I am solving this question. Suppose X is the set of the first 2n positive integer and S is any subset of X with n+1 elements. Show that S contains 2 integers such that one integer is divisible by the other. I checked the answer, it said that we let $$s=\{ x_1,x_2,x_3,...,x_{n+1}\}, x\in\{1,2,3...2n\}$$. Then, we might express the elements inside the set s to be $$x_i=2^{n_i}y_i, y_i\in\{1,2,3...,n\}$$
If $x_i$ is odd, then $n_i$ is 0. So, by Pigeonhole Principle, we would be able to find one $y_i$ that expresses two different $x_i$. Thus, one integer must be divisible by the other. However, I wonder how do we deal with the positive odd integers that between n and 2n , since we can't express them using the way that the answer did.
Thank you so much.
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Asaf Karagila
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Henry Cai
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4The representation should be $x_i = 2^{n_i} y_i$ where $y_i \in { 1, 3, 5, \dots, 2n-1 }$. – Greg Martin Oct 20 '20 at 08:02
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This makes sense. Thank you – Henry Cai Oct 20 '20 at 09:26