Compute a division with integer and fractional part I have a problem that I don't know how to solve:
Compute $[\frac{\sqrt{7}}{frac(\sqrt{7})}]$ 
Here's what I've tried: $[\sqrt{7}]=2 \rightarrow frac(\sqrt{7}) = \sqrt{7}-2 \rightarrow [\frac{\sqrt{7}}{frac(\sqrt{7})}] =[\frac{\sqrt{7}}{\sqrt{7}-2}] = [\frac{\sqrt{7}(\sqrt{7}+2)}{7-4}] = [\frac{7+2\sqrt{7}}{3}]$
But from here I don't know what to do anymore.
 A: It is immediate that $2<\sqrt7<3$, which yields
$$\frac{11}3<\frac{7+2\sqrt7}3<\frac{13}3$$ and the answer is one of $3$ or $4$.
So let us evaluate
$$\frac{7+2\sqrt7}3-4=\frac{2\sqrt7-5}3.$$
It is clear that the numerator is positive, by
$$2\sqrt7>5\iff 28>25.$$
Hence, $4$.

You can also start with a tighter bracketing,
$$2.5=\frac52<\sqrt7<3,$$ justified by $$\frac{25}4<7.$$
Then 
$$\frac{7+2\cdot\dfrac52}3=4<\frac{7+2\sqrt7}3<\frac{7+2\cdot3}3<5.$$
A: In some way you need to estimate $2\sqrt7$. First observe that $2\sqrt7=\sqrt{4\cdot 7}=\sqrt{28}$, thus $5=\sqrt{25}<\sqrt{28}<\sqrt{36}=6$, so
$$4=\left\lfloor\frac{7+5}3\right\rfloor\le\left\lfloor\frac{7+2\sqrt7}3\right\rfloor\le\left\lfloor\frac{7+6}3\right\rfloor=4$$
A: $$x\le \frac{7+2\sqrt{7}}{3}<x+1$$
$$3x\le 7+2\sqrt{7}<3x+3$$
$$3x-7\le 2\sqrt{7}<3x-4$$
$$\frac{3x-7}{2}\le \sqrt{7}<\frac{3x-4}{2}$$
$$\frac{9x^2-42x+49}{4}\le 7<\frac{9x^2-24x+16}{4}$$
$$9x^2-42x+49\le 28<9x^2-24x+16$$
where $x\in N$. There are integer solution is $x=4$.
A: Now, prove that
$$4<\frac{7+2\sqrt7}{3}<5$$
