# Ernesto's question at Yahoo! Answers regarding finding the simple continued fraction for sqrt(3)

#### MarkFL

Staff member
Here is the question:

Simple continued fraction of √3?

Find the simple ( all numerators have to be 1) continued fraction of √3 (square root of three)

Explain.

Thanks.
I have posted a link there to this thread so the OP can view my work.

#### MarkFL

Staff member
Hello Ernesto,

The minimal polynomial having $$\displaystyle 1+\sqrt{3}$$ as a root is:

$$\displaystyle x^2-2x-2=0$$

Using this as our characteristic equation, we obtain the recursion:

(1) $$\displaystyle A_{n+1}=2A_{n}+2A_{n-1}$$

Where we will have:

$$\displaystyle \lim_{n\to\infty}\left(\frac{A_{n+1}}{A_{n}} \right)=\sqrt{3}+1$$

Therefore, we may approximate $\sqrt{3}$ with:

$$\displaystyle \sqrt{3}\approx\frac{A_{n+1}}{A_{n}}-1=\frac{A_{n+1}-A_{n}}{A_{n}}$$

Using our recursion in (1), we may write:

$$\displaystyle \sqrt{3}\approx\frac{A_{n}+2A_{n-1}}{A_{n}}=1+\frac{1}{\dfrac{A_{n}}{2A_{n-1}}}$$

Now, since the recursion in (1) may be written in $A_{n}$ as:

$$\displaystyle A_{n}=2A_{n-1}+2A_{n-2}$$

We may write:

$$\displaystyle \sqrt{3}\approx1+\frac{1}{\dfrac{2A_{n-1}+2A_{n-2}}{2A_{n-1}}}=1+\frac{1}{1+\dfrac{1}{\dfrac{A_{n-1}}{A_{n-2}}}}$$

Using the recursion for $$\displaystyle A_{n-1}$$ we may state:

$$\displaystyle \sqrt{3}\approx1+\frac{1}{1+\dfrac{1}{\dfrac{2A_{n-2}+2A_{n-3}}{A_{n-2}}}}=1+\frac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{A_{n-2}}{2A_{n-3}}}}}$$

Hence, we will find that by allowing $n\to\infty$ and repeating the above process ad infinitum, we will obtain:

$$\displaystyle \sqrt{3}=1+\dfrac{1}{1+\dfrac{1}{2+ \dfrac{1}{1+\dfrac{1}{2+\cdots}}}}$$

Written in linear form, we have:

$$\displaystyle \sqrt{3}=\left[1;\overline{1,2} \right]$$

#### soroban

##### Well-known member
Hello, Erne4sto!

Here is a primitive solution.
It requires a calculator and some stamina.

Find the simple continued fraction for $$\sqrt{3}.$$

$$\sqrt{3} \;=\; 1 + 0.732050808$$

. . . .$$=\;1 + \frac{1}{1 + 0.366025404}$$

. . . .$$=\;1 + \frac{1}{1 + \dfrac{1}{2 +0.732050808}}$$

. . . .$$=\;1 + \frac{1}{1 + \dfrac{1}{2 + \dfrac{1}{1+0.366025404}}}$$

. . . .$$=\;1 + \frac{1}{1 + \dfrac{1}{2 + \dfrac{1}{1+\dfrac{1}{2 + 0.732050808}}}}$$

We note the repeating pattern: .$$1,1,2,1,2,1,2,\;.\;.\;.$$

Therefore: .$$\sqrt{3} \;=\;[1,\overline{1,2}]$$