Proving any number irrational How can I prove any number to be irrational (it must be irrational, of course). Specifically, which is a better method to prove that a given number is irrational: the contrapositive method or the rational zeroes theorem?
I am asking this question because I am having trouble proving various numbers (e.g. ${\sqrt 6}$, ${\sqrt 12}$) to be irrational by using the contrapositive method.
 A: Integers (greater than $1$) can be uniquely represented by products of integer powers of primes. That is, for any $n>1$
$$n=p_1^{q_1}p_2^{q_2}p_3^{q_3}\cdots p_n^{q_n}$$
where all $p_i$ are primes  and all the powers are (positive) integers.
If $n$ is a square of another number then all the $q_i$s are even numbers. Let $n$ be not the  square of another integer number. 
Assume that its square root is rational:
$$\frac mn=\sqrt n={p_1^{\frac{q_1}2}p_2^{\frac{q_2}2}p_3^{\frac{q_3}2}\cdots p_n^{\frac{q_n}2}}.$$
Since $n$ is not the square of any number, there will be at least one odd $q_j$ among the powers above. The half of that ${q_j}$ is not an integer then. This contradicts to the fact that in the same representation of $n$ and $m$ all the powers are integers. As a result the same is true for $\frac mn$; all the prime powers in $\frac mn$ are integers (some may be negative though but not a fraction).
A: Taking your example of $\sqrt{6}$
The following will be a proof by contradiction. For contradiction, assume $\sqrt{6} \in \mathbb{Q}$
Then we can write $\sqrt{6} = \frac{p}{q}$ with $hcf(p,q) = 1$ and $p,q \in \mathbb{Z}$ with $ q\neq 0$
$\Rightarrow 6 = \frac{p^{2}}{q^{2}} \Rightarrow 6q^{2} = p^{2}$
So we have that $6|p^{2}$. 
Let's consider all the possible cases for $p$:
If $p \equiv 0 \bmod 6 \Rightarrow p^{2} \equiv 0 \bmod 6$
If $p \equiv 1 \bmod 6 \Rightarrow p^{2} \equiv 1 \bmod 6$
If $p \equiv 2 \bmod 6 \Rightarrow p^{2} \equiv 4 \bmod 6$
If $p \equiv 3 \bmod 6 \Rightarrow p^{2} \equiv 3 \bmod 6$
If $p \equiv 4 \bmod 6 \Rightarrow p^{2} \equiv 1 \bmod 6$
If $p \equiv 5 \bmod 6 \Rightarrow p^{2} \equiv 1 \bmod 6$
Therefore, $p^{2} \equiv 0 \bmod 6 \Leftrightarrow p\equiv 0 \bmod 6$
Then we can write $p = 6k$ for some $k \in \mathbb{Z}$
$\Rightarrow 6q^{2} = 36k^{2} \Rightarrow q^{2} = 6k^{2}$
Then by the same argument, we can conclude that $6|q$, but this implies that $hcf(p,q) = 6$, which is a contradiction, hence $\sqrt{6}$ must be irrational.
A: Whole shebang:
$\sqrt {K} = m/n; m,n \in \mathbb Z; n \ne 0 \gcd(m, n) = 1 \implies$
$n^2K=m^2$
If we know the unique prime factorization theorem then if $p$ is a prime factor of $n$ then $p$ is a prime factor of $m$ which contradicts $\gcd(m,n) =1$ so $n$ has no prime factors so $n = 1$ and $K = m^2$ and $\sqrt{K} = m$. i.e. only perfect squares have integer square roots.  All other square roots of integers are irrational.
If you don't know the unique prime factorization theorem.. you are in a bit of a pickle.  
