows," see the reference:





The sequences (1) and (2) are called diagonal numbers and side numbers, respectively. Obviously the two sequences are monotone strictly increasing. In addition the sequences are generalized Fibonacci.

By induction,



Let's suppose that $\sqrt{2}$ is a rational number. Then we can write $\displaystyle \sqrt{2}=\frac{a}{b},$ where $a,b$ are positive integers. From (3),


$(bd_n-as_n)(bd_n+as_n)=b^2\cdot (-1)^n.$

As a matter of fact,

$\displaystyle \begin{align} (bd_n-as_n)(bd_n+as_n)&=(bd_n)^2-(as_n)^2\\ &=b^2d_n^2-a^2s_n^2=b^2\left(d_n^2-\frac{a^2}{b^2}s^2_n\right)\\ &=b^2(d_n^2-2s_n^2=b^2\cdot (-1)^n. \end{align}$

If $bd_n-as_n=0,$ then, successively, $bd_n=as_n,$ $\displaystyle d_n=\frac{a}{b}s_n,$ $\displaystyle d_n^2=\frac{a^2}{b^2}s_n^2,$ $d_n^2-2s_n^2=2s_n^2-2s_n^2=0,$ in contradiction with (3). Therefore, $bd_n-as_n\ne 0,$ and we can divide in (4):

$\displaystyle bd_n+as_n=\frac{b^2\cdot (-1)^n}{bd_n-as_n}.$

Since $bd_n+as_n\gt 0,$ $bd_n+as_n=|bd_n+as_n|$ and

$\displaystyle bd_n+as_n=\left|\frac{b^2\cdot (-1)^n}{bd_n-as_n}\right|=\frac{b^2}{|bd_n-as_n|}.$

In addition, from $bd_n-as_n\ne 0,$ it follows that $|bd_n-as_n|\ge 1,$ $\displaystyle \frac{1}{|bd_n-as_n|}\le 1,$ $\displaystyle \frac{b^2}{|bd_n-as_n|}\le b^2,$ or, $bd_n+as_n\le b^2,$ implying that the sequences $(d_n)$ and $(s_n)$ are bounded. A contradiction.


  1. Heath T., A History of Greek Mathematics, Oxford University Press, Vol I, 92
  2. Jacobson M., Williams H., Solving the Pell Equation, CMS books in Mathematics, Springer, 24- 25 (2009)
  3. Palmisani C., Side and diagonal numbers: additional results, to be appear in Periodico di Matematiche - Mathesis; No. 1 (2018)

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