An element not in a field extension 
Possible Duplicate:
Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? 

Consider the field extension $\mathbb{Q}(\sqrt2)$. I want to show that $\sqrt5 \notin \mathbb{Q}(\sqrt2)$. If this were not the case, then we could write $\sqrt5 = a + b\sqrt2$ for $a,b\in\mathbb{Q}$. However, I do not see the contradiction here. Is there a better/easier way to prove this?
 A: HINT $\ $ Since $\sqrt{2},\sqrt{5},\sqrt{10}\not\in\mathbb Q,\:$ it is an immediate consequence of this
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  $\rm\:\sqrt{b}  \not\in K\:,\:$  therefore it is sufficient to prove $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ But this fails only if  $\rm\:\sqrt{a} \in L = K(\sqrt{b})\ $ and then $\rm\ \sqrt{a}\ =\  r + s\ \sqrt{b}\ $  for $\rm\ r,s\in K\:.\:$ That's impossible: squaring yields $\rm(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\  \sqrt{b}\:,\: $ contra hypotheses as follows:  
$\rm\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\  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r \in K$ 
$\rm\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
Using the above as the inductive step one easily proves the following result of Besicovic.
THEOREM $\ $  Let $\rm\:Q\:$ be a field with $2 \ne 0\:,\:$ and $\rm\ L = Q(S)\ $ be an extension of $\rm\:Q\:$ generated by $\rm\: n\:$  square roots  $\rm\ S = \{ \sqrt{a}, \sqrt{b},\ldots \}$ of elts  $\rm\ a,\:b,\:\ldots \in  Q\:.\:$
If every nonempty subset of $\rm\:S\:$ has product not in $\rm\:Q\:$ then each successive 
adjunction  $\rm\ Q(\sqrt{a}),\  Q(\sqrt{a},\:\sqrt{b}),\:\ldots$ doubles the degree over $\rm\:Q\:,\:$ so, in total, $\rm\: [L:Q] \ =\ 2^n\:.\:$  Hence the $\rm\:2^n\:$ subproducts of the product of $\rm\:S\:$ form a basis of $\rm\:L\:$ over $\rm\:Q\:.$
A: Squaring both sides of $\sqrt5 = a + b\sqrt2$ will imply that $\sqrt2$ is rational, which it is not.
(Actually, you also need to consider the case $b=0$, which would imply $\sqrt5$ is rational, and the case $a=0$, which would imply $\sqrt{\dfrac52}$ is rational.)
