Exercise 2 (b) on page 100 of Analysis I by Amann and Escher asks me to show that $\mathbb Q$ is the smallest subfield of $\mathbb R$.
Wolfram MathWorld gives the following reasoning:
I don't know what a division algebra is, so I have the following alternative proof.
Assume we removed some rational $a/b$. Because the equation $c/d + x = a/b$ has a unique solution $x \in \mathbb Q$ for any $c/d \in \mathbb Q$ (proven on page 53), in order to maintain the additive group axioms we would have to remove $c/d$ and $x$. But since the equation holds for any $c/d \in \mathbb Q$, we would have to remove all elements of $\mathbb Q$ and therefore would end up with the empty set, which is not a field.
Is this valid? I appreciate any feedback.
It's not valid. In order to maintain the group axioms you would have to remove either $x$ or $c/d$, but not necessarily both. Indeed, if this were valid you would have shown that $\mathbb{Q}$ was the smallest additive subgroup of $\mathbb{R}$ - but this is clearly false, witness $\mathbb{Z}$, or for that matter $\{0\}$.
Trying to think about "removing" elements makes it unnecessarily complicated. Just suppose $K$ is an arbitrary subfield of $\mathbb{R}$, let $a/b$ be an arbitrary rational, and try to use the field axioms to show that $a/b$ has to be in $K$. (You could start by showing that $a$ and $b$ are each in $K$.)