Thread: Pigeon-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.

Originally Posted by alexmahone

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.

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 , , ,

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"?

Originally Posted by alexmahone

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"?

The pair (1,2n) is ordered set that contains two elements, one from [1,n] and one from [2n,3n-1].

(1,2n) - 1 is from closed interval [1,n] ; 2n is from closed interval [2n,3n-1].

Originally Posted by Also sprach Zarathustra

The pair (1,2n) is ordered set that contains two elements, one from [1,n] and one from [2n,3n-1].

(1,2n) - 1 is from closed interval [1,n] ; 2 is from closed interval [2n,3n-1].

But I thought you added "a" to all the numbers in the second step. So shouldn't we have (1+a, 2n+a) etc?

Originally Posted by alexmahone

But I thought you added "a" to all the numbers in the second step. So shouldn't we have (1+a, 2n+a) etc?

Okay!

The adding a needed only at the first stage, the purpose was eliminate closed interval [n+1,2n-1].