This is a follow-up question of a previous one.
Is it true ${}^{*}\mathbb{Q}={}^{*}\mathbb{R}$ ?
It seems to me it's true because for any $x \in{}^{*}\mathbb{R}$, there is a $y \in {}^{*}\mathbb{Q}$ such that $y < x$ and vice versa, and because for any $x' \in \mathbb{R}$, there is a $y' \in {}^{*}\mathbb{Q}$ such that $y' = x'$.
On
No, ${}^*\Bbb Q\ne{}^*\Bbb R$. It’s not true that $\Bbb R\subseteq{}^*\Bbb Q$. In the ultrapower construction this is especially easy to see: in ${}^*\Bbb R$, $\Bbb R$ is identified with the set of equivalence classes of constant sequences of reals, and the class of a constant sequence with irrational value is easily seen to be different from every element of ${}^*\Bbb Q$, though it’s also easy to see that there are elements of ${}^*\Bbb Q$ infinitely close to each such class.
On
By transfer, every element in $^*\mathbb Q$ is a quotient of two elements from $^*\mathbb Z$. By transfer, not every element in $^*\mathbb R$ is the quotient of two elements from $^*\mathbb Z$. So, the two enlargements are not equal.
What you are alluding to (I think) in the question is that both enlargements give rise the same completion of $\mathbb Q$, namely $\mathbb R$.
On
The simplest argument is to transfer the formula $\mathbb{Q}\not=\mathbb{R}$.
In more detail, the symbols "Q" and "R" in the language are interpreted as respectively the rationals $\Bbb Q$ and the reals $\Bbb R$ in the standard model, and as $^\star\Bbb Q$ and $^\star\Bbb R$ in the non-standard model. The fact that $\Bbb Q$ is PROPERLY contained in $\Bbb R$, and therefore distinct from it, is encoded by the formula Q$\not=$R. By the transfer principle, the same formula remains true when interpreted over the nonstandard model. Hence $^\star\Bbb Q\not= {}^\star\Bbb R$.
No. The theory of $\Bbb R$ is not the same first-order theory of $\Bbb Q$. If you define $^*\Bbb Q$ and $^*\Bbb R$ as ultrapowers then they have the same first-order theories as their standard counterparts.
In particular, $^*\Bbb Q$ satisfies $\forall x(x\cdot x\neq(1+1))$; whereas $^*\Bbb R$ satisfies $\exists x(x\cdot x=1+1)$, and that $x$ is the equivalence class of $f(n)=\sqrt2$ in the ultrapower.
You can do a bit better, e.g. since in $\Bbb R$ it is true that every odd-degree polynomial has a root, so we can find many algebraic numbers in $^\ast\Bbb R$ which do not exist in $^*\Bbb Q$.
One more thing on the use of the transfer principle. The transfer principle is really just Los theorem, that the ultrapower of a structure is elementary equivalent to it, i.e. a sentence is true in the ultrapower if and only if it is true in the structure.
Structures, and models come with a language. If your language is the language of ordered-fields then $\Bbb Q$ is not definable. This is because definable sets of real numbers are finite unions of intervals (possibly one-point intervals). The rationals are clearly not such set.
If, on the other hand, you are free to add whatever predicate that you choose to add and whatever function, constants or otherwise, then clearly you can add an unary predicate symbol $Q$ which is interpreted as true for $x$ if and only if $x\in\Bbb Q$. Since $\Bbb Q\models\forall x.Q(x)$ we have that $^*\Bbb Q$ must satisfy the same sentence, and by a similar reasoning we have that $Q^{^*\Bbb R}=\, ^*\Bbb Q$. So it's quite obvious now that $\exists x.\lnot Q(x)$ is true in $\Bbb R$ and therefore in $^*\Bbb R$, and so we have $^*\Bbb R\neq\, ^*\Bbb Q$.
But usually when we talk about $^*\Bbb R$ we talk about it as an ordered field, in which case there is no $Q$ predicate like that, and the rationals are not definable there. So we have that there is no formula which states $\Bbb Q\neq\Bbb R$.