Consider Q[sqrt(2)]. Does every element of Q[sqrt(2)] have a square root in Q[sqrt(2)]? Prove if true; and give a counterexample if false.
I'm thinking it is true, but I can't prove it. Please help!
Suppose $\displaystyle \sqrt{\sqrt{2}}\in\mathbb{Q}(\sqrt{2})\Longrightar row (a+b\sqrt{2})^2=\sqrt{2}$ , for some $\displaystyle a,b\in\mathbb{Q}$ , but then:
$\displaystyle a^2+2b^2+2\sqrt{2}ab=\sqrt{2}\Longrightarrow $ make some order here and show this would imply $\displaystyle \sqrt{2}\in\mathbb{Q}$ or some other harsh contradiction .
Tonio