Floor function of a product

Question

I'm reading a book about proofs and I'm currently stuck in this problem.

Prove that for all real numbers $x$ and $y$ we have that: $$\lfloor x\rfloor \lfloor y\rfloor \leq \lfloor xy\rfloor \leq \lfloor x\rfloor \lfloor y \rfloor + \lfloor x \rfloor + \lfloor y \rfloor$$

I though I could do it by cases, considering when the number is a pure integer and when is an integer plus some real number. But by doing this I end up with having a lot of cases to show. Is there any better, simpler and clever approach? Thank you!

PS: I end up with a lot of cases because there is a point where I will have to consider "subcases" of cases specifically when the integer part is multiplying with the positive "rest" less than one

Answer 1

I shall assume that $x, y \ge 0$. We also have $\lfloor x\rfloor, \lfloor y\rfloor \ge 0.$

Write $x = \lfloor x\rfloor + \{x\}$ and $y = \lfloor y\rfloor + \{y\}.$ Clearly, $0 \le \{x\}, \{y\} < 1$.

Note that $$xy =\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor\{y\} + \lfloor y\rfloor\{x\} + \{x\}\{y\}. \quad (*)$$

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Using the fact that all the rightmost three terms on the RHS of $(*)$ are nonnegative, we see that $$xy \ge \lfloor x\rfloor\lfloor y\rfloor.$$ Note that the RHS is an integer which is lesser than $xy$. By definition of floor, $\lfloor xy\rfloor$ must be the greatest such integer. Thus, we have $$\lfloor xy\rfloor \ge \lfloor x\rfloor\lfloor y\rfloor,$$ giving us the left inequality.

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Using $(*)$ and the fact that $\{x\}, \{y\} < 1$, we see that $$xy \le \lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y\rfloor + \{x\}\{y\}.$$ Since $\lfloor .\rfloor$ is an increasing function, we see that $$\lfloor xy\rfloor \le \lfloor\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \{x\}\{y\}\rfloor. \quad (**)$$ Note that $\lfloor n + z\rfloor = n + \lfloor z\rfloor$ for any $n \in \Bbb Z$ and $z \in \Bbb R$. Using this, the RHS of $(**)$ can be written as \begin{align} &\lfloor\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \{x\}\{y\}\rfloor\\ =&\lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor+ \lfloor\{x\}\{y\}\rfloor\\ =& \lfloor x\rfloor\lfloor y\rfloor + \lfloor x\rfloor + \lfloor y \rfloor, \quad (\because 0 \le \{x\}\{y\} < 1) \end{align}

giving us the right inequality.

Answer 2

It can help to write $x=a+r$ where $a$ is an integer and $0\leq r <1$, so that $\lfloor x \rfloor = a.$ Similarly, let $y=b+s$ where $b$ is an integer and $0\leq s <1.$

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

$$\lfloor x y \rfloor = \lfloor ab+as + br +sr \rfloor = ab +\lfloor as + br +sr \rfloor.$$

(Assuming $x$ and $y$ are positive,) the last term above has to be greater than $0$, so that's your first inequality.

Then take that last term, and since $r$ and $s$ are less than one:

$$\lfloor as + br +sr \rfloor \leq \lfloor a + b +sr \rfloor = a + b + \lfloor sr \rfloor.$$

The last term is $0$, so there's your second inequality.