A finite continued fraction is a fraction of the form \[ a_0+\cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{ \llap{\ddots} a_{n-2}+\cfrac{}{ a_{n-1} + \cfrac{1}{a_n}}}}} \] for some real numbers \(a_0,a_1,a_2,\ldots,a_n\) where \(a_i>0\) for \(i=1,2,\ldots,n\). It is denoted by \([a_0,a_1,a_2,\ldots,a_n]\). A continued fraction is called simple if \(a_0,a_1,a_2,\ldots,a_n\) are integers.

Example.

1. \[[5,2,3,2]=5+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{2}}}.\] We can verify that \([5,2,3,2]=\frac{87}{16}\). Similarly we can show that any finite simple continued fraction is a rational number.
   
   
2. Let us write \(\frac{1479}{17}\) as a finite simple continued fraction. We follow steps similar to the Euclidean Algorithm: \[ \begin{align*} \frac{1479}{272} &=5+\frac{119}{272}=5+\cfrac{1}{\frac{272}{119}} &&\implies \frac{1479}{272}=5+\cfrac{1}{\frac{272}{119}}\\ \frac{272}{119} &=2+\frac{34}{119}=2+\cfrac{1}{\frac{119}{34}} &&\implies \frac{1479}{272}=5+\cfrac{1}{2+\cfrac{1}{\frac{119}{34}}}\\ \frac{119}{34} &=3+\frac{17}{34}=3+\frac{1}{2} &&\implies \frac{1479}{272}=5+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{2}}} \end{align*} \] So \(\frac{1479}{272}=[5,2,3,2]\). Similarly we can show that any rational number can be written as a finite simple continued fraction.

Observation. A real number is a rational number if and only if it can be written as a finite simple continued fraction.

An infinite continued fraction is a fraction of the form \[ a_0+\cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{ \llap{\ddots} {\:\:}}}} \] for infinitely many real numbers \(a_0,a_1,a_2,\ldots\) where \(a_i>0\) for \(i=1,2,\ldots\). It is denoted by \([a_0,a_1,a_2,\ldots]\). An irrational number expressed by an infinite continued fraction \([a_0,a_1,a_2,\ldots]\) can be approximated by its truncated form \([a_0,a_1,a_2,\ldots,a_n]\) for some \(n\geq 0\) which is called the \(n\)th convergent of \([a_0,a_1,a_2,\ldots]\), denoted by \(C_n\). It can be proved that \(\displaystyle\lim_{n\to \infty} C_n\) exists and is equal to \([a_0,a_1,a_2,\ldots]\) using the following fact: \[ C_0< C_2< C_4<\cdots< C_{2n}<\cdots< C_{2n+1}<\cdots< C_5< C_3< C_1.\]

Example.

1. \(\pi=[3,7,15,1,292,1,1,1,2,1,3,1,\ldots]\).
   \(\pi\) can be approximated by \(C_0=[3]=3\), \(C_1=[3,7]=\frac{22}{7}\), \(C_2=[3,7,15]=\frac{333}{106}\) etc.
   
   
2. \(e=[2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots]\).
   Notice the interesting pattern.
   
   
3. \(\phi=[1,1,1,\ldots]=[\overline{1}]\).
   Note that \(x=[1,1,1,\ldots]=[1,x]=1+\frac{1}{x}\). Then \(x^2-x-1=0\implies x=(1\pm \sqrt{5})/2\). Since \(x>0\), \(x=(1+\sqrt{5})/2=\phi\), the golden ratio.
   
   
4. \(\sqrt{2}=[1,2,2,2,\ldots]=[1,\overline{2}]\).
   Note that \((\sqrt{2}-1)(\sqrt{2}+1)=2-1=1\implies \sqrt{2}=1+\cfrac{1}{1+\sqrt{2}}\). Repeating this process, we get \[\sqrt{2}=1+\cfrac{1}{1+\sqrt{2}} =1+\cfrac{1}{2+\cfrac{1}{1+\sqrt{2}}} =1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{1+\sqrt{2}}}} =\cdots=[1,2,2,2,\ldots].\]

We get the continued fraction of an irrational number \(n\) by the continued fraction algorithm:
With \(x_0=n\), recursively define \(x_{k+1}=\frac{1}{x_k-\left\lfloor x_k \right\rfloor}, \;k\geq 0\). Now define \(a_k=\left\lfloor x_k \right\rfloor\), \(k=0,1,2,\ldots\). Then \(n=x_0=[a_0,a_1,a_2,\ldots]\). For a rough justification, note that for \(k=0,1,2,\ldots\), \(x_{k+1}=\frac{1}{x_k-a_k} \implies x_k=a_k+\cfrac{1}{x_{k+1}}\). Thus \[ x_0 =a_0+\cfrac{1}{x_1} =a_0+\cfrac{1}{a_1+\cfrac{1}{x_2}} =a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{x_3}}}=\cdots \]

Example. Let us find the continued fraction of \(\sqrt{2}\). \[ \begin{array}{ll} x_0=\sqrt{2} &\implies a_0=\left\lfloor x_0\right\rfloor=1\\ x_1=\frac{1}{x_0-\left\lfloor x_0 \right\rfloor}=\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}+1 &\implies a_1=\left\lfloor x_1\right\rfloor=2\\ x_2=\frac{1}{x_1-\left\lfloor x_1\right\rfloor}=\frac{1}{(\sqrt{2}+1)-2}=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\sqrt{2}+1 &\implies a_2=\left\lfloor x_2\right\rfloor=2\\ \hspace{4pt}\vdots & \hspace{32pt}\vdots \end{array}\] So \(\sqrt{2}=[1,2,2,2,\ldots]\).

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