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Complete Induction

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Question asks to prove that all binary string that begins and ends with the same number(bit) has an even number of occurences of substrings from {0,1}

    Hint: you may find useful to combine claims about strings that start and end with different bit


    2. Relevant equations


    {0} = {} since 0 is a multiple of 2 it holds.
    {11} = {} since 0 is a multiple of 2 it holds.
    eg 010010 = (01), (10), (01) (10)

    3. The attempt at a solution


    My induction hypothesis was, for all natural number n, assume that p(j) holds for some integer j such that 0 <= j < n, that is a binary string with length j that begins and ends with the same bit has an even number of ooccurences of substrings from {0,1}

    so, let j = n - 1 and by I.H we know that p(j) holds.
    j, begins and ends with the same bit. and can be represented like so, P(j) = 2t for some integer multiple t. since 1st and last digit of j is the same, we know if n = jlast then there are no additional sets created so the equation will still be p(n) = 2t + 0;

    but if we were to add n != jlast, then there would be 1 extra subset at the end since ...01 or ....10 will take place. therefore, the equation will now be p(n) = 2t + 1; even number + odd = odd. hence we've shown that for all n, if first and last digit of string is the same, there are even number of substrings from {0,1}

    I feel like I have done too much explaining in english and not enough mathenatical proof.
    is there somewhere I went too far? help is appreciated.
     
    Last edited: Oct 3, 2012
  2. jcsd
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