Solution of Pell's Equation is a Convergent

Theorem

Let $x = a, y = b$ be a positive solution to Pell's equation $x^2 - n y^2 = 1$.

Then $\dfrac a b$ is a convergent of $\sqrt n$.

Let $a^2 - n b^2 = 1$.

Then we have:

$\paren {a - b \sqrt n} \paren {a + b \sqrt n} = 1$.

So:

$a - b \sqrt n = \dfrac 1 {a + b \sqrt n} > 0$

and so $a > b \sqrt n$.

Therefore:

$\ds \size {\sqrt n - \frac a b}$

$=$

$\ds \frac {a - b \sqrt n} b$

$\ds $

$=$

$\ds \frac 1 {b \paren {a + b \sqrt n} }$

$\ds $

$<$

$\ds \frac 1 {b \paren {b \sqrt n + b \sqrt n} }$

$\ds $

$=$

$\ds \frac 1 {2 b^2 \sqrt n}$

$\ds $

$<$

$\ds \frac 1 {2 b^2}$

$\blacksquare$