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T or F? The prime field of R=Q[sqrt(2)] then Frac(R)=Reals

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data

    If [itex]R=\mathbb{Q}[\sqrt{2}][/itex], then Frac(R)=[itex]\mathbb{R}[/itex]

    2. Relevant equations

    [itex]\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2} | a,b\in{\mathbb{Q}}\}[/itex]
    Frac(R) is the fraction field of R is basically [itex]\{\frac{a+b\sqrt{2}}{c+d\sqrt{2}} | a,b,c,d\in{\mathbb{Q}}\}[/itex].

    3. The attempt at a solution
    Back of the book says false...

    I tried to show [itex]\pi\not\in{Frac(R)}[/itex] by assuming by way of contradiction that there existed [itex]a,b,c,d\in{\mathbb{Q}}[/itex] s.t. [itex]\frac{a+b\sqrt{2}}{c+d\sqrt{2}}=\pi[/itex].

    That implies [itex]c\pi+d\pi\sqrt{2}=a+b\sqrt{2}[/itex] ... this led nowhere :(
     
    Last edited: Apr 5, 2013
  2. jcsd
  3. Apr 5, 2013 #2

    Dick

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    Frac(R) is countable, isn't it?
     
  4. Apr 5, 2013 #3
    Thanks for the response.

    If we could show Frac(R) is countable, we could conclude Frac(R)=/=reals because the reals are uncountable.

    How can we show Frac(R) is countable? My friend said a way to show countability is to find a bijection between Frac(R) and the natural numbers...

    Could we do something like this... http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function?
     
  5. Apr 6, 2013 #4

    Dick

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    That's the basis of it. That shows that |NxN|=|N|. You don't need to directly construct the bijection. If you can show that if there is a map from a countable set ONTO Frac(R) then you know Frac(R) is countable. If you pick a tuple of four rational numbers (a,b,c,d) there an obvious way to map that to an element of Frac(R), right? So that's a mapping from QxQxQxQ -> Frac(R). Now you know Q is countable, right?
     
    Last edited: Apr 6, 2013
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