1. ## Proof....rational numbers

Hey all,
how can i prove that if $\displaystyle a$ and $\displaystyle b$ are non-zero rational numbers, then $\displaystyle a + b\sqrt{2} \not\in \mathbb{Q}$

...using the fact that $\displaystyle \sqrt{2} \not\in \mathbb{Q}$

2. Does this need proof? It should be obvious that a rational times an irrational is irrational, so $\displaystyle \displaystyle b\sqrt{2}$ is irrational.

The sum of a rational and an irrational is also irrational, so $\displaystyle \displaystyle a + b\sqrt{2}$ is irrational.

If it requires proof, then you probably need to use the fact that the set of rationals is closed under addition and multiplication...

3. Originally Posted by Oiler
Hey all,
how can i prove that if $\displaystyle a$ and $\displaystyle b$ are non-zero rational numbers, then $\displaystyle a + b\sqrt{2} \not\in \mathbb{Q}$

...using the fact that $\displaystyle \sqrt{2} \not\in \mathbb{Q}$
Let $\displaystyle $$a \ne 0 and \displaystyle$$b \ne 0$ be rational and suppose $\displaystyle a+b\sqrt{2}$ is rational. Hence there exists a rational $\displaystyle$$c$ such that:

$\displaystyle c=a+b\sqrt{2}$

But that rationals are closed under addition so:

$\displaystyle d=c+a=b\sqrt{2}$

is rational. Also the rationals are closed under division (by non-zero elements anyway) so:

$\displaystyle e=d/b=\sqrt{2}$

is rational, which is a contradiction so our hypothesis that $\displaystyle a+b\sqrt{2}$ is rational fails, hence $\displaystyle a+b\sqrt{2}$ is irrational.

CB