T or F? The prime field of R=Q[sqrt(2)] then Frac(R)=Reals

[robertjordan]

Homework Statement

If $R=\mathbb{Q}[\sqrt{2}]$, then Frac(R)=$\mathbb{R}$

Homework Equations

$\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2} | a,b\in{\mathbb{Q}}\}$
Frac(R) is the fraction field of R is basically $\{\frac{a+b\sqrt{2}}{c+d\sqrt{2}} | a,b,c,d\in{\mathbb{Q}}\}$.

The Attempt at a Solution

Back of the book says false...

I tried to show $\pi\not\in{Frac(R)}$ by assuming by way of contradiction that there existed $a,b,c,d\in{\mathbb{Q}}$ s.t. $\frac{a+b\sqrt{2}}{c+d\sqrt{2}}=\pi$.

That implies $c\pi+d\pi\sqrt{2}=a+b\sqrt{2}$ ... this led nowhere :(

 

[Dick]

Homework Statement

If $R=\mathbb{Q}[\sqrt{2}]$, then Frac(R)=$\mathbb{R}$

Homework Equations

$\mathbb{Q}[\sqrt{2}]=\left{a+b\sqrt{2} | a,b\in{\mathbb{Q}}\right}$
Frac(R) is the fraction field of R is basically $\left{\frac{a+b\sqrt{2}}{c+d\sqrt{2}} | a,b,c,d\in{\mathbb{Q}}\right}$.

The Attempt at a Solution

Back of the book says false...

I tried to show $\pi\not\in{Frac(R)}$ by assuming by way of contradiction that there existed $a,b,c,d\in{\mathbb{Q}}$ s.t. $\frac{a+b\sqrt{2}}{c+d\sqrt{2}}=\pi$.

That implies $c\pi+d\pi\sqrt{2}=a+b\sqrt{2}$ ... this led nowhere :(

Frac(R) is countable, isn't it?

 

[robertjordan]

Frac(R) is countable, isn't it?

Thanks for the response.

If we could show Frac(R) is countable, we could conclude Frac(R)=/=reals because the reals are uncountable.

How can we show Frac(R) is countable? My friend said a way to show countability is to find a bijection between Frac(R) and the natural numbers...

Could we do something like this... http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function?

 

[Dick]

Thanks for the response.

If we could show Frac(R) is countable, we could conclude Frac(R)=/=reals because the reals are uncountable.

How can we show Frac(R) is countable? My friend said a way to show countability is to find a bijection between Frac(R) and the natural numbers...

Could we do something like this... http://en.wikipedia.org/wiki/Cantor_pairing_function#Cantor_pairing_function?

That's the basis of it. That shows that |NxN|=|N|. You don't need to directly construct the bijection. If you can show that if there is a map from a countable set ONTO Frac(R) then you know Frac(R) is countable. If you pick a tuple of four rational numbers (a,b,c,d) there an obvious way to map that to an element of Frac(R), right? So that's a mapping from QxQxQxQ -> Frac(R). Now you know Q is countable, right?