Proof....rational numbers

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March 4th 2011, 10:57 PM
Oiler

Proof....rational numbers

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

...using the fact that $\sqrt{2} \not\in \mathbb{Q}$
March 4th 2011, 11:02 PM
Prove It

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

The sum of a rational and an irrational is also irrational, so $\displaystyle a + b\sqrt{2}$ 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...
March 4th 2011, 11:48 PM
CaptainBlack

Quote:

Originally Posted by Oiler

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

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

Let $$$a \ne 0$ and $$$b \ne 0$ be rational and suppose $a+b\sqrt{2}$ is rational. Hence there exists a rational $$$c$ such that:

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

But that rationals are closed under addition so:

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

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

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

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

CB

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