Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
2  
Do you mean $\lfloor x\rfloor \lfloor y\rfloor$? –  Dan Oct 4 '13 at 9:24
3  
$x \ge \lfloor x\rfloor$ and $y\ge \lfloor y\rfloor$ imply $xy \ge \lfloor x\rfloor \lfloor y\rfloor$. –  njguliyev Oct 4 '13 at 9:25
    
@Dan yes, I edit it. –  user95733 Oct 4 '13 at 9:27

5 Answers 5

up vote 10 down vote accepted

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.

share|improve this answer
    
(+1) nice answer. –  Mhenni Benghorbal Oct 4 '13 at 11:48

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 $\lfloor x \rfloor \lfloor y \rfloor = ab$ and

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

share|improve this answer
1  
I would personally write $x\equiv\lfloor x \rfloor+b$ and $y\equiv \lfloor y \rfloor+d$ –  Mobius Pizza Oct 4 '13 at 9:38
    
@MobiusPizza I would go for what njguliyev wrote in comments, but this seemed a bit easier to comprehend to anyone having a problem with stuff like this, as it avoids getting expressions like $\lfloor \lfloor x \rfloor + b \rfloor$, i.e., $\lfloor \cdot \rfloor$ inside $\lfloor \cdot \rfloor$. That kind of stuff tends to confuse readers. –  Vedran Šego Oct 4 '13 at 9:59
    
(+1) nice solution. –  Mhenni Benghorbal Oct 4 '13 at 11:49

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 $n\le\lfloor z\rfloor$.

share|improve this answer
    
I am not sure I get the 'you only need'. Do you really mean that any $f:\mathbb R^+\rightarrow \mathbb R^+$ so that $f(x)\leq x$ will have $f(x)f(y)\leq f(xy)$? –  String Oct 4 '13 at 9:31
    
@String: It was overstated; I must have been thinking that the righthand side was $xy$. I’ve fixed it now. –  Brian M. Scott Oct 4 '13 at 9:39
    
@BrianMScott: Great! I was just wondering. –  String Oct 4 '13 at 9:53

$ \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}

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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