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Homework Help: Vector Space

  1. Feb 15, 2010 #1
    {∀ x ϵ ℝ+ : x>0}
    Define the operation of scalar multiplication, denoted ∘, by α∘x = x^α, x ϵ ℝ+ and α ϵ ℝ.
    Define the operation of addition, denoted ⊕, by x ⊕ y = x·y, x, y ϵ ℝ+.
    Thus, for this system, the scalar product of -3 times 1/2 is given by:
    -3∘(1/2)= (1/2)^-3 = 8 and the sum of 2 and 5 is given by:
    2 ⊕ 5 = 2·5 = 10.
    Is ℝ+ a vector space with these operations? Prove your answer.

    Axioms:
    1. x ⊕ y = x·y = y·x = y ⊕ x
    2. (x ⊕ y) ⊕ z = (x·y) ⊕ z = x·y·z = x·(y·z) = x·(y ⊕ z) = x ⊕ (y ⊕ z)
    3. There ∃ an element 0 ϵ V, x ⊕ 0 = x
    For Axiom 3, I obtain x ⊕ 0 = x·0 = 0; therefore, {∀ x ϵ ℝ+ : x>0} isn't a vector space. This makes sense but since axiom 3, says there exists, does that mean it has to be a zero element or can it be any element?
     
    Last edited: Feb 15, 2010
  2. jcsd
  3. Feb 15, 2010 #2
    I'm pretty sure axiom 3 means that there exists an element 0, such that for all elements x, x ⊕ 0 = 0. You can prove that this element is unique. But it must the be same element for all x.
     
  4. Feb 15, 2010 #3
    You're confusing the zero element in the vector space with the number 0. Call the zero element z. Then, if such an element exists, you know that x ⊕ z = xz = x. It should be clear what z has to be for this to work.
     
  5. Feb 15, 2010 #4

    vela

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    "0" is just a name for the additive identity element. It's not necessarily the real number zero, which you might notice isn't in your prospective vector space, R+.

    You're looking for an element e of R+ that satisfies the relationship e ⊕ x = x ⊕ e = x. If no such element exists, (R+, ⊕, ∘) isn't a vector space.
     
  6. Feb 15, 2010 #5
    Ok so the element 1 verifies 3 then, correct?
     
  7. Feb 15, 2010 #6
    4. ∀ x ϵ ℝ+, ∃ x ϵ ℝ +, x ⊕ (-x) = 0
    Since ⊕ is defined by multiplying the two elements, this axiom fails due to there isn't an element in ℝ+ that yields 0, correct or am I missing something here?
     
  8. Feb 15, 2010 #7
    What about instead of using the symbol "0" try using the symbol "e", like another poster did. The use of the symbol "0" in your axioms only means it's neutral in some way, not that it's the actual number half way between 1 and -1.
     
  9. Feb 15, 2010 #8
    Ok then for axiom 4 what is the desired outcome?
     
  10. Feb 15, 2010 #9
    Axiom 4 just says that each element has it's own opposite (inverse) element, in such a way that when you combine them you get back down to the neutral element.
     
  11. Feb 15, 2010 #10
    Then for example if the element is 2 the additive inverse would be 1/2 and that is what axiom for is asking?
     
  12. Feb 15, 2010 #11

    vela

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    Yes, exactly. The axiom requires that every element in R+ have an additive inverse.
     
  13. Feb 15, 2010 #12
    For axiom 4, this is what I would put: 4. x ⊕ (-x) = x·1/x = 1?
     
  14. Feb 15, 2010 #13
    You're working with the positive reals.. Looks okay to me.
     
  15. Feb 15, 2010 #14
    Thanks, now I am going to just check the rest of the axioms and see if I come across any more questions.
     
  16. Feb 15, 2010 #15
    Axiom 5 states: a(x ⊕ y) = ax ⊕ ay
    α·(x ⊕ y) = α·(x·y) = (α·x)·y = x·(α·y)
    Is this equivalent to what axiom 5 is asking?
     
  17. Feb 15, 2010 #16

    vela

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    No, you need to use scalar multiplication as you defined it above. You're using regular scalar multiplication when multiplying a into (x·y).
     
  18. Feb 15, 2010 #17
    Even though (x ⊕ y) was defined as x·y earlier?
     
  19. Feb 15, 2010 #18

    vela

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    That part's not the problem. It's when you say a(x·y)=(a·x)·y. That's not right.
     
  20. Feb 15, 2010 #19
    So this is the correct procedure then, α·(x ⊕ y) = (α·x ⊕ α·y) = α·x·α·y? Now is this equivalent to what the axiom is asking even though addition was defined as such?
     
  21. Feb 15, 2010 #20

    vela

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    No, you're missing the point. How is scalar multiplication defined for this problem?
     
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