# Subfield question

## 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

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

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

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)).

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

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
Staff Emeritus
Science Advisor
Gold Member
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?