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

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?

• 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. – John Omielan Feb 12 at 7:40
• @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, – superuser123 Feb 12 at 7:46

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