# Inequalities with floor function

I need some help with this exercise, I'm pretty new solving this exercises.

$$\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$

I know that I had to use the formal definition of the floor function, which is:

$$\lfloor x \rfloor = \max {\{ m \in \Bbb Z \mid m \le x\}}$$

• Hi @vonbrand, this a new topic to me, I just start seeing it yesterday. I tried to use the inequality property to prove $\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$ first. I use this property: $$a \le c; b \le d; a + b \le c+d$$ Commented Feb 17, 2016 at 17:33

Use that $x=\lfloor x\rfloor +\alpha,\,\alpha\in[0,1)$ and $y=\lfloor y\rfloor +\beta,\,\beta\in[0,1)$

$$\lfloor x+y\rfloor =\lfloor \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor$$ because $\lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\ge \lfloor x\rfloor+\lfloor y\rfloor$ and therefore $\lfloor x+y\rfloor$ is at least $\lfloor x\rfloor+\lfloor y\rfloor$

For the other inequality: $$\lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta < \lfloor x\rfloor+\lfloor y\rfloor+2\Rightarrow$$ From here you see that one integer $m\in\mathbb Z$ that satisfies $$(*)\quad\quad m\leq \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta$$ is $m=\lfloor x\rfloor+\lfloor y\rfloor$ and the greatest possible is $\lfloor x\rfloor+\lfloor y\rfloor+1$, because already $\lfloor x\rfloor+\lfloor y\rfloor+2>\lfloor x\rfloor+\lfloor y\rfloor+\alpha+\beta$ -see $(*)$. Therefore $$\lfloor x+y\rfloor =\lfloor \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\rfloor\leq \lfloor x\rfloor+\lfloor y\rfloor+1$$

$$\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$

Note that you can pull integers out from floor\ceil function

Hint: for three first cases

Case one: $x,y \in \mathbb{Z}$ so

$$x+y\le x+y \le x+y+1$$

Case two $x\in \mathbb{Z}$ but $y\in \mathbb{R}$ and $y\notin \mathbb{Z}$

$$x + \lfloor y \rfloor \le x+\lfloor y \rfloor \le x+\lfloor y \rfloor +1$$

Case three $y\in \mathbb{Z}$ but $x\in \mathbb{R}$ and $x\notin \mathbb{Z}$

$$y+\lfloor x \rfloor \le y +\lfloor x \rfloor \le y+1+\lfloor x \rfloor$$

Now, try to prove case four: $x,y\in \mathbb{R}$ and $x,y\notin \mathbb{Z}$

• Thank you for update and take your time to answer my question @Nehorai. Commented Feb 17, 2016 at 17:39

I do not know how rigorous or formal you need/want to be but it's straight forward that $[x]$ is the unique integer such that $[x] \le x < [x] + 1$[*]

Therefore $[x] \le x < [x] + 1$ and $[y] \le y < [y] + 1$ so $[x] + [y] \le x + y < [x] +[y] + 2$.

So there are two possible cases:

Case 1: $[x] + [y] \le x + y < [x] + [y] + 1$

This means $[x + y ] = [x] + [y]$. So $[x]+[y] = [x+y] < [x] + [y] + 1$.

Case 2: $[x] + [y] + 1 \le x + y < [x]+ [y] + 2$

This means $[x + y] = [x] + [y] + 1$. So $[x]+[y] < [x+y] = [x]+[y] + 1$.

And that's it.

[*] Rigor and formality??

By the archemedian principal, for every real $x$, there is a unique integer $n$ such that $n \le x < n+ 1$.

$n \in \mathbb Z$ and $n \le x$ so $n \in \{m \in \mathbb Z| m \le x\}$. If $m \in \mathbb Z; m > n$ then $m \ge n+1 > x$ so $m \not \in \{m \in \mathbb Z| m \le x\}$.

So $n = \max \{m \in \mathbb Z| m \le x\} = [x]$.

• Thank you for your time @fleablood. Commented Feb 17, 2016 at 17:37

As said by Nehorai, you can pull any integer from the floor function, so it suffices to validate with $x,y\in[0,1)$.

Then $$\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$

becomes

$$0\le \lfloor x + y \rfloor \le 1,$$

which is obviously true as

$$0\le x+y<2.$$