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Root 3 + root 2

  1. Feb 1, 2005 #1
    I need to show that this is an algebraic number.

    In other words,

    I need to show: an*x^n + an1*x^(n-1) + ... + a1 * x^1 + a0 * x^0 =
    where the a terms are not ALL 0 but some of them can be.

    Like for root 2 by itself,

    I have 1 * (root 2) ^ 2 + 0 * (root 2)^1 + -2 * (root 2) ^ 0
     
  2. jcsd
  3. Feb 2, 2005 #2
    How about a5=1, a3=-10, a1=1, all others zero?
     
  4. Feb 2, 2005 #3
    According to Maple, that gives -.30e-7.
     
  5. Feb 2, 2005 #4
    let [tex]x = \sqrt{3} + \sqrt{2}[/tex]

    then [tex](x - \sqrt{3})^2 = x^2 - 2\sqrt{3}x + 3 = 2[/tex]

    ... something like that. put one square root one one side & the rest on the other, square both sides, etc etc then you'll get a 4th-degree polynomial that has [tex]\sqrt{3} + \sqrt{2}[/tex] as a root. end of proof
     
  6. Feb 2, 2005 #5
    Rounding error.

    --J
     
  7. Feb 2, 2005 #6

    Curious3141

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    The easiest way is just to find an equation with integral coefficients that has the number as a root.

    What is the square of [tex]\sqrt{2} + \sqrt{3}[/tex] ? Express that in the form : integer plus a square root. Let's call this [tex]n_1 = a + \sqrt{b}[/tex]

    Next, find the conjugate of the number. [tex]n_2 = a - \sqrt{b}[/tex]

    Set up the equation [tex](y - n_1)(y - n_2) = 0[/tex] and expand it out. You know for certain that this equation is going to have integral coefficients because that's a property of conjugate surds.

    Once you have that equation in y, just put [tex]x^2 = y[/tex] and get a quartic in x. That equation will have [tex]\sqrt{2} + \sqrt{3}[/tex] as one of the roots, and you're done.
     
    Last edited: Feb 2, 2005
  8. Feb 2, 2005 #7
    You can look at this from the standpoint of Galois theory. Along with one surd we are going to get the conjugate of: [tex]\pm\sqrt2[/tex]. So putting both roots in a quadratic makes the coefficients rational. [tex]x^2-2=0[/tex].

    The same idea can be applied to this expression: [tex]\pm\sqrt{2}[/tex][tex]\pm\sqrt{3}[/tex]. We don't want the surds to appear in the coefficients.

    Maybe not as efficient as described in previous postings, but does offer some insight.
     
    Last edited: Feb 2, 2005
  9. Feb 2, 2005 #8

    Hurkyl

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    It shouldn't; Maple is capable of doing symbolic arithmetic. Don't ask it to approximate. :tongue2:

    By the way, please don't multiple post. :mad:
     
    Last edited: Feb 2, 2005
  10. Feb 2, 2005 #9

    HallsofIvy

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    Standard method: Use the fact that (a+ b)(a- b)= a2- b2 to get rid of the square roots.
    [tex](x- \sqrt{3}-\sqrt{2})(x-\sqrt{3}+ \sqrt{2})= (x- \sqrt{3})^2- 2= x^2- 2\sqrt{3}x+1[/tex]

    [tex](x^2+1-2\sqrt{3}x)(x^2+1+2\sqrt{3}x)= (x^2+1)^2- 12x^2= x^4+ 2x^2+ 1- 12x^2= x^4- 10x^2+ 1[/tex]

    [tex]\sqrt{3}+ \sqrt{2}[/tex]
    is algebraic because it satisfies
    [tex]x^4-10x^2+ 1[/tex]
     
    Last edited: Feb 2, 2005
  11. Feb 2, 2005 #10

    Hurkyl

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    You made a typo in there somewhere.
     
  12. Feb 2, 2005 #11
    I apologize for the multiple post...I was in a panic. Won't happen again.
     
  13. Feb 2, 2005 #12

    HallsofIvy

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    Yep, I dropped an "x" early on. I've edited it. Thanks, Hurkyl.
     
  14. Feb 2, 2005 #13
    Wow, I hope I never have to do these kind of problems. What course is this for?
     
  15. Feb 2, 2005 #14
    Modern Algebra course of some kind. I'm doing a similar one myself this semester. Actually, I solved that very exercise a couple of hours ago.

    Very nice course, by the way. Don't hope without knowing. :wink:
     
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