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Discrete Math Methods of proof 1

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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

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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

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**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

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**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

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**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

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**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.

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**Example: Disproving a 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

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**Example: Proving a 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 !!!

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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

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**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) ]

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**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

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**Example of Direct Proof**

∀a, ∃b, ∃c ( P(a,b)^Q(b,c)) → R(a,c)) ∃b, ∃c ( P(x,b)^Q(b,c) )→ R(x,c)) using Universal Instantiation ( P(x,y)^Q(y,z) → R(x,z) ) using Existential Instantiation Using the definition of divisible P(x,y) → ∃q y=qx where q is a positive integer Q(y,z) → ∃r z=ry where r is a positive integer z=ry=r(qx)=(rq)x where r and q are positive integers z=ry=r(qx)=(rq)x rq is a positive integer (because the product of two positive integers is a positive integer)

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**Example of Direct Proof**

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

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**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

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**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

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**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

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**Example of Indirect Proof**

¬R(a,b) → ¬Q(a,b) ¬R(a,b) Premise ¬ R(x,y) Universal Instantiation x|y from 2 y|x definition of universe y=rx and x=qy definition of divisible y=rx=r(qy)=(rq)y rq=1 so r=q= 1 and y=x ¬Q(x,y) ¬Q(a,b) Universal Generalization

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**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]

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**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 >= 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 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

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**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 1/101

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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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