I am trying to understand this question. This is from the book Essential Discrete Mathematics for Computer Science and Question is from Chapter one "the pigeonhole principle" So clearly we have to prove using pigeonhole principle
But lets say I choose the numbers $1, 2, 3,$ and $60$ these numbers are clearly within $1$ and $60$ inclusive but no two numbers differ by at most $19$. So what am I missing here?
In fact, since $\ 2-1=3-2=1\ $, and $\ 3-1=2\ $ there are three pairs of your numbers that differ by much less than $19$, and hence also by at most $19$.
Let $\ a,b,c,d\ $ be four distinct numbers between $1$ and $60$ arranged in increasing order:$\ a<b<c<d\ $. Then, since $\ d\le60\ $ and $\ a\ge1\ $, we must have $\ d-a\le59\ $. Let $\ m=\min(b-a,c-b,d-c)\ $. Then \begin{align} b-a&\ge m\ ,\\ c-b&\ge m\ \text{, and}\\ d-c&\ge m\ . \end{align} If we add these inequalities, we get $\ 59\ge d-a\ge3m\ $, which gives $\ m\le\frac{59}{3}<20\ $, and since $\ m\ $ must be an integer, it can be at most $\ 19\ $. It follows that at least one of the pairs $\ \{a,b\}, \{b,c\},\ $ and $\ \{c,d\}\ $ of numbers differs by at most $\ 19\ $.