1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted