Exercise 7.6.1 (Find some fractions $a/b$ such that $|\sqrt{2} - a/b|

Question

Find two rational numbers $a/b$ satisfying
$$\left | \sqrt{2} - \frac{a}{b} \right| < \frac{1}{\sqrt{5}\,  b^2}.$$

Richard Ganaye · Accepted Answer

\begin{proof} 
Since
\begin{align*}
&\left | \sqrt{2} - \frac{1}{1} ight| < 0.415<0.447 <  \frac{1}{\sqrt{5} \cdot 1^2},\
&\left | \sqrt{2} - \frac{3}{2} ight| < 0.086 <  0.111 < \frac{1}{\sqrt{5} \cdot 2^2},
\end{align*}
the fractions $1/1$ and $3/2$ are suitable.

For more with Sagemath:
\begin{verbatim}
sage: dfc = continued_fraction(sqrt(2)); dfc
[1; 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...]
sage: cvg = [dfc.convergent(i) for i in range(7)]; cvg
[1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169]
sage: for f in cvg:
....:     a,b = f.numer(), f.denom()
....:     if abs(sqrt(2) - f) < 1/(sqrt(5)* b^2):
....:         print(f)
....:         
1
3/2
7/5
17/12
41/29
99/70
239/169
\end{verbatim}
So all convergents up to $h_6/k_6 = 239/169$ are suitable. With further calculations, all convergents up to $h_{200}/k_{200}$ are convenient, so we may conjecture that all convergents $a/b$ of $\sqrt{2}$ satisfy $\left | \sqrt{2} - \frac{a}{b} ight| < \frac{1}{\sqrt{5}\,  b^2}$ (to be verified).
\end{proof}

Exercise 7.6.1 (Find some fractions $a/b$ such that $|\sqrt{2} - a/b| < 1/(\sqrt{5} b^2)$)

Answers

Comments

Exercise 7.6.1 (Find some fractions $a/b$ such that $|\sqrt{2} - a/b| < 1/(\sqrt{5} b^2)$)

Answers

Comments

Add answer