Show that $[x]$ is monotonic increasing

Question

$[x]$ is defined as the biggest integer which is $\leq x$. I don't know how I can prove this because I don't know how I can write the assignment rule. $x\rightarrow \text{ integer a for which: } x-1<a\leq x$.

I have to Show if $a>b$ then $[a]\geq[b]$ I have first tried a contradiction proof by assuming that $[a]<[b]$. But it resulted to nothing.

I have tried a constructive proof then:

We know $a>b$. Then we can directly say that $b-1<a-1<[a]\leq a$ Also we know that $ [b]<a$. I know that if $b-1<[b]\leq a-1$. Then $[b]<[a]$, but how can I proof that if $a-1<b<a$ then $[a]=[b]$? Is there an easy proof that there can not be two integers at the same time in this interval?

Answer 1

So, according to your definition, $$[x] = \max \{ n \in \Bbb{Z} : n \le x \}$$ Note that if $x \le y$, then $$\{ n \in \Bbb{Z} : n \le x \} \subseteq \{ n \in \Bbb{Z} : n \le y \}.$$ (Prove this for yourself!). Therefore, the max of the RHS, $[y]$, is an upper bound on the LHS. Since $[x]$ is the least upper bound on the LHS, we have $[x] \le [y]$.

Answer 2

Let $a,b \in \mathbb{R}$ with $a < b$. By definition, $[a]$ is the largest integer such that $[a] \leq a$. Because $a < b$, it immediately follows that $[a] < b$. Since $[a]$ is an integer with $[a] \leq b$, the very definition of $[b]$ ensures that $[a] \leq [b]$.

Answer 3

Let $x,y\in\mathbb R$ be such that $x\le y$.

We have $\lfloor x\rfloor\le x\le y$. However, $\lfloor y\rfloor$ is the biggest integer which is $\le y$. Since $\lfloor x\rfloor\le y$ and $\lfloor x\rfloor$ is an integer, we have $\lfloor x\rfloor\le\lfloor y\rfloor$.