Automorphisms of reals fixing rationals

Our goal today is to show that $\text{Aut}(\mathbb R /  \mathbb Q) = 1$. We follow the outline suggested in the exercises of section 14.1 of Dummit and Foote.

Let $\sigma$ be an automorphism of the reals fixing the rationals. Let $k \in \mathbb R$, and observe that
$$\sigma(k^2) = \sigma(k)\sigma(k) = \sigma(k)^2.$$
Any positive real can be written as a square in $\mathbb R$, so applying $\sigma$, we get another square and hence another positive real.

Next, suppose $a < b$ for any $a, b \in \mathbb R$. Then $0 < b - a$, so $0 < \sigma(b) - \sigma(a)$,
so $\sigma(b) > \sigma(a)$. Therefore $\sigma$ must be increasing.

Now, let $m$ be a positive integer. Then
$$-\frac 1m < b - a < \frac 1m \implies -\frac 1 m < \sigma(b) - \sigma(a) < \frac 1 m$$
by applying $\sigma$, so $\sigma$ is continuous. Every real number is a limit of a sequence of rationals, so since $\sigma$ is the identity on the rationals, it must also be the identity on every real number.