Does there exist an identity for multiplying two floor math expressions?

Question

If $e_1$ and $e_2$ are arbitrary real-valued expressions, then is there an algebraic identity relating $\lfloor e_1 \rfloor \lfloor e_2 \rfloor$ and $\lfloor e_1 e_2 \rfloor$? Is it true that $\lfloor e_1 \rfloor \lfloor e_2 \rfloor = \lfloor e_1 e_2 \rfloor ?$ I think no, but what is the proof?

I tried/attempted to google the answer for this question for days. I found many webpages that were talking about the definition, properties, identities and equivalents of the floor function, but none answered my question specifically.

Answer 1

In a technical sense there can be no identity relating these that does not use equivalently powerful notation. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\floor#1{\lfloor#1\rfloor} $ The reason is that any such identity would also give an identity relating $n \floor{x}$ and $\floor{nx}$ for integer $n$ and real $x$, but $\floor{nx}$ can attain $n$ possible values from $n \floor{x}$ to $n (\floor{x}+1) - 1$ as $x$ varies.

However, if we define $fr(x) = x-\floor{x}$ for any real $x$, then we can in fact get the following identity relating $\floor{xy}$ and $\floor{x}\floor{y}$:

$\floor{xy} = \floor{(\floor{x}+fr(x))(\floor{y}+fr(y))} = \floor{x}\floor{y}+\floor{fr(y)\floor{x}+fr(x)\floor{y}+fr(x)fr(y)}$
  for any reals $x,y$.

You may object that it is not a nice identity. That is true, which is likely why nobody bothered to write it down anywhere you have looked, until now. But at least it gives an algebraic explanation for the inequality:

$\floor{xy} \ge \floor{x}\floor{y}$ for any reals $x,y \ge 0$.

It also algebraically gives the other side of the inequality (since $fr(x),fr(y)<1$):

$\floor{xy} < \floor{x}\floor{y}+\floor{x}+\floor{y}+1$ and hence $\floor{xy} \le \floor{x}\floor{y}+\floor{x}+\floor{y}$.

Answer 2

You are asking in particular whether $\lfloor e_1 \rfloor \lfloor e_2 \rfloor = \lfloor e_1 e_2 \rfloor$. This is not true in general, as you suspected. You asked for a proof, but all that you need to see that this is not true in general is a counterexample, and it is easy to find one. For example

$$\lfloor 1.9\rfloor\cdot\lfloor 1.9\rfloor=1\cdot1=1\ne3=\lfloor 3.61\rfloor=\lfloor 1.9\cdot1.9\rfloor.$$ On the other hand, $$\lfloor 1.1\rfloor\cdot\lfloor 1.1\rfloor=1\cdot1=1=\lfloor 1.21\rfloor=\lfloor 1.1\cdot1.1\rfloor.$$ From this you see that no easy formula can be expected.