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Subfield question

  • #1

Homework Statement



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

Homework Equations


Q= rational numers, R= real numbers.


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.
 

Answers and Replies

  • #2
41
0
It looks like you have a mistake in the definitions...

What is the definition of Q[sqrt(2)]?
 
  • #3
236
0
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)).
 
  • #4
Q is defined as a+b(sqrt2). So subtraction is obviosly preserved, but I have no idea what to do with the inverses
 
  • #5
41
0
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].
 
  • #6
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
Q is defined as a+b(sqrt2). So subtraction is obviosly preserved, but I have no idea what to do with the inverses
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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