There is a number that occurs exactly once in $a_1/1, a_2/2, ...$ where $a_1,a_2,...$ is an increasing sequence in $(0,1)$. What I want to prove is that for every sequence between $1$ and $0$ where terms increase, that there is an $n$-th term that when divided by $n$ isn't equal to any kth term divided by $k$, where $k$ is not equal to $n$.
In other words, let $a_1<a_2<a_3<a_4<\cdots$ be an infinite sequence of real numbers in the interval $(0, 1)$. Show that there exists a number that occurs exactly once in the sequence $\frac{a_1}{1}, \frac{a_2}{2}, \frac{a_3}{3}, ... $
I've shown that this is true if there is a largest term in the second sequence that is equal to terms further down in the sequence, or, that there is a largest n such that $a_n=a_k$, where $k>n$. This is because this implies there is a number an that is equal to an infinite number of terms further down in the sequence. Because if $\frac{a_n}{n}=\frac{a_k}{k}, a_k=a_n+\frac{(k-n)(a_n)}{n}$, the terms further down in the sequence that are equal to $\frac{a_n}{n}$ will strictly increase with each term, and thus approach infinity, which is outside the range.
With this being said, how do I prove it for all sequences that satisfy the conditions?
 A: For two distinct indices $i$ and $j$, call $i$ and $j$ friends to each other if $a_i/i=a_j/j$.  We are asked to show there is an index that has no friends.
If there exists an index $i$ such that $i$ has infinitely many friends $f_1, f_2,\cdots$, then $a_{f_k}=f_k\frac{a_i}i$ goes to infinity when $f_k$ goes to infinity. However, all $a_*$ are smaller than $1$. Hence no index has infinitely many friends.
Towards contradiction, suppose every index has at least one friend.
Let us construct a sequence $q_0, p_1, q_1, p_2, q_2,\cdots$ such that $p_k$ and $q_k$ are friends, $k\le p_k< q_k$, and $q_{k-1}+1< p_{k+1}\le q_k+1$  for all $k\ge1$.

*

*Let $q_0=p_1=1$. Let $q_1$ be the largest friend of $p_1$.


*Suppose we have selected $q_0, p_1, q_1, \cdots, p_n, q_n$ such that

*

*$p_k< q_k$ are friends, $k\le p_k$ for $k=1,2,\cdots,n$,

*$q_{k-1}+1< p_{k+1}\le q_k+1$  for $k=1,2,\cdots, n-1$, and

*any friend of an index $i\le q_{n-1}+1$ must be $\le q_n$.

Consider set $$\{j\in\Bbb N\mid j\ge q_n+1, \exists i\in(q_{n-1}+1, q_n+1]\text{ such that } i\text{ and } j \text{ are friends}\},$$
which is not empty since

*

*either $q_n+1$ has a friend in $(q_{n-1}+1, q_n+1]$, meaning $q_n+1$ is in the set

*or any friend of $q_n+1$ is not in $(q_{n-1}+1, q_n+1]$. Since "any friend of an index $i\le q_{n-1}+1$ must be $\le q_n$", any friend of $q_n+1$ cannot $\le q_{n-1}+1$. So all friends of $q_n+1$ are $\ge q_n+1$, hence in the set.

Let $q_{n+1}$ be the maximum element in that set and $p_{n+1}$ be a friend of $q_{n+1}$ in $(q_{n-1}+1, q_n+1]$. (If there are several choices for $p_{n+1}$, pick any.)
Then we have selected $q_0, p_1, q_1, \cdots, p_{n+1}, q_{n+1}$ such that

*

*$p_k< q_k$ are friends, $k\le p_k$ for $k=1,2,\cdots,n+1$,

*$q_{k-1}+1< p_{k+1}\le q_k+1$  for $k=1,2,\cdots, n$ and

*any friend of an index $i\le q_{n}+1$ must be $\le q_{n+1}$.



Since $a_*$ is increasing and $q_{k-1}<q_{k-1}+1< p_{k+1}$, we know $\frac{a_{p_{k+1}}}{a_{q_{k-1}}}>1$.
$$\begin{aligned}
\frac{a_{q_{2n+1}}}{a_{p_1}}&=\prod_{i=1}^{n+1}\frac{a_{q_{2i-1}}}{a_{p_{2i-1}}}\prod_{i=1}^{n}\frac{a_{p_{2i+1}}}{a_{q_{2i-1}}}\ge\prod_{i=1}^{n+1}\frac{a_{q_{2i-1}}}{a_{p_{2i-1}}}
=\prod_{i=1}^{n+1}\frac{q_{2i-1}}{p_{2i-1}}\\
\frac{a_{q_{2n}}}{a_{p_2}}&=\prod_{i=1}^{n}\frac{a_{q_{2i}}}{a_{p_{2i}}}\prod_{i=1}^{n-1}\frac{a_{p_{2i+2}}}{a_{q_{2i}}}\ge\prod_{i=1}^{n}\frac{a_{q_{2i}}}{a_{p_{2i}}}
=\prod_{i=1}^{n}\frac{q_{2i}}{p_{2i}}\\
\left(\frac{a_{q_{2n}}}{a_{p_2}}\right)^5
&\ge\prod_{i=1}^n\left(\frac{q_{2i}}{p_{2i}}\right)^5
>\prod_{i=1}^n\frac{q_{2i}+1}{p_{2i}-1}
\ge\prod_{i=1}^n\frac{p_{2i+1}}{q_{2i-1}}\\
\frac{a_{q_{2n+1}}}{a_{p_1}}\left(\frac{a_{q_{2n}}}{a_{p_2}}\right)^5&\ge\prod_{i=1}^{n+1}\frac {q_{2i-1}}{p_{2i-1}}\prod_{i=1}^n\frac{p_{2i+1}}{q_{2i-1}}=\frac {q_{2n+1}}{p_1}
\end{aligned}$$
The strict inequality on the second line from last above comes from the lemma below.
The last inequality can be rearranged as
$$a_{q_{2n+1}}{a_{q_{2n}}}^5\ge \frac{a_{p_1}{a_{p_2}}^5}{p_1}q_{2n+1}.$$
While the LHS is bounded by $1\cdot1^5=1$, the RHS goes to infinity when $n$ go to infinity since $q_{2n+1}\ge2n+1$. This is a contradiction.

Lemma: If integers $y>x\ge2$, then $(\frac yx)^5\gt\frac{y+1}{x-1}$.
Proof: Multiply $(\frac yx)^2\ge(\frac y{y-1})^2>\frac{y+1}y$, $\ \ (\frac yx)^2\ge(\frac{x+1}{x})^2>\frac{x}{x-1}$ and $\ \frac yx=\frac yx$. $\quad\checkmark$
