Sections 1.6 & 1.7

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

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Backward reasoning - example • Prove that the square of every odd integer has the form 8k + 1 for some integer k: • Begin with some odd integer n, which by definition has the form n = 2i + 1 for some integer i. • Then n2 = (2i + 1)2 = 4i2 + 4i + 1 • We need to show that n2 has the form 8k + 1 • Reasoning backwards, this follows if 4i2 + 4i can be written as 8k • But 4i2 + 4i = 4i(i + 1) • i(i+ 1) is the product of 2 consecutive integers • Since every other integer is even, either i or i+1 is even • This means their product is even, and can be written 2k for some integer k • Therefore, 4i2 + 4i = 4i(i + 1) = 4(2k) = 8k; it follows that, since n2 = 4i2 + 4i + 1 and 4i2 + 4i = 8k, that n2 = 8k + 1