Could anybody help me to understand the solution?
$\textbf{Problem:}$Proof that $\sup(A) = \sqrt{2}$,$~~$where $A = \lbrace x \in \mathbb{Q}: x > 0, x^2 < 2 \rbrace$
For proving the right-hand side of the inequality, it says that $s$ is rational or irrational. In the irrational case, how does he find $w$?
If $s$ is irrational (in fact, also if $s$ is rational - so there is not really a need to distinguzish the cases), we can certainly find a rational number $w$ with $s<w<s+\frac 1n$. Indeed, if we only consider numbers of the form $\frac k{n+1}$, at least one of these will be in the desired interval (simply because the distance between consecutive candidates is shorter than the interval length). More formally, we wnat to find an integer $k$ with $s<\frac k{n+1}<s+\frac 1n$ or equivalently with $$(n+1)s<k<(n+1)s+\frac{n+1}n=(n+1)s+1+\frac1n.$$ One can simply let $k=\lfloor (n+1)s+1\rfloor =\lfloor (n+1)s\rfloor +1 $ because that makes $k$ an integer with, by definition, $$ (n+1)s<k\le (n+1)s+1.$$