Showing that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$ How can I prove that $\sqrt5$ is not in $\mathbb{Q}(\sqrt7)$ ?
I can only think of trying to write $\sqrt5 = a+b\sqrt7$ (where $a,b$ are in $\mathbb{Q}$), but I can't think of a good reason that shoes such $a$ and $b$ doesn't exist.
Any ideas ?
 A: Hint $\ $ Apply the following lemma, noting $\sqrt{5},\sqrt{7},\sqrt{35}$ all $\rm\not\in K = \mathbb Q$
LEMMA $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if  $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\:b}\ $  all $\rm\not\in K\:$ and $\rm\: 2 \ne 0\:$ in $\rm\:K\:.$
Proof $\ \ $  Let  $\rm\ L = K(\sqrt{b})\:.\:$ $\rm\:  [L:K] = 2\:$ by  $\rm\:\sqrt{b}  \not\in K,\:$  so it suffices to show $\rm\: [L(\sqrt{a}):L] = 2\:.\:$ This fails only if  $\rm\:\sqrt{a} \in L = K(\sqrt{b})$ $\:\Rightarrow\:$ $\rm \sqrt{a}\ =\  r + s\ \sqrt{b}\ $  for $\rm\ r,s\in K,\:$ which is false, because squaring yields $\rm\:(1):\ \ a\ =\ r^2 + b\ s^2 + 2\:r\:s\  \sqrt{b}\:,\: $ which is contra to hypotheses as follows:  
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \  \sqrt{b}\ \in\  K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\:,\:$ using  $\rm\:2 \ne 0$  
$\rm\qquad\qquad\  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\  r + s\ \sqrt{b}\ =\ r \in K$ 
$\rm\qquad\qquad\  r = 0\ \ \Rightarrow\ \  \sqrt{a\:b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\ \sqrt{b}\:,\: \ $times $\rm\:\sqrt{b}\quad\quad$ QED
See my post here for generalizations.
A: It suffices to find a field containing $\mathbb{Q}(\sqrt 7)$ in which $5$ is not a square. The field of $3$-adic numbers $\mathbb{Q}_3$ is such a field, since $7 \equiv1 \mod 3$ and thus $7$ is a square, but $5\equiv 2 \mod 3$ so $5$ is not a square. (For any odd prime $p$, an integer prime to $p$ is a square in $\mathbb{Q}_p$ if and only if it is a square modulo $p$.)
http://en.wikipedia.org/wiki/P-adic_number
A: If $\sqrt{5}\in\mathbb{Q}(\sqrt{7})$, then there exist rationals $a$ and $b$ such that $(a+b\sqrt{7})^2 = 5$. But 
$$(a+b\sqrt{7})^2 = a^2 + 7b^2 + 2ab\sqrt{7} = 5$$
implies $ab=0$ (since $1,\sqrt{7}$ are linearly independent over $\mathbb{Q}$, since $7$ is not a square), which implies $5=a^2$ or $5=7b^2$, both of which are impossible with $a$ and $b$ rational.
(In fact, not only are they different, they are not even isomorphic)
More generally, if $p$ and $q$ are distinct squarefree integers different from $1$, then $\mathbb{Q}(\sqrt{p})\neq\mathbb{Q}(\sqrt{q})$, since 
$$(a+b\sqrt{q})^2 = a^2+qb^2 + 2ab\sqrt{q}=p$$
implies $ab=0$, hence $p=a^2$ or $p=qb^2$, and the fact that $p$ is squarefree yields a contradiction.
See also this previous answer covering the same ground, or Bill Dubuque's answer to the same question which shows that for rationals $d$ and $d'$, $\mathbb{Q}(\sqrt{d})=\mathbb{Q}(\sqrt{d'})$ if and only if $d/d'$ is a (rational) square. 
