TITLE: When $\lfloor{ab}\rfloor = \lfloor{a}\rfloor\lfloor{b}\rfloor$
QUESTION [13 upvotes]: I was just thinking about the equality (for a bit of fun)

$\lfloor ab \rfloor = \lfloor a\rfloor\lfloor b \rfloor$

for non-integer $a$ and $b$. I was wondering if anyone could point me to some lecture notes or something such where this features, since I can't find anything on google. Alternatively, is there some trivial condition that I'm overlooking? I've been scribbling away for a while just writing some things down and was wondering if anybody could tell me anything about the conditions to place on that equality?
My thought process was as follows: let $a = a_0.a_1a_2a_3...$ and $b=b_0.b_1b_2b_3...$. From this, we can write
$$a = a_0 + \sum_{n=1}^k \frac{a_k}{10^n}\ \ \  \text{and}\ \ \ b = b_0 + \sum_{m=1}^j \frac{b_m}{10^m}$$
and then 
\begin{align}
ab &= \left(a_0 + \sum_{n=1}^k \frac{a_k}{10^n}\right) \left(b_0 + \sum_{m=1}^j \frac{b_m}{10^m}\right)\\
&= a_0b_0 + b_0\sum_{n=1}^k \frac{a_k}{10^n} + a_0\sum_{m=1}^j \frac{b_m}{10^m} + \left(\sum_{n=1}^k \frac{a_k}{10^n}\right)\left(\sum_{m=1}^j \frac{b_m}{10^m}\right)
\end{align}
and then we simply have that 

$\lfloor a b \rfloor = \lfloor a \rfloor \lfloor b\rfloor$ iff 
  $$0 < b_0\sum_{n=1}^k \frac{a_k}{10^n} + a_0\sum_{m=1}^j \frac{b_m}{10^m} + \left(\sum_{n=1}^k \frac{a_k}{10^n}\right)\left(\sum_{m=1}^j \frac{b_m}{10^m}\right) <1$$

... more concsely, see @MorganRogers' answer for some better use of notation.

REPLY [3 votes]: In the first quadrant, in the square tile $[A,A+1)\times[B,B+1)$, we have
$$\lfloor a\rfloor \lfloor b\rfloor=AB$$ and $$AB\le ab<(A+1)(B+1).$$
We are looking for the points such that
$$AB\le ab<AB+1,$$ which form a subset of the tile delimited by the equilateral hyperbola $$ab=AB+1.$$
For all tiles such that $AB=0$, the solutions cover the area $$0\le ab<1,$$
i.e. the space between the axis and $ab=1$.
For the other tiles, the intersections of the hyperbola with the edges are
$$\begin{align}
a&=A\to &Ab=AB+1\to &b=B+\frac1A,\\
a&=A+1\to&(A+1)b=AB+1\to &b=\frac{AB+1}{A+1}\le B,\\
b&=B\to &aB=AB+1\to &b=A+\frac1B,\\
b&=B+1\to &a(B+1)=AB+1\to &b=\frac{AB+1}{B+1}\le A.\\
\end{align}$$
So the solution area is the inside of the curvilinear triangle with straight sides $a=A,b=B$ and the branch of hyperbola $ab=AB+1$, from $(A,B+1/A)$ to $(A+1/B,B)$. The area of the triangle is roughly proportional to $1/AB$.
The study will be similar for the other quadrants.