Proving a specific vector space with two operations 
Prove that
  $$V = \{ (a,b) \in \Bbb R^2 : a,b > 0 \}$$
  is a vector space with the operations
  $$(a,b) \oplus(c,d) = (ac,bd)  \,\,\,\forall  (a,b),(c,d) \in V$$
  and
  $$\alpha(a,b) = (a^\alpha,b^\alpha)\,\,\,\forall\alpha \in \Bbb R\,\,\,\forall(a,b) \in V.$$

I know that I must show the 8 properties but this $(ac,bd)$ is confusing me in two properties and the $\alpha$ exponent also is driving me crazy to show the multiplication properties.
Could anyone help?
 A: We need closure first of all. Do the two operations in question produce vectors that are still in the space? The answer is yes; the multiplication of two positive numbers is positive, and a positive number raised to any real number is also positive. Check. Then we move on to:


*

*Associativity of addition:
\begin{align*}
(a,b)\oplus((c,d)\oplus(e,f))&=(a,b)\oplus(ce,df)\\
&=(a(ce),b(df))\\
&=((ac)e,(bd)f)\\
&=(ac,bd)\oplus(e,f)\\
&=((a,b)\oplus(c,d))\oplus(e,f),
\end{align*}
as required.

*Commutivity of addition. Perform a similar calculation as above; the result follows from the commutivity of real multiplication.

*Identity element of addition. I think you'll find this is $(1,1):$
\begin{align*}
(a,b)\oplus(1,1)&=(a\cdot 1, b\cdot 1)\\
&=(a,b)\\
&=(1\cdot a,1\cdot b)\\
&=(1,1)\oplus(a,b).
\end{align*}

*Inverse elements of addition. I think you'll find that, for vector $(a,b),$ its inverse is $(1/a,1/b).$ Note that $a,b>0,$ so this is defined. We'd have
\begin{align*}
(a,b)\oplus(1/a,1/b)&=(a/a,b/b)\\
&=(1,1),
\end{align*}
the additive inverse. And you can check that the other direction of multiplication works as well.

*Compatibility of scalar multiplication. This says that we need $a(b(c,d))=(ab)(c,d),$ where $a$ and $b$ are scalars. We have
\begin{align*}
a(b(c,d))&=a(c^b, d^b)\\
&=((c^b)^a,(d^b)^a)\\
&=(c^{ba},d^{ba})\\
&=(ba)(c,d)\\
&=(ab)(c,d),
\end{align*}
as required.

*Identity element for scalar multiplication. This'd be $1:$
$$1(a,b)=(a^1,b^1)=(a,b). $$

*Distribution over vector addition. We need
$$a((b,c)\oplus(d,e))=(a(b,c))\oplus(a(d,e)). $$
So, we have
\begin{align*}
a((b,c)\oplus(d,e))&=a(bd,ce)\\
&=((bd)^a,(ce)^a)\\
&=(b^ad^a,c^ae^a)\\
&=(b^a,c^a)\oplus(d^a,e^a)\\
&=(a(b,c))\oplus(a(d,e)),
\end{align*}
as required.

*I'll let you show that $(a+b)(c,d)=a(c,d)+b(c,d).$
A: Consider the map $V \to \mathbb{R} \oplus \mathbb{R}$ that sends $(a, b)$ to $(\ln a, \ln b$). This map converts coordinatewise products into coordinatewise sums and coordinatewise powers into scalar multiples, and is a bijection. It follows that since the codomain is a vector space, so is the domain, and the map is in fact an isomorphism of vector spaces.
