There is no rational number of $r^2=2$ 
The conventional proof of contradiction for: 
  $\nexists$ $r \in \mathbb{Q}$ such that
  $r^2=2$

I saw the proof for contradiction and I am a bit confused. In the proof it suggests that the rational number has no common factors. What I am confused on is that there may exist a fraction that is not fully reduced, even if the proof assumed a non-fully reduced fraction did exist, this does not imply to me there does not exist a non-fully reduced fraction. Can someone clarify this for me?
 A: Suppose $\displaystyle \frac{m}{n}=r=\frac{p}{q}$, for some integers $m,n,p,q$ and $r\in \mathbb{Q}$.
Suppose $\displaystyle \frac{m}{n}$ is fully reduced and $\displaystyle \frac{p}{q}$ isn't.
Whatever you say about $\displaystyle \frac{m}{n}$ you can also say about $\displaystyle \frac{p}{q}$ because they are $\textbf{equal}$.
Even if your intuition somehow tricks you into thinking that assuming that taking the fraction in its irreducible form doesn't provide a complete proof, the proof must be valid because $\displaystyle \frac{m}{n}=\frac{p}{q}$.
Edit: One could argue: "wait a minute, you can say that $\frac{m}{n}$ is fully reduced, but you can't say the same about $\frac{p}{q}$ even though they are equal". The fallacy in this argument lies in the fact that saying $\frac{m}{n}$ is fully reduced and $\frac{p}{q}$ isn't, is not a statement about the fractions. It's a statement about the order pairs $(m,n)$ and $(p,q)$ because saying $\frac{m}{n}$ is fully reduced is just short for $\gcd(m,n)=1$ (except when either $m$ or $n$ are $0$). Not only that, your question was in fact a question about the fractions, so there's no confusion here.
A: If there exists a non-fully reduced fraction, there must exist a reduced form of said fraction.  The proof shows that there is no fully reduced fraction and, implictly, no non-fully reduced fraction could exist because it would have to be equal to a fully reduced fraction that can't exist.
This is a fun concept to play with and it ends up that for such a number to exist, $$\color{Teal} {\underset{n \rightarrow \infty}{\lim}} \color{orchid}{\frac{2^{an}b}{2^{cn}}}$$
for some integers a, b, and c.  This does not make sense though it is a fun concept to play with.
A: Consider the fraction $\frac{a}{b}$, $b>0$. Then there exists integers $c,d$ with $d>0$, $\frac{a}{b}=\frac{c}{d}$, and $c$ and $d$ relatively prime.
Suppose, for contradiction, that this is not true for some $a,b$. Let $S$ be the set of positive integers $d$ for which there exists a $c$ with $\frac{a}{b}=\frac{c}{d}$. Since $b\in S$, $S$ is non-empty, and so it has a least element $d_0$, such that for some integer $c_0$, $\frac{a}{b}=\frac{c_0}{d_0}$. By assumption $c_0$ and $d_0$ are not relatively prime, and so $c_0 = kc_1$ and $d_0 = kd_1$ for $k>1$. In particular, $0<d_1<d_0$. Moreover, 
$$\frac{c_1}{d_1} = \frac{kc_0}{kd_0} = \frac{a}{b},$$
so $d_1\in S$, a contradiction.
So if there exists a rational number equal to $\sqrt{2}$, there must also exist a rational number in lowest terms equal to $\sqrt{2}$.
