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Proofs: If a|b then -a|b, a|-b, -a|-ab

  1. Feb 3, 2004 #1
    Hello,

    First I will post the question that I am working on.

    I am not good at proofs (even elementry proofs such as these ones). I was wondering if someone could take a look at my work and perhaps confirm whether my proofs are adequate and/or make some suggestions.

    First I will start off with a basic definition of a divisor:

    An integer, a, not equal to zero, is called a divisor of an integer b if there exists an integer c such the b = a c.

    i) If a|b then -a|b.

    Assume a|b. Then b = a c for some integer c by def.

    Let c = -k where -1, k are integers. Then

    b = a (-k) = - a k.

    Since k is an integer, then by def., if a|b then b = -a k.


    Similarly,

    ii) If a|b then a|-b.

    Assume a|b. Then b = a c for some integer c by def.

    Let c = -k where -1, k are integers. Then

    b = a c = a -k = -a k = -(a k)
    -b = --(a k)
    -b = a k

    Once again since k is an integer, then by def., if a|b then -b = a k.


    Also

    iii) If a|b then -a|-b.


    I am sort of stuck on this one. I am not yet sure how to show

    If a|b then -a|-b.

    I thought

    b = a c,
    -b = -a c

    By definition, -a|-b if -b = -a c for some integer c. Since c is an integer, then by def. if a|b then -a|-b.

    Part iii) seems pretty weak to me. In fact all look pretty weak now.

    Any help/insights are appreciated.

    Thankyou.
     
  2. jcsd
  3. Feb 4, 2004 #2
    this is fine but i would adjust it a bit:
    since a|b, b=ac for some c∈Z. then b=(-a)(-c), implying that -a|b as -c∈Z.

    (ie there is no need for the k)

    that's fine i think. here's how i would phrase it. since a|b, b=ac for some c∈Z. then -b=-ac, which implies that -a|-b.

    you want to find an integer q such that -aq=-ab. what might q be?
     
  4. Feb 4, 2004 #3
    Well, by cancellation I would say that q - b. But I don't understand your point.


    Let me try this again.

    iii)

    Prove if a|b then -a|-b.

    Assume

    a|b.

    Then by definition

    b = a c for some integer c.

    Let

    -b = -a c for some integer c.

    Then by definition

    -a|-b.

    Therefore

    If a|b then -a|-b.

    How would that be?

    It seems a little stronger than what I had. But it feels like I am missing something inbetween.
     
  5. Feb 4, 2004 #4
    that seems fine except for the word "let."

    q would be b.

    -ab=-ab implies that -a|-ab.
     
  6. Feb 4, 2004 #5
    That was a typo on my part. I meant q = b. 8)

    I am not sure I am getting this though.

    If you have

    -a q = -ab

    for some q in this case, I could then say

    -a|-ab by definition.

    I can see it better in the other direction:

    Say -a|g.

    Let g = -ab. Then

    -a|-ab. So by definition

    -ab = -a q from some q which is an element of the integers.

    If -ab = -aq, then b =q by cancellation.


    I feel like I am trying to run through a brick wall while the way through the brick wall is a door just a couple of feet to one side.

    8(
     
    Last edited by a moderator: Feb 4, 2004
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