If $N\in \mathbb N$ and $b$ irrational, why are there finitely many rationals $r=p/q$, with $p,q$ coprime & $0<q\leq N$, in the interval $(b-1,b+1)$?

Question

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.

Answer 1

There's a lot of obfuscation and apparently irrelevant information. But I think they put it in there to make the proof easier (in their minds....)

If $b$ is irrational than $b\pm 1$ are not integers and between $b$ and $b+1$ there is exactly $1$ integer (it is $\lceil b \rceil$) and between $b-1$ and $b$ there is exactly one integer (it is $\lfloor b \rfloor$), so between $b-1$ and $b+1$ there is exactly $2$ integers.

And if $b$ is irrational then $q(b-1)$ and $q(b+1)$ are not integers and between those two points there are exactly $2q$ integers. [Because between $bq$ and $bq+q$ there are $q$ integers ($\lceil bq \rceil$, $\lceil bq+1 \rceil, ..... \lceil bq + (q-1)\rceil$) and between $bq-q$ there are $q$ integers ($\lfloor bq \rfloor$, $\lfloor bq+1 \rfloor, ..... \lfloor bq + (q-1)\rfloor$

SO the total number of rational numbers of the form $\frac pq$ where $p, q: q\le N$ are relatively prime is less than or equal to the number of integers in each of the intervals $(q(b-1), q(b+1))$ and for each interval there are exactly $2q$ integers so the total number of integers is $\sum_{q=1}^N 2q$ and the number of rational numbers of the $p,1: q\le N$ are relatively prime are less that $\sum_{q=1}^N 2q$ which is clearly finite.

.....

But we don't need all that superfluous information.

Given any interval $(u, w)$ there are only a finite number of integers. Why? Actually how much detail one needs to go to depends on the level of mathematics. Argument one: The distance between two integers is $1$ so if any interval with $N$ integers must have a length of at least $N\times 1 = N$. So any interval with infinite integers in it must be infinite length. So any finite interval has a finite number if integers. Argument two: Let $(u,w)$ be an interval. Let $N_1 < u$ and $N_2 > w$. Then all integers in the interval are all between $N_1$ and $N_2$ and there are only a finite number of such intervals. Such arguments are not actually proofs as they make assumptions that must be proven.

So for any interval $(u,v)$ there are a finite number of integers in it. Call that number $M_1$.

For any $\frac pq$ so that $q \le$ then $u < \frac pq < w \iff qu < p < qw$. And in the interval $(qu,qw)$ there are only a finite number of integers in it. Call that number $M_q$.

So the total number of $p::q$ pairs so that $u < \frac pq < w$ and $q \le N$ is $M_1 + M_2 + ..... +M_N$ which is clearly finite. And we also need $p,q$ relatively prime there are even fewer.

Answer 2

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.)