Can you prove these floor and ceiling equations?

[Walshy1]

Hi, i need help with 2 proofs based off of floor and ceiling.
1.)For any real number x, if x is not an integer, then floor(x) + floor(-x) =- 1.2.For all real numbers x, floor(floor(x/2)/2) = floor(x/4).

Thanks.

 

[tkhunny]

floor(x) = x - a, where 0 <= a < 1

floor(-x) = -x - b, where 0 <= b < 1

You tell me why a + b = 1

 

[Walshy1]

floor(x) = x - a, where 0 <= a < 1

floor(-x) = -x - b, where 0 <= b < 1

You tell me why a + b = 1

I'm sorry that makes no sense to me.

 

[MarkFL]

I urge you to think more about the given hint...it leads immediately to the proof you seek.

Once this clicks, a very similar argument will work for the second proof.

 

[QuestForInsight]

$ \lfloor{x}\rfloor = \max\{m\in\mathbb{Z}\mid m < x\}$ when $x$ is not an integer, thus:

$ \begin{aligned} \lfloor{x}\rfloor+\lfloor{-x}\rfloor & = \max\{m\in\mathbb{Z}\mid m < x\}+\max\{m\in\mathbb{Z}\mid m < -x\} \\& = \max\{m\in\mathbb{Z}\mid 2m < 0\} = \max\{m\in\mathbb{Z}\mid m < 0\} \\& = -1.\end{aligned}$

This addition (of sets usually) is called Minkowski addition.

 

[Evgeny.Makarov]

For the second problem, you can consider cases when $x$ has the form $4n+y$ and $4n+2+y$ where $n\in\mathbb{Z}$ and $0\le y<2$. For example, suppose that $x=4n+y$. Then $x/2=2n+y/2$. Since $0\le y/2<1$, $\lfloor x/2\rfloor=2n$. Therefore, $\lfloor\lfloor x/2\rfloor/2\rfloor=n$. Now you show that $\lfloor (4n+y)/4\rfloor=n$. Also, show that $\lfloor\lfloor x/2\rfloor/2\rfloor=\lfloor x/4\rfloor$ when $x=4n+2+y$ where $n\in\mathbb{Z}$ and $0\le y<2$