Analysis

**heathrowjohnny** · February 16th 2008, 09:45 AM

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

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

**Plato** · February 16th 2008, 09:51 AM

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

**heathrowjohnny** · February 16th 2008, 10:31 AM

Because than $x$ is rational which is a contradiction.

**wingless** · February 16th 2008, 10:49 AM

Originally Posted by heathrowjohnny

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

You're right about $\frac{p^{2}}{q^{2}} = 12$

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

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

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

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

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

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