Proving that \sqrt{p} is irrational

[Mandelbroth]

I'm aware of the standard proof.

What I'm wondering is why we can't just do the following. Given, I haven't slept well and I'm currently out of caffeine, so this one might be trivial for you guys.

Suppose, by way of contradiction, that $\sqrt{p}=\frac{m}{n}$, for $m,n\in\mathbb{Z}$ coprime. Then, $p=\frac{m^2}{n^2}$. Because $p$ is an integer, $\frac{m^2}{n^2}$ must be as well. However, because $m$ and $n$ are coprime, so are $m^2$ and $n^2$. Thus, $n^2=1$ is necessary for $\frac{m^2}{n^2}$ to be an integer. But that means $m^2=p$, and $p$ is prime. Thus, a contradiction is met and we see that $\sqrt{p}$ is irrational.

Is this valid? I'll probably figure out my mistake (if I made one) by the time I get back with adequate caffeination, but until then, I'd like to make sure I figure it out.

Thank you.

Edit: Gee, I guess it might be important to mention that $p$ is prime. :facepalm:

 

[R136a1]

In my opinion, it's valid.

 

[jbunniii]

It looks OK to me.

 

[Curious3141]

Completely valid. The last step basically assumes that a prime cannot be a perfect square, which is true and fairly obvious. But if you really want to make the proof completely obvious, you can state: $m^2 = p$. Hence $m|p$ (m divides p), which is a contradiction.

 

[PeroK]

Edit: Gee, I guess it might be important to mention that $p$ is prime. :facepalm:

Actually, it doesn't matter that p is prime. It's valid for any +ve integer: the square root is either an integer or irrational; it cannot be a proper rational.

 

[Mandelbroth]

Thank you, all.