Square root of 2 divided by 0 is rational?

[PcumP_Ravenclaw]

Dear All,
Please help me understand how $\sqrt{2}$ divide by 0 is rational as stated in the excerpt from alan F beardon's book?

 

[phinds]

I don't follow his logic, but in any case it is actually undefined, and thus it is not meaningful to talk about it being rational or irrational. I think probably what he is saying is more complex than that "(sqrt 2) / 0 is rational"

 

[gopher_p]

He is saying that the multiplicative inverse of the element $x=a+b\sqrt{2}$ in the field $\mathbb{Q}(\sqrt{2})$ is the element $x^{-1}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}$.

There part where he mentions that $a^2-2b^2\neq 0$ is so that we can rest assured that the formula that he's given us for $x^{-1}$ is defined for all pairs of rationals $a$ and $b$.

 

[PeroK]

Dear All,
Please help me understand how $\sqrt{2}$ divide by 0 is rational as stated in the excerpt from alan F beardon's book?

You've misunderstood him. If $a^2 - 2b^2 = 0$ then $\sqrt{2} = \frac{a}{b}$ is rational. So, you know that $a^2 - 2b^2 \ne 0$

 

[PcumP_Ravenclaw]

Hey PeroK, I think this is what you mean??

$a^2 - 2b^2 = 0$
can be written as
$(a + \sqrt{2}b)(a - \sqrt{2}b) = 0$
here note that a and b are rational numbers ${a + b√2 : a, b ∈ Q},$
so either $(a + \sqrt{2}b)$ or $(a - \sqrt{2}b)$ must equal 0.
lets say $(a - \sqrt{2}b) = 0$ then $\sqrt{2} = a/b$ which is a contradiction so $(a - \sqrt{2}b) = 0$ cannot happen.

 

[PeroK]

Hey PeroK, I think this is what you mean??

$a^2 - 2b^2 = 0$
can be written as
$(a + \sqrt{2}b)(a - \sqrt{2}b) = 0$
here note that a and b are rational numbers ${a + b√2 : a, b ∈ Q},$
so either $(a + \sqrt{2}b)$ or $(a - \sqrt{2}b)$ must equal 0.
lets say $(a - \sqrt{2}b) = 0$ then $\sqrt{2} = a/b$ which is a contradiction so $(a - \sqrt{2}b) = 0$ cannot happen.

That's not quite the way I'd do it! I'd say:

$a^2 - 2b^2 = 0 \ \ \Rightarrow \ \ a^2 = 2b^2 \ \Rightarrow \ \ \frac{a^2}{b^2} = 2 \ \Rightarrow \ \ (\frac{a}{b})^2 = 2$