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Zakon Vol 1, Ch2, Sec-6, Prob-19 : Cardinality of union of 2 sets

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data


    Show by induction that if the fi nite sets A and B have m and n elements,
    respectively, then
    (i) A X B has mn elements;
    (ii) A has 2m subsets;
    (iii) If further A [itex]\cap[/itex] B = [itex]\varphi[/itex], then A [itex]\cup[/itex] B has m+ n elements.

    NOTE : I am only interested in the (iii) section of the problem. Section (i) and (ii), I was able to solve albeit with some help.

    2. Relevant equations



    3. The attempt at a solution
    Here is the attempt at the solution

    I am using mathematical induction to solve the query.


    P(n) : |A| = m, |B| = n; |A [itex]\cup[/itex] B| = m + n if A [itex]\cap[/itex] B = [itex]\varphi[/itex].

    Basis Step:

    P(n = 1) to be proven as true.

    Let A = { a1, a2,......am} => |A| = m
    Let B = {b1} => |B| = 1

    Now A [itex]\cup[/itex] B = {a1, a2,...am,b1}
    |A [itex]\cup[/itex] B| = m + 1 elements, since A and B are disjoint(given) Thus P(1) holds true.
    Basis Step is proven.

    Let us redefine B = {b1, b2, ....., bk, bk+1} => |B| = k+1.
    Also |A [itex]\cap[/itex] B| = [itex]\varphi[/itex]
    Now consider Bk = {b1, b2,..., bk} => Bk [itex]\subset[/itex] B, |Bk| = k

    Inductive hypothesis:
    Assume P(k) holds true for n = k i.e.,

    |A [itex]\cup[/itex] Bk| = m + k, when A [itex]\cap[/itex] Bk = [itex]\varphi[/itex]

    Inductive Step:

    Bk [itex]\subset[/itex] B => B = Bk [itex]\cup[/itex] {bk+1}
    A [itex]\cup[/itex] B = A [itex]\cup[/itex] (Bk [itex]\cup[/itex] {bk+1}) = (A [itex]\cup[/itex] Bk) [itex]\cup[/itex] {Bk+1} = m + k + 1 = m + (k + 1) => P(k+1) holds true. Thus Induction is complete.

    Somehow, I am not convinced with the solution that I have attempted. Please let me know if I am missing something.
     
  2. jcsd
  3. Mar 6, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is wrong. What you have written is 2 times m while, in fact, A has 2m subsets. If you don't want to use LaTeX (as you have below) use 2^m.

    That's correct but seems overly complicated. If A and B have no members in common, [itex]A\cup B[/itex] contains every member of A and B- thus has m+ n members.
     
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