Exercise 4.1.7 (Prove that $\lfloor x \rfloor \cdot \lfloor y \rfloor \leq \lfloor xy \rfloor$)

Question

For any positive real numbers $x$ and $y$ prove that $\lfloor x \rfloor \cdot \lfloor y \rfloor \leq \lfloor xy \rfloor$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} \begin{proof} We write $$\begin{array}{lll} x = n + u,& n \in \Z, &0 \leq u < 1,\ y = m + \mu,& m \in \Z, &0 \leq \mu < 1, \end{array} $$ so that $\lfloor x floor = n, \lfloor y floor = m$. Since $x>0$ and $y>0$, $n \geq 0$ and $m \geq 0$. Then $$\lfloor x floor \cdot \lfloor y floor = nm \leq nm + n\mu + m u + u \mu = xy.$$ By definition, $\lfloor xy floor = \max\, \{ k \in \Z \mid k \leq xy\}$, so we obtain $$ \lfloor x floor \cdot \lfloor y floor \leq \lfloor x y floor.$$ \end{proof}

Exercise 4.1.7 (Prove that $\lfloor x \rfloor \cdot \lfloor y \rfloor \leq \lfloor xy \rfloor$)

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Exercise 4.1.7 (Prove that $\lfloor x \rfloor \cdot \lfloor y \rfloor \leq \lfloor xy \rfloor$)

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