Finding Continued Fraction Expansion for Reals

To find a continued fraction expansion for real numbers, we use a similar technique to how we did so for rational numbers.

First let's try an example, $\sqrt{5}$. Well, we know that this is equivalent to $\frac{\sqrt{5}}{1}$. When we do the division, we have $$\sqrt{5}=2+(\sqrt{5}-2).$$

$sqrt\left\{5\right\}=2.23606=2+0.236...$ which is an integer plus irrational part less than .

Yes, we just took the the largest integer less than or equal to $\sqrt{5}$. We will deonte this by $\lfloor\sqrt{5}\rfloor=2$. So, we know that $$\sqrt{5}=2+\dfrac{1}{x},$$ for some $x$. Then, let's solve for $x$ to get $$x=\dfrac{1}{\sqrt{5}-2}=2+\sqrt{5}.$$ Now we repeat the process to get $$2+\sqrt{5}=4+\dfrac{1}{x},$$ where $x=2+\sqrt{5}$. However, note that we computed the continued fraction for $2+\sqrt{5}$ already! So, we know that it repeats. $$\sqrt{5}=2+\dfrac{1}{4+\dfrac{1}{4+\ddots}}=[2,4,4,4,4,\cdots].$$ We will express the repeating part of the continued fraction with a bar on top, like so: $$\sqrt{5}=[2,4,4,4,\cdots]=[2,\overline{4}].$$

Now let's generalize

If $n$ is irrational, by the division algorithm, we can write an equation in the following form: $$n=1q_1+r\quad\text{where }n-r_1<1.$$ However, note that $r_1=n-\lfloor n \rfloor$ is also irrational. Thus, we can write a similar equation: $$1=r_1q_2+r_2.$$ Since we will always get an irrational remainder, we can continue this process infinitely. Now let's rearrange the equations to get what we want. $$\begin{align} n=1q_1+r_1 &\implies n=q_1+\dfrac{1}{\frac{1}{r_1}}.\\ 1=r_1q_2+r_2 &\implies \frac{1}{r_1}=q_2+\dfrac{1}{\frac{r_1}{r_2}}\\ r_1=r_2q_3+r_3 &\implies \frac{r_1}{r_2}=q_3+\dfrac{1}{\frac{r_2}{r_3}}\\ &\vdots \end{align}$$ So, we can just repeatedly substitute into the previous equation to get the continued fraction $$\begin{align} n&=q_1+\dfrac{1}{\frac{1}{r_1}}\\ &=q_1+\dfrac{1}{q_2+\dfrac{1}{\frac{r_1}{r_2}}}\\ &=q_1+\dfrac{1}{q_2+\dfrac{1}{q_3+\dfrac{1}{\frac{r_2}{r_3}}}}\\ &=\qquad \vdots\\ &=[q_1,q_2,\cdots]. \end{align}$$