If $x\le y$, prove $\lfloor{x}\rfloor \le \lfloor{y}\rfloor$

Question

I've just started my infinitesimal course, and I would like to see if I got the essence of floor right.

Let $x, y \in \mathbb R$ and $x\le y$ , prove that $\left\lfloor x \right\rfloor\le \left\lfloor y \right\rfloor$.

So, we know from the definition of floor that:

A. $\left\lfloor x \right\rfloor\le x < \left\lfloor x \right\rfloor +1$

B. $\left\lfloor y \right\rfloor \le y < \left\lfloor y \right\rfloor +1$

Now, since $x\le y$:

$\left\lfloor x \right\rfloor \le x \le y \rightarrow \left\lfloor x \right\rfloor \le y$

Also from the definition, we understand:

A. $\left\lfloor x \right\rfloor = max\{n\in \mathbb Z | n\le x\} $

B. $\left\lfloor y \right\rfloor = max\{n\in \mathbb Z | n \le y\}$

Is it okay to derive from here since $x\le y$ and $\left\lfloor x \right\rfloor \le y$

That also $\left\lfloor x \right\rfloor \le \left\lfloor y \right\rfloor$?

Thanks!

Answer 1

I would go

$$\lfloor x\rfloor\le x\le y\lt\lfloor y\rfloor+1\implies-1\lt\lfloor y\rfloor-\lfloor x\rfloor\implies0\le\lfloor y\rfloor-\lfloor x\rfloor$$

where the final implication uses the fact that $\lfloor y\rfloor-\lfloor x\rfloor$ is an integer greater than $-1$.

Answer 2

Since floor is defined as a maximum of a bounded above set of numbers, you want to use the following property of maxima of bounded above subsets of $\mathbb R$:

When $A\subseteq B$ then $\max A \le \max B$.

Now note that when $x\le y$ you have $$ \{\,n\in\mathbb Z\mid n\le x\,\} \subseteq \{\,n\in\mathbb Z\mid n\le y\,\} $$ by transitivity of $\le$ and thus the above identity yields $\lfloor x\rfloor \le \lfloor y\rfloor$.

Answer 3

Another (perhaps more direct) way too see this is to split the problem into two cases: if $y \ge \lfloor x \rfloor + 1$ then $\lfloor y \rfloor \ge \lfloor x \rfloor + 1$ since this quantity is in $\mathbb{Z}$, hence $\lfloor y \rfloor \ge \lfloor x \rfloor $.

Otherwise, $x \le y < \lfloor x \rfloor + 1$ in which case $\lfloor y \rfloor = \lfloor x \rfloor$ since this is the biggest integer smaller than $y$.

In both case we conclude that $\lfloor x \rfloor \le \lfloor y \rfloor$.