415 lecture note #2

[Ch 1] Logic and Proofs (2)

1.4 Proofs

• Axiom -- a logical sentence/statement.
• Theorem -- a proposition that has been proved to be true.  e.g. "For all real number x, x² >= 0".
• Lemma -- a theorem that is not too interesting by itself but useful in proving other theorems.  e.g. "For all real number x, x² >= x".
• Corollary -- a theorem which follows from another theorem quickly.  e.g. "For all real number x, x² + 1 >= 0".

2) Various proof techniques

Example: 
To prove a theorem "If P(x₁,..,x_n), then Q (x₁,..,x_n)", where x₁ Î D₁,.., x_n Î D_n.  

1. Direct proof and counter example 
   • Show the theorem is true for all values of x₁ Î D₁,.., x_n Î D_n, or 
     show a counterexample (x1,..,xn) in which the theorem is false.

   • Example:

     Proposition:  If n is any even integer, then (-1)ⁿ = 1.

     Proof:  Suppose n is any even integer.  Then n = 2k for some integer k.  Hence, by laws of exponents of algebra,

      (-1)ⁿ = (-1)^2k = ((-1)²)^k = 1^k = 1.  Therefore, the proposition is true.

2. Proof by contradiction
   • To prove the theorem (to be true), show that assuming the negation of the theorem leads to a contradiction.  
   • Steps:  
     1. Start with ¬(P(x₁,..,x_n) ® Q (x₁,..,x_n)), which is equivalent to P(x₁,..,x_n) Ù ¬Q (x₁,..,x_n).  
     2. Then show there are no x₁,..,x_n which make that negated theorem true.
   • Example:

     Proposition: For all integers n, if n² is even, then n is even.

     Proof:  Suppose not.  That is, [Negation of the theorem] suppose there exists an integer n such that n² is even and n is odd.  Since n is odd, n = 2k + 1 for some integer k.  Then,

     n² = (2k + 1)² = (2k + 1)(2k + 1) = 4k² + 4k + 1 = 2(2k² + 2k) + 1

     by the laws of algebra.  Let m = 2k² + 2k.  Then, m is an integer because 2 and k are integers, and sums and products of integers are integers.  It follows by substitution that n² = 2m + 1 for an integer m, and thus n² is odd.  Thus, contradiction. {Hence, the supposition is false and the theorem is true.]

1.6 Mathematical Induction

• Another proof technique often used to prove theorems on values/objects that are composed of sequence (such as [1,2,3,4,..]).
• Suppose we want to show a theorem P(n) is true for all positive integer n (>= n₀).  Then we must show P(n₀), P(n₀+1), P(n₀+2),... are all true.  But instead of enumerating all possible (sometimes infinite number of) cases, we can show 2 cases, and generalize inductively.
  1. Show for the initial value of the sequence P(n₀) 
  2. Show for a general case "if P(n) is true, then P(n+1) is true."

• This scheme is analogous to dominos.

  
  P(1) P(2) P(3)

• Formal definition (slightly different from the textbook)

  Principle of (ordinary) Mathematical Induction: Let P(n) be a predicate that is defined for integer n, and let a be a fixed integer.  Suppose the following two statements are true: [Basic Step] P(a) is true. [Inductive Step] For all integers k >= a,  [Inductive Hypothesis] Assume P(k) is true.  [Inductive statement] Then P(k+1) is true. Then, the statement "For all integers n >= a, P(n)" is true.

   
• Example:

  Theorem:  for n = 1, 2, ..

  Proof:  We show by induction on n.

  Basic Step:  When n = 1,    .... (A)

  Inductive Step:  
  [Inductive Hypothesis] Assume is true for all integer n >= 1.  
  [Inductive statement] We show that the equation is true for n+1, that is, . 
  Using the inductive hypothesis, we derive it as follows.

  .... (B)

  By (A) and (B), the theorem is true (for all integers n >= 1).

• More examples:
  1. 1 + 3 + 5 + .. + (2n-1) = n²
  2. 1 + 2 + 2² + .. +  2ⁿ = 2ⁿ⁺¹ - 1
  3. n! >= 2^n-1
  4. 5ⁿ -1 is divisible by 4, for n = 1, 2,..
  5. For all integers n >= 4, n cents can be obtained using 2-cent and 5-cent coins.