Select $11$ different numbers from $f\{1,2,...,20\}$. Prove that two of your numbers, $a$ and $b$, will differ by two.
Clearly this is an application of the pigeonhole principle. However, I'm not sure how to write up a coherent proof.
2026-04-01 13:14:24.1775049264
Application of pigeonhole principle
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Consider the following sets: $$\{1, 3\}, \{2, 4\}, \{5, 7\}, \{6, 8\}, \{9, 11\}, \{10, 12\}, \{13, 15\}, \{14, 16\}, \{17, 19\}, \{18, 20\}$$
Together, these $10$ sets account for all of the integers $\{1, \ldots, 20\}$. When we pick 11 numbers, by the Pigeonhole Principle, we will pick both numbers from at least one of the sets. Hence, these two numbers (which we can denote $a$ and $b$) will differ by two.
Hope this helps. Cheers!