Can rationals be approximated by increasingly large-denominator rationals? For the purposes of this question, assume all rationals discussed in this post are in simplest form. Suppose $\frac{p}{q} \in \mathbb{Q}$ is in simplest form. The question is whether you can find $N \in \mathbb{N}$ so that there is another simplest form rational $\frac{a}{n}$ within $\varepsilon$ of $\frac{p}{q}$ for every $n \geq N$. Formally, the statement is proposed in the following way:
$$\forall \, \frac{p}{q} \in \mathbb{Q}, \gcd(p,q)=1, \forall \, \varepsilon >0, \exists \, N \in \mathbb{N}, \forall \, n \geq N, \exists \, \frac{a}{n} \in \Big( \frac{p}{q} - \varepsilon, \frac{p}{q} + \varepsilon\Big)$$
Now, if $A_n$ is the set of all relatively prime fractions with denominator $n$, 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 $A_1 = \mathbb{Z}, A_2 = \{\frac{1}{2}, \frac{3}{2}, \dots,\}$, $\dots$, $A_{56} = \{\frac{1}{56},\frac{3}{56}, \frac{5}{56}, \dots\}, \dots$
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?
 A: 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.
A: 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 $\infty\,.$
$$\tag*{$\Box
$}
\quad
$$
A: 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 $n>\frac{1}{2 \epsilon}$.
