Let $S\subseteq \{1,2,...,3n\}$ be a set of cardinality $2n+1$. Prove there exist $x,y \in S$ such that $x|y$.

Question

Not quite sure where to approach this from. I tried using the pigeonhole principle on equivalence classes of the relation $xRy \iff x|y$ but the problem is that this relation is not an equivalence relation.

Doesn't 1 | x for any x ? Perhaps your question is not posed as you wanted it to be. — 2ndYearFreshman, Oct 16 '23 at 17:03
It is posed as I wanted it to be, since $1$ doesn't have to be an element of $S$. — Kantig Shoter, Oct 16 '23 at 17:06
Note that, whenever you choose $2n+1$ numbers from ${1,\dots,3n}$, then at least $n+1$ chosen numbers will be in ${1,\dots,2n}$. Therefore, you can apply the result of this famous problem. — Mike Earnest, Oct 16 '23 at 17:49

score 2 · Answer 1 · edited Oct 16 '23 at 17:54

2

For each $l\in \{1,\dots,3n\}$ such that $l$ is odd, let $$ S_l:=\{l,2l,2^2l,\cdots,2^\alpha l\} $$ where $\alpha$ is the largest integer such that $2^\alpha l\le 3n$. That is, $\alpha =\left\lfloor \log_2 \left(\frac{3n}{l}\right)\right\rfloor$. Note that, for any $x, y\in S_l$, we either have $x\mid y$ or $y\mid x$. As $$\{1,2,\cdots,3n\}=\mathop{\bigcup}\limits_{1\le l\le 3n, \;l\text{ is odd}}S_l, $$ and there are fewer than $2n+1$ odd numbers in $\{1,\dots,3n\}$, there must be two elements $x\ne y\in S$ such that $\exists l, x,y\in S_l$.

You can see that the number $2n+1$ is far from optimal.

edited Oct 16 '23 at 17:54

Mike Earnest

84,902

answered Oct 16 '23 at 17:09

rushusuixing

595

3

Why "There must be two elements $x\ne y\in S$ such that $\exists l, x,y\in S_l$"? – Anne Bauval Oct 16 '23 at 17:18
@AnneBauval That's because of the Pigeonhole principle. – rushusuixing Oct 16 '23 at 17:41
1

@AnneBauval There are $2n+1$ elements of $S$, and only $\lceil (3/2)n\rceil $ sets of the form $S_l$ (one for each odd number between $1$ and $3n$). Since $2n+1>\lceil (3/2)n\rceil$, PHP implies there are two members of $S$ in the same $S_l$. – Mike Earnest Oct 16 '23 at 17:45
@MikeEarnest thanks for the edition. I'm poor in English. – rushusuixing Oct 16 '23 at 18:11

Let $S\subseteq \{1,2,...,3n\}$ be a set of cardinality $2n+1$. Prove there exist $x,y \in S$ such that $x|y$.

1 Answers1