I am stuck with this problem:
Consider $A:=\{x\in \mathbb R : x > 0\}$. Let $b\in A$ an irrational and $N\in\mathbb N$. Then there are finitely many rationals $r=p/q$, with $p,q$ coprime and $0<q\leq N$, in the interval $(b-1,b+1)$.
I understand it intuitively but I am not sure that my proof is correct.
Attempt of proof:
If $r=\frac{p}{q}\in(b-1+b+1)$ with $p,q$ coprime and $0<q\leq N$, then
$$b-1\leq \frac{p}{q}\leq b+1,$$
so
$$q(b-1)\leq p \leq q(b+1).$$
Question 1: Does this mean that $p$ is an integer with a finite number of values in the interval $\big(q(b-1), q(b+1)\big)$?
Question 2: Also since $q$ is such that $0<q\leq N$,can we conclude that there are finitely many rational $r=\frac{p}{q}\in(b-1,b+1)$ with $0<q\leq N$?
Can anybody help me please? Any feedback will be appreciated. Thanks in advanced.
Best Answer
One way to see that is by counting. Look at the numbers of the form $p/2$ in that interval. They can be ordered by size, and then the distance between two consecutive such numbers is $1/2$. So in an interval of length $2$ there are at most $4$ of them.
That reasoning can be applied to each $q < N$ instead of $q=2$. Since $N$ is finite, you can estimate the number of such rational numbers by counting.
(This has nothing to do with $b$ being irrational..)
(This has also nothing to do with choosing $p$ and $q$ coprime. For fixed $q\in \mathbb{N}$, the number of different rationals of the form $p/q$ within a given bounded interval $(a,b)$ is simply bounded by $(b-a) q$, and if you look at only a finite number of such $q$ then you get your claim.)