<s>[INST] $a, b$ are the integer part and decimal fraction of $\sqrt7$ find integer part of $\frac{a}{b}$ 
$a, b$ are the integer part and decimal fraction of $\sqrt7$
find integer part of $\frac{a}{b}$

using calculator :
$\sqrt 7$ = 2.645
$\frac{a}{b} = \frac{2}{0.6} = \frac{20}{6} = 3.333$
integer part of $\frac{a}{b} = 3$
how to do it without calculator? if i don't know $\sqrt7$ part
 [/INST] My solution without a calculator:
Since $2^2 = 4$ and $3^2=9$, we know that $\sqrt{7} = 2 +b $ and therefore
$$
\frac{a}{b} = \frac{2}{b} = \frac{2}{\sqrt{7}-2} = \frac{2}{(\sqrt{7}-2)} \frac{(\sqrt{7}+2)}{(\sqrt{7}+2)} = \frac{2(\sqrt{7}+2)}{7-4} = \frac{2}{3} \left(\sqrt{7}+2 \right)
$$
And since $2 < \sqrt{7} < 3$, we know that (plugging these into the expression)
$$
\frac{8}{3} < \frac{2}{3} \left(\sqrt{7}+2 \right) < \frac{10}{3}
$$
which indicates that the integer part of $a/b$ is $3$.
Edit: the lower bound is not tight enough. But this can be fixed.
First observation:
$
4\cdot 7 = 28 > 25 \Rightarrow \sqrt{7}> \frac{5}{2}
$
Then: 
$$
\left[ \frac{2}{3}\left( \sqrt{7}+2 \right) \right]^2 = \frac{4}{9} (11+4 \sqrt{7}) > \frac{4}{9} (11+4 \cdot \frac{5}{2}) = \frac{4}{9}\cdot 21 = \frac{28}{3}> \frac{27}{3} = 9
$$
so therefore $\frac{2}{3} \left(\sqrt{7}+2 \right) > 3$.
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