Proof that an algebraic structure is a vector space.

Question

Let R+ be the set of all real numbers that are strictly positive. Let x and y be two elements of R+ and let c be any real number. Define the sum of elements of R+ as

x "+" y = xy (multiplication of real numbers)

and the multiplication by a scalar as

c'*'x = x^c (exponentiation of real numbers)

Prove that R+ is a real vector space with these two definitions and find the element of R+ that acts as the 'zero' vector of the space.

Answer 1

Hey there, so to boil it down, to prove that a set under two binary operations is a vector space, we really just need to show 10 things. I'll list the first 5 which relate to the defined "+", and the second 5 relating to the defined "*".

1.) The set is closed under "+" (if you take any x,y in R⁺ and perform "+", your solution stays in R⁺

2.) The set is commutative under "+"

3.) The set is associative under "+"

4.) There exists a 0 vector (some call this the additive identity) with respect to "+". That is there is some vector (0) such that x "+" 0 = x for all x in the set.

5.) For each x in our set, there exists an additive inverse. So for any x, we have some number (x^-1) such that x + x^-1 = 0. (Important note, this is not the scalar zero, this is the additive identity that we defined in number 4.)

6.) The set is closed under "*"

7.) We have the distributive property of scalar over vectors

8.) We have the distributive property of vector over scalars

9.) We have the associative property of scalar multiplication

10.) We have scalar multiplication by "1" (the multiplicative identity element in our field). ie 1*x = x for any x in our set.

That's a lot, and it's cumbersome, but you need to go through all 10 of those points and show that they hold for our set, and for how we are defining + and *.

Hope that helps to get you started. If you're still stuck, feel free to reach out and we can set up an online session to discuss in further detail.