# Show that $\left< \mathbb{Q}\left[\sqrt 2\right], < \right> \simeq \left<\mathbb{Q}, < \right>$

Show that $$\left< \mathbb{Q}\left[\sqrt 2\right], < \right> \simeq \left<\mathbb{Q}, < \right>$$,

where $$\mathbb Q$$[$$\sqrt 2$$] is the smallest subfield of $$\mathbb R$$ containing $$\sqrt 2$$.

I do not really know where to go from here, i.e. start the proof.

• Welcome to Math.SE! Please update your question with the context in which you encountered the problem (e.g. a homework question) and your thoughts on how to approach it, as well as where you are getting stuck. We will be glad to guide you further. Mar 12, 2019 at 18:15
• i do not really know where to go from here. i.e. start the proof Mar 12, 2019 at 18:17
• For my part, I was uncertain what the symbol $\simeq$ was to denote. I guess, from the answer given below, that it’s to mean “order isomorphic”, but it will help readers for you to explain the context, as @gt6989b said. Mar 12, 2019 at 18:47

Any two "countable, dense, totally ordered sets without endpoints" are isomorphic (with respect to the order).

"Dense" refers to that for any $$a < b$$, there exists a $$c$$ inbetween; so there is a $$c$$ such that $$a < c < b$$. And "without endpoints" means that for any $$a$$, there is at least one larger and at least one smaller element (there exists $$b > a$$ and there exists $$c < a$$).

Your problem is an instance of this because $$\mathbb{Q}$$ and $$\mathbb{Q}[\sqrt{2}]$$ are both countable (prove this), dense (prove this also), and without endpoints (prove this finally).

Now to prove that any two countable, dense, totally ordered sets without endpoints are isomorphic: this is a classic result that we approach as follows. Let the sets be $$A$$ and $$B$$. Since they are countable, we can number them: \begin{align*} A &= \{a_1, a_2, a_3, \ldots \} \\ B &= \{b_1, b_2, b_3, \ldots \} \end{align*} (But be careful to note that these numberings are not in order, so it's not true that $$a_1 < a_2 < a_3 < \cdots$$. They could be in a completely random order.)

Anyway, the idea is then to match up $$A$$ and $$B$$ one element at a time: we build an isomorphism between the two:

• First pick the element $$a_1$$ of $$A$$. Match it with the first unmatched element of $$B$$, which is $$b_1$$.

• Now pick the first element of $$B$$ on the list that is not yet matched, that is, $$b_2$$. If $$b_2 > b_1$$, then pick some element of $$A$$ that is larger than $$a_1$$, say, $$a_{i_1}$$; otherwise pick some element of $$A$$ that is smaller than $$a_1$$. Match this element with $$b_2$$.

• Now pick the first element of $$A$$ on the list that is not matched. Maybe it's $$a_2$$, or $$a_3$$, for example. Consider how it compares to all the previous $$a$$'s in the ordering ($$<$$), and pick an element of $$B$$ that compares the same way. This is always possible due to "denseness" and "without endpoints".

Basically, we keep alternately picking the first element on the $$A$$ list that is not matched, and the first element on the $$B$$ list that is not matched; we compare the element with everything matched up so far, and we pick a corresponding element of $$B$$ (or $$A$$) to match it to which is consistent with the total orderings $$<$$ on $$A$$ and $$B$$.

This always preserves the ordering $$<$$ by construction -- and it actually eventually matches all elements of $$A$$ and all elements of $$B$$ , because we always match up the first element on the lists that is not matched, so we eventually get to every element of $$A$$ and every element of $$B$$. So it is an isomorphism.