Automorphisms in $\mathbb{R}$

Question

Let $\phi: \mathbb{R}\rightarrow \mathbb{R}$ be an automorphism. Suppose $p=\frac{m}{n}$ is a rational number. Then is it true that $\phi(p)=\frac{\phi(m)}{\phi(n)}$?

I got this problem while doing an algebra problem. In the original problem it has been asked to prove that $\phi(\mathbb{Q})=\mathbb{Q}$.

Answer 1

Hint:

Write down $\phi (p)=\phi(\frac m n)=\phi (m\cdot n^{-1})$

Now, since $\phi$ is an automorphism, in particular it is a homomorphism. What homomorphism properties can you use?

Answer 2

Let $\phi : \mathbb{R} \rightarrow \mathbb{R}$ be an automorphism. So $\phi (1) = 1.$ Thus, for each $n \in \mathbb{N}, \phi (n) = \phi(1 + 1 + \cdots + 1) = \phi (1) + \phi (1) + \cdots + \phi(1) = n \phi(1) = n.$ Also $\phi(-n) = - \phi(n) = -n.$ Hence $\phi(n) = n, \forall n \in \mathbb{Z}.$

Now let $\dfrac{m}{n} \in \mathbb{Q}.$ So $\phi (m) = m\phi (1) = m \Rightarrow \phi(n. \dfrac{m}{n}) = m \Rightarrow n\phi(\dfrac{m}{n}) = m \Rightarrow \phi(\dfrac{m}{n}) = \dfrac{m}{n}.$