If $r$ is a real number such that $r^2 = 2$, then $r$ is irrational. Prove that, If $r$ is a real number such that $r^2 = 2$, $r$ is irrational.

Proposition: If $r$ is a real number such that $r^2 = 2$, then $r$ is irrational.
Hypothesis: If $r$ is a real number such that $r^2 = 2$.
Conclusion: $r$ is irrational.
Using proof by contradiction, we negate the conclusion: $r$ is rational.

My textbook uses a long(er) proof, but I was wondering if it is valid to proceed as follows:
$r^2 = 2$
$ \implies r = \sqrt{2}$ where  $\sqrt{2}$ is irrational. 
This is a contradiction, since $r$ is assumed to be rational in the hypothesis. Although, it seems so trivial compared to the textbook proof that I am sceptical.

I would greatly appreciate it if people could please take the time to review this. Is this a valid proof by contradiction? If not, then where is the error in my reasoning?
 A: To demonstrate it in this way you must know (or assume) that $\sqrt2\notin \Bbb Q$. So actually you didn't demonstrate anything. The point of the proposition is to show that you can't write $\sqrt2$ as $\frac pq$ where $p,q \in \Bbb Z$.
If you need the usual demonstration I can write it, but I think it's the one written in your book.
A: That is not valid! You are assuming that $\sqrt{2}$ is irrational.
What you have to do is show that $\sqrt{2}$ is irrational. 
The most common way to do that is suppose that it is rational:
$$\sqrt{2}=\frac{p}{q} \Leftrightarrow p^2=2q^2$$
Let's take $\gcd(p,q)=1$. The above equation means that $p$ is even (once $p^2$ is even)  so, $p=2p_1$ and backing to the equation we have
$$4p_1^2=2q^2 \to q^2=2p_1^2$$
but that means that $q$ is also even what is a contradiction with $\gcd(p,q)=1$
A: The first thing you need to ask yourself is what $\sqrt 2$ is. This is not written neither in decimal form nor as a fraction, so we can't just say at a glance whether this is rational or not. Actually, we know nothing of it without precisely stating definition: 

$\sqrt 2$ is the positive root of polynomial $x^2 - 2$.

As you can see, you are back to your original question, whether $x^2 = 2$ has rational solution.
Your book probably proves this by contradiction, assuming that there are positive integers $p,q$ that are relatively prime and such that $\frac{p^2}{q^2} = 2$. From this point you use only facts about integers (specifically, divisibility properties of primes) to derive contradiction. 
If you use $\frac p q = \sqrt 2$ to conclude that there is contradiction, what you are saying in disguise is:

"Assume that solution $\alpha$ of $x^2 = 2$ is rational. But $\alpha$
  is irrational. Contradiction."

If you are looking for short and elegant proof, you can use rational root test, not hard to prove, but extremely useful theorem. Applied to polynomial $x^2 - 2$, it says that for any rational root $\frac p q$, where $p,q$ are relatively prime, it is necessary that $p$ divides $2$, and $q$ divides $1$. Thus, the only possible rational roots are $\{\pm 1,\pm 2\}$. You can easily check that these are not roots of $x^2 - 2$, and thus both roots of $x^2 - 2$ are irrational.
A: This proof is not valid because you are not proving that $\sqrt{2}$ is an irrational number. 
The one given in your book is correct. 
For your proof to be valid you must further prove that $\sqrt{2}$ is irrational which is again possible through contradiction method.
A: Let us suppose that $r$ is a rational number then $r=\frac{p}{q}$ where $\gcd(p,q)=1.$ Now, 
\begin{align*}
r^2 & = \frac{p^2}{q^2}\\
\implies 2& = \frac{p^2}{q^2}\implies p^2=2q^2\implies p^2\ \text{is an even number$\implies p$ is an even number. }\\
\implies p^2=(2k)^2 &= 2q^2 \implies q^2=2k^2\implies q\ \text{is an even number.} 
\end{align*}
This is a contradiction as $\gcd(p,q)=1.$
A: Proof by contradiction implies that you assume the opposite of what you propose, then prove that the new assumption is invalid, therefore the original proposition valid.
The only way to prove this is to assume that $\sqrt{2}$ is rational, then face a contradiction in the process.
Assumption: $\sqrt{2}$ is rational.
If it is rational, then you are able to write it as a fraction where the denominator and numerator different primes (the fraction cannot be simplified).
$$\sqrt{2}=\frac{p}{q}\iff2=\frac{p^2}{q^2}\iff2q^2=p^2$$
$p$ is an even number $(p = 2r)$.
$$2q^2=(2r)^2\iff2q^2=4r^2\iff q^2=2r^2$$
Which means that both $p$ and $q$ are even, therefore at least one of them is not a prime, which contradicts the assumption, thus proving that $\sqrt{2}$ is irrational.
What you did was simply state/assume the conclusion itself without actually making a valid proof.
A: Most of the answers here are at least somewhat misleading.  The only step of your proof that is definitely an error is that $r^2=2$ does not imply $r=\sqrt{2}$.  It implies either $r=\sqrt{2}$ or $r=-\sqrt{2}$, and so for your argument to work you would need to separately get a contradiction in both of these cases.  If you know that $\sqrt{2}$ is irrational, then you can get a contradiction in the case $r=-\sqrt{2}$ as well since if $-\sqrt{2}$ were rational then $\sqrt{2}=(-1)\cdot(-\sqrt{2})$ would be rational as well (the product of two rational numbers is rational).
With this correction, your proof is actually perfectly correct, assuming you already know that $\sqrt{2}$ is irrational.  However, probably your book is assuming you don't already know that $\sqrt{2}$ is irrational, so you can't use this fact in your proof.  That's why its proof is longer--it is essentially reproving the (not at all obvious!) fact that $\sqrt{2}$ is irrational, which you assumed to already be known.
