# If $x\lt y$ then prove that $\lfloor x\rfloor \leq \lfloor y\rfloor$.

If $x\lt y$ then we have to show $\lfloor x\rfloor \leq \lfloor y\rfloor$ for any real numbers $x,y$.

I have proceeded in the following way: for any real numbers $x,y$,

$x-1\lt \lfloor x\rfloor \leq x\lt \lfloor x\rfloor+1$ and

$y-1\lt \lfloor y\rfloor \leq y \lt \lfloor y\rfloor+1$. From these we have $\lfloor x\rfloor \lt y$ and $\lfloor y\rfloor \leq y$ .

Im stuck here. What can I do after this to prove the result?

By definition, $\lfloor y\rfloor$ is the largest integer $\leq y$, and similarly for $\lfloor x\rfloor$.

Then $\lfloor x\rfloor$ is an integer $\leq x\leq y$. Since $\lfloor y\rfloor$ is the largest such, we get the result.

• @LC thank you for your answer(+1) – user356595 Sep 19 '16 at 18:16

Let $x< y$ and let $n=\lfloor x\rfloor$. We have two cases:

1) if $y<n+1$ then $\lfloor x\rfloor=n=\lfloor y\rfloor$;

2) if $y\geq n+1$ then $\lfloor x\rfloor=n<n+1\leq \lfloor y\rfloor$.

• s thank you for your answer in different way(+1) – user356595 Sep 19 '16 at 18:19