0
$\begingroup$

I've got this exercise to prove :

if $x \leq y$ then $\lfloor x \rfloor \leq \lfloor y \rfloor$

And I want to know if logically my approach is fine:

Suppose for the sake of contradiction that $x \leq y$ and $\lfloor x \rfloor \gt \lfloor y \rfloor$ then if we assume $x = y$ then $\lfloor x \rfloor = \lfloor y \rfloor$ which is contradiction.

is this legal proof logically? Do I need to look for contradiction for the second case as well (where $ x < y$) or it's enough like that?

Thanks alot !

$\endgroup$
2
  • $\begingroup$ You only proved that if $x=y$ then $\lfloor x\rfloor\le \lfloor y\rfloor$, in other words , $\lfloor x\rfloor\le \lfloor x\rfloor$ which does not seem sufficient. $\endgroup$
    – user65203
    Commented Mar 17, 2021 at 18:10
  • $\begingroup$ But then I need to deduce that if $x < y$ then $⌊x⌋≤⌊y⌋$ which is not what I was trying to do, I was trying to solve it by contradiction $\endgroup$
    – Yarin
    Commented Mar 17, 2021 at 18:25

1 Answer 1

Reset to default
1
$\begingroup$

No, this is not sufficient as written. You need to handle the other case.

A recommendation: If $x<y$ then either $x< \left \lfloor{y}\right \rfloor $ or $\left \lfloor{y}\right \rfloor \leq x< y$. This splits your second case into two sub-cases. In my opinion, this is the easiest approach to finishing off your proof.

Edit: Just to be 100% clear, I did not finish the proof here. I am just recommending what's written be your next step.

Edit 2: I changed my two cases slightly. The earlier version was not wrong, but this way is better.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .