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

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

where a and b belong to $Q$ (the set of all rational numbers). Prove that $3^{1/2}$ does not belong to $Q[2^{1/2}]$ ."
March 7th 2011, 07:19 AM
FernandoRevilla

If $\sqrt{3}=a+b\;\sqrt{2}$ with $a.b\in\mathbb{Q}$ then, taking squares $3=a^2+2b^2+2\sqrt{2}\;ab$ . Now you can easily prove that $\sqrt{2}$ would be a rational number (contradiction) .

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

Thanks a lot! :)

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