Determine if the following is true or false and provide a proof:
$\forall x\in\mathbb{R},\exists y\in\mathbb{R}$ so that $\lfloor xy\rfloor = \lfloor x\rfloor \lfloor y\rfloor$
My attempt:
-The statement is false, the negation is as follows: $\exists x\in\mathbb{R}$, so that $\forall y\in\mathbb{R}$ $\lfloor xy\rfloor \neq \lfloor x\rfloor \lfloor y\rfloor$.
-Proof-
Let $x=10.99$ and suppose $y$ is an arbitrary but fixed real number. By the definition of floor, there exist unique integers $n$ and $m$, such that $n=\lfloor x\rfloor$ and $n\leq x < n+1$, and $m= \lfloor y\rfloor$ and $m \leq x < m+1$.
So, $10 = \lfloor 10.99\rfloor$ and $10 \leq 10.99 < 11$.
Now, $10y \leq 10.99y < 11y$ ........(stuck..??)
So I understand that what I am trying to prove is that: $\lfloor xy\rfloor \neq \lfloor x\rfloor \lfloor y\rfloor$. However I am unsure of how to correctly choose(suppose) one equation, and manipulate it enough to prove that it doesn't equate to the other. I assume the issue is with the sentence "By the definition of floor, there exists unique integers $n$ and $m$, such that $n= \lfloor x\rfloor$ and $n \leq x < n+1$, and $m= \lfloor y\rfloor$ and $m \leq x < m+1$", and whatever follows thereafter.
I am trying to figure out how to complete the proof. Any help is appreciated.
The statement you are trying to prove is true, so you are unable to disprove it.
Hint:
The statement says that for every $x$, there exists some $y$. So all you need to do is find some particular $y$ for which the equality holds.
Try to prove the statement by using very simple numbers for $y$....