Given $n_o \in \mathbb N$ , are there only finitely many rationals in the interval , such that denominator in reduced form does not exceed $n_o$? How to show that , given $n_o \in \mathbb N$ for any real $a,b$ with $0 \le a < b$  , there are only finitely many rationals $x \in (a,b)$ such that $x=m/n$ , g.c.d. $(m,n)=1$ , and $0<n \le n_0$ ? 
 A: The rational numbers of the form $\frac{a}{N}$ with $\gcd(a,N)=1$ in the interval $(0,1)$ are exactly $\varphi(N)$,
so the number of rational numbers in $(0,1)$ with a denominator $\leq N_0$ are given by:
$$ \sum_{N=2}^{N_0} \varphi(N) \approx \frac{3}{\pi^2} N_0^2, $$
see Farey sequence.
A: Without loss of generality, assume that m is an integer. Then all we have to do is show that for any value of n such that $0 < n \le n_0$, m must lie within a fixed range.

$x = \frac{m}{n}$ and $x \in (a,b)$, then

$a \le \frac{m}{n} \le b$

which implies,

$an \le m \le bn$
And as m is an integer, therefore only a finite number of values of m is possible within this range (an, bn), excluding also values which are not coprime with n. Also note that there is only a finite number of possible values of n as well.
Hence there is only a finite number of rationals of the form $\frac{m}{n}$ with the range (a,b) under the given conditions. 
A: There are clearly only finitely many denominators between 1 and $n_{0}$. For each of these denominators $n$ the numerator must be at least $a n$ and no smaller than $b n$.
