Equating Field of Elements via the Radix

[SubZir0]

Is there a way of calculating the field of elements in a system using the base? I expect that's the main way, I am just unlearned.

For Example base phi: (1+√5)/2 has the FoE Q[√5] = Q + [√5]Q

Is it just any non-rational elements are added to the field? what would base: (√2 / √7) be?

 

[HallsofIvy]

Yes, any rational elements are already in Q! If by [itex]\sqrt{2}/\sqrt{7}[/itex] you mean [itex]\sqrt{2/7}[/itex], any member of [itex]Q[\sqrt{2/7}[/itex] is of the form [itex]a+ b\sqrt{7/2}[/itex] for a and b rational numbers.

Note that [itex]Q[\sqrt{7/2}][/itex] is different from [itex]Q[\sqrt{7}, 1/\sqrt{2}]= Q[\sqrt{7}, \sqrt{2}][/itex] where the two roots can appear separately. Such numbers are of the form [itex]a+ b\sqrt{7}+ c\sqrt{2}+ d\sqrt{14}[/itex].

(We can use [itex]\sqrt{2}[/itex] rather than [itex]1/\sqrt{2}[/itex] because
[tex]\frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}[/tex])

 

[SubZir0]

Ok, excellent, I was wondering how to deal with more than one thing like you said:

[itex]a+ b\sqrt{7}+ c\sqrt{2}+ d\sqrt{14}[/itex]

So you just extend the equation until it's all in there then do a multiple of them? (14 at the end)