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