# Is an automorphism of the field of real numbers the identity map?

Is an automorphism of the field of real numbers $\mathbb{R}$ the identity map?
If yes, how can we prove it?

Remark An automorphism of $\mathbb{R}$ may not be continuous.

Here’s a detailed proof based on the hint given by lhf.

Let $\phi$ be an automorphism of the field of real numbers.
Let $x \gt 0$ be a positive real number.
Then there exists $y$ such that $x = y^2$.
Hence $\phi(x) = \phi(y)^2 \gt 0$.

If $a \lt b$, then $b - a \gt 0$.
Hence $\phi(b) - \phi(a) = \phi(b - a) \gt 0$ by the above.
Hence $\phi(a) \lt \phi(b)$.
This means that $\phi$ is strictly increasing.

If $n$ is a natural number, it can be written in the form $1 + \ldots + 1$, so $\phi(n) = n$. Now, any rational number is of the form $r = (a - b)c^{-1}$, for $a, b, c$ natural numbers, so it follows that $\phi(r) = r$ for any rational number.

Let $x$ be a real number.
Let $r, s$ be rational numbers such that $r \lt x \lt s$.
Then $r \lt \phi(x) \lt s$.
Since $s - r$ can be arbitrarily small, $\phi(x) = x$.
This completes the proof.