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: 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

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 xP(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:

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