ceiling-and-floor-functionsdiscrete mathematicsfunctionsinequality
Prove or disprove the statements below.
(a) For all positive real numbers x and y, $\lfloor x \cdot y\rfloor ≤ \lfloor x\rfloor \cdot \lfloor y\rfloor $.
(b) For all positive real numbers x and y, $ \lceil x \cdot y\rceil ≤ \lceil x\rceil\cdot \lceil y\rceil $ .
The brackets were not covered in my class and I can't seem to find them on google so I can't even start the problem as I can't figure out their meanings…
The first is the correct: you round "down" (i.e. the greatest integer LESS THAN $-0.8$).
In contrast, the ceiling function rounds "up" to the least integer GREATER THAN $-0.8 = 0$.
$$ \begin{align} \lfloor{-0.8}\rfloor & = -1\quad & \text{since}\;\; \color{blue}{\bf -1} \lt -0.8 \lt 0 \\ \\ \lceil {-0.8} \rceil & = 0\quad &\text{since} \;\; -1 \lt -0.8 \lt \color{blue}{\bf 0} \end{align}$$
In general, we must have that $$\lfloor x \rfloor \leq x\leq \lceil x \rceil\quad \forall x \in \mathbb R$$
And so it follows that $$-1 = \lfloor -0.8 \rfloor \leq -0.8 \leq \lceil -0.8 \rceil = 0$$
K.Stm's suggestion is a nice, intuitive way to recall the relation between the floor and the ceiling of a real number $x$, especially when $x\lt 0$. Using the "number line" idea and plotting $-0.8$ with the two closest integers that "sandwich" $-0.8$ gives us:
$\qquad\qquad$
We see that the floor of $x= -0.8$ is the first integer immediately to the left of $-0.8,\;$ and the ceiling of $x= -0.8$ is the first integer immediately to the right of $-0.8$, and this strategy can be used, whatever the value of a real number $x$.
Floor and ceiling functions are usually defined as
$$ \lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\} $$ and $$ \lceil x \rceil=\min\,\{n\in\mathbb{Z}\mid n\ge x\} $$ for $x\in\mathbb R$ (see Floor and ceiling functions for more details).
Best Answer
$a)$ is false (counter-example $(x,y)=(1.5,1.5)$).
$b)$ is true. Let $x=x_1-r_1,y=y_1-r_2, (x_1,y_1\in\mathbb Z^+), 0\le r_1,r_2< 1$.
$$\lceil xy\rceil=\lceil x_1y_1-r_2x_1-r_1y_1+r_1r_2\rceil\le \lceil x_1y_1 -r_1-r_2+r_1r_2\rceil\le \lceil x_1y_1\rceil=x_1y_1$$
This used $r_1+r_2\ge 2\sqrt{r_1r_2}\ge r_1r_2$.