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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$ .

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

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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.

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  • $\begingroup$ @LC thank you for your answer(+1) $\endgroup$ – user356595 Sep 19 '16 at 18:16
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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 $\lfloor x\rfloor=n<n+1\leq \lfloor y\rfloor$.

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

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