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)

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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