We chose n + 2 numbers from the set 1, 2, ..., 3n. Prove that there are always two among the chosen numbers whose difference is more than n but less than 2n.

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- August 5th 2011, 03:48 AMalexmahonePigeon-hole Principle
We chose n + 2 numbers from the set 1, 2, ..., 3n. Prove that there are always two among the chosen numbers whose difference is more than n but less than 2n.

- August 5th 2011, 06:37 AMAlso sprach ZarathustraRe: Pigeon-hole Principle
The solution is from the net, in "my words",http://www.worldscibooks.com/etextbo...177_chap01.pdf

page: 16, #11

First, denote the largest chosen number from as , a can be .

Second, we add to our all chosen numbers, that would not change the result:

Say, if and are chosen ones, ; then by adding to and we will still get .

So by adding a to the largest chosen, , we will get .

Now, we have two cases, one is if one of other chosen numbers is from ,

if so, we are done, if , then .

The second case is if the other n+1 chosen numbers are from and .

Now, consider the following pairs:

(If is a asome above pair then is from and is from )

So, we have pairs, but we choose numbers, then there must to be one pair with two chosen elements.

Say, that is , , ,

- August 5th 2011, 06:53 AMalexmahoneRe: Pigeon-hole Principle
How come "a" doesn't figure in the rest of the solution?

For example, how do you consider the pair (1, 2n)? Shouldn't the smallest number of the set now be "a"? - August 5th 2011, 07:03 AMAlso sprach ZarathustraRe: Pigeon-hole Principle
- August 5th 2011, 07:06 AMalexmahoneRe: Pigeon-hole Principle
- August 5th 2011, 07:20 AMAlso sprach ZarathustraRe: Pigeon-hole Principle