Is 1 considered the zero vector in this scenario?

[torquerotates]

I came across an interesting problem. I'm trying to determine if the following is a vector space, x+y=xy, kx=x(risen to power k), and I came across an interesting result. I used ax. 4 to show x+y+1=(x+y)+1=(xy)+1=1(xy)=xy=x+y. Doesn't that just seem strange that 1 is the zero vector. 1 is not even a vector let alone 0. Is there something wrong with my thinking?


let a,b,c be vectors and V is a vector space, then
1)a&b is in V then a+b is in V
2)a+b=b+a
3)a+(b+c)=(a+b)+c
4)0+a=a+0=a
5)a+(-a)=(-a)+a=0
6)a is in V implies ka is in V
7)k(a+b)=ka+kb
8)(k+m)a=ka+ma
9)k(ma)=(km)a
10) 1a=a

 

[Hurkyl]

I came across an interesting problem. I'm trying to determine if the following is a vector space, x+y=xy, kx=x(risen to power k), and I came across an interesting result.

What is the set of vectors? What is the field of scalars?

1 is not even a vector

Why not?

 

[torquerotates]

Well, the problem said, the set of all positive real numbers such that, x+y=xy, kx=x(risen to power k). But if we designate this as set of vectors, not scalars( something can be both simulataneously). Then I would think that I arrived at a contradiction. Because 1 is obviously a scalar. We can designate it as a vector. I have no problems with that. But then in this situation, what would be a scalar? Obviously it can't be any number k since k is a multiple of one and one is a vector. The thing that's getting me is that I think whenever there is a set of vectors, there is a field of scalars associated with it. Is this true?

 

[HallsofIvy]

You have completely confused yourself (well, at least me!). You say that you are using the set of real numbers as vectors (completely allowable) but then protest that 1 is a scalar not a vector! If you are thinking of the real numbers as a vector space over the real numbers, then any number, including 0 and 1, is both a scalar and a vector- you just have to keep track of which one you intend in a particular case. The real numbers, with ordinary addition as vector addition and ordinary multiplication as scalar multiplication, certainly does form a vector space over the real numbers. It has dimension 1 of course and isn't terribly interesting!

Now, back to this problem. Your set is the set of positive real numbers and you are defining vector addition, "x+ y", as xy, scalar multiplication, "kx", as x^k. Yes, the number 1 is now the "vector 0", the additive identity. A vector space must have all of the "group" properties- in particular every member must have an additive inverse. What is the "additive inverse" of a "vector"? (And do you see why you need positive real numbers and not all real numbers? What would happen if 0 were in the set?)

I think the crucial point here is the "distributive law": if a is a scalar and u and v are vectors, then a(u+ v)= au+ av. What does that say in terms of the operations defined here? Is it true?

Actually, you can construct a simple isomorphism between this space and the vector space of real numbers with ordinary addition and multiplcation, based on the fact that e^x+y= e^xe^y and e^kx= (e^x)^k.