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Vector Space

  • Thread starter Dustinsfl
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  • #1
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{∀ 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?
 
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Answers and Replies

  • #2
127
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{∀ 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?
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.
 
  • #3
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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.
 
  • #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.
 
  • #5
699
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Ok so the element 1 verifies 3 then, correct?
 
  • #6
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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?
 
  • #7
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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?
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.
 
  • #8
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Ok then for axiom 4 what is the desired outcome?
 
  • #9
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Ok then for axiom 4 what is the desired outcome?
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.
 
  • #10
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Then for example if the element is 2 the additive inverse would be 1/2 and that is what axiom for is asking?
 
  • #11
vela
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Yes, exactly. The axiom requires that every element in R+ have an additive inverse.
 
  • #12
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For axiom 4, this is what I would put: 4. x ⊕ (-x) = x·1/x = 1?
 
  • #13
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For axiom 4, this is what I would put: 4. x ⊕ (-x) = x·1/x = 1?
You're working with the positive reals.. Looks okay to me.
 
  • #14
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Thanks, now I am going to just check the rest of the axioms and see if I come across any more questions.
 
  • #15
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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?
 
  • #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).
 
  • #17
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Even though (x ⊕ y) was defined as x·y earlier?
 
  • #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.
 
  • #19
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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?
 
  • #20
vela
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No, you're missing the point. How is scalar multiplication defined for this problem?
 
  • #21
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Oh yeah, I forgot about that. α∘(x ⊕ y) = x^α ⊕ y^α Now do I stop here or do I continue with addition definition and carry the operation further?
 
  • #22
vela
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Well, you can't directly say that α∘(x ⊕ y) = x^α ⊕ y^α. You have to prove that you can distribute ∘ across ⊕.

What you want to do is start with α∘(x ⊕ y), and then using the definitions you have and what you know about real numbers, turn it into α∘x ⊕ α∘y.
 
  • #23
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α∘(x ⊕ y) = α∘x ⊕ α∘y = x^α ⊕ y^α
Should I continue simplifying to x^α · y^α and then onto (xy)^α or is stopping at x^α ⊕ y^α more appropriate?
 

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