Proof that a given set is a field

Question

I am solving the following exercise (linear algebra):

show that: $\ \mathbb{Q}\lbrack\sqrt{2}\rbrack = \{ a + b\cdot\sqrt{2} \ \vert \ a,b \in \mathbb{Q}\} \subset \mathbb{R} \ $ is a field, with the addition and multiplication of the real numbers.

Background: First of all, i am a first semester mathematics student and we had not covered a lot so far. In our lectures we had the definition of a group, abelian group and of a field so far.

Problem: The sentence "with addition and multiplication of reall numbers" confuse me because I dont understand if i have to apply the field axioms to an element of the set $\mathbb{Q}\lbrack\sqrt{2}\rbrack$ and one of $\mathbb{R}$ or just to element of the set. Further I have no clue how to start that proof. If anyone could give me some helpfull piece of advice i would be very happy about.

Answer 1

Hint

The proof in some steps:

• The set $\Bbb Q[\sqrt 2]$ isn't empty
• if $u=a+b\sqrt2$ and $u'=a'+b'\sqrt2$ are in $\Bbb Q[\sqrt2]$ then $u-u'\in \Bbb Q[\sqrt2]$
• if $u=a+b\sqrt2$ and $u'=a'+b'\sqrt2$ are in $\Bbb Q[\sqrt2]$ , $u'\ne0$ then $u(u')^{-1}\in \Bbb Q[\sqrt2]$

so if you prove these points then you proved that $\Bbb Q[\sqrt 2]$ is a sub-field of $(\Bbb R ,+,\times)$.