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

2. Methods of Proof Many theorems are implications
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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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4. Example of Direct Proof
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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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6. Indirect Proof Example
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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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7. Vacuous Proof Suppose p is false - if so, then p  q is true
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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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8. Vacuous Proof Example Show that P(0) is true where P(n) is:
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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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10. Example of Trivial Proof
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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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11. Proof by Contradiction
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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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12. Example of proof by contradiction
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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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13. Example of proof by contradiction
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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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14. Example of proof by contradiction
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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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15. Proof by contradiction and indirect proof
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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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16. Proof by Cases To prove (p1 p2 …  pn)  q, can use the tautology:
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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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17. Proof by Cases To prove an equivalence (p  q), can use the tautology:
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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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18. Theorems & Quantifiers
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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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19. Types of Existence Proofs
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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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20. Constructive Existence Proof Example
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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: 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
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21. Non-constructive Existence Proof Example
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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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22. Proof by Counter-example
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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 xP(x) is true
Many examples exist - 3, 6 and 7 are all candidates
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23. Choosing a method of proof
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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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24. Forward reasoning Start with the hypothesis
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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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26. Backward reasoning - example
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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
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