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Equating Field of Elements via the Radix

  1. Sep 2, 2011 #1
    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. jcsd
  3. Sep 2, 2011 #2


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    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])
    Last edited by a moderator: Sep 2, 2011
  4. Sep 2, 2011 #3
    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)
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