I'm trying to solve the following problem.
Consider a set of $n+1$ positive integers, each less than or equal to $2n$. Show there must always exist a pair of integers in the set, one dividing the other.
Here is my attempt.
Let $S = \{s_i \mid 1 \leq i \leq n+1\}$ be the set of $n+1$ positive integers, each of which is less than or equal to $2n$. By the fundamental theorem of arithmetic, we can write each $s_i$, uniquely, in the form $s_i = 2^{a_i - 1} \cdot (2b_i - 1)$ for $a_i, b_i \in \mathbb{N}$. But there are only $n$ odd numbers less than or equal to $2n$, so there must exist distinct $i,j$ such that $2b_i - 1 = 2b_j - 1$. Without loss of generality, we can assume that $a_i - 1 \geq a_j - 1$. Then $2^{a_j - 1} \mid 2^{a_i - 1}$. Specifically, $2^{a_i - 1} \cdot 2^{a_j - a_i} = 2^{a_j - 1}$, so $$ s_i \cdot 2^{a_j - a_i} = 2^{a_i - 1} \cdot (2b_i - 1) \cdot 2^{a_j - a_i} = 2^{a_j - 1} \cdot (2b_i - 1) = 2^{a_j - 1} \cdot (2b_j - 1) = s_j, $$ so $s_i \mid s_j$, as required.
How does this look?
Your answer looks perfectly fine. Just a minor typo: if you are assuming that $a_i \geq a_j$, then in your last computation you should swap $i$ and $j$ if you want to work with integers all the time.