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We have n+1 numbers from the set {1...2n} and prove that there are 2 such that one divides the other... I just could not understand the solutions that already exist some

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Well, we can divide your set to two separate sets: $ A=\{1,\,3\,,\,..., \,n\} $ and $ B=\{2,\,4\,,\,..., \,2n\} $.

Providing that $ n\in\mathbb{N}$ those two sets have equal sizes, and we notice that each element from $ B $ is some element of $ A $ multiplied by $ 2 $ or some other element of $B$ multiplied by $2$, hence if we remove powers of $ 2 $ from each element (that means, we divide each element named $a$ by $2^k$ where $k$ is $\lfloor\log_2a\rfloor$ and take rest) of $A$ and $B$ we will receive 2 sets, $A'$ and $B'$ such that $A'\cup B'=A'$.

Now we use pigeonhole principle - $A'$ has $n$ elements, we "mark" all of them, +1 addional, but since we have no other elements that those of $A'$ and $B'$ we have to mark one that we already have, QED.

Edit: Sets were not divided well, thanks @Henry Swanson

Annisar
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  • If $n$ is $10$, $11 \in B$, but $11$ is certainly not an element of $A$ multiplied by $2$. – Henry Swanson Nov 08 '13 at 02:26
  • that was a great observation thanks Mr. @HenrySwanson –  Nov 08 '13 at 02:35
  • could you expand a little on "... of them, +1 addional, but ..." –  Nov 08 '13 at 02:50
  • $A'$ has $n$ elements. You have to choose $n+1$ elements from $A\cup B$, but if we think about what $A'$ consists of, we see that we have only $n$ elements that do not divide one another in whole $A\cup B$. – Annisar Nov 08 '13 at 02:58
  • now I totally understand this question. thank you very very much Mr. @KamilSołtysik –  Nov 08 '13 at 03:01
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Do what the hint says.

Write each number as $2^k m$ for some odd number $m$ between $1$ and $2n - 1$.

How many odd numbers ($m$) are there here? How many total numbers are in your set?

Now apply the pigeon hole principle.

Deven Ware
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