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Phi is irrational

  1. Oct 4, 2007 #1
    The golden ratio is irrational. Do you know any clever proofs for this fact? I put this here, because it's not homework--only more of a discussion.
     
  2. jcsd
  3. Oct 4, 2007 #2

    matt grime

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    Clever? Does the fact that it is an algebraic number that is not an integer count as clever?
     
  4. Oct 4, 2007 #3

    CRGreathouse

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    Once you show that sqrt(5) is irrational it's pretty easy. You can use the standard proof for that -- suppose a/b = sqrt(5) with a/b in lowest terms, then consider [itex]a^2=5b^2[/itex] mod 25.

    The root of 4x-3=0 is algebraic but rational.
     
  5. Oct 4, 2007 #4
    Yeah, but phi is the root of a *monic* polynomial. Matt meant to say it is an algebraic integer which is not an integer.
     
  6. Oct 4, 2007 #5
    How about

    [tex]\frac{F_{2n}}{F_{2n-1}} < Phi < \frac{F_{2n+1}}{F_{2n}}[/tex]

    If you assume [tex]phi = a/b[/tex] then the above inequality conflicts with that.
     
    Last edited: Oct 4, 2007
  7. Oct 5, 2007 #6

    CRGreathouse

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    Nice proof, ramsey2879.
     
  8. Oct 6, 2007 #7
    I don't get it. There has to be a gazillion rationals between those two fractions, no matter how big n is or how close to the limit you are.
     
  9. Oct 6, 2007 #8

    matt grime

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    Not a gazillion, Dodo. Infinitely many in fact.
     
  10. Oct 6, 2007 #9
    your right
    But [tex] phi^{n} = F_{n-1} + F_{n}phi[/tex]

    Then

    [tex]1+phi = phi^2[/tex]

    so we have phi is a root of x^2- x -1 but the discriminate is [tex]\sqrt{5}[/tex] so phi is irrational.
     
  11. Dec 13, 2007 #10
    supose sqrt(5) = p/q in lowest terms ==> p,q integer such that gcd(p,q)=1 ==>

    ==> 5 = p^2/q^2 ==> 5q^2 = p^2

    if gcd(5,p) [tex]\neq[/tex] 1 ==> q^2 = 5x^2 ==> q = sqrt(5)x ==> contradiction

    if gcd(5,q) [tex]\neq[/tex] 1 ==> 125y^2 = p^2 ==> 25y = p ==> gcd(5,p) [tex]\neq[/tex] 1 ==> contradiction

    ** the two contradictions shows up because gcd(p,q)=1

    so gcd(5,p) = gcd(5,q) = 1 ==> gcd(p,q) [tex]\neq[/tex] 1 ==> contradiction

    sqrt(5) is irrational
     
  12. Dec 13, 2007 #11
    could someone prove that irrational OP integer = irrational, OP = operations +, -, / and *
     
  13. Dec 13, 2007 #12

    CRGreathouse

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    Let z be an integer, n be a positive integer, and x be an irrational number.

    x + z is irrational (else a/b - z = (a-bz)/b which is rational)

    x - z is irrational by the above.

    x * n is irrational (else a/b / n = a/(bn) which is rational)

    x / n is irrational (else a/b * n = (an)/b which is rational)

    x * 0 is rational

    x / 0 is undefined
     
  14. Dec 13, 2007 #13

    HallsofIvy

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    Those are "two fractions". Those are two sequences of fraction. Phi is between every pair of corresponding numbers in those sequences.
     
  15. Dec 13, 2007 #14

    Gib Z

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    That seems to be the most common definition for phi.
     
  16. Dec 13, 2007 #15

    CRGreathouse

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    How about the continued fraction form for phi?
     
  17. Dec 17, 2007 #16
    Thank you, really very simple.

    Another simple proof: proves that if the nth-root of a positive whole number will not be a positive whole number**, also will not be a rational number.

    This should generalize our results.

    consider gcd(p,q) = 1 (p/q in lowest terms)

    k^1/n = p/q ==> kq^n = p^n ==> k | p ==> kq^n = (k^n)*(x^n) ==>

    ==> q^n = k^(n-1)*x^n ==> k | q ==> k | p and q ==> contradiction

    ** note that if k^1/n is a whole number ==> p/q will not be in lowest terms

     
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