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Sections 1.6 1.7

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**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