- #1
Dustinsfl
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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?
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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