Can axioms of the Euclidean space be proven in the Real space? I'm not a mathematical logic student so my question can be naïve. I'm thinking if we identify $\Bbb{E}^2$ with $\Bbb{R}^2$( as a normed space with the Euclidean norm), Then can we prove all Euclidean axioms (Hilbert's axioms) by the axioms of the real number system?
If so what does it say about the completeness of the real analysis?
 A: I will attempt to give you a vague outline what needs to be done to answer your first question. The second one I don't really understand, so I'll leave it for now.
For your first question, I understand that you mean the first-order axioms of real number system; you can't really do that for the original Euclid's axioms (they're too vague), and you can't for Hilbert's axioms either, because they're not first-order (though you can with some, most likely rather considerable effort show all the axioms except the last two). You can do it, however, for Tarski's axioms.
To do that in a formal sense, you need to somehow intepret the primitive notions of this axiomatisation, that is, those of a point, betweenness and congruence of segments. This is not hard: a point is a triple of numbers, and betweenness and congruence are definable because the real field can calculate the euclidean norm and it knows what a convex combination of two points is.
Once you have that, the Tarski's axioms follow from the axioms of the real field, since by Godel's completeness theorem any true statement has a proof, and the theory of real numbers is complete and is satisfied by the usual $({\bf R},+,\cdot,0,1)$, which also has all the properties postulated by Tarski under our interpretation (because, of course, ${\bf R}^3$ certainly is an euclidean space!). Alternatively, you can with some effort prove each axiom, but this can be a little tricky because of the axiom schema of continuity, which actually is an infinite set of axioms, but not all that hard.
