# Homework Help: Vector Space {∀ x ϵ ℝ+ : x>0}

1. Feb 16, 2010

### Dustinsfl

I am not sure if my #4 holds and I don't know how to approach #7. My Axioms are below the general axioms.
{∀ 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.

Vector Space Axioms
1. x + y = y + x
2. (x + y) + z = x + (y + z)
3. x + 0 = x
4. x + (-x) = 0
5. α(x + y) = α·x + α·y
6. (α + β)x = α·x + β·x
7. (αβ)·x = α·(βx)
8. 1·x = x

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. x ⊕ 1 = x·1 = x
4. -x = -1∘x = x^-1 = 1/x ⇒x ⊕ (-x) = x·1/x = 1
5. α∘(x ⊕ y) = α(x·y) = (x·y)^α = x^α·y^α = x^α ⊕ y^α = α∘x ⊕ α∘y
6. (α + β)∘x = x^(α + β) = x^α·x^β = x^α ⊕ x^β = α∘x ⊕ β∘x
7. (α·β)∘x =
8. 1∘x = x^1 = x

2. Feb 16, 2010

### vela

Staff Emeritus
For #4, you can just note that 1/x is in R+, and that it adds with x to give the identity element. You really don't have to show how you figured out what the additive inverse is, just that it's in the space.

For #7, if you can't see it starting from the lefthand side, try seeing what happens if you start with the righthand side.

3. Feb 16, 2010

### Dustinsfl

7. (α·β)∘x = x^(α·β) = x^(β·α) = (β∘x)^α = α∘(β∘x)
So this is what I obtained. Is it correct?

4. Feb 16, 2010

### vela

Staff Emeritus

$$x^{\beta\alpha} = (x^\beta)^\alpha = (\beta\circ x)^\alpha$$

5. Feb 16, 2010

### Dustinsfl

Do you know how to do superscript in Maple with the greeks, by any chance? I can only do it with standard letters.

6. Feb 16, 2010

### Dustinsfl

Here is the completed Vector Space problem: Can you tell me if you see any issues? Thanks.
{∀ 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.

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. x ⊕ 1 = x·1 = x
4. -x = -1∘x = x^-1 = 1/x ⇒x ⊕ (-x) = x·1/x = 1
5. α∘(x ⊕ y) = α(x·y) = (x·y)^α = x^α·y^α = x^α ⊕ y^α = α∘x ⊕ α∘y
6. (α + β)∘x = x^(α + β) = x^α·x^β = x^α ⊕ x^β = α∘x ⊕ β∘x
7. (α·β)∘x = x^(α·β) = x^(β·α) = (x^β)^α = (β∘x)^α = α∘(β∘x)
8. 1∘x = x^1 = x

Closure Properties:
1. If x ϵ ℝ+ and α is a scalar, then α∘x ϵ ℝ+.
α∘x = x^α > 0 ∴ α∘x ϵ ℝ+
2. If x,y ϵ ℝ+, then x ⊕ y ϵ ℝ+.
x ⊕ y = x·y > 0 ∴ x ⊕ y ϵ ℝ+

Yes, ℝ+ is a vector with these operations.

7. Feb 16, 2010

### vela

Staff Emeritus
Nope, sorry. I've never used Maple. Someone else can probably answer your question, hopefully.

8. Apr 6, 2011

### SpY]

Hi, I don't mean to necro topics but I just have a quick question on the same title:

For the above Vector Space {∀ x ϵ ℝ+ : x>0} to be a vector space, is it necessary to contain the zero vector? (Which it doesn't, since x>0 so I'm thinking it's not a vector space).

9. Apr 6, 2011

### vela

Staff Emeritus
The zero vector is the identity element for vector addition. What it is depends on your definition for vector addition. It doesn't have to actually equal 0, the identity element of the usual addition of real numbers.

10. Apr 6, 2011

### HallsofIvy

It is necessary that a vector space be non-empty. But once you have a vector, v, in the vector space, (-1)v= -v is in the set and so v+ (-v)= 0 is in the set. So just saying that the vector space is non-empty is equivalent to saying that it contains the 0 vector.

11. Apr 7, 2011

### SpY]

Hmm I might be missing the point here, but say we use ordinary addition and scalar multiplication. My question is that for a VS defined to be only positive reals, there is no number you can add to a number x such that you obtain x. So there is no zero element since zero is non positive (not in the set) so R+ is not a vector space!

12. Apr 7, 2011

### vela

Staff Emeritus
Yes, with regular addition and multiplication, the set of positive reals does not satisfy the requirements of a vector space. But that's not the question that was originally asked in this thread.

The point you seem to be missing is that a vector space isn't simply a set. It's a set and two binary operators which define scalar multiplication and vector addition. x=1 is the "zero" vector, the identity element for vector addition as it was defined in this problem. R+ and the given operators do satisfy the axioms of a vector space.