The answers above seem to suggest there is a homomorphism from this interesting little line group to $\mathbb{R}$ sending the cartesian product to multiplication. But we have $A^1 \times A^2 =A^3$ when the answer we want is surely $A^2$. Cartesian product is additive on dimensions so perhaps eg. Andre has found a value for $ \frac12 A$. My guess as to how to get a multiplicative structure would be $X*Y:=hom(X,Y)$ in some suitable category, for example if $A^i=\mathbb{R}^i$ then linear maps would do the job. But now $\sqrt{\mathbb{R}}=[Y, hom(Y,Y)=\mathbb{R}] =\mathbb{R}$ and the real (and probably impossible) fun comes in finding '$\sqrt{\mathbb{R}^2}$'
[Edit: re-read question- didn't realise those were powers of $A$- thought it was just an index! Oh well- will leave this here as I think it's an interesting reformulation...]
Edit2: Bit of a fiddle but you can make this work in the category of $\mathbb{Q}$ Vector spaces with continuous $\mathbb{Q}$-linear maps where $\sqrt{\mathbb{Q}^2}=\mathbb{Q}[\sqrt{m}]$ for any non-square m and $\sqrt{\mathbb{Q}^n}=\mathbb{Q}[^n\sqrt{m},^n\sqrt{m}^2...]$ for any non-nth power m, we've lost the opportunity for fractions and a cube root seems unlikely, but it's sort of neat...
