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

If yes, how can we prove it?

RemarkAn automorphism of R may not be continuous.

**Answer**

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

Let ϕ be an automorphism of the field of real numbers.

Let x>0 be a positive real number.

Then there exists y such that x=y2.

Hence ϕ(x)=ϕ(y)2>0.

If a<b, then b−a>0.

Hence ϕ(b)−ϕ(a)=ϕ(b−a)>0 by the above.

Hence ϕ(a)<ϕ(b).

This means that ϕ is strictly increasing.

If n is a natural number, it can be written in the form 1+…+1, so ϕ(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 ϕ(r)=r for any rational number.

Let x be a real number.

Let r,s be rational numbers such that r<x<s.

Then r<ϕ(x)<s.

Since s−r can be arbitrarily small, ϕ(x)=x.

This completes the proof.

**Attribution***Source : Link , Question Author : Makoto Kato , Answer Author : Aleks Kissinger*