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Discrete Math

Methods of proof

Proofs A proof is a valid argument that establishes the

truth of a theorem. The statements used in a proof include

axioms (or postulates) which are statements we assume to be true

Premises of the theorem, Previous proven theorems, and lemmas (sometimes

prove a part of the theorem) Rules of inference Definitions, All terms used in a proof must be defined

Theorems A theorem is a statement that can be shown to

be true (using a proof) The statement of a theorem is a conjecture, a

statement that is proposed to be true (usually based on some evidence)

A conjecture becomes a theorem after it is formally proven to be true

A theorem may be stated as a quantification of a conditional statement

A statement that is somewhat important will become a theorem when it is proven, Simpler statements (propositions) will become facts, or results

Types of Proof Direct Proof Indirect Proof or Proof by Contraposition Proof by contradiction Proof using cases Exhaustive proofs Proof by mathematical induction Other methods of proof

Definition (to be used in proofs) Consider any two integers a and b. We

say that a divides b if and only if there exists an integer q such that b=qa. We write this symbolically as a|b When a divides b then a is a factor of b When a divides b then b is a multiple of a When a divides b then a is divisible by b If no such q exists then a does not divide b

a b

Another Definition (used in proofs)

Let p>1 be an integer. p is prime when it is divisible by only 1 and

itself (by 1 and p) Otherwise p is called a composite integer

Disproving a proposition Sometimes it is easy to disprove a

proposition or theorem by finding a counter example for which the statement is not true.

Example: Disproving a proposition Proposition

Every positive integer is a prime number x universe of discourse positive integers P(x) x is a prime number ∀x P(x)

It is easy to find a counterexample that will disprove the proposition ∀x P(x) Consider the positive integer 8 2|8, the definition of a prime number states that a prime

number is divisible only by 1 and by itself 2 is not 1 or 8, there is a contradiction and the proposition

is disproved

Example: Proving a proposition? Proposition

Every positive integer is a prime number x universe of discourse positive integers P(x) x is a prime number ∀x P(x)

It is easy to find examples that will satisfy the proposition ∀x P(x) Consider the positive integers 2, 3, and 11 We can show that 2, 3 and 11 are all prime We cannot say that all positive integers are prime based

on those particular examples Proof by example is no more convincing when the

proposition is true !!!

Direct Proofs A direct proof shows that a conditional

statement p → q is true. Proving the theorem or proposition requires that We begin with p and show step by step that q is true We show the case P true Q false never occurs We justify each step of our proof with an axiom or

proven proposition (like a rule of inference) We have already seen direct proofs of a

number of propositions while we were studying the rules of inference

Example of Direct Proof Proposition (Theorem) to be proven

If a|b and b|c then it follows that a|c a, b, c are positive integers P(a,b) a|b Q(b,c) b|c R(a,c) a|c ∀a, ∃b, ∃c [ ( P(a,b)^Q(b,c) ) → R(a,c) ]

Example of Direct Proof Proposition (Theorem) to be proven

∀a, ∃b, ∃c ( P(a,b)^Q(b,c) ) → R(a,c) ) Replace a, by arbitrary positive integers x x, is arbitrarily chosen from the universe of

positive integers (may represent any positive integer)

(P(x,b)^Q(b,c) ) → R(x,c) This is an example of Universal Instantiation

of x

Example of Direct Proof1. ∀a, ∃b, ∃c ( P(a,b)^Q(b,c)) → R(a,c))

2. ∃b, ∃c ( P(x,b)^Q(b,c) )→ R(x,c)) using Universal Instantiation

3. ( P(x,y)^Q(y,z) → R(x,z) ) using Existential Instantiation

4. Using the definition of divisible1. P(x,y) → ∃q y=qx where q is a positive integer

2. Q(y,z) → ∃r z=ry where r is a positive integer

5. z=ry=r(qx)=(rq)x where r and q are positive integers

6. z=ry=r(qx)=(rq)x rq is a positive integer (because the product of two positive integers is a positive integer)

Example of Direct Proof1. Using the definition of divisible

x|z means ∃s=rq z=sx where s is a positive integer

So if P(x,y)^Q(y,z) is true then R(x,z) is true P(x,y)^Q(y,z) → R(x,z) is true

We have constructed a y and a z for which this is true we can use Existential Generalization to give ∃b, ∃c ( P(a,b)^Q(b,c) ) → R(a,c) )

Since x is arbitrary (any element of the universe) P(a,b)^Q(b,c) ) → R(a,c) by Universal Generalization

Indirect Proof Also called Proof by Contraposition The contrapositive of a proposition or theorem is

proven rather than the proposition or theorem itself The contrapositive is logically equivalent to the

original statement, so the same thing is being proved

Sometimes the contrapositive is easier to prove than the original statement

Example Problem Proposition (Theorem) to be proven

if a ≠ b then it follows that b ∤ a a, b are positive integers for which a|b Q(a,b) a ≠ b R(a,b) b ∤ a Q(a,b) → R(a,b) ¬ R(a,b) → ¬Q(a,b) Contrapositive

Example Problem Proposition (Theorem) to be proven

If a|b and a ≠ b then it follows that b ∤ a ¬ R(a,b) → ¬ Q(a,b) Contrapositive a, b are positive integers for which a|b ¬ Q(a,b) a = b ¬ R(a,b) b∣a

Prove the contrapositive: demonstrate that when a|b and b∣a then a = b

Example of Indirect Proof ¬R(a,b) → ¬Q(a,b)

1. ¬R(a,b) Premise

2. ¬ R(x,y) Universal Instantiation3. x|y from 2 4. y|x definition of universe5. y=rx and x=qy definition of divisible6. y=rx=r(qy)=(rq)y rq=1 so r=q= 1 and y=x 7. ¬Q(x,y)8. ¬Q(a,b) Universal Generalization

Proof by contradiction We want to prove a statement p is true We can prove p is true if we can

demonstrate that for some proposition r that contradiction q

Because ¬p → q is true we conclude that ¬p is false (p is true)

This approach to proving p is true is called proof by contradiction

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Example: proof by contradiction Prove that is irrational

P(x) is irrational

Suppose that ¬P(x) is true (p is rational) Then there exist two integers, with no common factors, a and b

such the a/b = Squaring both sides gives 2 = a2/b2

2b2 = a2 so a2 is even by the definition of even By the definition of even a=2c So 2b2 = 4c2 or b2 = 2c2, so b2 is even, and b is even But both a and b are even so they have a common factor 2 This contradiction shows ¬P(x) is false so P(x) must be true

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Proof by cases Sometimes cannot prove a theorem or

proposition using a single argument. In these situations you can often divide

the problem into cases, then demonstrate the validity of the proposition or theorem using a different argument for each case

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Proof by cases If P ↔ P1 v P2 v … v Pk then

P → Q iff P1 v P2 v … v Pk → Q

or

P → Q iff P1 → Q ^ P2 → Q ^ … ^ Pk → Q

So we can prove by demonstrating that each of the following statements is true

P1 → Q (one case)

P2 → Q (another case)

… Pk → Q (last case

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Proof by cases: Example Prove ∀n ∈ Z [n/2] [n/2] = [n2/4] Case 1: n is even n= a*2

[2a/2] [2a/2] = [4a2/4]= a2

[n2/4] = [(2a*2a)/4] = a2

Case 2: n is odd n =a*2+1 [(2a+1)/2] [(2a+1)/2] = ([a+ ½ ] [a+ ½ ])

= a(a+1) + ¼ [n2/4] = [(2a+1)2/4] = [(4a2+4a+1)/4]

= a(a+1) + ¼ 23

Your turn Prove by cases

If x,y ∈ ℝ and x+y >= 100 then x >= 50 or y >= 50

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Your turn: If x,y ∈ ℝ and x+y >= 100 then

x >= 50 or y >= 50

Case 1: for x,y ∈ ℝ y<50 x+y >= 100

x >= 100 – y If y>=50 then y = 50 + r for r an arbitrary non negative real

number x >= 100 – (50+r) = 50 – r x >= 50-r So x may be <50 , 50, or >50, all three possibilities are

consistent with the conclusion x >= 50 or y >= 5025

Your turn: If x,y ∈ ℝ and x+y >= 100 then

x >= 50 or y >= 50

Case 1: for x,y ∈ ℝ y>=50 x+y >= 100

x >= 100 – y If y<50 then y = 50 - r where r is some positive real number x >= 100 – (50 - r) = 50 + r x >= 50+r So x must be >50, If x>50 then x>=50 So if y<50 then x>=50 and the conclusion is true

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Your turn: Indirect proof

If x,y ∈ ℝ and x+y >= 100 then x >= 50 or y >= 50

Contrapositive for x,y ∈ ℝ If x<50 and y<50 then x+y < 100

If x<50 then x = 50 – q where q is some positive real number

If y<50 then y = 50 – r where r is some positive real number

Since q and r are positive number q+r is a positive number

So x+y = 100 – (q+r) = 100 – positive real # < 100 27

Proof by Exhaustion When we prove by cases for all possible

cases. Usually break down into a few cases and

prove them all Sometimes will need to prove for many

different cases

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Existence proofs Many theorems are assertions that some

particular type of object exists We can prove these types of theorems by

constructive proofs in which we construct an example of the object, since we have an example we know the type of object exists

We can also use non-constructive proofs that show the type of object exists without demonstrating a particular example of that type of object

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Example: Constructive Prove that there exists an integer solution

to the equation x * y = z2

Proof: The integers x=2 y=8 and z=4 satisfy the

equation. We have shown that an integer solution to

the equation exists

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Another example: constructive There exists a positive rational number less than

1/100. By the definition of a rational number x, the rational

number can be expressed as the ratio of two integers a/b

Let a=1 and b=100 To construct a smaller positive rational number we

must increase b or decrease a. But a is already the smallest possible positive rational number so we must increase b

Thus a smaller rational number is 1/10131

Uniqueness proofs Theorems often assert that there is

exactly one object with particular properties

Uniqueness proofs show That an object with the particular properties

exists We show that only one object with the

particular properties exists

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