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CSC 236            Sample Solutions for Week  4 Tutorial          Fall 2007
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A.  Recall the formal principles of well ordering, simple induction, and
    complete induction, then review the proof that all three principles are
    equivalent, make sure you understand it, and bring your questions to
    tutorial.

    { See my summary of lecture notes.  Ask if anything is still unclear. }

B.  Exercise 13 on page 45 of the textbook.

    The inductive hypothesis holds for all j, 4 <= j < i, but the inductive
    step relies on the value i-4.  While it is true that i-4 < i, it is NOT
    the case that 4 <= i-4, since all we know of i is i >= 4.  Hence, the
    IH does NOT apply to i-4.

C.  Prove the following statements using well ordering

     -  \-/ n >= 4, n^2 <= 2^n

        Q:  How do we get started?
        A:  For a contradiction, suppose that -] n >= 4, n^2 > 2^n.

        Q:  What does the principle of well ordering allow us to conclude?
        A:  By the principle of well ordering, there must be a smallest
            such n; call it k.

        Q:  Now what?
        A:  Similar to proof by induction: eliminate k = 4, then reason
            from k-1 to k to get contradiction.

        Q:  How do we show k =/= 4?
        A:  Direct proof: 4^2 = 16 = 2^4.  This means k > 4.

        Q:  What next?
        A:  Since k > 4, k-1 >= 4, and since k is smallest such that
            k^2 > 2^k, we know that (k-1)^2 <= 2^{k-1}.

        Q:  How do we get a contradiction?
        A:  Express k^2 as a function of (k-1)^2:
                k^2 = (k-1+1)^2
                    = (k-1)^2 + 2(k-1) + 1
                   <= (k-1)^2 + 3(k-1)      [since k-1 >= 4]
                   <= (k-1)^2 + (k-1)^2     [since k-1 >= 4]
                   <= 2^{k-1} + 2^{k-1}     [by IH]
                   <= 2^k.
            This contradicts k^2 > 2^k.  Hence, \-/ n >= 4, n^2 <= 2^n.

     -  \-/ n, 12^n - 1 is divisible by 11

        For a contradiction, suppose -] n, \-/ k, 12^n - 1 =/= 11 k.
        Then there must be a smallest such n by well ordering, call it m.

        Because 12^0 - 1 = 0 = 0 * 11, we know m > 0.  This means m-1 >= 0.
        Since m is smallest such that 12^m - 1 not divisible by 11, we know
        12^{m-1} - 1 = 11 * k for some k.  But then,
            12^m - 1 = 12 * 12^{m-1} - 1
                     = 11 * 12^{m-1} + 12^{m-1} - 1
                     = 11 * (12^{m-1} + k).          [by IH]
        This contradicts 12^m - 1 =/= 11 k for all k.  Hence, 12^n - 1 is
        divisible by 11 for all n.