Prove $\sqrt6$ is irrational Suppose $\sqrt6 = \frac pq$ where $p$ and $q$ have no common factors.
$$6 = \frac {p^2}{q^2}$$
$$6p^2 = q^2$$
So $q^2$ and therefore $q$ is divisible by $6$. 
$$p^2 = \frac {q^2}{6}$$
So $p^2$ and therefore $p$ is divisible by $6$. 
S0, $p$ and $q$ have a common factor $6$. Contradiction.
Therefore $\sqrt6$ is irrational.
Does the proof work?
 A: As others have pointed out, it is important to justify that $6 \mid q^2 \Rightarrow 6\mid q$. And it isn't totally clear from your proof.
I suggest the following.
First show that $\sqrt{6}$ is not an integer. It's not difficult to do that. Since $4<6<9$, it follows that $2<\sqrt{6}<3$ and that means that $\sqrt{6}$ is not an integer.
Now assume that $\sqrt{6}$ is a rational number, $\frac{p}{q}$ where $p$ and $q$ are co-prime positive integers and $q>1$.
Now you can write
$$6=\frac{p^2}{q^2}$$
$$\Rightarrow 6q=\frac{p^2}{q}$$
It is clear that the left hand side is an integer. But the right hand side isn't since $p^2$ and $q$ share no common factors.
So this equality can not hold. And $\sqrt{6}$ can not equal $\frac{p}{q}$.
So it has to be an irrational number.
There's an incredibly short proof of this if you know the rational root theorem. Just notice that $\sqrt{6}$ is a root of the monic polynomial $x^2-6$. The proof is almost immediate.
EDIT: Here's a messy justification of why $q$ does not divide $p^2$. Let $p=\prod {p_i}^{x_i}$ and $q=\prod {p_j}^{y_j}$ such that $x_i$ and $y_j$ are positive integers. This notation is incredibly informal but it gets the message across.
Now since $p$ and $q$ are co-prime, $p_i\neq p_j$ for any $i$ & $j$. Now $p^2=\prod {p_i}^{2{x_i}}$. Notice that $p^2$ has the same prime divisors as $p$. Since $p$ and $q$ share no common prime factors, it follows that $p^2$ and $q$ share no common prime factors.
That means $q\nmid p^2$.
A: If you don't mind using some stronger weapons, then Eisenstein's criterion can help.
Choose $p=2$ and take the polynomial $1x^2+0x-6$. Since 
$$2\not\mid1 ,\quad 2\mid0 ,\quad 2\mid -6 ,\quad 2^2\not\mid-6 ,$$
we have that the polynomial is irreducible. It has $\sqrt6$ as a root. Therefore $\sqrt6$ is quadratic algebraic and not linear (=rational).
A: Using the prime factorization theorem (every positive integer has a unique prime factorization), we can write, for any rational number $r$:
$$
r = {p\over q} = {{2^{a_1}\cdot 3^{a_2}\cdot 5^{a_3} ...} \over {2^{b_1}\cdot 3^{b_2}\cdot 5^{b_3} ...}}
= {2^{c_1} \cdot3^{c_2}\cdot 5^{c_3} ...}
$$
where $a_i$ and $b_i$ are non-negative integers, and $c_i = a_i - b_i$ are integers (possibly negative).  Since the factorizations of $p$ and $q$ are unique, the factorization of $r$ must also be unique.
Thus:
$$
r^2 = 2^{2 c_1}\cdot 3^{2 c_2} \cdot5^{2 c_3}...
$$
If $r^2 = 6 = 2^1\cdot3^1$, then we have $2c_1 = 1$ and $2c_2 = 1$. This is impossible.
This technique generalizes to any number that has any prime factor an odd number of times.  It also generalizes to other roots.  For example: $^3\sqrt 4$.
A: Let us assume that $\sqrt{6}$ is rational number. Then $\exists p, q\in \mathbb{Z}, \,\, q>1$ and $gcd(p, q)= 1$ i.e they have no common factors.
$\dfrac{p}{q}= \sqrt{6} \Rightarrow p^2= 6q^2$
Hence $p^2$ is even i.e $2|p^2 \Rightarrow 2|p$
$\therefore \hspace{1cm} p= 2r \hspace{1cm}[r\in \mathbb{Z}]$
$\therefore \hspace{1cm} 4r^2= 6q^2 \Rightarrow 3q^2= 2r^2$
Again $3q^2$ is even i.e $2|3q^2 \Rightarrow 2|q^2 \Rightarrow 2|q$
Hence we have $2|p$ and $2|q$, which contradicts our supposition of $gcd(p, q)= 1$.
Hence $\sqrt{6}$ can't be rational.
