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Math Help - Help with Assignment

  1. #1
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    Help with Assignment

    I should serve her until Sunday
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  2. #2
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    Quote Originally Posted by sbsite View Post
    I should serve her until Sunday
    Problem 1 is fairly easy. Suppose there in a non-integral rational solution
    X=a/b where b!=1, and a and b coprime.

    Then multiply the equation through by b^(n-1). The first term is then not
    an integer, and all the other terms are integers, and the sum is zeros, which
    is impossible as the sum cannot be an integer.

    Therefore if there is a rational solution it is integral.

    RonL
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  3. #3
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    Problem 2 is also easy.

    The leading coeffienct is 1 and the last coefficent is 1. By the rational roots test a solution is either 1 or -1.

    If the solution is 1 then:
    1^n+p_{n-1}1^{n-1}+...+p_11^n+1=2+p_1+...+p_{n-1}=0
    But that cannot be.

    If the solution is -1 then:
    1+p_1(-1)+p_2(-1)^2+...+p_{n-1}(-1)^{n-1}=0
    Which is it not.


    Thus there are no solutions.
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  4. #4
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    Little request???

    Quote Originally Posted by CaptainBlack View Post
    Problem 1 is fairly easy. Suppose there in a non-integral rational solution
    X=a/b where b!=1, and a and b coprime.

    Then multiply the equation through by b^(n-1). The first term is then not
    an integer, and all the other terms are integers, and the sum is zeros, which
    is impossible as the sum cannot be an integer.

    Therefore if there is a rational solution it is integral.

    RonL

    can you writh that in matematics .
    i have problems with my english/???
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Problem 2 is also easy.

    The leading coeffienct is 1 and the last coefficent is 1. By the rational roots test a solution is either 1 or -1.

    If the solution is 1 then:
    1^n+p_{n-1}1^{n-1}+...+p_11^n+1=2+p_1+...+p_{n-1}=0
    But that cannot be.

    If the solution is -1 then:
    1+p_1(-1)+p_2(-1)^2+...+p_{n-1}(-1)^{n-1}=0
    Which is it not.


    Thus there are no solutions.


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  6. #6
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    Talk normally. And stop with the smiley abuse.
    -=USER WARNED=-
    ~~~
    I got that because 1 or -1 are solutions.
    If I substitute -1 into that polynomial
    Then I get (if you look at the powers only):
    (-1)^n+(-1)^{n-1}+...+(-1)^2+(-1)^1
    That is alternating,...
    1,-1,1,-1,...
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  7. #7
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    Quote Originally Posted by sbsite View Post

    can you writh that in matematics .
    i have problems with my english/???
    Suppose \exists\ a,b \in \mathbb{Z} such that gcd(a,b)=1, and that X=a/b is a root of:

    <br />
X^n + p_{n-1}X^{n-1}+\ ...\ +p_1X +1=0<br />

    Then:

    <br />
b^{n-1}X^n =-[p_{n-1}b^{n-1}X^{n-1}+\ ...\ +p_1b^{n-1}X +b^{n-1}]<br />

    Now:

    <br />
-[p_{n-1}b^{n-1}X^{n-1}+\ ...\ +p_1b^{n-1}X +b^{n-1}]\in \mathbb{Z}<br />

    but:

    <br />
b^{n-1}X^n \notin \mathbb{Z}<br />

    a contradiction.

    Therefor no such a, b exist and so if the given equation has a rational root that root must be integral.

    RonL
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