false proof of $\root \of 4$ is irrational. 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?

 A: You begin your proof saying that $\sqrt4=\frac pq$ with $p$ and $q$ natural coprimes greater than $1$. That works in the case of $\sqrt2$, since $\sqrt2$ is not a natural number, and then, yes, if it could be written as $\frac pq$ with $p$ and $q$ natural coprime numbers, then both of them would have to be greater than $1$. But $\sqrt4$ is a natural number. And the assumption that $\sqrt4$ can be written as $\frac pq$ with $p$ and $q$ coprime and $p,q>1$ is false.
A: You seem to be trying to understand the Proof by contradiction method used for the irrationality of numbers.

You can do the right steps as follows:
Let $\gcd(m,n)=1$, then you have
$$\sqrt 4=\frac mn$$
$$4=\frac {m^2}{n^2}$$
$$m^2=4n^2$$
This implies, $m=2k$. Hence,
$$4k^2=4n^2$$
$$k=n$$
This means, $$m=2k=2n$$
Finally,
$$\sqrt 4=\frac mn=\frac{2n}{n}=2.$$

Proof - verification

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 $q > 1$.

Your error strarts here:
You claim that $p,q> 1$. This claim is wrong.  The correct claim should be as follows:
$$\gcd(p,q)=1.$$

Obviously , $4q$ is an integer but $\frac{p^2}{q}$ is not a integer.

This claim is also wrong. Because, $q=1$ is a counter-example.
A: 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.
A: The restriction that $p,q$ should be $>1$ is unwarranted in either proofs, unnecessary in the first one and false for the second one. It should have been stated for what it was: that $p,q\in\Bbb Z$ coprime numbers such that $q\ge1$. Then the proof for $\sqrt 2$ goes more or less $$[2q^2=p^2]\Rightarrow [2\mid p]\Rightarrow [q^2=2\left(p/2\right)^2]\Rightarrow [2\mid q]$$
But then $2\mid\operatorname{gcd}(p,q)=1$.
Your second attempt at a proof relies on a false theorem, therefore the rest is hardly worth worrying about.
