Exercise 7.4.1 (Some expansions of quadratic numbers)

Question

Expand each of the following as infinite simple continued fractions: $\sqrt{2},\ \sqrt{2} - 1,\ \sqrt{2}/2,\ \sqrt{3},\ 1/\sqrt{3}$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
Put $\xi = \sqrt{2} - 1$. Then 
$$\xi = \sqrt{2} - 1= \frac{1}{\sqrt{2}+1} = \frac{1}{2 + (\sqrt{2} - 1)} = \frac{1}{2 + \xi}.$$
The algorithm of section 7.4 shows that
$$\xi = \sqrt{2} - 1 =\langle 0,2,2,2,\ldots angle,$$
so
$$\sqrt{2} =1 + \xi = \langle 1,2,2,2,2,\ldots angle = \langle 1, (2)* angle.$$
For a more formal proof, put $\eta = 1/\xi = \sqrt{2} + 1$. Then $\eta= 2 + \frac{1}{\eta}$. If $\eta = \langle a_0,a_1,\ldots,a_n,\ldots angle$, then
\begin{align*}
\langle a_0,a_1,\ldots,a_n,\ldots angle &= 2 + \frac{1}{\langle a_0,a_1,\ldots,a_n,\ldots angle} \
&= \langle 2, \langle a_0,a_1,\ldots,a_n angle angle\
&= \langle 2, a_0,a_1,\ldots,a_n,\ldots angle.
\end{align*}
The unicity of the expansion gives $2 = a_0 = a_1 = a_2 = \cdots$, so 
$$\sqrt{2} + 1 = \langle 2 ,2,2,2, \ldots angle.$$
This gives
$$\sqrt{2} = \langle 1,2,2,2,2,\ldots angle = \langle 1, (2)* angle,$$
and
$$\sqrt{2}- 1 =  \langle 0,2,2,2,2,\ldots angle = \langle 0, (2)* angle.$$
Moreover
$$\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} = 0 + \frac{1}{\langle 1,2,2,2,2,\ldots angle  }  = \langle 0,1,2,2,2,2,\ldots angle =  \langle 0,1, (2)* angle.$$

\bigskip
Put $\zeta = \sqrt{3} + 1$. Then
\begin{align*}
\zeta &= \sqrt{3} + 1\
&=2 + (\sqrt{3} - 1)\
&= 2 + \frac{1}{\left(\frac{\sqrt{3}+1}{2} ight)}\
&=2 + \frac{1}{1 + \frac{\sqrt{3}-1}{2}}\
&= 2 + \frac{1}{1 + \frac{1}{\sqrt{3} + 1}}\
&= 2 + \frac{1}{1 + \frac{1}{\zeta}}.
\end{align*}
If $\zeta = \langle a_0,a_1,\ldots,a_n,\ldots angle$, then
\begin{align*}
\langle a_0,a_1,\ldots,a_n,\ldots angle &= 2 + \frac{1}{1 + \frac{1}{\langle a_0,a_1,\ldots,a_n,\ldots angle}}\
&=2 + \frac{1}{\langle 1,a_0,a_1,\ldots,a_n,\ldots angle}\
&=\langle 2,1,a_0,a_1,\ldots,a_n,\ldotsangle.
\end{align*}
The unicity of the expansion gives $a_{2k} = 1$ and $a_{2k+1} = 1$ for all $k \in \N$. Therefore
$$\sqrt{3}+ 1 = \langle 2,1,2,1,\ldots angle = \langle (2,1)* angle,$$
so
$$\sqrt{3} = \langle 1,1,2,1,2,\ldots angle = \langle 1, (1,2)* angle,$$
and
$$\frac{1}{\sqrt{3}}= 0 + \frac{1}{\langle 1,1,2,1,2,\ldots angle} = \langle 0, 1, 1, 2, 1,2,\ldots, angle = \langle 0, 1, (1,2)*angle.$$

Verification:
\begin{verbatim}
sage: continued_fraction(1/sqrt(3))
[0; 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...]
\end{verbatim}
\end{proof}

Exercise 7.4.1 (Some expansions of quadratic numbers)

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Exercise 7.4.1 (Some expansions of quadratic numbers)

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