$n$ is also given to be an even number.
So I want to prove this by using the pigeon-hole principle.
I would partition the numbers $\{1, 2, . . . , 2n\}$ into $n$ boxes with numbers in each as follows : $$\{1,2\}, \{3,4\},……,\{2n-1,2n \}$$
if $n+1$ numbers are to be chosen from these $n$ boxes, How to show that there will be remaining numbers that differ by 2 ?
Let's say we have $n=4$ then we would have
$$\{1,2\} ,\{3,4\}, \{5,6\}, \{7,8 \}$$
Now if we $5$ numbers are chosen from the above, then we want to show that there are two of the three numbers remaining that differ by $2$
Here I have $4$ holes and $5$ pigeons, so one of these pairs must be chosen when choosing the $5$ numbers , But still I can't argue that we will be left with 2 numbers that differ by 2.
ANy suggestions ?
Best Answer
Why not instead consider: $$\{1,3\},\{2,4\},\{5,7\},\{6,8\},\ldots$$
Alternately:
Pick an element from the set. Now there is at least one element that is ineligible to be picked next, as it is 2 away from the first element. Proceed by induction.