Sections 1.6 1.7

1. Sections 1.6 & 1.7 Methods of Proof & Proof Strategies

2. 2 Methods of Proof Many theorems are implications Recall that an implication (p ? q) is true when both p and q are true, or when p is false; it is only false if q is false To prove an implication, we need only prove that q is true if p is true (it is not common to prove q itself)

3. 3 Direct Proof Show that if p is true, q must also be true (so that the combination of p true, q false never occurs) Assume p is true Use rules of inference and theorems to show q must also be true

4. 4 Example of Direct Proof Prove “if n is odd, n2 must be odd” Let p = “n is odd” Let q = “n2 is odd” Assume n is odd; then n = 2k + 1 for some integer k (by definition of an odd number) This means n2 = (2k + 1)2 = 2(2k2 + 2k) + 1 Thus, by definition, n2 is odd

5. 5 Indirect Proof Uses the fact that an implication (p ? q) and its contrapositive ?q ? ?p have the same truth value Therefore proving the contrapositive proves the implication

6. 6 Indirect Proof Example Prove “if 3n + 2 is odd, then n is odd” Let p = “3n + 2 is odd” Let q = “n is odd” To prove ?q ? ?p , begin by assuming ?q is true So n is even, and n = 2k for some integer k (by definition of even numbers) Then 3n + 2 = 3(2k) +2 = 6k + 2 = 2(3k + 1) Thus, 3n + 2 is even, ?q ? ?p and p ? q

7. 7 Vacuous Proof Suppose p is false - if so, then p ? q is true Thus, if p can be proven false, the implication is proven true This technique is often used to establish special cases of theorems that state an implication is true for all positive integers

8. 8 Vacuous Proof Example Show that P(0) is true where P(n) is: “if n > 1, then n2 > n” Let p = n>1 and q = n2 > n Since P(n) = P(0) and 0>1 is false, p is false Since the premise is false, p ? q is true for P(0) Note that it doesn’t matter that the conclusion (02 > 0 ) is false for P(0) - since the premise is false, the implication is true

9. 9 Trivial Proof If q can be proven true, then p ? q is true for all possible p’s, since: T ? T and F ? T are both true

10. 10 Example of Trivial Proof Let P(n) = “if a >= b then an >= bn” where a and b are positive integers; show that P(0) is true so p = a >=b and q = a0 >= b0 Since a0 = b0, q is true for P(0) Since q is true, p ? q is true Note that this proof didn’t require examining the hypothesis

11. 11 Proof by Contradiction Suppose q is false and ?p ? q is true This is possible only if p is true If q is a contradiction (e.g. r ? ?r), can prove p via ?p ? (r ? ?r)

12. 12 Example of proof by contradiction Prove ?2 is irrational Suppose ?p is true - then ?2 is rational If ?2 is rational, then ?2 = a/b for some numbers a and b with no common factors So (?2 )2 = (a/b)2 or 2 = a2/b2 If 2 = a2/b2 then 2b2 = a2 So a2 must be even, and a must be even

13. 13 Example of proof by contradiction If a is even, then a = 2c and a2 = 4c2 Thus 2b2 = 4c2 and b2 = 2c2 - which means b2 is even, and b must be even If a and b are both even, they have a common factor (2) This is a contradiction of the original premise, which states that a and b have no common factors

14. 14 Example of proof by contradiction So ?p ? (r ? ?r) where ?p = ?2 is rational, r = a & b have no common factors, and ?r = a & b have a common factor r ? ?r is a contradiction so ?p must be false thus p is true and ?2 is irrational

15. 15 Proof by contradiction and indirect proof Can write an indirect proof as a proof by contradiction Prove p ? q by proving ?q ? ?p Suppose p and ?q are both true Go through direct proof of ?q ? ?p to show ?p is also true Now we have a contradiction: p ? ?p is true

16. 16 Proof by Cases To prove (p1? p2? … ? pn) ? q, can use the tautology: ((p1? p2? … ? pn) ? q) ? ((p1 ? q) ? (p2 ? q) ? … ? (pn ? q)) as a rule for inference In other words, show that pi ? q for all values of i from 1 through n

17. 17 Proof by Cases To prove an equivalence (p ? q), can use the tautology: (p ? q) ? ((p ?q) ? (q ? p)) If a theorem states that several propositions are equivalent (p1 ? p2 ? … ? pn), can use the tautology: (p1 ? p2 ? … ? pn) ? ((p1 ? p2) ? (p2 ? p3) ? … ? (pn ? p1))

18. 18 Theorems & Quantifiers Existence proof: proof of a theorem asserting that objects of a particular type exist, aka propositions of the form ?xP(x) Proof by counter-example: proof of a theorem of the form ?xP(x)

19. 19 Types of Existence Proofs Constructive: find an element a such that P(a) is true Non-constructive: prove ?xP(x) without finding a specific element - often uses proof by contradiction to show ??xP(x) implies a contradiction

20. 20 Constructive Existence Proof Example For every positive integer n, there is an integer divisible by >n primes Stated formally, this is: ?n?x(x:x is divisible by >n primes) Assume we know the prime numbers and can list them: p1, p2, … If so, the number p1 * p2 * … * pn+1 is divisible by >n primes

21. 21 Non-constructive Existence Proof Example Show that for every positive integer n there is a prime greater than n This is ?xQ(x) where Q(x) is the proposition x is prime and x > n Let n be a positive integer; to show there is a prime > n, consider n! + 1 Every integer has a prime factor, so n! + 1 has at least one prime factor When n! + 1 is divided by an integer <= n, remainder is 1 Thus, any prime factor of this integer must be > n Proof is non-constructive because we never have to actually produce a prime (or n)

22. 22 Proof by Counter-example To prove ?xP(x) is false, need find only one element e such that P(e) is false Example: Prove or disprove that every positive integer can be written as the sum of 2 squares We need to show ?x?P(x) is true Many examples exist - 3, 6 and 7 are all candidates

23. 23 Choosing a method of proof When confronted with a statement to prove: Replace terms by their definitions Analyze what hypotheses & conclusion mean If statement is an implication, try direct proof; If that fails, try indirect proof If neither of the above works, try proof by contradiction

24. 24 Forward reasoning Start with the hypothesis Together with axioms and known theorems, construct a proof using a sequence of steps that leads to the conclusion With indirect reasoning, can start with negation of conclusion, work through a similar sequence to reach negation of hypothesis

25. 25 Backward reasoning To reason backward to prove a statement q, we find a statement p that we can prove with the property p ? q The next slide provides an example of this type of reasoning

26. 26 Backward reasoning - example