if we choose 14 different numbers from the following set 1, 2, 3, 4,...,20, then ...

Question

Using the pigeonhole principle, prove that if we choose 14 different numbers from the following set {1, 2, 3, 4,...,20}, then definitely there are two numbers such as a and b (among our 14 selected numbers) which their difference is at least 7 (i.e. |a-b| ≥7)

Answer 1

Assume there exist $14$ different numbers that we can choose such that the smallest difference is $6$, or that there are $6$ numbers between the largest and smallest. Then, by the pidgeonhole princple, we must have chosen at least one number twice, since $14 > 6$. This is a contradiction, since all of our numbers are unique. Hence, we must have a difference of at least $7$.

Redo the phrasing since it's a bit convoluted but this is the main concept of the pidgeonhole principle through a contradiction.