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Subfield question

  1. Mar 7, 2010 #1
    1. The problem statement, all variables and given/known data

    proof that Q(sqrt(2)) is a subfield of R

    2. Relevant equations
    Q= rational numers, R= real numbers.


    3. The attempt at a solution
    Clearly (sqrt(2)) is a subgroup of R. Then a[sqrt2]. b[sqrt2] are elements of Q[sqrt2] if a and b are eleemnts of Q. Therefore Q[sqrt2] contains at least two elements.
    2. a[sqrtb]-b[sqrt2] is an element of Q[sqrt2] since a-b is an element of Q since Q is closed under subtraction.
    From there I have to prove that a[sqrt2]*[b[sqret2]^-1] are elements of Q[sqrt2] but i don't know how to do this. Any help would be appreciated.
     
  2. jcsd
  3. Mar 7, 2010 #2
    It looks like you have a mistake in the definitions...

    What is the definition of Q[sqrt(2)]?
     
  4. Mar 7, 2010 #3
    This could depend a lot on what you're studying right now, and what you are/aren't allowed to use. But I'd just note that Q is a subfield of R, and sqrt(2) is in R, and look at the definition of Q(sqrt(2)).
     
  5. Mar 8, 2010 #4
    Q is defined as a+b(sqrt2). So subtraction is obviosly preserved, but I have no idea what to do with the inverses
     
  6. Mar 10, 2010 #5
    All right. Do you remember how to divide complex numbers? The same trick can be applied here to find the inverses in [tex] \mathbb{Q}[\sqrt2] [/tex].

    Just note that
    [tex] a^2 - 2b^2 \neq 0 [/tex]
    whenever [tex]a, b \in \mathbb{Q} [/tex] unless [tex]a = b = 0[/tex].
     
  7. Mar 10, 2010 #6

    Hurkyl

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    Try solving a linear equation!

    Wait a moment -- what do you need to do with inverses? Why do they need to be considered specially at all?
     
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