Let $\phi: \QQ \longrightarrow \QQ$ be an automorphism of $(\QQ, +)$. Show that
for any $q \in \QQ$, we have $\phi(q) = q\phi(1)$.
Proposed as exercise 4.55 in Abstract Algebra: An Introductory Course, by Gregory T. Lee.
The proof of this result is very similar to that of problem
21.
Notice that if $q \in \ZZ$ then $$ \phi(q) = \phi(\underbrace{1 + 1 + \cdots + 1}_{q\text{ times}}) = \underbrace{\phi(1) + \phi(1) + \cdots + \phi(1)}_{q\text{ times}} = q\phi(1), $$ whereas if $q = r/s \in \QQ$ it follows that $$ \begin{align*} &r\phi(1) = \phi(r) = \phi\underbrace{\left(\frac rs + \frac rs + \cdots + \frac rs\right)}_{s\text{ times}} = s\phi\left(\frac rs\right) = s\phi(q)\\ \implies &\phi(q) = \frac rs \phi(1) = q\phi(1), \end{align*} $$ concluding the proof.
Notice that if $q \in \ZZ$ then $$ \phi(q) = \phi(\underbrace{1 + 1 + \cdots + 1}_{q\text{ times}}) = \underbrace{\phi(1) + \phi(1) + \cdots + \phi(1)}_{q\text{ times}} = q\phi(1), $$ whereas if $q = r/s \in \QQ$ it follows that $$ \begin{align*} &r\phi(1) = \phi(r) = \phi\underbrace{\left(\frac rs + \frac rs + \cdots + \frac rs\right)}_{s\text{ times}} = s\phi\left(\frac rs\right) = s\phi(q)\\ \implies &\phi(q) = \frac rs \phi(1) = q\phi(1), \end{align*} $$ concluding the proof.