TITLE: false proof of $\root \of 4$ is irrational.
QUESTION [9 upvotes]: There is a proof in my math textbook about the fact that $\root \of 2$ is irrational.
Proof:
Lets assume that $\root \of 2$ is rational.Then there will exist $2$ coprime natural numbers $p$ , $q > 1$ such that ,
$$ \root \of 2 = \frac{p}{q} \to 2 = \frac{p^2}{q^2} \to 2q = \frac{p^2}{q}$$
Obviously , $2q$ is an integer but $\frac{p^2}{q}$ is not a integer , because $p$ and $q$ are natural numbers , coprime and $q > 1$. So,
$$2q \neq \frac{p^2}{q} \to \root \of 2 \neq \frac{p}{q}$$
So , $\root \of 2$ is an irrational number.$\square$

But the confusion to me is , it seems like I can use this argument to show that $\root \of 4$ is an irrational number.

Proof:
Lets assume that $\root \of 4$ is rational.Then there will exist $2$ coprime natural numbers $p$ , $q > 1$ such that ,
$$ \root \of 4 = \frac{p}{q} \to 4 = \frac{p^2}{q^2} \to 4q = \frac{p^2}{q}$$
Obviously , $4q$ is an integer but $\frac{p^2}{q}$ is not a integer , because $p$ and $q$ are natural numbers , coprime and $q > 1$. So,
$$4q \neq \frac{p^2}{q} \to \root \of 4 \neq \frac{p}{q}$$
So , $\root \of 4$ is an irrational number.$\square$

Can someone tell me what is wrong with this proof?

REPLY [3 votes]: If $p,\,q$ are coprime natural numbers, $q|p^2$ iff $q=1$. The condition $p/q=\sqrt{2}$ contradicts $q=1$ because $\sqrt{2}$ isn't an integer, but $p/q=\sqrt{4}$ doesn't because $\sqrt{4}$ is. So while $q>1$ finishes the first proof, it can't be deduced in the second. In particular, we need only check $1^2<2<2^2=4$ to see $\sqrt{4}$, but not $\sqrt{2}$, is natural.