sections 1.6 & 1.7 methods of proof & proof strategies
TRANSCRIPT
Sections 1.6 & 1.7
Methods of Proof & Proof Strategies
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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))
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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)
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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
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Constructive Existence Proof Example
• For every positive integer n, there is an integer divisible by >n primes
• Stated formally, this is: nx(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
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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)
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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 xP(x) is true– Many examples exist - 3, 6 and 7 are all
candidates
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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
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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
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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
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Backward reasoning - example
Prove that the square of every odd integer has the form 8k + 1 for some integer k:
1. 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
2. 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
3. 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