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$\forall x \in \Bbb R , \exists y \in \Bbb Z$ so that $\lfloor xy \rfloor = \lfloor x \rfloor \lfloor y \rfloor $.

Assume that $x \in \Bbb R, y \in \Bbb Z.$

Let $\lfloor x \rfloor = n$ for some integer $n$ such that $n \le x \lt n+1,$ and let $\lfloor y \rfloor = y$ since $y$ is an integer.

By multiplying all sides of the inequality by $y$, you get:

$ny \le xy \lt y(n+1)$

$\lfloor xy \rfloor = ny = \lfloor x \rfloor \lfloor y \rfloor$, since $n = \lfloor x \rfloor$ and $y = \lfloor y \rfloor$


As I worked through these next two, I ended doing the exact same process as the one above. Is that correct or am I making a mistake in my interpretation of one of these?


$\forall x \in \Bbb R , \exists y \in \Bbb Z$ so that $\lfloor xy \rfloor \neq \lfloor x \rfloor \lfloor y \rfloor $ rewritten as a negation is:

$\exists x \in \Bbb R$, such that $\forall y \in \Bbb Z \lfloor xy \rfloor = \lfloor x \rfloor \lfloor y \rfloor $


$\exists y \in \Bbb Z $ so that $ \forall x \in \Bbb R, \lfloor xy \rfloor \neq \lfloor x \rfloor \lfloor y \rfloor $ rewritten as a negation is:

$\forall y \in \Bbb Z, \exists x \in \Bbb R$, such that $\lfloor xy \rfloor = \lfloor x \rfloor \lfloor y \rfloor $

Travis
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  • Note that $y(n+1)\neq yn+1$ unless $y=1$. So the fact that $ny\leq xy\leq y(n+1)$ only shows that $\lfloor x\rfloor y \leq \lfloor xy\rfloor<y(\lfloor x\rfloor +1)$ – Thomas Andrews Oct 03 '14 at 01:28
  • Never mind, I misread the problem. Your arguments are still wrong, but the (first) statement is true, since it is true when $y=1$, so $\exists y$ is true. – Thomas Andrews Oct 03 '14 at 01:41
  • So for that first argument should I also assume that y = 1, or Let y = 1? Would that fix the first problem? – Travis Oct 03 '14 at 01:46

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