1. ## Floor function

$\displaystyle \forall x,y \in \mathbb{R}$

Show that:

$\displaystyle \lfloor x+y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor$

.................................................. .................................................. .

where $\displaystyle \lfloor x \rfloor$ is the floor function, the greatest
integer less than $\displaystyle x$.

2. Let: $\displaystyle x = \lfloor x \rfloor + a$ and $\displaystyle y = \lfloor y \rfloor + b$ for some $\displaystyle 0 \leq a,b < 1$.

Fact: $\displaystyle \lfloor m + n \rfloor = \lfloor m \rfloor + n$ where $\displaystyle m \in \mathbb{R}, \ n \in \mathbb{Z}$

Consider: $\displaystyle x + y = \lfloor x \rfloor + \lfloor y \rfloor + (a + b)$

Take the floor function of both sides and with the fact, the conclusion follows.

3. Originally Posted by Sea
$\displaystyle \forall x,y \in \mathbb{R}$
Show that: $\displaystyle \lfloor x+y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor$.
The rigor required by the instructor/text may well vary.
But this is the general idea.
$\displaystyle \left\lfloor x \right\rfloor \leqslant x\;\& \,\left\lfloor y \right\rfloor \leqslant y\, \Rightarrow \,\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor \leqslant x + y$.
Because, $\displaystyle \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor$ is an integer and $\displaystyle \left\lfloor {x + y} \right\rfloor$ is the greatest integer which does not exceed $\displaystyle x + y$ it follows that $\displaystyle \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor \leqslant \left\lfloor {x + y} \right\rfloor$.

4. Thanks a million.......

I understand....

And... Can you help me?

__________________________________________________

Show that:

$\displaystyle \lfloor m + n \rfloor = \lfloor m \rfloor n$ where $\displaystyle m \in \mathbb{R}, \ n \in \mathbb{Z}$

__________________________________________________ _

Show that:

$\displaystyle \lfloor \frac{\lfloor x \rfloor }{n} \rfloor = \lfloor \frac{x}{n} \rfloor +$ where $\displaystyle x \in \mathbb{R}, \ n \in \mathbb{Z^+}$

5. Originally Posted by Sea
Show that:

$\displaystyle \lfloor m + n \rfloor = \lfloor m \rfloor n$ where $\displaystyle m \in \mathbb{R}, \ n \in \mathbb{Z}$
I think this is supposed to be $\displaystyle \lfloor m+n \rfloor=\lfloor m \rfloor +n$. A quick counterexample to what you have written is $\displaystyle \lfloor \sqrt{2}+3 \rfloor=4\ne\lfloor\sqrt{2}\rfloor\cdot 3=3$.

So let $\displaystyle m=m_1+\varphi\quad m_1\in\mathbb{Z},\varphi\in[0,1)$. So $\displaystyle \lfloor m+n\rfloor=\lfloor m_1+n+\varphi\rfloor$, and since $\displaystyle m_1,n\in\mathbb{Z}\wedge\varphi\in[0,1)$ we can conclude by the defintion of the floor function that $\displaystyle \lfloor m_1+n+\varphi\rfloor=m_1+n$. Now lets look at the left side. Defining $\displaystyle m$ the same as above we can see that $\displaystyle \lfloor m\rfloor+n=\lfloor m_1+\varphi\rfloor+n$ and once again because $\displaystyle m_1\in\mathbb{Z}\wedge\varphi\in[0,1)$ we can conclude that $\displaystyle \lfloor m\rfloor+n=\lfloor m_1+\varphi\rfloor+n=m_1+n$. Finally $\displaystyle \lfloor m+n\rfloor=m_1+n=\lfloor m\rfloor +n$

Show that

$\displaystyle \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\ right\rfloor\quad x\in\mathbb{R}\wedge n\in\mathbb{Z}^+$
Similar to before define $\displaystyle x=x_1+\varphi\quad x\in\mathbb{Z}\wedge\varphi\in[0,1)$. So once again by the definition of the floor function $\displaystyle \lfloor x_1+\varphi\rfloor=x_1$. So $\displaystyle \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{\lfloo r x_1+\varphi\rfloor}{n}\right\rfloor=\left\lfloor\f rac{x_1}{n}\right\rfloor$

Now for the second part notice that $\displaystyle \left\lfloor\frac{x}{n}\right\rfloor=\left\lfloor\ frac{x_1}{n}+\frac{\varphi}{n}\right\rfloor=\left\ lfloor\frac{x_1}{n}\right\rfloor+\left\lfloor\frac {\varphi}{n}\right\rfloor$. Now conclude the answer by noticing that since $\displaystyle n\in\mathbb{Z}^+\implies\frac{\varphi}{n}\in[0,1)$

6. I write false....... But you write true......

I'm sorry for this and I thank you so much... for answering my
question after all...

Thanks a million......