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 $\displaystyle \sqrt{2} \not\in \mathbb{Q}$

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- Mar 4th 2011, 10:57 PMOilerProof....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 $\displaystyle \sqrt{2} \not\in \mathbb{Q}$ - Mar 4th 2011, 11:02 PMProve It
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... - Mar 4th 2011, 11:48 PMCaptainBlack
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