Continued fraction for $\sqrt{14}$

Question

I'm referencing this page: An Introduction to the Continued Fraction, where they explain the algebraic method of solving the square root of $14$.

$$\sqrt{14} = 3 + \frac1x$$

So, $x_0 = 3$, Solving for $x$, we get

$$x = \frac{\sqrt{14} + 3}5$$

However, in the next step, how do we get the whole number $x_1$ = 1?

$$\frac{\sqrt{14} + 3}5 = 1 + \frac1x$$

My understanding is we would substitute for $x$ in the original equation for $\sqrt{14}$ whereas $$\sqrt{14} = 3 + \frac1{\frac{\sqrt{14} + 3}5}$$

Then substitute the $\sqrt{14}$ again here for $x = \frac{\sqrt{14} + 3}5$ to get the $x_1$ of the continued fraction? Am I just getting the algebra at this point wrong or am I botching steps?

Answer 1

$\sqrt{14}=3+\sqrt{14}-3=3+\frac{1}{\frac{\sqrt{14}+3}{5}}\implies x_0 = 3$

$\frac{\sqrt{14}+3}{5}=\frac{6+\sqrt{14}-3}{5}=1+\frac{\sqrt{14}-2}{5}=1+\frac{1}{\frac{\sqrt{14}+2}{2}} \implies x_1 = 1$

$\frac{\sqrt{14}+2}{2}=\frac{5+\sqrt{14}-3}{2}=2+\frac{\sqrt{14}-2}{2}=2+\frac{1}{\frac{\sqrt{14}+2}{5}} \implies x_2 = 2$

etc.

Answer 2

It's really easy to compute $\left\lfloor \frac{\sqrt a + b}{c}\right\rfloor$ for integers $a,b,c$. Just use the fact that

$$\left\lfloor \frac{r + b}{c}\right\rfloor = \left\lfloor \frac{\lfloor r \rfloor + b}{c}\right\rfloor$$

for real $r$ and integers $b,c$.

Here, $\lfloor\sqrt{14}\rfloor = 3$, so $x = \left\lfloor\frac{6}{5}\right\rfloor = 1$.