1. ## set and relation

Define the the relation R on the integers Z by xRy if and only if either x/y or y/x.
For each n E Z, define An to be the set of integers related to n. An={xEZ:xRn}
Proving the quality below
Intersection An= {-1,0,1}

So here my prove
An is a subset of {-1,0,1}, so if xEAn then XE{-1,0,1} ==> Intersection An is a subset of {-1,0,1}.

Suppose that m is not belong of {-1,0,1} then m is not belong to Intersection An ===> {-1,0,1} is a subset of Intersection An

I a little confusing , i know im missing something out there,
Any help or suggestions would be greatly appreciated. Thanks

2. Originally Posted by logglypop
Prove
An={xEZ:xRn}
Intersection An= {-1,0,1}
nEZ

An is a subset of {-1,0,1}, so if xEAn then XE{-1,0,1} ==> Intersection An is a subset of {-1,0,1}.

Suppose that m is not belong of {-1,0,1} then m is not belong to Intersection An ===> {-1,0,1} is a subset of Intersection An

I a little confusing , i know im missing something out there,
Any help or suggestions would be greatly appreciated. Thanks
I'm not sure what the question is actually asking. Could you type it in LaTeX or at least more carefully?

3. I edited abit
i hope u understand

4. Originally Posted by logglypop
I edited abit
i hope u understand
What is the relation?

5. What is the relation?
sorry there no relation.
I just have to prove the Intersection of the set An = {-1,0,1}

6. Suppose you and me are talking and I ask, "By the way, did you know that Peterson is a bullplooper?" You would naturally ask, "Who is Peterson? And what is a bullplooper?"

It's the same way when you say "An={xEZ:xRn}" but don't say what R is.

7. Define the the relation R on the integers Z by xRy if and only if either x/y or y/x.

8. Originally Posted by logglypop
Define the the relation R on the integers Z by xRy if and only if either x/y or y/x.
I presume you mean $xRy \Leftrightarrow x/y \in \mathbb{Z} \text{ or } y/x \in \mathbb{Z}$.

I will write $a\sim b$ instead of $aRb$ as this is easier to read.

For your proof, firstly note that $0\sim n$ for all $n \in \mathbb{Z}$ as $0/n = 0 \in \mathbb{Z}$. Also, $\pm1 \sim n$ as $n/\pm 1 = \pm n \in \mathbb{Z}$. Therefore, $\{-1, 0, 1\}$ is in the intersection.

(It's late, so this bit is a bit rough around the edges.) To prove that this is precisely the intersection, choose $a \in \mathbb{Z} \setminus \{-1, 0, 1\}$. Then there exists a number, $b$, such that $a$ and $b$ are coprime (for example, if a is positive choose a+1, while if a is negative choose a-1). Clearly $a \not\sim b$, and so $a$ is not contained in the intersection. As a is arbitrary, we have the result.

9. OK. Does "x / y" mean x divides y? Usually this is denoted by "x | y".

So here my prove
An is a subset of {-1,0,1},
If you prove this, you solve the whole problem. The whole difficulty of the problem is to show that for every integer $n$, $A_n\subseteq\{-1,0,1\}$.

Now, $x\in\bigcap_{n\in\mathbb{Z}}A_n$ iff $x\in A_n$ for every $n\in\mathbb{Z}$. In turn, this happens iff $xRn$ for all $n\in\mathbb{Z}$, i.e., for every $n$, either $n$ divides $x$ or vice versa.

You can check that $x = -1$, $0$ or $1$ satisfy this property. This means that $\{-1,0,1\}\subseteq\bigcap_{n\in\mathbb{Z}}A_n$. On the other hand, when $x$ is none of the three, it is easy to find an $n$ such that neither $x\mid n$ nor $n\mid x$. This means that $x\notin\{-1,0,1\}$ implies $x\notin A_n$ (where $n$ was found in the previous sentence) and so $x\notin\bigcap_{n\in\mathbb{Z}}A_n$, which is the same thing to say as $\bigcap_{n\in\mathbb{Z}}A_n\subseteq\{-1,0,1\}$.