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Prove/disprove that $\lfloor x\rfloor \leq t \iff x\leq\lfloor t\rfloor +1$

Playing around I can see why this is true, but I have no idea how to prove that, any ideas?

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  • $\begingroup$ This is most likely a typo, but note that if $x = \lfloor t \rfloor + 1$, then $\lfloor x \rfloor \le t$ won't be true. In other words, you should change the $\le$ to a $\lt$ on the right side part. Also, please show what you've already tried, and in particular have had any trouble with. Thanks. $\endgroup$ – John Omielan Feb 12 at 7:40
  • $\begingroup$ @JohnOmielan well it's not some kind of exercise, I saw that in some note somewhere and couldn't find any source for that claim, $\endgroup$ – superuser123 Feb 12 at 7:46
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I think (as noted in the comment) that it should read:

$\lfloor x\rfloor \leq t \iff x <\lfloor t\rfloor +1$.

Let $m =\lfloor x\rfloor$, then $m \in \mathbb Z$ and $m \le x <m+1.$ If $m \le t$, then $\lfloor t\rfloor \ge m$, hence $\lfloor t\rfloor+1 \ge m+1 >x.$

It is your turn to prove the reversed implication.

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