# Equating Field of Elements via the Radix

1. Sep 2, 2011

### 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?

2. Sep 2, 2011

### HallsofIvy

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

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

(We can use $\sqrt{2}$ rather than $1/\sqrt{2}$ because
$$\frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$$)

Last edited: Sep 2, 2011
3. Sep 2, 2011

### SubZir0

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

$a+ b\sqrt{7}+ c\sqrt{2}+ d\sqrt{14}$

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