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
I presume you mean .
I will write instead of as this is easier to read.
For your proof, firstly note that for all as . Also, as . Therefore, 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 . Then there exists a number, , such that and are coprime (for example, if a is positive choose a+1, while if a is negative choose a-1). Clearly , and so is not contained in the intersection. As a is arbitrary, we have the result.
OK. Does "x / y" mean x divides y? Usually this is denoted by "x | y".
If you prove this, you solve the whole problem. The whole difficulty of the problem is to show that for every integer , .So here my prove
An is a subset of {-1,0,1},
Now, iff for every . In turn, this happens iff for all , i.e., for every , either divides or vice versa.
You can check that , or satisfy this property. This means that . On the other hand, when is none of the three, it is easy to find an such that neither nor . This means that implies (where was found in the previous sentence) and so , which is the same thing to say as .