Proof Q(sqrt(2)) is Subfield of R

[tyrannosaurus]

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.

 

[Mathnerdmo]

It looks like you have a mistake in the definitions...

What is the definition of Q[sqrt(2)]?

 

[Tinyboss]

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)).

 

[tyrannosaurus]

Q is defined as a+b(sqrt2). So subtraction is obviosly preserved, but I have no idea what to do with the inverses

 

[Mathnerdmo]

All right. Do you remember how to divide complex numbers? The same trick can be applied here to find the inverses in \mathbb{Q}[\sqrt2].

Just note that
a^2 - 2b^2 \neq 0
whenever a, b \in \mathbb{Q} unless a = b = 0.

 

[Hurkyl]

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?