# Complete ordered field is an Archimedean field that cannot be extended to an Archimedean field

As a bonus problem, our professor of real analysis asked us to prove that the real numbers (a complete ordered field) cannot be extended into an Archimedean field, with no definition of what he meant by extending.

I have tried using proof by contradiction to show that if we have some set, $\mathbb{R}^{*}$ such that $\mathbb{R}$ is a proper subset of $\mathbb{R}^{*}$, and that $\mathbb{R}^{*}$ forms an Archimedean field, then because for there to be new elements in $\mathbb{R}^{*}$ as opposed to $\mathbb{R}$, they would have to be larger (or smaller) than all the elements of R. But then we would have reached a contradiction with the Archimedean property.

Professor returned this solution and said it's not the right solution (without any further comment). Can anyone offer some enlightenment on what I have done wrong or what I could try now?

All I have found on Google are mentions in textbooks that go along the lines of "every Archimedean field is isomorphic to a subfield of real numbers". That would imply that we cannot extend reals into an Archimedean field, but how can one go about proving that?

I can only use basic definition of an ordered (Archimedean) field and other "basics", we have not yet covered sequences, etc.

Outline: Let $F$ be a proper Archimedean extension field of $\mathbb{R}$. Let $\eta$ be an element of $F$ which is not in $\mathbb{R}$.

By taking the additive inverse if necessary, we can assume that $\eta$ is positive. By the assumed Archimedean property of $F$, there is an integer greater than $\eta$.

Thus the set of $x$ in $\mathbb{R}$ which are $\lt \eta$ is bounded above, and hence has a least upper bound $b$.

Now comes the key fact, that is being left for you to prove: There is no rational number between $0$ and $|\eta-b|$.

From this you can conclude that $\frac{1}{|\eta-b|}$ is greater than any integer, contradicting the assumption that $F$ is Archimedean.

Remark: In a sense, this fleshes out your notion that any "new" element is bigger or smaller than every element of $\mathbb{R}$. That part was not correct. However, from a new element $\eta$, assumed positive and smaller than some integer, we produce an element $\frac{1}{|\eta-b|}$ which is larger than any integer.