# Any Automorphism of the Field of Real Numbers Must be the Identity Map

## Problem 507

Prove that any field automorphism of the field of real numbers $\R$ must be the identity automorphism.

Contents

- Problem 507
- Proof.
- Claim 1. For any positive real number $x$, we have $\phi(x)>0$.
- Claim 2. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.
- Claim 3. The automorphism $\phi$ is the identity on positive integers.
- Claim 4. The automorphism $\phi$ is the identity on rational numbers.
- Claim 5. The automorphism $\phi$ is the identity on real numbers.

## Proof.

We prove the problem by proving the following sequence of claims.

Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

**Claim 1**. For any positive real number $x$, we have $\phi(x)>0$.**Claim 2**. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.**Claim 3**. The automorphism $\phi$ is the identity on positive integers.**Claim 4**. The automorphism $\phi$ is the identity on rational numbers.**Claim 5**. The automorphism $\phi$ is the identity on real numbers.

Let us now start proving the claims.

Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

**Claim 1**. For any positive real number $x$, we have $\phi(x)>0$.

Since $x$ is a positive real number, we have $\sqrt{x}\in \R$ and

\[\phi(x)=\phi\left(\sqrt{x}^2\right)=\phi(\sqrt{x})^2 \geq 0.\]

Note that since $\phi(0)=0$ and $\phi$ is bijective, $\phi(x)\neq 0$ for any $x\neq 0$.

Thus, it follows that $\phi(x) > 0$ for each positive real number $x$.

Claim 1 is proved.

**Claim 2**. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.

Since $x > y$, we have $x-y > 0$ and it follows from Claim 1 that

\[0<\phi(x-y)=\phi(x)-\phi(y).\]
Hence, $\phi(x)> \phi(y)$.

**Claim 3**. The automorphism $\phi$ is the identity on positive integers.

Let $n$ be a positive integer. Then we have

\begin{align*}

\phi(n)=\phi(\underbrace{1+1+\cdots+1}_{\text{$n$ times}})=\underbrace{\phi(1)+\phi(1)+\cdots+\phi(1)}_{\text{$n$ times}}=n

\end{align*}

since $\phi(1)=1$.

**Claim 4**. The automorphism $\phi$ is the identity on rational numbers.

Any rational number $q$ can be written as $q=\pm m/n$, where $m, n$ are positive integers.

Then we have

\begin{align*}

\phi(q)=\phi\left(\, \pm \frac{m}{n} \,\right)=\pm \frac{\phi(m)}{\phi(n)}=\pm \frac{m}{n}=q,

\end{align*}

where the third equality follows from Claim 3.

**Claim 5**. The automorphism $\phi$ is the identity on real numbers.

In this claim, we finish the proof of the problem.

Let $x$ be any real number.

Seeking a contradiction, assume that $\phi(x)\neq x$.

There are two cases to consider:

\[x < \phi(x) \text{ or } x > \phi(x).\]

First, suppose that $x < \phi(x)$.
Then there exists a rational number $q$ such that
\[x< q < \phi(x).\]
Then we have
\begin{align*}
\phi(x) &< \phi(q) && \text{by Claim 2 since $x < q$}\\
&=q && \text{by Claim 4 since $q$ is rational}\\
&<\phi(x) && \text{by the choice of $q$},
\end{align*}
and this is a contradiction.
Next, consider the case when $x > \phi(x)$.

There exists a rational number $q$ such that

\[\phi(x) < q < x.\]
Then by the same argument as above, we have
\[\phi(x) < q =\phi(q) < \phi(x),\]
which is a contradiction.
Thus, in either case we reached a contradiction, and hence we must have $\phi(x)=x$ for all real numbers $x$.
This proves that the automorphism $\phi: \R \to \R$ is the identity map.

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