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Finding the zero vector

  1. Feb 6, 2008 #1
    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
    6)a is in V implies ka is in V
    10) 1a=a
  2. jcsd
  3. Feb 6, 2008 #2


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

    Why not?
  4. Feb 6, 2008 #3
    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 thats 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?
  5. Feb 7, 2008 #4


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    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 xk. 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 ex+y= exey and ekx= (ex)k.
    Last edited by a moderator: Feb 7, 2008
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