Prove $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$

Question

Prove the following for all real $x$

i. $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$

ii. $⌊x⌋-2⌊x/2⌋$ is equal to either $0$ or $1$

For ($ii$.) I attempted to split it into cases of whether the fraction part {$x$} is $≥.5$ or $<5$ but that ended up being too tedious and I know there must be a more elegant, simpler method.

For ($i$) I tried cases like in part ($ii$) but since there are 2 variables that would lead to 4 cases.

An elegant and easy solution will be much appreciated

Answer 1

For (i), define $\{x\} = x - \lfloor x \rfloor$. Then \begin{align*} \lfloor 2x \rfloor + \lfloor 2y \rfloor = 2\lfloor x\rfloor + 2\lfloor y \rfloor + \lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor \end{align*} and \begin{align*} \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x + y \rfloor = 2\lfloor x\rfloor + 2\lfloor y \rfloor + \lfloor \{x\} + \{y\} \rfloor \end{align*} So the inequality to be proven is equivalent to \begin{align*} \lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor \ge \lfloor \{x\} + \{y\} \rfloor \end{align*} But this is true, since \begin{align*} \lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor \ge \lfloor 2\max(\{x\}, \{y\})\rfloor \ge \lfloor \{x\} + \{y\} \rfloor \end{align*}

Answer 2

Part ii:

In general: $$\lfloor z\rfloor \leq z< \lfloor z\rfloor+1.$$

For ii, set $z=x/2$ then we get: $$\left\lfloor\frac{x}{2}\right\rfloor\leq \frac{x}{2}<\left\lfloor\frac{x}{2}\right\rfloor +1$$

Double and you get:

$$2\left\lfloor\frac{x}{2}\right\rfloor\leq x<2\left\lfloor\frac{x}{2}\right\rfloor +2$$

Taking the floor gives:

$$2\left\lfloor\frac{x}{2}\right\rfloor\leq\lfloor x\rfloor<2\left\lfloor\frac{x}{2}\right\rfloor +2$$

Subtracting gets:

$$0\leq \left\lfloor x\right\rfloor-2\left\lfloor\frac{x}{2}\right\rfloor<2.$$

Which is the result you want.

Part i

You can actually prove the same way that $\lfloor x+y\rfloor-\lfloor x\rfloor -\lfloor y\rfloor$ is always either $0$ or $1.$ More specifically, you can show that $\lfloor x+y\rfloor-\lfloor x\rfloor -\lfloor y\rfloor=1$ if and only if $\{x\}+\{y\}\geq 1.$

In particular, when $y=x,$ $\lfloor 2x\rfloor -2\lfloor x\rfloor=1$ if and only if $\{x\}\geq \frac{1}{2}.$

Now, if $\{x\}+\{y\}\geq 1$ then one or both of $\{x\}$ and $\{y\}$ are $\geq \frac{1}{2}$, so one or both of $\lfloor 2x\rfloor -2\lfloor x\rfloor$ and $\lfloor 2y\rfloor-2\lfloor y\rfloor$ are one.

So that means, doe any $x,y:$

$$\lfloor x+y\rfloor -\lfloor x\rfloor -\lfloor y\rfloor\leq \left(\lfloor 2x\rfloor -2\lfloor x\rfloor\right)+\left(\lfloor 2y\rfloor -2\lfloor y\rfloor\right)$$

This is because when the left side is $0,$ we know the right side is at least $0,$ and when the left side is $1$, then the right side is either $1$ or $2.$

Adding $2\lfloor x\rfloor + 2\lfloor y\rfloor$ to both sides gives you:

$$\lfloor x+y\rfloor +\lfloor x\rfloor +\lfloor y\rfloor\leq \lfloor 2x\rfloor +\lfloor 2y\rfloor$$

Answer 3

Let's denote with $\{ x \}$ the fractional part: $$ x = \left\lfloor x \right\rfloor + \left\{ x \right\} $$

Then I would suggest that you first master the addition $$ \eqalign{ & \left\lfloor {x + y} \right\rfloor = \left\lfloor {\left\lfloor x \right\rfloor + \left\{ x \right\} + \left\lfloor y \right\rfloor + \left\{ y \right\}} \right\rfloor = \cr & = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr & = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \cr} $$ where $[P]$ denotes the Iverson bracket $$ \left[ P \right] = \left\{ {\begin{array}{*{20}c} 1 & {P = TRUE} \\ 0 & {P = FALSE} \\ \end{array} } \right. $$

Thereafter you simply have

i)
$$ \left\lfloor {2x} \right\rfloor = \left\lfloor {x + x} \right\rfloor = 2\left\lfloor x \right\rfloor + \left[ {1/2 \le \left\{ x \right\}} \right] $$ so $$ \left\{ \matrix{ \left\lfloor {2x} \right\rfloor + \left\lfloor {2y} \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1/2 \le \left\{ x \right\}} \right] + \left[ {1/2 \le \left\{ y \right\}} \right] \hfill \cr \left\lfloor {x + y} \right\rfloor + \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2\left\lfloor x \right\rfloor + 2\left\lfloor y \right\rfloor + \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \hfill \cr} \right. $$ and clearly $$ \left[ {1 \le \left\{ x \right\} + \left\{ y \right\}} \right] \le \left[ {1/2 \le \left\{ x \right\}} \right] + \left[ {1/2 \le \left\{ y \right\}} \right] $$

ii)
$$ \eqalign{ & \left\lfloor x \right\rfloor = \left\lfloor {x/2 + x/2} \right\rfloor = \cr & = 2\left\lfloor {x/2} \right\rfloor + \left[ {1/2 \le \left\{ {x/2} \right\}} \right] \cr} $$