Today, I have to write a proof to the following arguement:
For every $x,y \in \mathbb{R} \hspace{0.1cm}$ , $x\le y \implies \lfloor x\rfloor \le \lfloor \:y\rfloor \:$
So, I tried to write something
The proof:
Let $x,y\in \mathbb{R}, \hspace{0.1cm} x\le y$.
According to the properties of the floored value (I can reference them like this) :
$\hspace{0.1cm} x-1 < \lfloor x \rfloor \leq x$
$\hspace{0.1cm} y-1 < \lfloor y \rfloor \leq y$
$x\le y \iff x-1\le y-1$
So, I don't have much clue what to do next from here. I see that the right side of inequality no.2 $\ge $ the right side of inequality no.1. $(x\le y)$
And that the left side of inequality no.2 $\ge $ the left side of inequality no.1 $(x-1\le y-1)$
So I think I can "infer" that the middle term of ineqality no.2 $\ge$ the middle term of inequlity no.1 ($x\le y \rightarrow \lfloor x\rfloor \le \lfloor \:y\rfloor\:$).
But of course, I can't "infer" in a proof, and I don't think I could say such a thing (if it's even correct).
Is there something I miss?
Thank you for reading.
$\lfloor x \rfloor \le x \le y$
$\lfloor x \rfloor$ is an integer that is less than or equal to $y$
The greatest integer that is less than or equal to $y$ must be greater than or equal to $\lfloor x \rfloor$
$\implies \lfloor y \rfloor$ must be greater than or equal to $\lfloor x \rfloor$
$\implies \lfloor x \rfloor \le \lfloor y \rfloor$