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Proof about polynomials.

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex] A_n [/itex] be the algebraic numbers obtained as roots of polynomials with integer
    coeffiecients that have degree n. Using the fact that every polynomial has a finite number of roots. Show that [itex] A_n [/itex] is countable.
    3. The attempt at a solution
    So an nth degree polynomial has n roots. And (n+1) coefficients.
    so for the nth degree polynomial I will have the first coefficient go to the first prime and then the next one to the second prime and then (n+1) coefficient to the n+1 prime.
    And we take the absoulte value of the integers before we raise them to a prime.
    For example the first coefficient will go to [itex] 2^x [/itex]
    now to take care of the same nth degree polynomial but might have negative coefficients.
    we will have the first coefficient go to the (n+2) prime and then keep continuing this.
    We will just keep continuing this and shifting down the list of primes to make sure we got every polynomial. Will this work or Am I way off.
     
  2. jcsd
  3. Feb 4, 2012 #2

    Dick

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    That is complete garbage. It has nothing to do with primes. Are there a countable number of polynomials with degree n? Can you show that? Start with n=0 or n=1 or n=2. Are the roots then countable? Or not?
     
  4. Feb 4, 2012 #3
    Its a good thing we have the forums and people like you to straighten me out. I wasn't really implying that it had something to do with primes, I was just using primes to generate unique numbers. But any way's that's complete nonsense. Could I do something like add up their coefficients to get an integer and use this somehow to show that they are a countable number of polynomials. Their roots would seem to be countable because I could factor that nth degree polynomial into n factors.
     
  5. Feb 5, 2012 #4

    Dick

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    Take n=1. The polynomials are n1*x+n0. That corresponds to an ordered pair (n1,n0). That's in Z x Z. Is Z x Z countable? Some of the polynomials might have the same root. Does that matter? Some might have no root. You are probably too worried about making a unique correspondence. You don't have to.
     
  6. Feb 5, 2012 #5
    ok, so ZxZ would be countable because we could construct (n1,n0) by
    writing the integers across the top and then a column of the integers going down vertically. the pair (n1,n0) would be n1 across the top and n0 would be the vertical place.
    like a multiplication table and we could write the naturals in a diagonal fashion by going back and fourth so this would be countable. is ZxZxZ countable. we could just group it like this
    (ZxZ)x(Z) and do the same thing so this will cover all polynomials of degree n.
    So all the polynomials of degree n are countable.
    A polynomial of degree n has n roots, so the roots are countable because if we took the union of an infinite amount of countable things, it is still countable.
     
  7. Feb 5, 2012 #6

    Dick

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    Yes, there are a countable number of polynomials of degree n, and each polynomial has a countable (finite) number of roots. So the union of the set of all roots of polynomials of degree n is countable. Now you take the union over n, right?
     
  8. Feb 5, 2012 #7
    SO now I just need to take the union of all polynomials of degree n, so this union over all n will be countable because each set in the union is countable.
     
  9. Feb 5, 2012 #8

    Dick

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    Yes, but you want to state that a little more clearly. You want to show the set of ROOTS is countable, finally.
     
  10. Feb 5, 2012 #9

    Deveno

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    it may help to recall that a polynomial of degree n can have at most n roots.
     
  11. Jun 20, 2012 #10
    why is it incorrect to map the stuff to primes like I originally tried to do. Like when they prove that the rationals are countable they map them to prime bases
    like [itex] 2^p 3^q [/itex] because each natural has a unique prime factorization.
     
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