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Algebraic and Transcendental numbers proof

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Let α be a transcendental number and β an algebraic number. Prove that α+β is transcendental.


    3. The attempt at a solution

    It's known that adding two algebraic numbers results in an algebraic number.

    Since β is algebraic, it is a root of a polynomial with integer coefficients.

    That is p(x)=Ʃbix^i i=0 to n, p(β)=0.

    So,

    p(x)= b+b1x1+b2x2+b3x3+...+b(n-1)xn-1+bnxn

    I'm stuck, I'm not sure where to go from here...

    Any help is appreciated.

    Thanks.
     
  2. jcsd
  3. Sep 27, 2012 #2
    By contradiction. Assume the sum is algebraic.
     
  4. Sep 27, 2012 #3
    Okay, assuming the sum is algebraic, that means the root of p(x) added to α will give me a number that can be expressed as a root of another polynomial. How would I proceed?
     
  5. Sep 27, 2012 #4
    You have polynomial R(x) with integer coefficients whose root is α+β. Now consider R(x - β).
     
  6. Sep 27, 2012 #5
    Would x represent any real number? I'm sorry, it's not really clicking with me...
     
  7. Sep 27, 2012 #6
    Are you implying that I should take x=α+β? So that R(x-β) would be R(α+β-β)=R(α)?
     
  8. Sep 27, 2012 #7
    x is the argument of the polynomial R(x - β). What do its coefficients look like?
     
  9. Sep 27, 2012 #8
    R(x-β)=b0+b1(x-β)+b2(x-β)2+....+bn(x-β)n

    But since β is a transcendental number, we can further distribute the bn coefficients right? So for b1(x-β)=b1x-b1β
     
  10. Sep 27, 2012 #9
    β is algebraic according to your original post.
     
  11. Sep 27, 2012 #10
    Oh yeah, sorry I made a mistake. It is algebraic, but I am still trying to figure out what this implies.
     
  12. Sep 27, 2012 #11
    Are the coefficients of R(x - β) integer, rational, algebraic, transcendental? What does that mean with regard to its roots?
     
  13. Sep 27, 2012 #12
    The coefficients are algebraic, and the roots must also be algebraic.
     
  14. Sep 27, 2012 #13
    To prove the statement, you need to consider R(x + β) in the same way. Can you now connect all the pieces?
     
  15. Sep 27, 2012 #14
    So, it R(x-β) and R(x+β) have algebraic coefficients then R(x) mus also have algebraic coefficients right?
     
  16. Sep 27, 2012 #15
    R(x) is defined in #4. Its definition follows from the assumption that α+β is algebraic.
     
  17. Sep 27, 2012 #16

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You started by assuming that [itex]\alpha+ \beta[/itex] was algebraic so that it was a root of a polynomial equation [itex]R(x)= 0[/itex] where R is a polynomial with integer coefficients. There is no point is now saying that they must be algebraic!
     
  18. Sep 27, 2012 #17
    I know that, I think I worded what I was trying to say wrong. I meant that algebraic coefficients yield algebraic roots. So R(x) would be a contradiction to this, or am I completely off?
     
  19. Sep 27, 2012 #18
    R(x) does not contradict anything. You assume that ##\alpha + \beta ## is algebraic, so there must be a polynomial with integer coefficients it is a root of. This is R(x). Then you consider another polynomial, which is ##R(x + \beta) ##. ##\alpha ## obviously is its root. What does that mean, if ## \beta ## is algebraic?
     
  20. Sep 27, 2012 #19
    Do you mean α is a root of R(x+β)? Sorry, I know the solution must be obvious and I'm totally missing it. I'm not really familiar with algebraic and transcendental numbers.
     
  21. Sep 27, 2012 #20
    ## P(x) = R(x + \beta) ##. What coefficients does P have? What are its roots (algebraic, transcendental)? Is ## \alpha ## its root? Why?
     
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