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CSC 236                  Examples of Well Ordering              Winter 2008
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Proof that \-/ n, \sum_{0<=i<=n} 2^i = 2^{n+1}-1, using well ordering.

    For a contradiction, suppose ! \-/ n, \sum_{0<=i<=n} 2^i = 2^{n+1}-1,
    i.e., -] n, \sum_{0<=i<=n} 2^i != 2^{n+1}-1.
        Consider the set S = { n (- N : \sum_{0<=i<=n} 2^i != 2^{n+1}-1 }.
        By assumption, S is not empty, so by the well ordering principle,
        S contains a smallest element -- call it k.  At this point, we know
        that:
          (i)   \sum_{0<=i<=k} 2^i != 2^{k+1}-1  (because k (- S),
          (ii)  \sum_{0<=i<=j} 2^i = 2^{j+1}-1  \-/ j (- {0,...,k-1}
                (because k is the smallest element in S).
        However, \sum_{0<=i<=0} 2^i = 2^0 = 1 = 2-1 = 2^{0+1}-1, so k != 0.
        Then k > 0 so k-1 >= 0 and \sum_{0<=i<=k-1} 2^i = 2^{(k-1)+1}-1, by
        (ii), which means that
            \sum_{0<=i<=k} 2^i = \sum_{0<=i<=k-1} 2^i + 2^k
                               = (2^k-1) + 2^k = 2*2^k - 1 = 2^{k+1} - 1.
        But this contradicts (i)!
    Hence, there is no value of n (- N for which \sum_{0<=i<=n} 2^i !=
    2^{n+1}-1, i.e., \-/ n (- N, \sum_{0<=i<=n} 2^i = 2^{n+1}-1.

Note:  Proof contains content of base case and inductive step from
    induction proof -- not surprising because we are proving same statement
    using equivalent principle.  Only difference is in structure of proof.

Template for proof of \-/ n, P(n) using well ordering:

    For a contradiction, suppose -] n, ! P(n).
        Then set S = { n (- N : ! P(n) } is not empty so by well ordering,
        S contains a smallest element k.  At this point we know:
          (i)   ! P(k)  (because k (- S)
          (ii)  P(i)  \-/ i (- {0,...,k-1}  (because k is smallest in S).
        Use both facts to derive a contradiction -- standard techniques:
        prove (\-/ i < k, P(i)) -> P(k) to contradict (i) (same as writing
        proof by complete induction) or prove ! P(k) -> -] i < k, ! P(i) to
        contradict (ii) (equivalent to first technique but sometimes easier
        to do).
    Hence, by contradiction, \-/ n, P(n).

Comparison of induction and well ordering:  Induction often more direct,
but sometimes well ordering easier to use.  For example, consider statement
\-/ m, \-/ n > 0, -] r, -] q, r < n /\ m = q n + r.  Read proof by
induction in textbook (Example 1.5, pp.27-28), and compare with following
proof by well ordering -- this is *direct* proof by well ordering, not just
rewrite of induction proof.  Notice how proof by well ordering is much
shorter (and more direct).

    Suppose m (- N, n (- N and n > 0.
        Consider the set R = { r (- N : -] q, m = q n + r }.
        R is not empty because m (- R (m = 0 n + m), so by well ordering,
        R contains a smallest element r, i.e.,
          (i)   -] q, m = q n + r
          (ii)  \-/ r' (- {0,...,r-1}, \-/ q, m != q n + r'
        Claim:  r < n.
        Proof:  For a contradiction, suppose r >= n and
            let q (- N such that m = q n + r, by (i).
            Then m = q n + n + r - n = (q+1) n + (r-n),
            so r-n (- R (because r-n >= 0) and r-n < r (because n > 0),
            contradicting (ii).
        Hence, -] r, -] q, r < n /\ m = q n + r.
    Therefore, \-/ m, \-/ n > 0, -] r, -] q, r < n /\ m = q n + r.