Negative x Negative = Positive?

[Xerxes]

I'm fairly sure if you were to tweak the distributive property in the fashion I suggested, there'd really be no-one calling it the Reals.

Well, then, what would they be calling "it"?

 

And, I'm sorry, I really don't understand what you mean by "tweaking the distributive property". Do we have a choice here? Surely, distributivity is what it is (i.e. no tweaking allowed)?

[the tree]

I think I'll call it the Freaky Numbers, that sounds a good name.

 

Axioms can be tweaked to your heart's content so long as they don't wind up contradicting each other, although you'd probably change the names just to avoid confusion. People tend not to because like I said, there's often no point at all.

 

Sometimes changing the axioms can be genuinely useful, for instance when it was found that the existing ones for geometry only usefully described flat surfaces.

[D H]

[math]\mathbb{R}[/math] can be viewed as:

 

a Set

 

Absolutely not. The reals are not a set. They are an algebraic structure (and a very special one at that), [math](S,+,\cdot)[/math]. The operations of addition and multiplication are a part of what constitutes the definition of the reals. Take these away and you do not have the reals.

 

a Monoid

a Group

an Abelian Group

Getting closer. However, these labels pertain to an algebraic structure with one operation.

a Field

a Ring

The reals of course form a ring because any field is also a ring. However, there are many rings that are not fields. Similarly, there are many fields that are not complete or orderable. On the other hand, any complete ordered Archimedean field is isomorphic with the reals.

 

For example, suppose you define some algebraic structure [math](S,+,\cdot)[/math] in which the members of [math]S[/math] can be classified as "nonnegative" or "negative", but in which the product of two negative members is negative. That algebraic structure is not isomorphic to the reals.

a Vector Space

a Topological Space

a Manifold

 

Wrong again. A vector space is a space over a field. Just as a group is more than just a set (it is a set plus an operation that is closed with respect to the set), a space is more than just a field.

 

When trying to make proofs in [math]\mathbb{R}[/math] we have to be clear under which of the above we are working, all have different axioms, and therefore different allowable proofs.

 

No. The reals are one thing, and one thing only (within isomorphism).

[Xerxes]

D.H.: Maybe you missed this bit

R [b']can be viewed as[/b]

I never meant to imply that the reals are merely a set, monoid, group, etc, but they most certainly can be viewed as all these things.

 

 

I respectfully invite you to open any text on Abstract Algebra, and you will see what I mean.

 

A vector space is a space over a field. Just as a group is more than just a set (it is a set plus an operation that is closed with respect to the set), a space is more than just a field.

And in your text you will see that the field over which a vector space is defined is itself the prototypical vector space.

 

Or, if you have no such text, maybe you'd like to look here:

http://en.wikipedia.org/wiki/Real_number or here

http://en.wikipedia.org/wiki/Examples_of_vector_spaces just for starters

 

I am not wrong, don't be so rude.

[redmafiya]

The best way to understand it is to solve some real life word problems that lead to a negative times a negative equals a positive.

 

View an example:

http://www.idealmath.com/algebra/FAQ/negxneg.htm

 

Practices

http://www.idealmath.com/algebra/Problems/negativenumbers/multiply1.htm