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?

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    $\begingroup$ Hint : The rationals form a dense set. So, there are ininfite many rational numbers closer to the given rational number than $\epsilon$. The rest should be easy. $\endgroup$
    – Peter
    Jan 15, 2023 at 15:58
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    $\begingroup$ @peter : The nontrivial part is the large denominator when the approximating fraction is required to be in lowest terms. You don't want to approximate $\frac pq$ and then find you've approximated it with $\frac12$. $\endgroup$
    – MJD
    Jan 15, 2023 at 16:29
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    $\begingroup$ @Peter The "rest" is the whole game. You want to show that for every $a<b$ and sufficiently large $n$, there exists $m\in(an,bn)$ relatively prime with $n$. That is true but I wouldn't say it is very easy despite the fact that there are short proofs ;-) $\endgroup$
    – fedja
    Jan 15, 2023 at 16:46
  • $\begingroup$ I assume that the $M$ in your formal statement should be $N$. $\endgroup$ Jan 15, 2023 at 16:50
  • $\begingroup$ Consider Dirichlet approximation theorem $\endgroup$
    – TravorLZH
    Jan 15, 2023 at 17:34

3 Answers 3


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.

  • $\begingroup$ Thank you for this thoughtful response. I will come back to look at this argument later this evening when I am available. $\endgroup$ Jan 15, 2023 at 18:43
  • $\begingroup$ Nice. I hope there is a proof that needs less than the prime number theorem. I thought about convergents for continued fraction expansions of irrationals near the target, but got nowhere. $\endgroup$ Jan 15, 2023 at 20:00
  • $\begingroup$ @EthanBolker Well, „prime“ is the simplest form of the „coprime“ we actually need. If $n$ has many small prime divisors, it may be hard to find a coprime number in the required range by other means. - Farey sequences may be more promising than continued fractions. $\endgroup$ Jan 16, 2023 at 0:11
  • $\begingroup$ My answer in mathoverflow.net/questions/434707/… contains an alternative proof of the density (actually, uniform distribution) of fractions with denominator exactly $n$ as $n\to\infty$ and the remark by Fedor Petrov makes that argument even more elementary, but still far from trivial. $\endgroup$
    – fedja
    Jan 16, 2023 at 3:32

Given $n$, we have $a \approx \frac {pn}{q}$; the best $a$ is that value rounded to the nearest integer. So $E=|a - \frac{pn}{q}|\le 0.5$.

$|\frac{a}{n} - \frac {p}{q}| = \frac {1}{n}\left|a-\frac {pn}{q}\right|=\frac{E}{n}\le\frac{0.5}{n}$

If I've understood your statement correctly, then: $\frac{0.5}{n}< \epsilon$ means $n>\frac{1}{2 \epsilon}$.


Let $(x_n)$ be a sequence of rationals that converges to $x\not\in\{x_1,x_2,x_3,...\}\,.$ (This $x$ can also be irrational but it should not be part of the sequence.)

Claim. Every $x_n$ is of the form $$ x_n=\frac{p_n}{q_n}\quad\text{ with }\operatorname{gcd}(p_n,q_n)=1\quad\text{ and }\quad q_n\to+\infty\,. $$ Proof. Because the limit of $(x_n)$ is not part of the sequence the following must hold: $$ \forall n\;\exists M\;\forall m\ge M\;: x_m\not=x_n\,. $$ Otherwise, we could find a subsequence $(x_{m_j})$ that is identically equal to some $x_n$ and therefore converges to an element of the sequence.

Because $(x_n)$ must also be a Cauchy sequence we have $$\tag{1} \forall \varepsilon>0\;\exists N\;\forall m,n\ge N:|x_n-x_m|<\varepsilon\,. $$ Clearly, $x_m\not=x_n$ is equivalent to $|p_nq_m-p_mq_n|\ge 1\,$ since this is a natural number or zero. Therefore we have $$ \forall\varepsilon>0\;\exists N\;\forall n\ge N\;\exists M\;\forall m\ge M:\frac{1}{|q_nq_m|}\le \frac{|p_nq_m-p_mq_n|}{|q_nq_m|}=\Bigg|\frac{p_n}{q_n}-\frac{p_m}{q_m}\Bigg|<\varepsilon\, $$ from (1). This implies that $\frac{1}{|q_nq_m|}$ must converge to zero when $n\to\infty$ or $m\to\infty\,.$ In other words, $q_n$ must diverge to $\infty\,.$ $$\tag*{$\Box $} \quad $$

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    $\begingroup$ This gives “arbitrarily large”denominators, but not “all sufficiently large” denominators $\endgroup$ Jan 15, 2023 at 18:30
  • $\begingroup$ @HagenvonEitzen Yes. Sounds like OP asks for more in the body than in the title only. Thanks for feedback and for your other answer. $\endgroup$
    – Kurt G.
    Jan 15, 2023 at 18:39
  • $\begingroup$ This proof shows that $q_n \to \infty$ but it seems to assume from the start that each $x_n = \frac{p_n}{q_n}$. In other words, I'm not seeing how this proof proves that $x_n=\frac{p_n}{q_n}$ $\endgroup$ Jan 16, 2023 at 14:49
  • $\begingroup$ @DavidC.Huang $x_n$ is a sequence of rationals, and I write them just as you wrote your rationals in OP. $\endgroup$
    – Kurt G.
    Jan 16, 2023 at 14:52
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    $\begingroup$ @DavidC.Huang I inserted one more expression to clarify this. It is from the Cauchy condition. $\endgroup$
    – Kurt G.
    Jan 16, 2023 at 15:20

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