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A Quick, Disjointed Survey of Mathematical Techniques

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Title: A Quick, Disjointed Survey of Mathematical Techniques


1
A Quick, Disjointed Survey of Mathematical
Techniques
  • Reading for this lecture Rosen 3.1 - 3.2
  • Be sure you know the material from 1.4-1.5,
    1.7-1.8, 4.1-4.4 (OR CLRS 2/e A.1, B.1, C.1)

2
The Nature of Proof
  • We usually want to prove implications (for
    example, If n is even, then n2 is also even.)
    -- usually truths that depend on other truths
  • We try to show that knowing the if part is true
    leads logically to certain truths that make up
    the then part
  • If each step is logically sound, then you must
    accept the end result

3
Four Ways to Prove pq
  • Direct Proof Prove pq by creating an
    implication chain
  • Transitive law ((ab) Ù (bc)) (ac)
  • Indirect proof A direct proof of ØqØp
  • Proof by contradiction Prove pq by assuming p
    Ù Øq and showing a contradiction that arises from
    it
  • Inductive proof Proof of "(ngtc) P(n) over the
    universe of discourse of integers

4
The Direct Proof
  • Proving that the statement pq is true by showing
    that p logically leads to q
  • Predicate calculus applies here for true
    correctness
  • Example If n is even, then n2 is even.
  • P(n) n is even
  • Q(n) n2 is even
  • The statement is really "n (P(n) Q(n)) where
    the universe of discourse is (nonnegative)
    integers

5
Direct Proof Example
  • Let n be an element in the universe of discourse
  • Assume n is even, show its square is even
  • n even 2 must be one of its prime factors
  • Thus, n n n2 has 22 in its prime
    factorization
  • Since 2 is one of n2s prime factors, n2 is even
  • Logically, we show pq by following a chain of
    implications (pa, ab, bq, thus pq)
  • Usually, the sub-implications are small steps
    leading toward the big truth

6
Indirect Proof
  • Since pq is logically equivalent to ØqØp
    (contrapositive law!), you can prove the former
    by directly proving the latter this is called an
    indirect proof
  • Lets prove the statement we just examined
    indirectly -- in other words, show that "n (ØQ(n)
    ØP(n))

7
Indirect Proof Example
  • Let n be an element in the universe of discourse
  • Assume n2 is not even
  • Then 2 is not a prime factor of n2
  • Therefore, 2 cannot appear in the prime
    factorization of n (or else it would have in
    n2s)
  • So, n is not even

8
Proof by Contradiction
  • Show that pq is true by initially assuming that
    pÙØq is true
  • Because Øq logically leads to Øp since pq is
    actually true, we get pÙØp
  • Thus pÙØq is false, and its negation must be
    true, but its negation is exactly equivalent to
    the original implication
  • In general contradiction proof, assume the
    negation of what you are trying to prove

9
Proof By Contradiction Ex.
  • Proof by contradiction is similar to an indirect
    proof
  • Ex. If n is even, n2 is even
  • Assume n is even, but n2 is odd
  • Since n2 is odd, 2 is not a prime factor of n2,
    so 2 is not a prime factor of n either
  • Thus, n is odd -- contradicts assumption
  • NOTE for pq, dont assume ØpÙq and then show Øq
    -- this proves nothing!

10
Another Contradiction Ex.
  • Prove that Ö (2) is not a rational number
  • Assume that it is, and so it can be expressed as
    a lowest-terms fraction of two integers a/b (a
    and b have no common factors)
  • So 2 a2/b2, and 2b2 a2, so a2 is even
  • Since a2 is even, a is even (exercise for reader)
  • Thus, a 2c where c is some integer, and
    therefore 2b2 4c2 2(2c2)
  • So b is also even, thus a and b are both
    divisible by 2, but the fraction was in lowest
    terms, so there is a contradiction -- assumption
    was false

11
The Inductive Proofs
  • Used to show truth over a universe of discourse
    consisting of the integers larger than some base
    number
  • Example Prove that for all integers n, the sum
    of the integers from 1 to n is n(n1)/2
  • Let P(m) be Sum of integers from 1 to m is
    m(m1)/2
  • Want to show "ngt1 P(n)

12
Induction Methodology
  • To show P(n) true for all ngtc
  • Show P(c), called the basis step
  • Show for any arbitrary m, P(m)P(m1), called the
    inductive step
  • Since it works for c, it must work for (c1), and
    since it works for (c1), it must work for (c2),
    and so on and so forth
  • Most common form of inductive proof

13
Induction Example
  • Prove that for any positive integer n, a 2n x 2n
    chessboard with one square removed can be tiled
    entirely with three-square, L-shaped tiles
  • Basis Prove P(1). A 2 x 2 chessboard with one
    square removed looks exactly like an L-shaped
    tile
  • Inductive Step Assume P(n), show P(n1). For
    the 2n1 x 2n1 chessboard, split it into four
    quadrants each of size 2n x 2n
  • For the quadrant with the missing square,
    inductive hypothesis P(n) applies -- we assume we
    can tile it
  • For the quadrants without the missing square,
    note that we can temporarily remove the middle
    adjacent corners to make another L, tile whats
    left, and then tile the missing corners
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