Proof square root of 4 is not irrational.

Question

I was recently working on a question essentially worded in the following way:

Where does a proof of $\sqrt{4}$ being irrational fall apart when we try to apply the same principles used for proving that $\sqrt{2}$ is irrational.

I attempted by making the same (in this case, intuitively correct) assumptions that led to a contradiction in the case of $\sqrt{2}$:

1. $\sqrt{4}$ is a rational number and can be written as $\dfrac{m}{n}$ where $n\neq0$

2. $\dfrac{m}{n}$ is in lowest reduced terms; i.e. $m$ and $n$ are co-prime due to definition of rational numbers

Then I took the following steps:

$$m^2 = 4n^2$$

$$m^2 = 2(2n^2)$$

Thus, $m^2$ is even $\implies$ $m$ is even and can be written as $2k$.

$$m^2 = 4k^2 = 4n^2$$

$$k = n$$

Thus, $k$ is a factor of both $m$ and $n$, contradicting the second assumption that I made ($m$ and $n$ are co-prime).

Although I understand intuitively that this is not the case, doesn't this show that $\sqrt{4}$ is irrational?

Answer 1

You have proven that $n = k$ and $m = 2k$. In the case that $m$ and $n$ are coprime, set $k = 1$.

Answer 2

All you can prove with this strategy is that $m=2n$, but the HCF is $1$ if we take $m=2,\,n=1$. By contrast, the analogous treatment of $\sqrt{2}$ shows $m,\,n$ must both be even.