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Proof Techniques ICS141 (H1.6) 1 p ! 8/23/94 Vacuous Proof q is true if p is false. Thus, we can prove ) p q by establishing that is false. p Example 1 Theorem: If 2x is odd for some integer x, x2 + 1 is prime. Proof: By the denition of odd and even integers, it is not the case that 2x is odd. 2 p ! Trivial Proof q is true if q is true. Thus, we can prove p ) q by establishing that q is true. Example 2 Theorem: If x = p + 1 > 0 for some prime integer p, then 2x + 1 is odd. Proof: By the denition of odd integers, 2x + 1 is obviously odd. 3 2 2 Direct Proof Apply inference rules in order to deduce the truth of q from the truth of p. Example 3 Theorem: If 6x + 9y = 101, then either x or y is not an integer. Proof: Assume that 6x + 9y = 101. This can be rewritten as 3(2x + 3y ) = 101. But 101=3 is not an integer. Hence 2x + 3y is not an integer. Therefore, either x or y must not be an integer. 2 Example 4 Theorem: If S is a set of one- and two-digit integers such that each of the digits 0 through 9 occurs exactly once in the set S , then the sum of the elements of S is divisible by 9. Proof: Assume that the hypothesis of the above assertion is true. Note that the digits 0 through 9 sum to 45. In any set S , some of the digits will occur in the 10's position and the reminder will occur in the 1's position. Let k denote the sum of the digits which occur in the 10's position. Then the sum of the elements of S can be expressed as follows. 10k + (45 0 k) = 9k + 45 = 9(k + 5) Since it has a factor 9, the sum is divisible by 9 regardless of the value k. 4 2 Proof of the Contrapositive Show that the contrapositive : !: q p of p ! q is true, by using other proof techniques. Example 5 A perfect number is an integer which is equal to the sum of all its divisors except the number itself. For example, 6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14. Theorem: A perfect number is not a prime number. Proof: The contrapositive is that a prime number is not a perfect number. We use direct proof to show the contrapositive. Suppose that k is a prime number. Then k 2 and k has exactly two divisors: 1 and k. The sum of all its divisors except k itself is 1. Since k 2, k 6= 1. Thus k is not a perfect number. 2 1 5 Proof by Contradiction Assume that p ! q were false, i.e., r ^ :r from this assumption. p were true and q were false. Derive a contradiction such as General Principle of Proof by Contradiction Assume that p were false. Derive a contradiction from this assumption. Then we can conclude that p is true. Example 6 Theorem: There is no largest prime number. Proof: Assume that there were a largest prime number, say K . Since all prime numbers are greater than 1 and none of prime numbers are greater than K , there must be only a nite number of prime numbers. Consider the product of all those prime numbers and call it j . j = 2 1 3 1 5 1 7 1 1 1K If we divide j + 1 by any prime number between 2 and K , the remainder is 1. It means that j + 1 cannot be expressed as a product of any two integers other than (j + 1) and 1. Thus j + 1 must be a prime number. Since j > K , j + 1 is a prime number greater than K . This contradicts the assumption that K is the largest prime number. Therefore, there is no largest prime number. 2 6 Reduction to Absurdity Assume that p ! q were false. That is, p is true and q is false. Derive :r for some assertion r which is known to be true, by using inference rules. Example 7 Theorem: If x2 + x 0 2 = 0, then x 6= 0. 2 2 2 Proof: Assume that x + x 0 2 = 0 and x = 0. Then x + x 0 2 = 0 + 0 0 2 = 02 must be equal to 0. It is obvious that 02 6= 0. Thus the assumption is not valid and hence the theorem is proved. 2 7 Proof by Cases (p1 _ p2 _ 1 1 1 _ pn) ! q is equivalent to (p1 ! q) ^ (p2 ! q) ^ 1 1 1 ^ (pn (1 i n). ! q ). Show pi ) q for each i Example 8 Let max(a; b) be the operation to choose the maximum of two integers a and b. Theorem: max(a; max(b; c)) = max(max(a; b); c) Proof: For any three integers a, b and c, one of the following 6 cases must hold. Case 1: a b c Then max(b; c) = b and max(a; b) = a and max(max(a; b); c) = max(a; c) = a. Therefore the equality holds. Case 2: a c b max(a; max(b; c)) = max(a; c) = a and max(max(a; b); c) = max(a; c) = a. Thus the equality holds. Case 3: b a c max(a; max(b; c)) = max(a; b) = b and max(max(a; b); c) = max(b; c) = b. Thus the equality holds. 2 Case 4: b c a max(a; max(b; c)) = max(a; b) = b and max(max(a; b); c) = max(b; c) = b. Thus the equality holds. Case 5: c a b max(a; max(b; c)) = max(a; c) = c and max(max(a; b); c) = max(a; c) = c. Thus the equality holds. Case 6: c b a max(a; max(b; c)) = max(a; c) = c and max(max(a; b); c) = max(b; c) = c. Thus the equality holds. Since the equality holds in all cases, the equality holds. 3 2