Compute a division with integer and fractional part

Question

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.

Answer 1

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

$$4=\left\lfloor\frac{7+5}3\right\rfloor\le\left\lfloor\frac{7+2\sqrt7}3\right\rfloor\le\left\lfloor\frac{7+6}3\right\rfloor=4$$

Answer 2

Now, prove that $$4<\frac{7+2\sqrt7}{3}<5$$

Answer 3

$$x\le \frac{7+2\sqrt{7}}{3}<x+1$$ $$3x\le 7+2\sqrt{7}<3x+3$$ $$3x-7\le 2\sqrt{7}<3x-4$$ $$\frac{3x-7}{2}\le \sqrt{7}<\frac{3x-4}{2}$$ $$\frac{9x^2-42x+49}{4}\le 7<\frac{9x^2-24x+16}{4}$$ $$9x^2-42x+49\le 28<9x^2-24x+16$$ where $x\in N$. There are integer solution is $x=4$.

Answer 4

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, $4$.

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You can also start with a tighter bracketing,

$$2.5=\frac52<\sqrt7<3,$$ justified by $$\frac{25}4<7.$$

Then

$$\frac{7+2\cdot\dfrac52}3=4<\frac{7+2\sqrt7}3<\frac{7+2\cdot3}3<5.$$