If $\displaystyle (n + 1)$ integers are chosen from integers $\displaystyle 1,2,3,\mbox{...},2n$, show that, among the chosen integers, there are two, such that one of them divides the other.

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- May 27th 2009, 09:37 AMfardeen_genCombinatorics - show result?
If $\displaystyle (n + 1)$ integers are chosen from integers $\displaystyle 1,2,3,\mbox{...},2n$, show that, among the chosen integers, there are two, such that one of them divides the other.

- May 27th 2009, 05:23 PMDeadstar
I cant remember exactly how to do this but what you have to note is that any integer can be written in the form $\displaystyle a \cdot 2^k$ where $\displaystyle k \geq 0$ and a is odd. Note also that in this case a has n values since there is n odd and n even integers between 1 and 2n.

Now you have to use the pigeonhole principle. How exactly i cant remember. But im guessing since there n values of a and you choose n+1 integers you'll end up at the fact that one 'hole' has 2 'pigeons' in it. - May 27th 2009, 05:24 PMSampras
You could probably use induction.

*Proof.*Base case: For $\displaystyle n=1 $, we choose $\displaystyle 2 $ integers of $\displaystyle 1,2 $. And $\displaystyle 1|2 $. Inductive step: Suppose that $\displaystyle k+1 $ integers are chosen from the integers $\displaystyle 1,2,3, \dots, 2k $ where $\displaystyle k \in \mathbb{N} $ and there are two, such that one divides the other. Now append $\displaystyle 2k+1 $ and $\displaystyle 2k+2 $. Then if we choose $\displaystyle k+2 $ of the integers we know have to use the Pigeonhole principle. You could have a case where you choose the $\displaystyle k+2 $ integers such that the integer in the inductive hypothesis are contained. Or could have both numbers not contained. Finally you could have one contained and the other not. $\displaystyle \square $