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

Question

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?

Answer 1

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.