Assume Q[sqrt(2)] is a subfield of the Real field R, and prove that it is the smallest subfield of R that contains sqrt(2)
Q[sqrt(2)]= a + b(sqrt(2))
The way $\displaystyle \mathbb{Q}(\sqrt{2})$ is usually defined is to be the smallest subfield containing $\displaystyle \mathbb{Q}$ and $\displaystyle \sqrt{2}$. Look Here.
So I will assume you are defining $\displaystyle \mathbb{Q}(\sqrt{2}) = \{ a+b\sqrt{2}|a,b\in \mathbb{Q}\}$ and you want to show it is the smallest such field. Let $\displaystyle \mathbb{Q}\subseteq K\subseteq \mathbb{R}$ be a field containing $\displaystyle \mathbb{Q}$ and $\displaystyle \sqrt{2}$. We need to show $\displaystyle \mathbb{Q}(\sqrt{2})\subseteq K$. Let $\displaystyle x\in \mathbb{Q}(\sqrt{2})$ then $\displaystyle x=a+b\sqrt{2}$ but then $\displaystyle a+b\sqrt{2}\in K$ because it is closed under sums and products, thus, $\displaystyle x\in K$.
Because a field has to be closed under sums and products (that is basically the definition of "field"). Meaning if $\displaystyle a,b\in F$ then $\displaystyle a+b,ab\in F$.This is JCIR you answered my question on fields and subfields, can you tell my why is it closed under sums and products ( i need to prove that too)
Thanks alot for your help.