Demonstrate by induction Let $p,q \in \mathbb{N}$, be prime among them and $p<q$.
Proposition : $\exists{n} \in \mathbb{N}\;|\; \frac{1}{n+1} \le \frac{p}{q} < \frac{1}{n}$
Demonstration :

And I'm stuck there!
Can you help me?


Hint: Start with a concrete example, $p = 7, q = 1999$. Note that $q = 285p + 4$ so 
$$
\dfrac{p}{q} = \dfrac{7}{285\cdot 7 + 4}
$$
and so, since $$q - 4 \leq q \leq q + 3,$$ you have $$285 \cdot 7 + 4 - 4 \leq q \leq 285\cdot 7 + 4 + 3$$ and thus
$$
\dfrac{7}{285\cdot 7 + 7} \leq  \dfrac{p}{q} \leq \dfrac{7}{285 \cdot 7}.
$$
What does $n$ need to be in this case?
Now adapt to a general $p,q$.
