Can rationals be approximated by increasingly large-denominator rationals

Question

Lets denote by $A_n$ the set of all relatively prime fractions with denominator $n\,.$ Then one should observe that the members of $A_n$ become more densely populated in the real line as $n \to \infty$. To see this, note that $$A_1 = \mathbb{Z}, A_2 = \Big\{\frac{1}{2}, \frac{3}{2}, \dots,\Big\},\dots,A_{56} = \Big\{\frac{1}{56},\frac{3}{56}, \frac{5}{56}, \dots\Big\},\; \dots$$

Thus, it would seem intuitively true that the statement I have proposed is true, but seeing how I could find $N \in \mathbb{N}$ so that this is true is elusive to me. Any suggestions here?

Answer 1

The way you formalised it, the answer is “yes”, but for way less trivial reasons than one might expect.

Let $x=\frac pq$ and $\epsilon$ be given. Then $\frac an$ is a valid (=in shortest terms and within $\epsilon$) approximation for $x$ iff $\frac{a-kn}n$ is a valid approximation for $x-k$, $k\in \Bbb Z$. Therefore, wlog $0\le x<1$, i.e., $0\le p<q$.

Under certain circumstances, let us replace $x$ and $\epsilon$ with “better” $x’$ and $\epsilon’$ with $0<x’<1$ and $(x’-\epsilon’, x’+\epsilon’)\subseteq (x-\epsilon,x+\epsilon)$. Then a valid approximation for $x’$ is also a valid approximation for $x$. Obviously, it suffices to specify suitable $x’$ with $x<x’<\min\{1,x+\epsilon\}$ as $\epsilon’=\epsilon +x-x’$ will work.

• If $p=0$ (i.e., $x=0$), pick $k>\frac1\epsilon$ and let $x’=\frac2{2k+1}$.
• If $p=1$, then $q\ge2$. Pick $k>\max\{q(q-1),\frac1\epsilon\}$ and let $x^\prime=x+\frac1k=\frac{q+k}{qk}$ (not necessarily in shortest terms). Then indeed $x<x^\prime<x+\epsilon$ and $q>\frac1{x^\prime}>\frac1{\frac1q+\frac1{q(q-1)}}=q-1$ so that the numerator of $x^\prime$ is $>1$ (and also $x^\prime<1$).

By this, we may assume wlog that $p\ge2$. Then $\frac1x$ is between two naturals, $k<\frac1x<k+1$. Then we may assume wlog that $\frac1{k+1}<x-\epsilon<x+\epsilon<\frac1k$. In particular, our $\epsilon$-interval now contains no number with numerator $1$.

Let $\delta=\frac\epsilon x>0$. By the Prime Number Theorem, there exists $m_0$ such that for all $m\ge m_0$, there exists a prime between $m$ and $(1+\delta)m$. Pick $N>\max\{\frac {m_0}x,\frac1\epsilon\}$. We want to show that for every $n\ge N$, there exists a valid approximation with denominator $n$: Let $n\ge N$ and set $m=\lfloor xn\rfloor$. Then $m\ge m_0$ and there exists a prime $a$ between $m$ and $(1+\delta)m$. Then $x-\epsilon<x-\frac1n\le\frac an<(1+\delta)x=x+\epsilon$, as desired. If $\frac an$ is not in lowest terms, we can cancel the prime $a$ completely and end up with a reciprocal of a natural, which cannot be in our $\epsilon$-interval. Hence $\frac an$ is in lowest term as desired, thus completing the proof.