Square Root of 2 as a Continued Fraction

To represent the \(\sqrt{2}\) as a continued fraction we start with the obvious

\(\sqrt{2}=1+(\sqrt{2}-1)=1+\frac{1}{1+\sqrt{2}}\).

What is worth observing is that \(\sqrt{2}\) appears on the two sides of the equality, making it possible to replace it recursively. The first step gives

\( \begin{align} \sqrt{2} &= 1+\frac{1}{1+\sqrt{2}} \\ &=1 + \frac{1}{1+(1+\frac{1}{1+\sqrt{2}})} \\ &=1 + \frac{1}{2+\frac{1}{1+\sqrt{2}}}. \end{align} \)

Naturally, we may continue:

\( \begin{align} \sqrt{2} &= 1+\frac{1}{1+\sqrt{2}} \\ &=1 + \frac{1}{2+\frac{1}{1+\sqrt{2}}} \\ &=1 + \frac{1}{2+\frac{1}{1+(1+\frac{1}{1+\sqrt{2}})}} \\ &=1 + \frac{1}{2+\frac{1}{2+\frac{1}{1+\sqrt{2}}}}. \end{align} \)

The process never ends so that \(\sqrt{2}\) comes out to be an infinite comtinued fractions:

\( \begin{align} \sqrt{2} &= 1+\frac{1}{1+\sqrt{2}} \\ &=1 + \frac{1}{2+\frac{1}{1+\sqrt{2}}} \\ &=1 + \frac{1}{2+\frac{1}{2+\frac{1}{1+\sqrt{2}}}} \\ &=1 + \frac{1}{2+\frac{1}{2+\frac{1}{1+(1+\frac{1}{1+\sqrt{2}})}}} \\ &=1 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\ldots }}}, \end{align} \)

which is written compactly as \([1;2,2,2,\ldots ]\). Being an inifinite continued fraction, \(\sqrt{2}\) is irrational.

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