The Pigeonhole Principle

Theorem 1.1. If n + 1 objects are put into n boxes, then at least one box contains two or more objects. Proof. Trivial. Example 1.1. Among 13 people there are two who have their birthdays in the same month. Example 1.2. There are n married couples. How many of the 2n people must be selected in order to guarantee that one has selected a married couple? Other principles related to the pigeonhole principle: • If n objects are put into n boxes and no box is empty, then each box contains exactly one object. • If n objects are put into n boxes and no box gets more than one object, then each box has an object. The abstract formulation of the three principles: Let X and Y be finite sets and let f : X −→ Y be a function. • If X has more elements than Y , then f is not one-to-one. • If X and Y have the same number of elements and f is onto, then f is one-to-one. • If X and Y have the same number of elements and f is one-to-one, then f is onto. Example 1.3. In any group of n people there are at least two persons having the same number friends. (It is assumed that if a person x is a friend of y then y is also a friend of x.) Proof. The number of friends of a person x is an integer k with 0 ≤ k ≤ n− 1. If there is a person y whose number of friends is n− 1, then everyone is a friend of y, that is, no one has 0 friend. This means that 0 and n− 1 can not be simultaneously the numbers of friends of some people in the group. The pigeonhole principle tells us that there are at least two people having the same number of friends. Example 1.4. Given n integers a1, a2, . . . , an, not necessarily distinct, there exist integers k and l with 0 ≤ k < l ≤ n such that the sum ak+1 + ak+2 + · · ·+ al is a multiple of n. Proof. Consider the n integers a1, a1 + a2, a1 + a2 + a3, . . . , a1 + a2 + · · ·+ an. Dividing these integers by n, we have a1 + a2 + · · ·+ ai = qin + ri, 0 ≤ ri ≤ n− 1, i = 1, 2, . . . , n. If one of the remainders r1, r2, . . . , rn is zero, say, rk = 0, then a1 + a2 + · · · + ak is a multiple of n. If none of r1, r2, . . . , rn is zero, then two of them must the same (since 1 ≤ ri ≤ n− 1 for all i), say, rk = rl with k < l. This means that the two integers a1+a2+· · ·+ak and a1+a2+· · ·+al have the same remainder. Thus ak+1+ak+2+· · ·+al is a multiple of n.