Floor function

**Sea** · December 21st 2008, 01:13 PM

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

Show that:

$\lfloor x+y \rfloor \geq \lfloor x \rfloor<br /> + \lfloor y \rfloor$

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

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

**o_O** · December 21st 2008, 02:25 PM

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

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

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

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

**Plato** · December 21st 2008, 02:32 PM

Originally Posted by Sea

$\forall x,y \in \mathbb{R}$
Show that: $\lfloor x+y \rfloor \geq \lfloor x \rfloor<br /> + \lfloor y \rfloor$ .

The rigor required by the instructor/text may well vary.
But this is the general idea.
$\left\lfloor x \right\rfloor \leqslant x\;\& \,\left\lfloor y \right\rfloor \leqslant y\, \Rightarrow \,\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor \leqslant x + y$ .
Because, $\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor$ is an integer and $\left\lfloor {x + y} \right\rfloor$ is the greatest integer which does not exceed $x + y$ it follows that $\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor \leqslant \left\lfloor {x + y} \right\rfloor$ .

**Sea** · December 21st 2008, 04:25 PM

Thanks a million...

....

I understand..

..

And... Can you help me?

__________________________________________________

Show that:

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

__________________________________________________ _

Show that:

$\lfloor <br /> \frac{\lfloor x \rfloor }{n} \rfloor = \lfloor <br /> \frac{x}{n} \rfloor +$ where $x \in \mathbb{R}, \ n \in \mathbb{Z^+}$

**Mathstud28** · December 21st 2008, 05:48 PM

Originally Posted by Sea

Show that:

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

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

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

Show that

$\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\ right\rfloor\quad x\in\mathbb{R}\wedge n\in\mathbb{Z}^+$

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

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

**Sea** · December 21st 2008, 10:13 PM

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...

...

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