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

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

Remark An 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 ba>0.
Hence ϕ(b)ϕ(a)=ϕ(ba)>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=(ab)c1, 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 sr can be arbitrarily small, ϕ(x)=x.
This completes the proof.

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

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