Let $Q$ follow subspace topology from $R$ Then I think it is true that $Q^n$ and $Q^m$ (with product topology) are not homeomorphic.I also think it will be possible to define "rational" homotopy groups by considering $[0,1]\cap Q$ etc. But I can't find any reference regarding topology of product spaces of rationals in general . So I was wondering if someone can give a sketch of a proof and give some references.

Edit: The following is a second attempt to repair problems in earlier proposed solutions, as pointed out by Gerald Edgar in comments. Hopefully this time I've gotten it right this time. It's classical (by a backandforth argument; see for instance A Shorter Model Theory by Hodges) that any two countable dense unbounded linear orders without endpoints are isomorphic. For example the linear order $\mathbb{Q}$ is isomorphic to $L = \mathbb{Z}_{(2), (5)} \cap (0, 1)$, referring here to the integers localized at the primes 2 and 5 (i.e., rationals whose decimal expansions don't have an infinite tail of 9's or 0's  this is for technical reasons). $L$ and $\mathbb{Q}$ are homeomorphic under their order topologies. Let $f: L \times L \to \mathbb{Q} \cap (0, 1)$ be the map that takes a pair of elements $\alpha, \beta \in L$ and forms a rational number by interleaving their decimal expansions, with the decimal expansion of $\alpha$ appearing in odd places and that of $\beta$ in the even places. Let $I$ be the image of $f$. The map $f: L \times L \to I$ is a homeomorphism, and $I$ is again a countable dense linear order without endpoints, hence homeomorphic to $\mathbb{Q}$ again. Then the evident composite $$\mathbb{Q} \times \mathbb{Q} \cong L \times L \stackrel{f}{\to} I \cong \mathbb{Q}$$ is a homeomorphism $\mathbb{Q} \times \mathbb{Q} \cong \mathbb{Q}$. 

