Is $V$ a vector space under these operations? Let $$V=\{x \in \mathbb{R} : x>0\}$$
For $x,y,a \in \Bbb{R}$, define $x\oplus y=xy$ and $x\odot a=x^a$. Is $V$ a vector space under these operations? Justify your answer.
 A: Define $\phi:V \to \mathbb{R}$ by $\phi(v) = \log v$. Note that $\phi$ is bijective.
We see that $\phi( x\oplus y) = \phi(xy)=\log x + \log y = \phi(x)+\phi(y)$, and $\phi( x \odot a) =\log(x^a) = a \log x = a \phi(x)$.
Since $(\mathbb{R}, +,\cdot)$ is a vector space, it follows that $(V,\oplus, \odot)$ is a vector space.
A: To be a vector space, you must first check that it satisfies the abelian group axioms: For all $x,y,z\in V$:


*

*$x\oplus y\in V$

*$x\oplus y=y\oplus x$

*$(x\oplus y)\oplus z=x\oplus (y\oplus z)$

*There is an element $e\in V$ such that $x\oplus e=x$

*$x\oplus x^*=e$ for some $x^*\in V$


Furthermore, it must "play nicely" with the underlying field (in this case, $\mathbb{R}$). Rigorously, this means satisfying all of these axioms, for all $x,y\in V$ and all $a,b\in\mathbb{R}$:


*

*$x\odot a\in V$

*$x\odot 1=x$

*$x\odot (ab) = (x\odot a)\odot b$

*$(x+y)\odot a = (x\odot a)+(y\odot b)$

*$x\odot(a+b) = (x\odot a)+(x\odot b)$


where $+$ is the usual addition.
To not be a vector space, all you need is one set of $x,y,z,a,b$ that fails just one of these axioms (or a proof that no such $e$ can exist). So if one of them seems unlikely to you, you might start by looking at simple examples to see if they hold up.
A: It's not hard to verify $\oplus$ and $\odot$ are closed operations. Now $\langle V,\oplus\rangle$ is associative, commutative, has an identity $1$, and inverse is possible $x^{-1}$ (we say it's an Abelian group). For $\langle V,\odot\rangle$, it satisfies two distributive laws, $(x^b)^a=x^{ab}$, and $x^1=x$. Hence, $\langle V,\mathbb{R},\oplus,\odot\rangle$ satisfies the definition of a vector space.
