Floor function inequality, for which x<y real is $\lfloor x \rfloor < \lfloor y \rfloor$?

Question

Floor function inequality, for which $x<y$ real is $\lfloor x\rfloor < \lfloor y\rfloor$ ?

Maybe I'm not thinking about this correctly but would x have to negative and y positive?

Answer 1

$0 < .5$ but $0=\lfloor 0\rfloor \not < \lfloor .5 \rfloor = 0$

So in general we need that their integer part is not the same (the part before the decimal)

Put another way, we say that $x< y$ if and only if $y-x>0$ (this works for positive and negative numbers and integers).

Answer 2

Let $\epsilon_x,\epsilon_y\in R$, and $0\leq\epsilon_x,\epsilon_y<1$, and also $a_x,a_y\in Z$ ($a_x,a_x$ are integer valued: positive, negative or zero), then:

Define: $x=a_x+\epsilon_x$, and $y=a_y+\epsilon_y$, by definition of the floor function we have: $\lfloor x \rfloor=a_x$ and $\lfloor y \rfloor=a_y$,

Now if $x<y$, then $a_x+\epsilon_x<a_y+\epsilon_y$, now we prove $\lfloor x\rfloor \leq \lfloor y\rfloor$, by contradiction:

let's assume: $\lfloor x\rfloor > \lfloor y\rfloor$, then equivalently $a_x>a_y$ (while based on the problem statement: $a_x+\epsilon_x<a_y+\epsilon_y$).

$a_x>a_y$ implies $a_x-a_y\geq 1$ (*) (because $a_x$ and $a_y$ are integers.), therefore:

$a_x+\epsilon_x<a_y+\epsilon_y$

$\Rightarrow a_x-a_y<\epsilon_y-\epsilon_x$

$\Rightarrow 1\leq a_x-a_y<\epsilon_y-\epsilon_x$

$\Rightarrow \epsilon_y-\epsilon_x\geq 1$, which is a contradiction, because $0\leq\epsilon_x,\epsilon_y<1$ and their difference can not be greater than or equal to $1$.

Therefore if $x<y$ then $a_x=\lfloor x \rfloor\leq \lfloor y \rfloor=a_y$