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Degree of this number

  1. Mar 6, 2005 #1
    Does anyone know how to find the degree over Q of this number:
    [tex]\sqrt{3} + \sqrt[3]{4}[/tex]

    In fact i'm having trouble finding any generic polynomial that this satisfies! Please help :biggrin:
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  3. Mar 6, 2005 #2


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    I don't know what the question is that you're asking so I'm probably being really stupid here, but I assume it would help if you had that expressed as the rote of some polynomial equation with integer coefficients:

    [tex]x = \sqrt{3} + \sqrt[3]{4}[/tex]

    [tex]x - \sqrt{3} = \sqrt[3]{4}[/tex]

    [tex]x^3 - 3\sqrt{3}x^2 + 9x - 3\sqrt{3} = 4[/tex]

    [tex]x^3 + 9x - 4 = 3\sqrt{3}x^2 + 3\sqrt{3}[/tex]

    [tex]x^6 + 6x^4 - 8x^3 + 81x^2 - 72x - 16 = 27x^4 + 54x^2 + 27[/tex]

    [tex]x^6 - 21x^4 - 8x^3 + 27x^2 - 72x - 43 = 0[/tex]

    That help at all? (still probably worth checking my steps haha)
    Last edited: Mar 6, 2005
  4. Mar 6, 2005 #3


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    Or, if you'd like to do it a different way, just keep raising your number to powers until you get a linearly dependant set. (Treating each distinct irrational number as a basis vector)

    For example, for a = √2 + √3:

    a^0 = 1
    a^1 = √2 + √3
    a^2 = 5 + 2 √6
    a^3 = 11 √2 + 9 √3
    a^4 = 49 + 20 √6

    So, Q(a) is a vector space over Q, with basis vectors 1, √2, √3, and √6. We have 5 different vectors, so 1, a, a^2, a^3, and a^4 form a linearly dependant set...
  5. Mar 6, 2005 #4


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    Of course, the fact that you have to go to 5th power to get a dependent set means that the degree is 4?

    (And notice that Hurkyl was using [itex]\sqrt{2}+\sqrt{3}[/itex], NOT the number of the original question. I like Zurtex's method: [itex]\sqrt{2}+ ^3\sqrt{3}[/itex] is algebraic of order 6 and so must have degree 6 over Q.
  6. Mar 6, 2005 #5


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    Zurtex's approach is certainly easier to execute, but sometimes it's nonobvious how to manipulate things so that radicals don't proliferate. For example, if the cube roots of both 2 and 4 are in the number, when you cube to get rid of the cube root of 2, the cube root of 4 will just introduce more cube roots of 2.
  7. Mar 6, 2005 #6
    Great, thanks a lot guys!
    Following Zurtex's method, i calculated:
    [tex]x^6-9x^4-8x^3+27x^2-72x-11=0[/tex] for [tex]x=\sqrt{3}+\sqrt[3]{4}[/tex].
    (There was a little error with zurtex's calculation i think)
    I managed to prove that this was irreducible tediously......but (this might be a dumb question) how did you know x was algebraic of order 6 to begin with? That would save me a lot of trouble heh
  8. Mar 7, 2005 #7


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    Because [itex][\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2[/itex] and [itex][\mathbb{Q}(\sqrt[3]{4}}) : \mathbb{Q}] = 3[/itex]. Therefore, both 2 and 3 must divide [itex][\mathbb{Q}(\sqrt{3}, \sqrt[3]{4}}) : \mathbb{Q}][/itex].

    This strongly suggests that the degree of your number must be 6... but more work would be needed to prove it.

    Oh, and a bit of shamelessness. :smile: If you used my approach to come up with a polynomial your number satisfies, you can also use it directly to prove that it is the minimum polynomial.

    For example, for √2 + √3, I just have to prove that {1, a, a^2, a^3} is linearly independent, which is a fairly straightforward task.
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