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

1. Apr 5, 2013

### robertjordan

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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 :(

Last edited: Apr 5, 2013
2. Apr 5, 2013

### Dick

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

3. Apr 5, 2013

### robertjordan

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?

4. Apr 6, 2013

### Dick

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?

Last edited: Apr 6, 2013