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Show that negative divisors of an integer, are just the negatives of the pos divisors

  1. Jun 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that negative divisors of an integer, are just the negatives of the positive divisors.


    3. The attempt at a solution

    having an integer [itex]n[/itex] and a negative divisor [itex]k[/itex], we get the positive divisor by multiplying [itex](-1)k[/itex], thus each negative divisor of an integer, is the negative of the positive divisor of [itex]n[/itex].

    I have the idea but don't know if this kind of proof is correct.
     
  2. jcsd
  3. Jun 14, 2012 #2

    Mark44

    Staff: Mentor

    Re: Show that negative divisors of an integer, are just the negatives of the pos divi

    It looks to me like what you showed is that if k is a negative number, then -k is a positive number. There's nothing in your proof that is related to division. What does it mean to say that a number is a divisor of some other number.

    If k is a negative number that divides n, how do you know that -k also divides n? That's the part thats missing.
     
  4. Jun 15, 2012 #3
    Re: Show that negative divisors of an integer, are just the negatives of the pos divi

    By a theorem, if [itex]m|p[/itex], and [itex]n|p[/itex], while [itex]gcd(m,n) = 1[/itex], then [itex]mn|p[/itex]. Since [itex](-1)|n[/itex] and [itex]k|n[/itex], and [itex]gcd(-1,k) = 1[/itex], then [itex](-1)k|n[/itex].

    Is this correct?
     
  5. Jun 15, 2012 #4

    Mark44

    Staff: Mentor

    Re: Show that negative divisors of an integer, are just the negatives of the pos divi

    Looks OK, but is more complicated than it needs to be. What I was thinking was more along these lines: if k|n, then for some integer a, n = ak. Using this basic concept, it's easy to show that if k|n, then -k|n as well, without the need to invoke any other theorem.
     
  6. Jun 15, 2012 #5
    Re: Show that negative divisors of an integer, are just the negatives of the pos divi

    Hmmm what about this one:
    if [itex]k|n[/itex], then for some integer [itex]a[/itex], [itex]n = ak = ak1 = ak(-1)(-1) = a(-k)(-1)[/itex] which shows that [itex](-k)|n[/itex]

    So if [itex]k[/itex] is a negative divisor, then [itex](-k)[/itex] which is positive, also divides [itex]n[/itex]. So for each negative divisor [itex]k[/itex], we have the positive [itex](-k)[/itex] integer. Now if [itex]k[/itex] is positive, then [itex](-k)[/itex] is negative, which shows that for each positive divisor [itex]k[/itex], [itex](-k)[/itex] which is a negative integer also divides [itex]n[/itex]. So the converse is also true, and the statement holds.
     
  7. Jun 15, 2012 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Show that negative divisors of an integer, are just the negatives of the pos divi

    It seems to me easier just to note that n= ak= (-a)(-k). Since so a is an integer, so is b= -a and you have shown that n= b(-k).

     
  8. Jun 15, 2012 #7
    Re: Show that negative divisors of an integer, are just the negatives of the pos divi

    yeah, I just wrote it the way it came on my mind. Thanks for noting it
     
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