Confused with proof exercise real analysis I'm sure it's easy, but I can't see why. Let $p,q,n\in\mathbb{N}$, if $p/q\in(0,1/n)$, then $1/q\in(0,1/n)$.

If $p/q\in(1,1+1/n)$, show that $1/q\in(0,1/n)$.

Attempt: Since $q\in\mathbb{N}$, $0<1/q$. Now, if $p/q\in(1,1+1/n)$, then we can write $p/q$ as 
$$
p/q = 1 + a/b,
$$
where $a/b\in(0,1/n)$, and $a/b$ is rational. By hypothesis, $1/b\in(0,1/n)$. How does one show that $1/q<1/n$? I thought in using contradiction for $1/n\leq 1/q$ but I can't find where everything breaks down.
 A: Since $1 < \frac{p}{q} < 1 + \frac{1}{n}$, we have $0 < \frac{1}{q} \le \frac{p - q}{q} < \frac{1}{n}$.
A: Might be easier to use inequality notation.  Remember $k \in (a,b)$ means nothing more or less than $a < k < b$.
ANd if we keep in mind that if $p \in \mathbb N$ that means $0 < 1 \le p$.
And if we keep in mind that if $a< b$ and $c > 0$ then $ac < bc$.
And that if $q > 0$ then $\frac 1q > 0$  then everything falls into place!
....
$\frac pq \in (0, \frac 1n)\implies$
$0 < \frac pq < \frac 1n$.
And $0 < 1 \le p$ means
$0 < 1*\frac 1q \le p*\frac 1q$ so
$0 < \frac 1q \le \frac pq$ and $0 < \frac pq < \frac 1n$ so
$0 < \frac 1q < \frac pq < \frac 1n\implies$
$0 < \frac 1q < \frac 1n\implies$
$\frac 1q \in (0, \frac 1n)$.
That's it!
And 
$\frac pq \in (1, 1 + \frac 1n) \implies$
$1 < \frac pq < 1 + \frac 1n \implies$
$0 < \frac pq - 1 < \frac 1n \implies$
$0 < \frac pq - \frac qq < \frac 1n \implies$
$0 < \frac {p-q}q < \frac 1n$.
And $0 < \frac {p-q}q$ means $0*q < \frac {p-q}q*q$ means $0 < p-q$.  But $p-q$ is an integer so $p-q \ge 1$.
So $0 < \frac 1q \le \frac {p-q}q < \frac 1n$ so
$0 < \frac 1q < \frac 1n$ so 
$\frac 1q \in (0, \frac 1n)$.
That's it.
