<s>[INST] Continued fraction for $\sqrt{14}$ 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? 
 [/INST] 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 $x = \left\lfloor\frac{6}{5}\right\rfloor = 1$.
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