Results 1 to 4 of 4

Math Help - Divisibility

  1. #1
    Newbie Demandeur's Avatar
    Joined
    Jun 2010
    Posts
    14

    Divisibility

    Question: If a\mid{b} and c\mid{d}, must (a+c)\mid(b+d)?

    Counterexample: 2\mid{4} and 3\mid{9}, but [tex]5\nmid{13}[/Math].

    However, I want to argue along the lines of if a\mid{b} and c\mid{d}, then b = aq and d = cq'
    for some integers [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] . How should we proceed from there?
    Last edited by Demandeur; July 22nd 2010 at 03:49 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408
    I don't understand your question. The counterexample is proof enough that the theorem is false. Are you seeking a way of finding solutions?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Demandeur's Avatar
    Joined
    Jun 2010
    Posts
    14
    Quote Originally Posted by Media_Man View Post
    I don't understand your question. The counterexample is proof enough that the theorem is false. Are you seeking a way of finding solutions?
    Is the counterexample the only way to show that the statement is false? I was hoping that if we proceed as if we were to prove it,
    we could derive a contradiction. Perhaps I'm expecting the wrong thing, but I do of course realise that a counterexample is enough
    to be prove it false. Also, now that you mention it, how can we find the set of values for which it holds?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408

    Finding Solutions

    how can we find the set of values for which it holds?
    Problem: Choose arbitrary a,c and find values b,d fitting the following three criteria: a|b, c|d, and a+c|b+d

    Solution: Restate the three criteria as b=aq, d=cq', b+d=(a+c)n (as you have correctly done in your first post). Substitute (1) and (2) into (3) to get aq+cq'=(a+c)n. This can be rearranged as a/c=(n-q')/(q-n). The left and right are both equal rational numbers, therefore the numerators and denominators are scalar multiples: as=n-q', cs=q-n. So q=n+cs and q'=n-as. Substituting back into (1) and (2), b=an+sca and d=cn-sca. So given 4 independent parameters a,c,n,s, b and d can be gotten by this formula to satisfy your three criteria. Proving this captures ALL solutions is probably trickier, but someone else might like to tackle it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Divisibility 11
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: December 20th 2008, 03:41 AM
  2. Divisibility (gcd) 10
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 19th 2008, 05:44 PM
  3. Divisibility (gcd) 9
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 19th 2008, 02:12 PM
  4. Divisibility (gcd) 8
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: December 19th 2008, 04:53 AM
  5. Divisibility
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 14th 2008, 10:24 AM

Search Tags


/mathhelpforum @mathhelpforum