# Prove that 3^(1/2) does not belong to the set Q[2^(1/2)]

• Mar 7th 2011, 06:23 AM
mbmstudent
Prove that 3^(1/2) does not belong to the set Q[2^(1/2)]
Hi, I have this problem which I do not know how to solve... thanks for any help you can give me! And please, if you can solve it, I would really appreciate it if you could tell me the theorem you have used since I am feeling so lost in abstract algebra :(

"Let $\displaystyle Q[2^{1/2}]$ be the set consisting of all numbers of the form
$\displaystyle a+b(2)^{1/2}$

where a and b belong to $\displaystyle Q$ (the set of all rational numbers). Prove that $\displaystyle 3^{1/2}$ does not belong to $\displaystyle Q[2^{1/2}]$."
• Mar 7th 2011, 07:19 AM
FernandoRevilla
If $\displaystyle \sqrt{3}=a+b\;\sqrt{2}$ with $\displaystyle a.b\in\mathbb{Q}$ then, taking squares $\displaystyle 3=a^2+2b^2+2\sqrt{2}\;ab$ . Now you can easily prove that $\displaystyle \sqrt{2}$ would be a rational number (contradiction) .

P.S. There another way to prove it considering $\displaystyle \mathbb{Q}[\sqrt {2}]=\mathbb{Q}[x]/(x^2-2)$
• Mar 7th 2011, 07:56 AM
mbmstudent
Thanks a lot! :)