Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 3 / e

Set $k=\lfloor x\rfloor$. The definition of the floor function gives $k\le x<k+1$. Adding the integer $m$ yieldsSince $k+m$ is an integer, it is exactly the floor of $x+m$. Thus

Put $k=\lfloor y^{1/n}\rfloor$. Because $y>0$, $k$ is a nonnegative integer, andThe function $t\mapsto t^n$ is strictly increasing for $t\ge 0$, so $k^n\le y<(k+1{)}^n$. Both powers are integers. Taking the floor function therefore givesTaking nonnegative $n$th roots now gives $k\le \lfloor y{\rfloor }^{1/n}<k+1$. ConsequentlyThe argument also covers $0<y<1$, when both outer floors are zero.