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

Methods of Proof Many theorems are implications
4/21/2017 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) ch1.5&3.1

4/21/2017 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 ch1.5&3.1

Example of Direct Proof
4/21/2017 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 ch1.5&3.1

4/21/2017 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 ch1.5&3.1

Indirect Proof Example
4/21/2017 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 ch1.5&3.1

Vacuous Proof Suppose p is false - if so, then p  q is true
4/21/2017 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 ch1.5&3.1

Vacuous Proof Example Show that P(0) is true where P(n) is:
4/21/2017 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 ch1.5&3.1

4/21/2017 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 ch1.5&3.1

Example of Trivial Proof
4/21/2017 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 ch1.5&3.1

Proof by Contradiction
4/21/2017 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) ch1.5&3.1

Example of proof by contradiction
4/21/2017 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 ch1.5&3.1

Example of proof by contradiction
4/21/2017 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 ch1.5&3.1

Example of proof by contradiction
4/21/2017 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 ch1.5&3.1

Proof by contradiction and indirect proof
4/21/2017 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 ch1.5&3.1

Proof by Cases To prove (p1 p2 …  pn)  q, can use the tautology:
4/21/2017 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 ch1.5&3.1

Proof by Cases To prove an equivalence (p  q), can use the tautology:
4/21/2017 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)) ch1.5&3.1

Theorems & Quantifiers
4/21/2017 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) ch1.5&3.1

Types of Existence Proofs
4/21/2017 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 ch1.5&3.1

Constructive Existence Proof Example
4/21/2017 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 ch1.5&3.1

Non-constructive Existence Proof Example
4/21/2017 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) ch1.5&3.1

Proof by Counter-example
4/21/2017 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 ch1.5&3.1

Choosing a method of proof
4/21/2017 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 ch1.5&3.1

Forward reasoning Start with the hypothesis
4/21/2017 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 ch1.5&3.1

4/21/2017 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 ch1.5&3.1

Backward reasoning - example
4/21/2017 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 ch1.5&3.1

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