It is obvious that Q_r is topologically isomorphic to Q_s while r and s denote different primes.But I really don't know whether it is true in the aspect of algebra.As I failed to prove it,I think that it is false,but I can't give a counterexample. Last I'm quite sorry that I'm new to MathJax and I don't know how to use it properly.Thanks for reading and I would appreciate it if you could solve my problem.

closed as too localized by Alex Bartel, Felipe Voloch, Chandan Singh Dalawat, GH from MO, Igor Rivin Dec 23 '11 at 15:00
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For a prime $p \neq 2$, the multiplicative group $\mathbb{Q}_p^\times$ is isomorphic to $\mathbb Z \times \mathbb Z_p \times \mathbb Z/(p1)$, while $\mathbb Q_2^\times$ is isomorphic to $\mathbb Z \times \mathbb Z_2 \times \mathbb Z/2$, cf. Serre, A course in arithmetic, Theorem II.3.2 (p. 17). Since a field isomorphism $\varphi\colon \mathbb Q_p \to \mathbb Q_r$ will preserve the torsion subgroup of the multiplicative subgroup, this shows that $\mathbb Q_p \not\cong \mathbb Q_r$ whenever $\{p,r\} \neq \{2,3\}$. The remaining case is taken care of by the fact that $\mathbb{Q}_p^\times$ / $\mathbb {Q}_p ^{\times 2}$ $ \cong \mathbb Z/2 \times \mathbb Z/2$ whenever $p \neq 2$, while $\mathbb Q_2^\times$ / $\mathbb Q_2^{\times 2} \cong \mathbb Z/2 \times \mathbb Z/2 \times \mathbb Z/2$, cf. loc. cit., p.18. 

