I'm trying to prove that a particular $V$ is a $\Bbb{Q}$-vector space. The question says to take the element $0_V = 1$, the function $+_V : V \times V \to V$ given by the function $[x +_V y = xy]$, and the function ${}_V: \Bbb{Q} \times V \to V$ given by $[a{}_Vx = x^a]$.
This notation isn't used in lecture or anywhere in the text, what does it mean?
The point of abstract algebra is to recognise when two apparently different situations are "essentially the same". In studying linear algebra you have probably already seen many examples of vector spaces, including pairs $(x,y)$ and triples $(x,y,z)$ of real numbers, matrices of a fixed size, polynomials and so on. We can call all of these "vectors", and they are "essentially the same" because in each case we can add vectors, multiply them by numbers, and they satisfy the same basic laws. For example, $${\bf u}+{\bf v}={\bf v}+{\bf u}$$ for all ${\bf u},{\bf v}$ in $\mathbb R^3$, and $$p+q=q+p$$ for all $p,q$ in the set of real polynomials.
To come to the question you are asking, the idea is that if we can change the meaning of the word "vector", we may as well also change the meaning of "addition", "scalar multiplication" and "zero", as long as we do so in such a way that the basic axioms are still true. In fact, we have done this already - addition of triples is not the same as addition of polynomials, though they are related - but the question you ask seeks to change them more radically.
You didn't mention what $V$ is but I suspect it's $\mathbb R^+\!$, the set of positive real numbers. The notation $$x\mathbin{+_V}y=xy$$ means, look at your vector space axioms, and whenever you see addition of vectors, replace it by multiplication. So the associative law $(x+y)+z=x+(y+z)$ now actually means $$(xy)z=x(yz)\ ,$$ and you have to decide whether this is true or false. (Easy, of course!) Similarly, a product $\alpha x$ is to be replaced by exponentiation, $x^\alpha$. So the distributive law $\alpha(x+y)=\alpha x+\alpha y$ becomes $$(xy)^\alpha=x^\alpha y^\alpha$$ and once again you have to decide if it's true. Likewise, you will replace the zero vector by $1$, and the negative of $x$ by... but I think you can work this out for yourself. Good luck!
