Pigeonhole proof of the existence of two numbers with given sum

Question

Let $|W|=m+1$ and $W$ be a subset of $X=\{1,2,3,\dots ,2m\}$ ($m$ is any natural number). Prove there exists two numbers in $W$ whose sum is $2m+1$.

Can anyone give me a hint to prove this? I know I should be using the pigeonhole principle.

Answer 1

There are $m$ disjoint pairs of elements $a,b\in{X}$ such that $a+b=2m+1$:

• $1,2m-0$
• $2,2m-1$
• $3,2m-2$
• $\dots$
• $m,m+1$

Since $W$ consists of $m+1$ elements, at least one of these pairs must be in it.