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I need help with a proof for the following:

Show that the subset sum property (below) holds for 11, for example, given 6 distinct natural numbers between 1 and 10, there must exist a pair which sum to 11. Then propose and prove a generalization of the subset sum property that holds for an arbitrary n. [Note that the generalized result is vacuous for n<3.]

Subset sum property: Given any six distinct natural numbers between 1 and 9, there must be a pair that sums to 10. To see this, we divide the numbers from 1 to 9 into five sets: {1,9}, {2,8}, {3,7}, {4,6}, {5}. As we have six distinct numbers, at least two of them must belong to the same set. But all of the sets in this partition containing at least two distinct numbers consist of two distinct numbers that sum to 10.

We were given a hint to 1) State a generalization of the subset property 2) List subsets of {1,..,n} and 3) Apply the Pigeonhole Principle and conclude my generalization.

My generalization is that for an arbitrary number n, (besides n<3), the # of distinct natural numbers one must pick from 1 to (n-1) is (the floor of n/2) + 1. But I'm not sure how to prove this by applying the PHP

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    Where are you stuck? Have you listed out the subsets of {1 ... 10} that sum to 11? – 2017-01-30
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    Well, just write out more pairs. {1,10}{2,9}{3,8}{4,7}{5,6}. As you have six distinct, two in the same pair. To extend to sum is 12 do the following pairs {1,11}{2,10}{3,9},{4,8}{5,7}{6} but now you *can* find six distinct. But you can't find 7. Can you generalize this? – 2017-01-30

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