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Analysis/ Abstract Algebra

  1. Jan 20, 2009 #1
    Def: A polynomial f(x) with coefficients in Q (the rationals) is called a "numerical polynomial" if for all integers n, f(n) is an integer also.

    I have to use induction to prove that for k > 0

    that the function f(x) := (1/k!)*x*(x-1)...(x-k+1) is a numerical polynomial

    I checked that this is true for k=1, but to be honest I'm not even sure what the dot dot dot means. If k=5 for say, I interpreted the dot dot dot as (1/5!)*x*(x-1)*(x-2)*(x-3)*(x-4). Is this the correct interpretation? If so it is indeed true for k=1, but nonetheless I don't know how to show that it is true for k+1.

    Thanks so much.

    P.S My professor sucks and it is really discouraging as a recently declared math major. So I will be on here often! Loves.
  2. jcsd
  3. Jan 20, 2009 #2


    Staff: Mentor

    ... is an ellipsis, and means continuing in the same pattern. Your interpretation above is correct.

    For an induction proof, you need to show that the statement is true for some base case (n = 1 will do), assume that the statement is true for n = k, and then use that statement to show that the statement is true for n = k + 1. (You'll notice that I changed your k to n.)

    The induction hypothesis is that f(x) = 1/k!*x*(x -1)*(x - 2)* ... *(x - k + 1) is a numerical polynomial.

    What can you say about 1/(k + 1)!*x*(x -1)*(x - 2)* ... *(x - k)?
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