Divisibility 11

**Sea** · December 19th 2008, 11:23 PM

Show that: $a, k\in \mathbf{Z},$

$a=2k \Leftrightarrow a-2\| \frac{a}{2}\| = 0$

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

Symbol...
Example:

$\| 5,3\|=5$

$\| -2,2\|=-3$

**CaptainBlack** · December 20th 2008, 02:07 AM

Originally Posted by Sea

Show that: $a, k\in \mathbf{Z},$

$a=2k \Leftrightarrow a-2\| \frac{a}{2}\| = 0$

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

Symbol...
Example:

$\| 5,3\|=5$

$\| -2,2\|=-3$

In more familiar notation:

Show that: $a, k\in \mathbb{Z},$

$a=2k \Leftrightarrow a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0$

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

CB

**CaptainBlack** · December 20th 2008, 02:14 AM

Originally Posted by CaptainBlack

In more familiar notation:

Show that: $a, k\in \mathbb{Z},$

$a=2k \Leftrightarrow a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0$

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

CB

Suppose $a=2k$ for some $k \in \mathbb{Z}$ then:

$<br /> a-2\left\lfloor \frac{a}{2}\right\rfloor=a-2 \lfloor k \rfloor=a-2k=0<br />$

Which proves

$a=2k \Rightarrow a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0$

To complete the proof assume that: $a-2 \left\lfloor \frac{a}{2}\right\rfloor = 0$ and show that this implies that there exists a $k \in \mathbb{Z}$ such that $a=2k$

CB

**Sea** · December 20th 2008, 02:41 AM

I didn't know this words means in english...I'm sorry for this and I thank you so much... for answering my question after all...

Thanks a million...

...

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