Proofs1 Elementary Discrete Mathematics Jim Skon.

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Presentation transcript:

Proofs1 Elementary Discrete Mathematics Jim Skon

Proofs2 §Why proofs? l Careful examination to determine if mistake has been made. l Convince someone else about proposition.

Proofs3 §Proofs based on systems of rules. §A set of rules should be: l consistent - can't prove anything invalid l complete - can prove anything that is true. §Problem: Gödel has proved that any system of consistent rules is incomplete!

Proofs4 §Proofs must be based on some underlying set of truths which, in general, everyone believes. l Axioms or Postulates l Definitions

Proofs5 Proofs - Axioms §Axioms or Postulates - set of assumptions which are believed to be fundamentally true - no proof is given. §Examples of Axioms: l Given two distinct points, there is exactly one line that contains them. l for all real numbers xy = yx

Proofs6 Proofs - Definitions §Definitions - set of statements used to define new concepts in terms or existing ones. No proof needed. §Examples of Definitions: l Two lines are parallel if they are on the same plain and never meet l The absolute value |x| of a real number x is defined to be x if x is positive and -x otherwise.

Proofs7 Proofs - Definitions §Example definitions: x is even   a:x = 2a x is odd   a:x = 2a + 1

Proofs8 Valid reasoning in proofs §A mathematical proof is a sequence of statements, such that each statement: 1. is an assumption, or 2. is a proposition already proved, or 3. Follow logically from one or more previous statements in proof.

Proofs9 Valid reasoning in proofs §Logically follows: l A proposition Q follows logically from propositions P 1, P 2,..., P n if Q must be true whenever P 1, P 2,..., P n are true.

Proofs10 Valid reasoning in proofs §Example: modus ponens P  (P  Q)  Q modus ponens P: The car is running Q: The car has gas. If we know that the car is running (P), we can prove that (Q) it has gas. Q  (P  Q)  P non-logical implication

Proofs11 Rules of Inference §Rules of Inference - used in proofs, or arguments, to move from what is known to what we want to prove. §modus ponens is a valid rule of inference.

Proofs12 Argument §Argument - consists of a collection of statements, called premises of the argument, followed by a conclusion statement. A 1 A 2 : A n  A } Premises Conclusion

Proofs13 Valid Argument §An argument is said to be valid if whenever all the premises are true, the conclusion is also true. § If the premises are true, but the conclusion false, the argument is said to be invalid.

Proofs14 Example (modus ponens): §Prove: l If I have a cold, then I will not go to the game l I have a cold l Therefore, I will not go to the game p  q p q p  q p T T T  q T F F F T T F F T

Proofs15 Example §Prove: l If I do my homework, I will get a passing grade on the test. l I passed the test. l Therefore I did all my homework. p  q q  p §Not Valid! A fallacy! §Called the fallacy of affirming the conclusion.

Proofs16 Addition §Prove: l It is windy outside. l Therefore it is either windy outside or cloudy outside. p  p  q

Proofs17 Simplification §Prove: l It is sunny and it is cold. l Therefore it is sunny. p  q  p

Proofs18 Modus tollens §Prove: l It is not cold today. l If is was clear last night, then it will be cold today l Therefore it was not clear last night.  q p  q  p

Proofs19 Hypothetical syllogism §Prove: l If Dr. Fairbanks is speaking in chapel today, I will not skip l If I don’t skip, I will have perfect attendance. l Therefore If Dr. Fairbanks is speaking in chapel today, then I will have perfect attendance. p  q q  r  p  r

Proofs20 Disjuctive syllogism §Prove: l I will work in the library today or I go fishing. l I did not work in the library today l Therefore I went fishing. p  q  p  q

Proofs21 Hypothetical syllogism §Prove: l If you love me you will keep my commandments. (Jn 14:15) l If you keep my commandments, you will abide in my love. (Jn 15:10) l Therefore, if you love me then you will abide in my love. §This argument valid by the law of hypothetical syllogism.

Proofs22 Proofs §Three Techniques l Show true using logical inference l Show true by showing that no way exists to make all premises true but conclusion false l Show false by finding a way to make premises true but conclusion false.

Proofs23 Proof Example §Consider:  q   p q  r p  s  r

Proofs24 Proof Example §Consider: p  (q  r) q  p  r  Show no way to make all premises true but conclusion false

Proofs25 Proof by contradiction §Consider: r  s p  s r  q  p  q §Use proof by contradiction

Proofs26 Proof Example §If the law is sufficient, then Christ died in vain §The law is sufficient §Therefore Christ died in vain.

Proofs27 Sorites §Using proof by induction, we can show that the law of syllogism may be extended to more than two premises. This argument is called a sorites. p 1  p 2 p 2  p 3 : p n-1  p n  p 1  p n

Proofs28 Sorites §Romans 10: ) Whoever will call upon the name of the Lord will be saved. 2) They must believe to call on the Lord. 3) They must hear the Gospel to believe. 4) They must have the word preached to them to hear the Gospel. 5) A person must be sent for the word to be preached.

Proofs29 Sorites § p - He is saved § q - He calls on the Lord § r - He believes § s - He hears the Gospel § t - He has the word preached to him § u - A person is send to preach

Proofs30 Sorites  u   t  t  s  s  r  r  q  q  p  u   p If no one is sent to preach the gospel, then no one will be saved!

Proofs31 Example §Babies are illogical §Nobody is despised who can manage a crocodile §Illogical persons are despised §Therefore, babies cannot manage crocodiles

Proofs32 Types of proof:  Vacuous Proof of P  Q The truth value of P  Q is true if P is false. If P can be shown false, then P  Q holds. Thus prove P  Q by showing P is false.

Proofs33 Trivial Proof of P  Q l If it is possible to establish that Q is true, then only the first and third lines of the truth table below apply. PQ P  Q TTT TFF FTT FFT Thus prove P  Q by showing Q true.

Proofs34 Direct Proof of P  Q §Prove Q, using P as an assumption.  Thus prove P  Q by showing Q is true whenever P is true.

Proofs35 Indirect Proof of P  Q  Prove the contrapositive, e.g.  Q   P is true, using one of the other proof methods.

Proofs36 Proof by contradiction §Assume the negation of the proposition is true, then derive a contradiction.  Thus to prove of P  Q, assume P   Q is true, then derive a contradiction.

Proofs37 Proof by cases of P  Q  To prove P  Q, find a set of propositions P 1, P 2,..., P n, n  2, in which at least one P j must be true for P to be true. P  P 1  P 2 ...  P n  Then prove the n propositions P 1  Q, P 2  Q,..., P n  Q.

Proofs38 Vacuous Proof §Consider the proposition: If you your grandfather dies as a baby then you will get an A in this class. §Proof of this statement: Your grandfather didn’t die, thus thus the premise must be false. Thus P  Q must be true.

Proofs39 Trivial Proof §Consider the proposition: l If 3n 2 + 5n -2  2n 2 + 7n - 16 then n = n 2. P(n). §Proof of P(0): l 0 = 0 2, thus P(0) is trivially true. QED.

Proofs40 Direct Proof §Consider: The sum of two even numbers is even. §Restate as:  x:  y: (x is even and y is even)  x + y is even §Proof: 1. Remember: x is even   a:x = 2a (definition) 2. Assume x is even and y is even (assume hypothesis) 3. x + y = 2a + 2b (from 1 and 2) 4. 2a + 2b = 2·(a+b) 5. By 1, 2·(a+b) is even - QED.

Proofs41 Direct Proof §Consider: Every multiple of 6 is also a multiple of 3.  Rewrite:  x  z  y:(6·x = y  3·z = y) §Proof: 1. Assume 6x = y (hypothesis) 2. 6x = y can be rewritten as 3 · 2x = y 3. Let z = 2x, then 3·z = y holds. QED.

Proofs42 Indirect Proofs §Prove the contrapositive, e.g. Prove that:  Q   P is true

Proofs43 Indirect Proofs §Prove: If x 2 is even, then x is even.  Rewrite:  x : (EVEN(x 2 )  EVEN(x))

Proofs44 Indirect Proofs §Prove: If x 2 is even, then x is even 1.  x : (ODD(x)  ODD(x 2 )) (contrapositive) 2. Assume 1 ODD(n) true for some n (hypothesis) 3. x is odd   a:x = 2a + 1 (definition) 4. n = 2a + 1 for some a (2 & 3) 5. n 2 = (2a + 1) 2 (substitution) 6. (2a + 1) 2 = (2a + 1)(2a + 1) = 4a 2 + 4a + 1 = 2 (2a 2 + 2a) (2a 2 + 2a) + 1 is odd(3 & 6) QED

Proofs45 Proof by contradiction  To prove of P  Q, assume  P  Q , derive a contradiction.  Recall that: P  Q   P  Q  Then:  P  Q  P  Q)  P   Q (Demorgan’s)  Thus to prove P  Q we assume P   Q and show a contradiction.

Proofs46 Proof by contradiction §Consider Theorem: There is no largest prime number. §This can be stated as "If x is a prime number, then there exists another prime y which is greater"  Formally:  x  y: (PRIME(x)  PRIME(y)  x < y)

Proofs47 Proof by contradiction §There is no largest prime number l Assume largest prime number does exist. Call this number p. l Restate implication as p is prime, and there does not exist a prime which is greater. 1. Form a product r = 2 · 3 · 5 ·... p) (e.g. r is the product of all primes) 2. If we divide r+1 by any prime, it will have remainder 1 3. r+1 is prime, since any number not divisible by any prime which is less must be prime. 4. but r+1 > p, which contradicts that p is the greatest prime number.QED.

Proofs48 Proof by cases  To prove P  Q, find a set of propositions P 1, P 2,..., P n, n  2, in which at least one P j must be true for P to be true. P  P 1  P 2 ...  P n Then prove the n propositions P 1  Q, P 2  Q,..., P n  Q.  Thus: P  (P 1  P 2 ...  P n ) and (P 1  Q)  (P 2  Q) ...  (P n  Q)  (P  Q)

Proofs49 Proof by cases §Consider: For every nonzero integer x,x 2 > 0. §Let: P = "x is a nonzero integer” Q = x 2 > 0  We want to prove P  Q

Proofs50 Proof by cases §If: P = "x is a nonzero integer” Q = x 2 > 0  Prove P  Q §P can be broken up into two cases: P 1 = x > 0 P 2 = x < 0  Note that P  (P 1  P 2 ).

Proofs51 Proof by cases §For every nonzero integer x,x 2 > 0. §Prove each case - Prove P 1  Q: If x > 0, then x 2 > 0, since the product of two positive numbers is always positive. Prove P 2  Q: If x 0, since the product of two negative numbers is always positive. QED.