Prove the following statement:
In any set of 26 integers chosen from the set of (1,2,3, ...50) there must be at least a pair of numbers such that one of them divides the other.
(In reply to re(2): @ Steve - toward reasonable proof
by Ady TZIDON)
We are not quite on the same page. My "inelegant" proof is already solid. I have shown that it is impossible to pick 26 numbers without picking more than once from one of my sets. That is all that is needed for a solid proof.
I am not claiming, and I do not need to claim that "it is possible to choose one number from each of those subsets, creating a new set of 15 numbers, none of them being divisor of any other."
As a small bonus, it follows that any valid set of 25 numbers must contain the 10 numbers that are not in any of my sets, namely (29,31,33,35,37,39,43,45,47). If I had constructed my sets differently I could have come up with more numbers that must be in a valid set of 25. For instance, if my sets included (14,42) and (13,39) instead of (14,28) and (13,26) instead of then it would follow that that 26 and 28 must be in any valid set. So there are at least 12 numbers that must be in any valid set, namely (26, 28,29,31,33,35,37,39,43,45,47). I would not be surprised if there were more.