$a, b$ 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
 A: 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 $\frac{2}{3} \left(\sqrt{7}+2 \right) > 3$.
A: Note that $$ 2.5<\sqrt 7<3$$
$$ \frac {a}{b} = \frac {2}{\sqrt 7 -2} = \frac {2(\sqrt 7+2)}{3}$$
$$3= \frac {2( 2.5+2)}{3} <        \frac {2(\sqrt 7+2)}{3} <\frac {2( 3+2)}{3}<4$$
Thus the integer part of $\frac {2}{\sqrt 7 -2}$ is $3$
A: $4<7<9$, so the integer part is $a=2$ and thus $b=\sqrt7-2=\frac3{\sqrt7+2}$. With that
$$
\frac ab=\frac{2(\sqrt7+2)}3
$$
Now also $25<4\cdot 7=28<36$, so that one can further simplify
$$
...=\frac{4+5}3+\frac{\sqrt{28}-5}3=3+\frac1{\sqrt{28}+5}
$$
Here you can again see what the integer part is.
A: $2.6^2=6.76<7<7.29=2.7^2.$
So $a=2$ and $0.6<b<0.7.$ 
So $4>10/3=2/0.6>a/b>2/0.7=20/7>2.$ 
So the integer part of $a/b$ is $2$ or $3.$
Now $$3\le a/b\iff 3b\le a\iff  3b\le 2 \iff b\le 2/3 \iff$$ $$\iff  \sqrt 7=a+b=2+b\le 2+2/3=8/3\iff$$ $$\iff 7\le (8/3)^2=64/9 =7+1/9.$$ Since we used "$\iff$" throughout, and since it is true that $7\le (8/3)^2,$ therefore $3\le a/b.$ And we already have $a/b<4$. So we have $$3\le a/b<4.$$
