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?

To demonstrate it in this way you must know (or assume) that $\sqrt2\
otin \Bbb Q$. 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.