prove that $\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$

Question

How can I prove that if $x$ and $y$ are positive then

$$\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$$

Answer 1

First note that $f:\mathbb R\rightarrow \mathbb Z$ given by $f(x)=\lfloor x\rfloor$ is an increasing function that is the identity on the integers. Then note that for positive $x$ we have $0\leq f(x)\leq x$. With this we get $$ f(x)f(y)\leq xy $$ and applying the increasing function $f$ on both sides above noting that the left hand side is an integer we then get: $$ f(x)f(y)=f(f(x)f(y))\leq f(xy) $$ which proves the claim.

Answer 2

Assuming that you mean $\lfloor x \rfloor \lfloor \color{red}{y} \rfloor \le \lfloor xy \rfloor$, write

$$x := a + b, \quad y = c + d, \quad a,c \in \mathbb{N} \cup \{0\}, \quad b,d \in [0,1\rangle.$$

Then $\lfloor x \rfloor \lfloor y \rfloor = ac$ and

$$\lfloor xy \rfloor = \lfloor (a+b)(c+d) \rfloor = \dots$$

Answer 3

HINT: For this you need only a little more than the fact that if $x\ge 0$, then $0\le\lfloor x\rfloor\le x$ and basic facts about manipulating inequalities. Specifically, you need to realize that if $z\in\Bbb R$, $n\in\Bbb Z$, and $n\le z$, then $n\le\lfloor z\rfloor$.

Answer 4

I am using the notation $\{x\}$ to mean the fractional part of $x$. $$ x = \lfloor x \rfloor + \{x\}, \;\; y = \lfloor y \rfloor + \{y\} \\ xy = \lfloor x\rfloor \lfloor y \rfloor + \lfloor x\rfloor \{y\} + \{x\}\lfloor y\rfloor + \{x\} \{y\} \\ \lfloor xy\rfloor=\lfloor x\rfloor \lfloor y\rfloor + \lfloor\lfloor x\rfloor\{y\}\rfloor + \lfloor\{x\}\lfloor y\rfloor\rfloor\\ \text{but } x\ge0,\;\;y\ge0\\ \lfloor\lfloor x\rfloor\{y\}\rfloor + \lfloor\{x\}\lfloor y\rfloor\rfloor\ge0\\ \therefore \lfloor xy\rfloor\ge\lfloor x\rfloor\lfloor y\rfloor$$

Answer 5

$ \newcommand{floor}[1]{\left\lfloor #1 \right\rfloor} $Here is how I would write down this proof, essentially the proof from njguliyev's comment.

For all real $\;x,y \ge 0\;$, \begin{align} & \floor x \times \floor y \;\le\; \floor{x \times y} \\ \equiv & \qquad \text{"basic property of $\;\floor{\cdot}\;$, using that the left hand side is integer"} \\ & \floor x \times \floor y \;\le\; x \times y \\ \Leftarrow & \qquad \text{"arithmetic: the simplest possible strengthening, using $\;x,y \ge 0\;$"} \\ & \floor x \le x \;\land\; \floor y \le y \\ \equiv & \qquad \text{"basic property of $\;\floor{\cdot}\;$, twice"} \\ & \text{true} \\ \end{align}