Square root of 6 proof rationality I was proving $\sqrt 6 \notin \Bbb Q$, by assuming its negation and stating that: $\exists (p,q) \in \Bbb Z \times \Bbb Z^*/ \gcd(p,q) = 1$, and $\sqrt 6 = (p/q)$.
$\implies p^2 = 2 \times 3q^2 \implies \exists k \in \Bbb Z; p = 2k \implies 2k^2 = 3q^2$ and found two possible cases, either $q$ is even or odd, if even we get contradiction that $\gcd(p, q) \neq 1$, if odd we get contradiction that $2k^2 = 3q^2$.
Is it a right path for reasoning it? 
 A: How about a proof by descent?
First show that $2^2<6<3^2$.  Then if $\sqrt{6}$ is to be rational it must have a form $p/q$ where $p,q\in \mathbb{Z}, q>0, 2q<p<3q$.  By simple algebra the square root is also equal to $6q/p$, thus
$p/q=6q/p\text{.....Eq. 1}$.
Now if $a/b=c/d$ then also
$a/b=(ma+nc)/(mb+nd)$
for any coefficients $m,n$ where the denominator is nonzero.  In particular, Eq. 1 implies
$p/q=(3p-6q)/(3q-p)\text{.....Eq. 2}$
where we already have $2q<p<3q$ and thus $0<3q-p<q$.  So the proposed rational fraction $p/q$ must be equal to an alternative rational fraction with a smaller positive denominator.  This causes an infinite descent contradiction forcing the assumption of a rational value to be false.
We can form a similar proof for the square root of any natural number that is not a squared integer.  "Not a squared integer" is needed because the square root must be strictly between two adjacent integers to obtain a descent of positive denominators.
A: Assume $\sqrt{6} = {a \over b}$ with $a, b \in \mathbb{Z}$ . This implies 
$$2 \cdot 3 b^2 = a^2$$
Use the fundamental theorem of algebra to decompose both sides into a unique prime factorization.  There are an even number of factors of $2$ on the rhs and an odd number of factors of $2$ on the lhs... a contradition.  Same for factors of $3$. 
Thus $\sqrt{6} \neq {a \over b}$, i.e., is irrational.
A: I'll play more generally
and see what happens.
Suppose
$\sqrt{m}
=\dfrac{u}{v}
$
where
$m = \prod_{p \in P} p^{m_i},
u = \prod_{p \in P} p^{u_i},
v = \prod_{p \in P} p^{v_i}
$.
Then
$\prod_{p \in P} p^{m_i}
=\dfrac{u^2}{v^2}
=\dfrac{ \prod_{p \in P} p^{2u_i}}{\prod_{p \in P} p^{2v_i}}
$
so
$\prod_{p \in P} p^{m_i}\prod_{p \in P} p^{2v_i}
=\prod_{p \in P} p^{2u_i}
$
or
$\prod_{p \in P} p^{m_i+2v_i}
=\prod_{p \in P} p^{2u_i}
$.
By unique factorization,
$m_i+2v_i
=2u_i
$,
so
$m_i
=2u_i-2v_i
=2(u_i-v_i)
$.
Therefore
$m
=\prod_{p \in P} p^{m_i}
=\prod_{p \in P} p^{2(u_i-v_i)}
=\left(\prod_{p \in P} p^{u_i-v_i}\right)^2
$
so $m$ is a perfect square.
Therefore
the square root of an integer
is rational
only if
it is a square.
I'll now try to generalize this
to $k$-th roots,
with as much cut-and-paste
as possible.
Suppose
$\sqrt[k]{m}
=\dfrac{u}{v}
$
where
$m = \prod_{p \in P} p^{m_i},
u = \prod_{p \in P} p^{u_i},
v = \prod_{p \in P} p^{v_i}
$.
Then
$\prod_{p \in P} p^{m_i}
=\dfrac{u^k}{v^k}
=\dfrac{ \prod_{p \in P} p^{ku_i}}{\prod_{p \in P} p^{kv_i}}
$
so
$\prod_{p \in P} p^{m_i}\prod_{p \in P} p^{kv_i}
=\prod_{p \in P} p^{ku_i}
$
or
$\prod_{p \in P} p^{m_i+kv_i}
=\prod_{p \in P} p^{ku_i}
$.
By unique factorization,
$m_i+kv_i
=ku_i
$,
so
$m_i
=ku_i-kv_i
=k(u_i-v_i)
$.
Therefore
$m
=\prod_{p \in P} p^{m_i}
=\prod_{p \in P} p^{k(u_i-v_i)}
=\left(\prod_{p \in P} p^{u_i-v_i}\right)^k
$
so $m$ is a perfect $k$-th power.
Therefore
the $k$-th root of an integer
is rational
only if
it is a $k$-th power.
A: When you get $2p^2 = 3q^2$ that means that $2|q^2$ so $2|q$ and that $3|p^2$ so $3|p$.  
Then if you replace $p = 3p'$ for some integer $p'$ and $q=2q'$ for some integer $q'$ you get
$2(3p')^2 = 3(2q')^2$
$18p'^2 = 12q'^2$
$3p'^3 = 2q'^2$.
Can you finish from there?
This assumes you know Euclid's lemma that 1) prime numbers exist and 2) if $p$ is prime and $p|a*b$ then either $p|a$ or $p|b$ (or both).
