Elementary fact about positive integers and rational numbers I am puzzled by the following result: Let $a$ be irrational number. Let $\epsilon > 0$ be given. We know by the Archimedean property that we can choose $q \in \mathbb{N}$ so that $ \frac{1}{q} < \epsilon $. Let $I = (0,1)$.  Then why does it follow that in any open interval centered at $a$ and contained within $I$ there are fewer than $ 1 + 2 + ... + (q-1) = \frac{q(q-1)}{2} $ positive rational numbers less than $1$ and of the form $\frac{m}{n} $ where $ 0 < m,n , n < q$
I am really not understanding why this must be true. Would be really helpful if someone can help see through it. 
 A: (By request of the OP.)
Suppose that $a\in(0,1)$; what’s the largest open interval $B_a$ centred at $a$ and contained in $(0,1)$? If $a\le\frac12$ it’s $(0,2a)$, and if $a\ge\frac12$ it’s $(1-2a,1)$. Let 
$$Q=\left\{\frac{m}n:0<m<n\le q\right\}\;;$$
we’re interested in the size of $B_a\cap Q$. Moreover, we want a bound that’s independent of $a$, so we might as well look at the case in which $B_a$ is as large as possible: when $a=\frac12$, $B_a=(0,1)$, so $B_{1/2}=Q$. Thus, we really just want an upper bound on $|Q|$.
Fix a possible denominator $n$, i.e., one satisfying $2\le n\le q$. The associated numerators are the integers $1,2,\ldots,n-1$, so there are $n-1$ fractions $\frac{m}n$ in $Q$. Summing over the possible values of $n$, we get
$$\sum_{n=2}^q(n-1)=\sum_{k=1}^{q-1}k=\frac{q(q-1)}2\;.\tag{1}$$
That would be the size of $Q$ if all of these fractions $\frac{m}n$ represented distinct rational numbers, but for $q\ge 4$ they don’t, because $\frac12=\frac24$. And of course if $a\ne\frac12$ the interval $B_a$ need not contain all of $Q$. Still, we can be sure from $(1)$ that
$$|B_a\cap Q|\le|Q|\le\frac{q(q-1)}2\;,$$
which is what we wanted to prove.
A: There are exactly $1+2+\ldots +(q-1)=\frac{q(q-1)}{2}$ ordered pairs $(m.,n)$ with $0<m<n\le q$. Of course each fraction $\frac mn$ with $0<\frac mn<1$ and $n\le q$ gives rise to such a pair, hence thare are at most $\frac{q(q-1)}{2}$ such fractions in the interval $(0,1)$ or any subinterval thereof.
We make a transition form at most to less than as soon as either the subinterval leaves out one of the fractions ($\frac1q$ and $\frac{q-1}q$ are the first candidates to get dropped this way), or as soon as two ordered pairs correspond to the same rational; this happens for the first time with $(1,2)$ and $(2,4)$ both representing $\frac12$, i.e. it happens as soon as $q\ge 4$. Indeed for $q=3$, there are not less than, but precisely $\frac{q(q-1)}2=3$ fractions $\frac13,\frac12,\frac23$ in $(0,1)$  and all sufficiently large subintervals.
