I have seen in many textbooks on analysis that the Archimedean property of reals is a consequence of the completeness axiom. However I am not convinced that we need to use such a powerful axiom (as the completeness axiom) to prove a very basic property like Archimedean Property. To me it looks simple enough as follows:


1) If a and b are natural numbers (like 1, 2, 3, ..., but not 0) and there is another natural number c such that b = a + c, then we write b > a. Using this we can see that given any natural number we can find another greater than it.


2) If a & b are positive rationals then a/b is rational and we can write a/b = p/q (p, q positive integers) and by division algorithm we have a non-negative integer p = qx + r with 0 <= r < q. Then clearly we have a positive integer (x + 1) > p/q = a/b. So that field of 
rationals possesses the Archimedean property.


3) If a, b are positive reals then a/b is also real. Any definition of real numbers (Dedekind's or Cauchy's for example) will lead to the fact that given a real number there is a rational greater than it and a rational less than it. So we have rational c > a/b. And clearly by Archimedean Property of rationals (point 2 above) we have a positive integer 'n' greater than 'c'. Thus we finally have a/b < n or a < nb.


I am really not sure why textbooks try to prove Archimedean property via completeness axiom. If it was really the case Archimedean property is a consequence of completeness property, then how come rational possess Archimedean property but not completeness axiom.


Please clarify if there is any mistake in above reasoning or the textbooks are treating this in high handed manner.



Archimedean Property of Real Numbers
One cannot prove the Archimedean property for the reals by appealing only to first-order algebraic truths of the ordered real field and the subring of the integers sitting inside it. The reason is that those statements are all first-order expressible in the structure $\langle\mathbb{R},{+},{\cdot},0,1,\lt,\mathbb{Z}\rangle$, and there are nonstandard models $\langle\mathbb{R}^\ast,{+},{\cdot},0,1,\lt,\mathbb{Z}^\ast\rangle$, which are not Archimedean and yet which satisfy exactly the same first-order truths as the standard model. In this sense, there cannot be a truly elementary proof of the Archimedean property. 


What this observation shows is that the Archimedean property is fundamentally a second-order property of the reals, and to establish it one must appeal to the fact that one is using the actual standard integers instead of merely some first-order property of the integers and how they relate to the reals. 


Your argument, which is fine, does this by appealing to the explicit constructions of the reals, say, as Dedekind cuts, rather than to its characterization by the completeness property. 


So while I agree that there are other proofs of the Archimedean property that don't appeal to completeness---and as you point out there certainly are incomplete Archimedean ordered fields---nevertheless I don't mind the proof from completeness, since this is a defining property characterizing the reals, and the proof from completeness is both easy and clear.


