Proving $\sqrt{6}$ is not part of a field

Suppose $\sqrt{6}=a+b\sqrt{2}+c\sqrt{3}$ with $a,b,c$ rational. Then also $$ \sqrt{6}-a=b\sqrt{2}+c\sqrt{3} $$ and when you square both sides of this, the only surd around will be $\sqrt{6}$, which makes your life a lot easier. You should be able to manipulate the result to show that $$ (b^2-3)(c^2-2)=0 $$ which contradicts the original rationality assumption.

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Real Analysis

Field Theory

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