1
$\begingroup$

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...

$\endgroup$
  • 1
    $\begingroup$ So $bxc=\lfloor x\rfloor$? $\endgroup$ – Thomas Andrews Mar 31 '15 at 2:37
  • 1
    $\begingroup$ That would be the floor function $\endgroup$ – ASKASK Mar 31 '15 at 2:38
  • $\begingroup$ The d and e would be the ceiling function $\endgroup$ – ASKASK Mar 31 '15 at 2:38
  • 1
    $\begingroup$ The "ceiling function" basically rounds up (as opposed to the floor function, which rounds down). For example, $\lceil3.142\rceil=4$, $\lceil2.718\rceil=3$, $\lceil1.618\rceil=2$, $\lceil6\rceil=6$, etc. Note that, if $k$ is an integer, then $\lfloor k\rfloor=\lceil k\rceil$! If $k$ isn't an integer, than $\lfloor k\rfloor+1=\lceil k\rceil$. The ceiling function can be defined as "the smallest integer $k$ such that $k\ge x$." $\endgroup$ – Akiva Weinberger Mar 31 '15 at 3:04
  • 1
    $\begingroup$ It gets annoying with negative numbers. $\lfloor-2.7\rfloor=-3$, even though it looks like $-2$ should be the integer part. This can be explained by going back to the definition: "the greatest integer $k$ such that $k\le x$." Since $-3<-2.7<-2$, the floor of $-2.7$ is $\lfloor-2.7\rfloor=-3$, and the ceiling is $\lceil-2.7\rceil=-2$. $\endgroup$ – Akiva Weinberger Mar 31 '15 at 3:06
2
$\begingroup$

$a)$ 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 $r_1+r_2\ge 2\sqrt{r_1r_2}\ge r_1r_2$.

$\endgroup$
  • $\begingroup$ sorry but I'm completely lost with what just happened. I just "learned" what floor and ceiling functions are so excuse me if I got this wrong. For the first one, how can you use a counter example of 1.5 if you don't know the function? I mean 1.5 * 1.5 would give you 3, and on the other side it would give you 1? is that how you went about it? $\endgroup$ – Marc-Andre Leclair Mar 31 '15 at 2:59
  • $\begingroup$ And for b), how did you come to use r1 and r2? also why would it be important to highlight the fact that x1 and y1 are part of the positive integer set? Are you doing so to make x and y rational numbers? $\endgroup$ – Marc-Andre Leclair Mar 31 '15 at 3:01
  • $\begingroup$ @Marc-AndreLeclair $\lfloor 1.5\cdot 1.5\rfloor =\lfloor 2.25\rfloor=2\not\leq \lfloor 1.5\rfloor \lfloor 1.5\rfloor = 1\cdot 1 = 1$, so the inequality $a)$ doesnt hold for every $x,y\in\mathbb R^+$. The statement is false and citing a counter-example is enough to disprove it. $\endgroup$ – user26486 Mar 31 '15 at 3:03
  • $\begingroup$ I meant 2.25, sorry, in my tired state I just did a mental addition. $\endgroup$ – Marc-Andre Leclair Mar 31 '15 at 3:05
  • $\begingroup$ @Marc-AndreLeclair If $x=x_1-r_1,y=y_1-r_2, (x_1,y_1\in\mathbb Z^+), 0\le r_1,r_2< 1$, then $\lceil x\rceil =x_1$ and $\lceil y\rceil =y_1$. The answer used the fact that $x_1,y_1\ge 1$ by stating $-r_2x_1-r_1y_1\le -r_2-r_1$, and the answer also used the fact that $r_1+r_2\ge 2\sqrt{r_1r_2}\iff (\sqrt{r_1}-\sqrt{r_2})^2\ge 0$ and the fact that $2\sqrt{r_1r_2}\ge r_1r_2$, which uses $0\le r_1r_2<1$. $\endgroup$ – user26486 Mar 31 '15 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.