Results 1 to 3 of 3

Math Help - Fraction in Lowest terms

  1. #1
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288

    Fraction in Lowest terms

    Prove \frac{n^3+2n}{n^4+3n^2+1} is a fraction in its lowest terms

    Rewriting
    \frac{n(n^2+2)}{(n^2+1)^2+n^2}

    The numerator is a product, so we shall consider them separately

     \frac{n^2+2}{(n^2+1)^2+n^2}
    Let z=n^2+1

    \frac{z+1}{z^2+z-1}. Numerator and denominator share no common factors

    Consider
     \frac{n}{(n^2+1)^2+n^2}
    This also has no common factors

    Hence
    \frac{n^3+2n}{n^4+3n^2+1}
    has no common factors
    QED

    Is this proof valid?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774
    It believe it is correct, but whether it is acceptable to an instructor depends on your course. You may have studied an algorithm to prove similar claims, and then you need to go over the steps without justifying them. However, for this to look like a proof to an outsider, it should include not only computational steps, but also invocation of some lemmas that prove that the computations indeed show what you claim they do.

    Quote Originally Posted by I-Think View Post
    Rewriting
    \frac{n(n^2+2)}{(n^2+1)^2+n^2}

    The numerator is a product, so we shall consider them separately
    Here one should say that n and n^2 + 1 are irreducible (why?) and so, if some polynomial divides n(n^2+2), then it is either n or n^2+2, using something like Euclid's lemma for polynomials. Therefore, it is sufficient to check whether the denominator is divisible by these two polynomials. Then one should say a word or two why z+1 does not divide z^2+z-1 and n does not divide (n^2+1)^2+n^2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    I think that's correct. Here is another way to approach it:

    Write n^3+2n = n(n^2+2). Since n^2+2 is irriducible over \mathbb{Q}, either  n or n^2+2 must be a factor of
    f(n) = n^4+3n^2+1 for the fraction not to be on the lowest form. It's obvious that n isn't a factor of
    f(n). For n^2+2 to be a factor, we must have n^4+3n^2+1 = (n^2+2)(an^2+bn+c), for some
    a, b, c \in\mathbb{R}. Expanding the RHS we have n^4+3n^2+1 = an^4+bn^3+(c+2a)n^2+2bn+2c.
    Comparing the coefficients we have (from the coefficient of [LaTeX ERROR: Convert failed] ) a = 1, and (from the coefficient of n^2)
    c+2a = 3, so c = 3-2a = 1. But (from the coefficient of n^0) 2c = 1, so c = \frac{1}{2}. A contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: June 6th 2010, 04:42 PM
  2. Lowest terms
    Posted in the Algebra Forum
    Replies: 7
    Last Post: April 8th 2009, 12:16 AM
  3. Reduce fraction to lowest terms
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 23rd 2009, 12:11 AM
  4. Replies: 1
    Last Post: April 26th 2008, 11:28 PM
  5. lowest terms
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 8th 2007, 05:03 PM

Search Tags


/mathhelpforum @mathhelpforum