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Math Help - Automorphisms of R/Q

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by robeuler View Post
    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
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