1. ## Analysis

1. If $\displaystyle r$ is rational ($\displaystyle r \neq 0$) and $\displaystyle x$ is irrational, prove that $\displaystyle r+x$ and $\displaystyle rx$ are irrational. So $\displaystyle r = \frac{p}{q}$ where $\displaystyle p,q \in \bold{Z}$. So then $\displaystyle \frac{p}{q} + x$ and $\displaystyle \frac{p}{q}x$ somehow have to be irrational.

2. Prove that there is no rational number whose square is $\displaystyle 12$. So maybe use a proof by contradiction (i.e. assume that $\displaystyle \frac{p^{2}}{q^{2}} = 12$)?

2. Say that $\displaystyle r + x = s$ where s is rational then $\displaystyle x = s - r$.
Do you see the contradiction?

3. Because than $\displaystyle x$ is rational which is a contradiction.

4. Originally Posted by heathrowjohnny
2. Prove that there is no rational number whose square is $\displaystyle 12$. So maybe use a proof by contradiction (i.e. assume that $\displaystyle \frac{p^{2}}{q^{2}} = 12$)?
You're right about $\displaystyle \frac{p^{2}}{q^{2}} = 12$

$\displaystyle \frac{p^{2}}{q^{2}} = 12$

$\displaystyle \left| \frac{p}{q} \right| = \sqrt{12}$

$\displaystyle \frac{p}{q} = \sqrt{12},~~ \frac{p}{q} = -\sqrt{12}$

$\displaystyle \sqrt{12}$ and $\displaystyle -\sqrt{12}$ are irrational. Irrational numbers cannot be expressed as rational numbers. So there's no such $\displaystyle \frac{p}{q}$

Well, we could also start with "$\displaystyle a$ is a rational number" instead of $\displaystyle \frac{p}{q}$.