I'm trying to learn how to express a square root as continued fraction, but I can't get one thing.
The following example of $\sqrt{14}$ is from this page (click the image to see it at full size):
In the 2nd row of the table, can anyone please tell me where the 1 (in red) comes from? For instance, why can't it be 2, or 3, or 4?
This is my only doubt, which I've been trying hard to understand, but unable to.
On
More generally, if $x$ is real and $n$ is a natural number, then $$\left\lfloor \frac{x}{n}\right\rfloor = \left\lfloor \frac{\lfloor x\rfloor}{n}\right\rfloor$$
So in computing the continued fraction expansion of a square root of $D$, all the $x_i$ are of the form $$\frac{\sqrt D + A}{B}$$ and to find the integer part of this, we only need to find the integer part of:
$$\frac{\lfloor\sqrt D \rfloor+ A}{B}$$
That greatly simplifies the process of computing the integer parts at each step - the only "approximation" we need for $\sqrt{D}$ to compute this step is $\lfloor \sqrt{D}\rfloor$. Consider how much harder it would be if the $x_i$ could be of the form:
$$\frac{A_i\sqrt{D}+B_i}{C_i}$$
This essentially requirs a "better" estimate for $\sqrt{D}$ to find the integer part of this expression.
Because $1$ is the greatest integer that is $\le \dfrac{\sqrt{14}+3}{5}$.
Remark: Recall the general continued fraction procedure. At any stage, we are looking at a number $x_n$, which is $\ge 1$ except possibly at the beginning. We compute $a_n=\lfloor x_n\rfloor$. If $a_n\ne x_n$, we define $x_{n+1}$ by $x_{n+1}=\dfrac{1}{x_n-a_n}$. Then $$x_n=a_n +\frac{1}{x_{n+1}}.$$ (There is a "better" algorithm for square roots of integers, and related numbers.)