1. ## Automorphisms of R/Q

I am doing a series of exercises with the goal of proving that the automorphism group of R/Q is trivial.

I have proven that any automorphism, g, takes positive reals to positive reals.

I haven't made progress on any of the following steps:

b) If a<b then g(a)<g(b) for any real numbers a and b.

c) Prove that -1/m<a-b<1/m implies that -1/m<g(a)-g(b)<1/m for every positive integer m.

d) It follows from c) that g is continuous (this is clear to me)

e) And finally prove that any continuous map on R which fixes Q is the identity map.

I think I know how to do e). Since Q is dense in R, for any real r pick 'nearby' rationals p and q such that p<r<q. Since p and q are fixed by g, and g(r) has to be between p and q, we know that g is the identity, because we can pick p and q as close to r as we like.

2. Originally Posted by robeuler
I am doing a series of exercises with the goal of proving that the automorphism group of R/Q is trivial.

I have proven that any automorphism, g, takes positive reals to positive reals.

I haven't made progress on any of the following steps:

b) If a<b then g(a)<g(b) for any real numbers a and b.

c) Prove that -1/m<a-b<1/m implies that -1/m<g(a)-g(b)<1/m for every positive integer m.

d) It follows from c) that g is continuous (this is clear to me)

e) And finally prove that any continuous map on R which fixes Q is the identity map.

I think I know how to do e). Since Q is dense in R, for any real r pick 'nearby' rationals p and q such that p<r<q. Since p and q are fixed by g, and g(r) has to be between p and q, we know that g is the identity, because we can pick p and q as close to r as we like.
this seems to be a Frequently Asked Problem! see this thread: http://www.mathhelpforum.com/math-he...m-problem.html