TITLE: How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$?
QUESTION [22 upvotes]: How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field.
I want to know the detail about the proof. Thanks in advance.
Actually I know any two of them and three of them are linearly independent.

REPLY [15 votes]: HINT $\ $ Specialize the Lemma below to $\rm\ K = \mathbb Q,\ a,b\ =\ 2,3\:. $
LEMMA $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K]\ =\ 4\ $ if  $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $  all are not in $\rm\:K\:$ and $\rm\: 2\: \ne\: 0\:$ in $\rm\:K\:.$
Proof $\ \ $  Let  $\rm\ L = K(\sqrt{b})\:.\:$ Then $\rm\:  [L:K] = 2\:$  via  $\rm\:\sqrt{b}  \not\in K\:,\:$  so it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ It fails only if  $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\  r + s\ \sqrt{b}\ $  for $\rm\ r,s\in K\:.\:$ But that is impossible since squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\  \sqrt{b}\:,\: $ which contradicts hypotheses as follows:  
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad rs \ne 0\ \ \Rightarrow\ \  \sqrt{b}\ \in\  K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using  $\rm\:2 \ne 0$  
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r \in K$ 
$\rm\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\  r = 0\ \ \Rightarrow\ \  \sqrt{a\:b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED
REMARK $\ $ By induction, the lemma easily generalizes to algebraic extensions generated by adjoining $\rm\:n\:$ square-roots, see my post here on Besicovic's Theorem, which includes references to generalizations by Mordell and Siegel. These results are elementary special cases of the Galois theory of Kummer extensions.