Results 1 to 2 of 2

Thread: Prove that there is no complex root such that both the Re and Im parts are rational.

  1. #1
    Junior Member
    Jun 2008

    Prove that there is no complex root such that both the Re and Im parts are rational.

    Let $\displaystyle P$ be a polynomial with integer (real) coefficients such that; (1) - the coefficient of the leading term and the coefficient of the independent term (counted as the coefficient of $\displaystyle x^0, 0$ degree) are odd; (2) -The total number of odd coefficients is odd.

    Like in
    $\displaystyle P(x) = x^3 - 5x^2 + 2x -7$
    $\displaystyle P(x) = 9x^3 - 6 x^2 + 3x -5$
    $\displaystyle P(x) = x^4 +5x^3 + 7x^2 + x +1$
    $\displaystyle P(x) = 7x^5 + 2x^4 - x^3 + 2x^2 - 8x -3
    Prove that $\displaystyle P$ has no root such that both the real and the imaginary parts are rational. In other words, if $\displaystyle a + bi$ is a root of $\displaystyle P$, then at least one of the numbers $\displaystyle a$ and $\displaystyle b$ is irrational
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    Let $\displaystyle P(x) = a_nx^n + \ldots+a_1x+a_0$, and suppose that $\displaystyle x = \textstyle\frac{p+iq}r$ is a root, where p, q and r are integers with no common factor. Then

    . . . . . $\displaystyle a_n(p+iq)^n + a_{n-1}(p+iq)^{n-1}r + \ldots + a_1(p+iq)r^{n-1} + a_0r^n = 0$.. . . . . (*)

    If r is even then 2=(1+i)(1-i) divides $\displaystyle a_n(p+iq)^n$. Since 1ħi are Gaussian primes which do not divide a_0, they must divide p+iq. Thus p and q are both even, contradicting the assumption that p, q and r have no common factor. Therefore r is odd.

    If 1+i divides p+iq then it divides r (since it cannot divide the odd number a_0). But then 1-i will also divide r, and so r is even. This is another contrdiction, and we conclude that 1+i does not divide p+iq.

    Next, notice that 1+i divides the Gaussian integer m+in if and only if m and n have the same parity (both even or both odd). If we call a Gaussian integer "even" if it is a multiple of 1+i, and "odd" if it is not, then "even" and "odd" have the same additivity and multiplicativity properties as they do for real integers. That is, the sum of two "odd" Gaussian integers is "even", as is the sum of two "even" integers, whereas the sum of an "odd" integer and an "even" integer is "odd". Also, the product of two Gaussian integers is "even" unless both of them are "odd", in which case the product is "odd".

    Coming back to (*), you see that the left side contains an odd number of "odd" terms, and is therefore "odd". But the right side is 0, which is "even". That is the final contradiction, showing that the original assumption (that (*) has a solution with rational real and imaginary parts) is false.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove rational raised to a rational is rational.
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Feb 15th 2011, 08:12 PM
  2. Replies: 6
    Last Post: May 5th 2009, 06:49 AM
  3. Replies: 1
    Last Post: Mar 23rd 2009, 06:01 PM
  4. Replies: 12
    Last Post: Nov 22nd 2008, 12:41 PM
  5. Rational Root Theorem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Oct 13th 2008, 05:07 PM

Search Tags

/mathhelpforum @mathhelpforum