Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 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.

Similar presentations


Presentation on theme: "1 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."— Presentation transcript:

1 1 Discrete Math Methods of proof

2 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

3 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

4 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

5 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

6 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

7 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.

8 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

9 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 !!!

10 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

11 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) ]

12 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

13 Example of Direct Proof 1. ∀ 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 divisible 1. 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)

14 Example of Direct Proof 1. 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

15 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

16 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

17 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

18 Example of Indirect Proof  ¬ R(a,b) → ¬ Q(a,b) 1. ¬ R(a,b) Premise 2. ¬ R(x,y) Universal Instantiation 3. x|y from 2 4. y|x definition of universe 5. y=rx and x=qy definition of divisible 6. 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

19 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 19

20 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 = a 2 /b 2  2b 2 = a 2 so a 2 is even by the definition of even  By the definition of even a=2c  So 2b 2 = 4c 2 or b 2 = 2c 2, so b 2 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 20

21 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 21

22 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 22

23 Proof by cases: Example  Prove ∀ n ∈ Z [n/2] [n/2] = [n 2 /4]  Case 1: n is even n= a*2  [2a/2] [2a/2] = [4a 2 /4]= a 2  [n 2 /4] = [(2a*2a)/4] = a 2  Case 2: n is odd n =a*2+1  [(2a+1)/2] [(2a+1)/2] = ([a+ ½ ] [a+ ½ ]) = a(a+1) + ¼  [n 2 /4] = [(2a+1) 2 /4] = [(4a 2 +4a+1)/4] = a(a+1) + ¼ 23

24 Your turn  Prove by cases  If x,y ∈ ℝ and x+y >= 100 then x >= 50 or y >= 50 24

25 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, all three possibilities are consistent with the conclusion x >= 50 or y >= 50 25

26 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 and the conclusion is true 26

27 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 # <

28 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 28

29 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 29

30 Example: Constructive  Prove that there exists an integer solution to the equation x * y = z 2  Proof:  The integers x=2 y=8 and z=4 satisfy the equation.  We have shown that an integer solution to the equation exists 30

31 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 31

32 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 32


Download ppt "1 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."

Similar presentations


Ads by Google