How are the following inequalities resolved? (Using Properties of floor function)
$$a \in \mathbb{R}$$ $$\lfloor x \rfloor > a \\ \lfloor x \rfloor < a \\ \lfloor x \rfloor \geq a \\ \lfloor x \rfloor \leq a$$
I know that :
$$x-1< \lfloor x \rfloor\leq x \\\lfloor x \rfloor \leq x<\lfloor x \rfloor+1 \\\ 0\leq x- \lfloor x \rfloor<1$$
I can not use it . please help me !
Let's find the values of $x$ that satisfy $\lfloor x \rfloor > a$. We consider two case: $a$ is an integer, and $a$ is not an integer (and use $\lceil a \rceil=\lfloor a \rfloor+1$ when $a$ is not an integer).
When $a$ is an integer then $\lfloor x \rfloor > a \implies x \ge a+1$. On the other hand, when $a$ is not an integer then $\lfloor x \rfloor > a \implies x \ge \lceil a \rceil=\lfloor a \rfloor+1$. Combining both these results, we get $$\lfloor x \rfloor > a \implies x \ge \lfloor a \rfloor+1. \tag 1$$
I leave the calculations for the remaining inequalities to you. In particular, show that $$\lfloor x \rfloor < a \implies x < \lceil a \rceil, \tag 2$$ $$\lfloor x \rfloor \ge a \implies x \ge \lceil a \rceil, \tag 3$$ and $$\lfloor x \rfloor \le a \implies x < \lfloor a \rfloor+1. \tag 4$$