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Proving that a set is a Vector Space - addition is defined very oddly

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Let V = {0,1} with addition defined modulo 2 (i.e. the remainder upon division by 2), and scalar multiplication given by ku = u^k for all k in the real numbers and u in V. Is the set V a vector space?

    2. Relevant equations

    The 10 axioms!

    3. The attempt at a solution

    I understand that to determine if V is a vector space you need to establish the ten axioms, but I don't understand what my teacher means when she says addition is defined modulo 2. Does that mean that u + v = u%2? Or u + v = v%2? Or something else?
     
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  3. Oct 6, 2012 #2

    Dick

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    It means that u+v (+ meaning the sum in your possible vector space) is equal to (u+v)%2 (+ here meaning the usual sum). You probably have more to worry about with your scalar multiplication.
     
    Last edited: Oct 6, 2012
  4. Oct 6, 2012 #3
    Thanks for the response. I've got a couple more questions:

    Does the axiom that says u + (v + w) must equal (u + v) + w apply in this case as there are only two vectors in the set?

    This set seems to fail the test when you try to show that -u + u = 0. Using it on the first vector in the set gets you -0 = 0^-1 = 1/0 which is undefined. Therefore, this is not a vector space. Is that right?

    Also, how do you make a distinction between a member of a set that happens to be zero and the zero vector? In this case they are one and the same but I imagine there will be some cases where that isn't true.
     
    Last edited: Oct 6, 2012
  5. Oct 6, 2012 #4

    Dick

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    Yes, you still have to show u + (v + w)=(u + v) + w. Associativity doesn't require that all three vectors have to be different. And -0=0^-1 is completely off the mark. So that doesn't work. And right on the third point. The element you label '0' doesn't have to be the additive identity of your vector space. But here it is. Can you show that? And to reiterate my previous hint, the problem is going to be with scalar multiplication, not with addition. Can you find it?
     
  6. Oct 7, 2012 #5
    What do you mean by this?
     
  7. Oct 7, 2012 #6

    HallsofIvy

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    When you say "-0= 0^{-1}" are you talking about the scalar 0 or the vector 0?
    There is no "vector multiplication" defined on vector spaces so it makes no sense to talk about "0^{-1}" for the vector 0.

    If you are talking about scalar multiplication, in "modulo 2", while it is true that "-0= 0" (that's true in any field), 0^{-1} is NOT 0- it does not exist. In "mod 2" arithmetic, we have the two members, 0 and 1 with addition defined by 0+ 0= 0, 1+ 1= 0, and 0+ 1= 1+ 0= 1. Multiplication is defined by 0*0= 0*1= 1*0= 0, and 1*1= 1. There is no x such that 0*x= 1 so 0 has no multiplicative inverse.
     
  8. Oct 7, 2012 #7
    I'm sorry; I'm still not understanding. Maybe I need to be more clear on what it is I was doing.

    I'm testing this axiom: For every v ∈ V, there exists an element −v ∈ V, called the additive inverse of v, such that v + (−v) = 0.

    In this case, there are only two vectors in V: 0 and 1. So I started with 0. If the axiom holds, then 0 + -0 = the zero vector. But because scalar multiplication was defined as ku = u^k, -0 is undefined. Therefore, this axiom cannot hold, and V is not a vector space.
     
  9. Oct 7, 2012 #8

    Dick

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    Yes, that could be problem if you phrase it correctly. There is really no problem with additive inverses. -u is the additive inverse of u. Since 0+0=0 and 1+1=0. -0=0 and -1=1. Since you don't necessarily know you have a vector space it may not be true that -u=(-1)*u. That is a problem if u=0. You might also think about what 0*u should be. What happens if u=1?
     
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