Is the scalar multiplication of (R>0)^n over Q associative?

[ilyas.h]

∴

Homework Statement

Let ℝ>0 together with multiplication denote the reals greater than zero, be an abelian group.

let (R>0)^n denote the n-fold Cartesian product of R>0 with itself.

furthermore, let a ∈ Q and b ∈ (ℝ>0)^n

we put a⊗b = $(b_1)^a + (b_2)^a + ... + (b_n)^a$

show that the abelian group (R>0)^n together with scalar multiplication

Q x (R>0)^n = (R>0)^n,
(a, b) = (a⊗b)

be a vector space over Q.

The Attempt at a Solution

proof of associativity:

p,q in Q
b in (R>0)^n

p(qb) = (pq)b

===> p(q ⊗ b) = (pq)⊗b

LHS:

p(q⊗b) = p⊗$((b_1)^q + (b_2)^q + ... + (b_n)^q)$

= $((b_1)^{pq} + (b_2)^{pq} + ... + (b_n)^{pq})$

∴associativity true.

 

[Dick]

∴

Homework Statement

Let ℝ>0 together with multiplication denote the reals greater than zero, be an abelian group.

let (R>0)^n denote the n-fold Cartesian product of R>0 with itself.

furthermore, let a ∈ Q and b ∈ (ℝ>0)^n

we put a⊗b = $(b_1)^a + (b_2)^a + ... + (b_n)^a$

show that the abelian group (R>0)^n together with scalar multiplication

Q x (R>0)^n = (R>0)^n,
(a, b) = (a⊗b)

be a vector space over Q.

The Attempt at a Solution

proof of associativity:

p,q in Q
b in (R>0)^n

p(qb) = (pq)b

===> p(q ⊗ b) = (pq)⊗b

LHS:

p(q⊗b) = p⊗$((b_1)^q + (b_2)^q + ... + (b_n)^q)$

= $((b_1)^{pq} + (b_2)^{pq} + ... + (b_n)^{pq})$

∴associativity true.

So far so good. What's your question?

 

[ilyas.h]

So far so good. What's your question?

just wanted to clarify if my logic is correct, I struggle on these sorts of Q's. Thanks.