Math Is Fun Forum

I think that by stating that a,b∈ Z, we can know for certain that
(-a)(b) = -(ab), and
(-1)(-1)=1  since we have that 1 is the multiplicative identity and we know that no integer squared has a negative value...

Well, yes...it is fairly obvious
I, for one, would assume that there was such a set...
It should be possible to derive a contradiction...
But I don't know how just yet...
I'll be back...**said in Arnold's Austrian accent**

hmmm...I believe the exact same argument would work for the entire set of real numbers, wouldn't it?

And I just re-read your explanation, Ricky, and I'm having a little bit of trouble with your statement of the definition of an open set as
"A set O is open if for all points a∈O there exists an ∈-neighborhood of a."

What did you mean by that?...You can create an epsilon neighborhood around any point, whether the set is open or closed.  I think that you meant to say something about the intersection of that neighborhood and the set...

The definitions of open and closed that I remember have to do with inclusion of boundary points.
A boundary point of a set S is defined as a point, say x (in R), such that for any epsilon>0, the neighborhood N(x;epsilon)
intersected with S is nonempty, and that same neighborhood N intersected with R\S is nonempty...

Now the definition that I know for a closed set is a set that contains all of those boundary points.  (And thus, an open set is one whose complement is closed)

For example, consider the set of integers Z.
the set of boundary points of Z (denoted bd Z) is actually all of the integers.  [Does that make sense?...because any neighborhood that you take about an integer includes both an integer (in Z), and all of the other real numbers between integers (in R\Z)]
So since Z contains bd Z (in fact, it IS bd Z), the set Z of integers is closed.

I hope I'm making sense...
So then, for this particular question, I would say (R="the set of real numbers")
bd R = empty set (because there are no points in R that would fit the definition of a boundary point of R)
then, bd R is a subset of R  ----> R is closed
but bd R is also a subset of R\R (which is the empty set) -----> R is open.
Thus, we have that R is both open and closed....

I don't know's on third  (I think...I can't remember too well)

I don't know...the only thing I can think is that
x+x+x+.....+x  (x times) is only defined for x in Z+

So that is isn't continuous anywhere or differentiable, really...

Who's alfred?...
and only "I" was going to St. Ives, so....ONE

Love it 

In that case....give us the proof please, krassi!!!!
It's been bugging me since you posted this and I'm having the same problem...I've gone through all my notes to try to find some long lost number theory theorem that would make it all come together, but I can't....and I'm a bit obsessive so I need to see the proof...and I need to see it fast!

When you use the absolute value symbols, you mean the order of those sets (number of elements in the set), right?

If so, then I'm not sure that you need to prove that they are subsets both ways, because actually each of those terms would just represent a number.