To determine a fraction

Question

How to calculate the fraction $\frac{a}{b}$ with $0<b\leq 7,$ such that $\vert \sqrt 3-\frac{a}{b}\vert\leq \frac{1}{8b}$

Answer 1

Multiplying by $b$ throughout we get $$ \left| b\sqrt{3} - a \right| \le \frac18 \iff -\frac18 \le b\sqrt{3} - a \le \frac18 $$ Can you finish this now?

Answer 2

Truncated continued fractions provide good rational approximations.

The continued fraction for $\sqrt3$ is $1+\dfrac1{1+\dfrac1{2+\dfrac1{1+\dfrac1{2+\dfrac1{1+...}}}}},$

so good rational approximations to $\sqrt3$ are

$1, 1+1, 1+\dfrac1{1+\dfrac12}, \color{blue}{1+\dfrac1{1+\dfrac1{2+1}}}, \color{purple}{1+\dfrac1{1+\dfrac1{2+\dfrac1{1+1}}}}$, etc.

It can be seen that $\color{blue}{\dfrac74}$ and $\color{purple}{\dfrac{12}7}$ are good enough rational approximations to meet the desired criterion.