Results 1 to 10 of 10

Math Help - Contradiction of Galios theory

  1. #1
    Newbie
    Joined
    Jun 2011
    Posts
    15

    Contradiction of Galios theory

    Hello.
    Galios theory tell us x^5-6x+3 is not solvable by radical but every equation lower than fifth can solve by radical.
    If G is solvable and H is solvable too G*H are solvable . for x^5-6x+3 we can use newton’s method and find one root of this equation .
    we obtian x=1.4 and factor this equation .
    x^5-6x+3=(x-1.4)(ax^4+px^3+qx^2+sx+t)
    a=1
    p=1.4
    q=(1.4)^2
    s=(1.4)^3
    t=-6+(1.4)^4
    x^5-6x+3=(x-1.4)(x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4)
    (x-1.4) is solvable and (x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4) is solvable too so (x-1.4)*(x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4) are solvable .
    why did Galios say that x^5-6x+3 is not solvable by radical?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2

    Re: Contradiction of Galios theory

    Quote Originally Posted by footmath View Post
    Hello.
    Galios theory tell us x^5-6x+3 is not solvable by radical but every equation lower than fifth can solve by radical.
    If G is solvable and H is solvable too G*H are solvable . for x^5-6x+3 we can use newton’s method and find one root of this equation .
    we obtian x=1.4 and factor this equation .
    x^5-6x+3=(x-1.4)(ax^4+px^3+qx^2+sx+t)
    a=1
    p=1.4
    q=(1.4)^2
    s=(1.4)^3
    t=-6+(1.4)^4
    x^5-6x+3=(x-1.4)(x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4)
    (x-1.4) is solvable and (x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4) is solvable too so (x-1.4)*(x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4) are solvable .
    why did Galios say that x^5-6x+3 is not solvable by radical?
    As I understand it, and I'm not really an algebraist, Galois theory tells you that the general quintic polynomial is not solvable by radicals. As a trivial example, the polynomial

    (x-1)(x-2)(x-3)(x-4)(x-5)=0

    is quite easily solved by radicals, although even that's not necessary. So, the existence of special cases of quintic polynomials that are solvable does not negate the result that the general quintic, which is

    ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0,

    is not solvable by radicals.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,572
    Thanks
    1412

    Re: Contradiction of Galios theory

    By the "rational root theorem", any rational root of a polynomial equation with integer coefficients must be of the form a/b where b evenly divides the leading coefficient and a evenly divides the constant term. Here, the leading coefficient is 1 and the constant term is 3 so the only possible rational roots are 1, -1, 1/3, and -1/3. 1.4= 7/5 is NOT a root of this equation.

    You could just as easily calculated that (1.4)^5- 6(1.4)+ 3= -0.02176, NOT 0.

    Your basic assumption, that 1.4 is a root, is wrong.

    I did my own Newton's method solution and got
    1.401618792653142 after about 17 iterations, starting at x= 1, but that still is only an approximation to the solution The equation x^5- 6x+ 3= 0 has NO rational roots.
    Last edited by HallsofIvy; August 5th 2011 at 06:23 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2010
    Posts
    193

    Re: Contradiction of Galios theory

    Quote Originally Posted by HallsofIvy View Post
    By the "rational root theorem", any rational root of a polynomial equation with integer coefficients must be of the form a/b where b evenly divides the leading coefficient and a evenly divides the constant term. Here, the leading coefficient is 1 and the constant term is 3 so the only possible rational roots are 1, -1, 1/3, and -1/3. 1.4= 7/5 is NOT a root of this equation.

    You could just as easily calculated that (1.4)^5- 6(1.4)+ 3= -0.02176, NOT 0.

    Your basic assumption, that 1.4 is a root, is wrong.
    This, and

    Quote Originally Posted by Ackbeet View Post
    As I understand it, and I'm not really an algebraist, Galois theory tells you that the general quintic polynomial is not solvable by radicals. As a trivial example, the polynomial

    (x-1)(x-2)(x-3)(x-4)(x-5)=0

    is quite easily solved by radicals, although even that's not necessary. So, the existence of special cases of quintic polynomials that are solvable does not negate the result that the general quintic, which is

    ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0,

    is not solvable by radicals.
    this. The general quintic is not solvable by radicals because the Galois group of a polynomial of degree 5 or larger is not necessarily solvable (the symmetric groups S_n,n\geq 5 are not solvable). Galois theory doesn't say that EVERY quintic of degree 5 or larger is insolvable by radicals. This example by Ackbeet is the simplest example.

    For another slightly less trivial example (but only slightly less), take p(x)=x^5-1\in \mathbb{Q}[x]. It is fairly easy to see that the Galois group of this polynomial is isomorphic to \mathbb{Z}_4, a solvable group. This implies that the polynomial is solvable by radicals.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2011
    Posts
    15

    Re: Contradiction of Galios theory

    Quote Originally Posted by HallsofIvy View Post
    By the "rational root theorem", any rational root of a polynomial equation with integer coefficients must be of the form a/b where b evenly divides the leading coefficient and a evenly divides the constant term. Here, the leading coefficient is 1 and the constant term is 3 so the only possible rational roots are 1, -1, 1/3, and -1/3. 1.4= 7/5 is NOT a root of this equation.

    You could just as easily calculated that (1.4)^5- 6(1.4)+ 3= -0.02176, NOT 0.

    Your basic assumption, that 1.4 is a root, is wrong.

    I did my own Newton's method solution and got
    1.401618792653142 after about 17 iterations, starting at x= 1, but that still is only an approximation to the solution The equation x^5- 6x+ 3= 0 has NO rational roots.
    Thank you.
    Can you prove that The equation x^5- 6x+ 3= 0 has NO rational roots ?
    When we want to calculate the root of a equation for examplex^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546 $ we cannot obtain a rational root but we say this equation is solvable by radical . $ \sqrt[5]{5.40985+0i} $ is the root of x^5- 6x+ 3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Contradiction of Galios theory

    Quote Originally Posted by footmath View Post
    Thank you.
    Can you prove that The equation x^5- 6x+ 3= 0 has NO rational roots ?
    When we want to calculate the root of a equation for examplex^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546 $ we cannot obtain a rational root but we say this equation is solvable by radical . $ \sqrt[5]{5.40985+0i} $ is the root of x^5- 6x+ 3
    It is a common fact that \overline{\mathbb{Z}}\cap\mathbb{Q}=\mathbb{Z} where \overline{\mathbb{Z}} is the ring of algebraic integers (look it up). From there you can check that f(x)=x^5-6x+3 strictly decreases for x<-2 and strictly increases for x>2. Thus, to check that f has no rational roots you must merely check that f doesn't vanish at \{-2,-1,0,1,2\}.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,572
    Thanks
    1412

    Re: Contradiction of Galios theory

    Quote Originally Posted by footmath View Post
    Thank you.
    Can you prove that The equation x^5- 6x+ 3= 0 has NO rational roots ?
    By the "rational root theorem" that I quoted above, the only possible rational roots are 1, -1, 3, and -3.
    1^5- 6(1)+ 3= -2, not 0.
    (-1)^5- 6(-1)+ 3= 8, not 0.
    (3)^5- 6(3)+ 3= 243- 18+ 3= 228, not 0.
    (-3)^5- 6(-3)+ 3= -243+ 18+ 3=-222 , not 0.

    When we want to calculate the root of a equation for examplex^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546 $
    Since you say the solution is not a rational number, I presume you understand that number, which is rational, once again, is an approximation, not an exact solution. The "cubic formula" is rather complicated so I may well be wrong but I get
    \sqrt[3]{-\frac{1}{2}+i\sqrt{\frac{5}{108}}}+\sqrt[3]{\frac{1}{2}-i\sqrt{\frac{5}{108}}}

    Notice those "i"s? It is the fact that, even for cubics with all real roots, you may require complex numbers midway through the formula that led to the acceptance of complex numbers.

    we cannot obtain a rational root but we say this equation is solvable by radical . $ \sqrt[5]{5.40985+0i} $ is the root of x^5- 6x+ 3
    Where did you get that "5.40985"? I strongly suspect that is also an approximation. If so that is NOT a solution in radicals because it is not in terms of radicals of exact numbers.
    And, in fact, for that value of x, x^5- 6x+ 3 is -8.7899913273470831550756050870503e-7, or -0.000000087899913273470831550756050870503 very, very close to 0 but NOT 0. Again, this is an approximation, not a solution.
    Last edited by HallsofIvy; August 5th 2011 at 05:45 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jun 2011
    Posts
    15

    Re: Contradiction of Galios theory

    Quote Originally Posted by HallsofIvy View Post
    By the "rational root theorem" that I quoted above, the only possible rational roots are 1, -1, 3, and -3.
    1^5- 6(1)+ 3= -2, not 0.
    (-1)^5- 6(-1)+ 3= 8, not 0.
    (3)^5- 6(3)+ 3= 243- 18+ 3= 228, not 0.
    (-3)^5- 6(-3)+ 3= -243+ 18+ 3=-222 , not 0.


    Since you say the solution is not a rational number, I presume you understand that number, which is rational, once again, is an approximation, not an exact solution. The "cubic formula" is rather complicated so I may well be wrong but I get
    \sqrt[3]{-\frac{1}{2}+i\sqrt{\frac{5}{108}}}+\sqrt[3]{\frac{1}{2}-i\sqrt{\frac{5}{108}}}

    Notice those "i"s? It is the fact that, even for cubics with all real roots, you may require complex numbers midway through the formula that led to the acceptance of complex numbers.


    Where did you get that "5.40985"? I strongly suspect that is also an approximation. If so that is NOT a solution in radicals because it is not in terms of radicals of exact numbers.
    And, in fact, for that value of x, x^5- 6x+ 3 is -8.7899913273470831550756050870503e-7, or -0.000000087899913273470831550756050870503 very, very close to 0 but NOT 0. Again, this is an approximation, not a solution.
    What is your opinion about x^5 + 20x^3 + 20x^2 + 30x + 10 =0
    Are you accept that the one root of this equation is \sqrt[5]{2}-\sqrt[5]{2^2}+\sqrt[5]{2^3}-\sqrt[5]{2^4} ? if you substitute this answer in the equation you will see very, very close to 0 but NOT 0.
    \sqrt[5]{5.40985+0i} is close to 0 for the equation x^5- 6x+ 3
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,572
    Thanks
    1412

    Re: Contradiction of Galios theory

    My opinion is wrapped up in your "If you substitute this answer in the equation you will see very, very close to 0 but NOT 0". A solution to an equation is, by definition, a value that makes the equation exactly true, not "very, very close". If you know about Galois theory, surely you must know what a "solution to an equation" means.

    Galois theory says that, for n greater than 4, there exist polynomial equations of degree n with solutons that cannot be written in terms of radicals. Of course, any number can be approximated, to any desired accuracy, by radicals.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5

    Re: Contradiction of Galios theory

    The question has been answered, and answered well, several times now. Thread closed.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proof by contradiction
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 22nd 2011, 01:17 AM
  2. contradiction help
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: March 1st 2010, 11:05 PM
  3. Galios groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 18th 2009, 08:42 AM
  4. contradiction
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 18th 2008, 08:52 AM
  5. Reed Solomon & Galios
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 1st 2007, 06:13 PM

Search Tags


/mathhelpforum @mathhelpforum