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Course Summary

Course Summary. CS 202 Aaron Bloomfield. Outline. A review of the proof methods gone over in class Interspersed with the most popular demotivators from CS 101 Course summary. Demotivator winners!. Methodology 1 st place vote counted for 3 points 2 nd place vote counted for 2 points

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Course Summary

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  1. Course Summary CS 202 Aaron Bloomfield

  2. Outline • A review of the proof methods gone over in class • Interspersed with the most popular demotivators from CS 101 • Course summary

  3. Demotivator winners! • Methodology • 1st place vote counted for 3 points • 2nd place vote counted for 2 points • 3rd place vote counted for 1 point • Have not tallied the Toolkit ones yet • I won’t get those for a few weeks • Will then buy two demotivators and hang them in my office…

  4. Proof methods learned so far • Logical equivalences • via truth tables • via logical equivalences • Set equivalences • via membership tables • via set identities • via mutual subset proof • via set builder notation and logical equivalences • Rules of inference • for propositions • for quantified statements • Pigeonhole principle • Combinatorial proofs • Ten proof methods in section 1.5: • Direct proofs • Indirect proofs • Vacuous proofs • Trivial proofs • Proof by contradiction • Proof by cases • Proofs of equivalence • Existence proofs • Constructive • Non-constructive • Uniqueness proofs • Counterexamples • Induction • Weak mathematical induction • Strong mathematical induction • Structural induction

  5. Using Logical Equivalences Original statement Definition of implication DeMorgan’s Law Associativity of Or Re-arranging Idempotent Law

  6. In 12th place (11 votes)

  7. Set equivalences (§1.7) • Prove that A∩B=B-(B-A) Definition of difference Definition of difference DeMorgan’s law Complementation law Distributive law Complement law Identity law Commutative law

  8. Proof by set builder notation and logical equivalences 2 Original statement Definition of difference Negating “element of” Definition of difference DeMorgan’s Law Distributive Law Negating “element of” Negation Law Identity Law Definition of intersection

  9. In 11th place (1/3) (12 votes)

  10. Rules of inference (§1.5) • “It is not sunny this afternoon and it is colder than yesterday” • “We will go swimming only if it is sunny” • “If we do not go swimming, then we will take a canoe trip” • “If we take a canoe trip, then we will be home by sunset” • Does this imply that “we will be home by sunset”? • “It is not sunny this afternoon and it is colder than yesterday” • “We will go swimming only if it is sunny” • “If we do not go swimming, then we will take a canoe trip” • “If we take a canoe trip, then we will be home by sunset” • Does this imply that “we will be home by sunset”? • “It is not sunny this afternoon and it is colder than yesterday” • “We will go swimming only if it is sunny” • “If we do not go swimming, then we will take a canoe trip” • “If we take a canoe trip, then we will be home by sunset” • Does this imply that “we will be home by sunset”? • “It is not sunny this afternoon and it is colder than yesterday” • “We will go swimming only if it is sunny” • “If we do not go swimming, then we will take a canoe trip” • “If we take a canoe trip, then we will be home by sunset” • Does this imply that “we will be home by sunset”? • “It is not sunny this afternoon and it is colder than yesterday” • “We will go swimming only if it is sunny” • “If we do not go swimming, then we will take a canoe trip” • “If we take a canoe trip, then we will be home by sunset” • Does this imply that “we will be home by sunset”? • Example 6 of Rosen, section 1.5 • We have the hypotheses: • “It is not sunny this afternoon and it is colder than yesterday” • “We will go swimming only if it is sunny” • “If we do not go swimming, then we will take a canoe trip” • “If we take a canoe trip, then we will be home by sunset” • Does this imply that “we will be home by sunset”? ¬p q r → p ¬r → s s → t t p q r s t

  11. Rules of inference (§1.5) • ¬p  q 1st hypothesis • ¬p Simplification using step 1 • r → p 2nd hypothesis • ¬r Modus tollens using steps 2 & 3 • ¬r → s 3rd hypothesis • s Modus ponens using steps 4 & 5 • s → t 4th hypothesis • t Modus ponens using steps 6 & 7

  12. In 11th place (2/3) (12 votes)

  13. Rules of inference (§1.5) • Rosen, section 1.5, question 10a • Given the hypotheses: • “Linda, a student in this class, owns a red convertible.” • “Everybody who owns a red convertible has gotten at least one speeding ticket” • Can you conclude: “Somebody in this class has gotten a speeding ticket”? C(Linda) R(Linda) x (R(x)→T(x)) x (C(x)T(x))

  14. Rules of inference (§1.5) • x (R(x)→T(x)) 3rd hypothesis • R(Linda) → T(Linda) Universal instantiation using step 1 • R(Linda) 2nd hypothesis • T(Linda) Modes ponens using steps 2 & 3 • C(Linda) 1st hypothesis • C(Linda)  T(Linda) Conjunction using steps 4 & 5 • x (C(x)T(x)) Existential generalization using step 6 Thus, we have shown that “Somebody in this class has gotten a speeding ticket”

  15. In 11th place (3/3) (12 votes)

  16. Proof methods (§1.5) • Ten proof methods: • Direct proofs • Indirect proofs • Vacuous proofs • Trivial proofs • Proof by contradiction • Proof by cases • Proofs of equivalence • Existence proofs • Uniqueness proofs • Counterexamples

  17. In 10th place (1/5) (13 votes)

  18. Direct proof example • Rosen, section 1.5, question 20 • Show that the square of an even number is an even number • Rephrased: if n is even, then n2 is even • Assume n is even • Thus, n = 2k, for some k (definition of even numbers) • n2 = (2k)2 = 4k2 = 2(2k2) • As n2 is 2 times an integer, n2 is thus even

  19. Indirect proof example • If n2 is an odd integer then n is an odd integer • Prove the contrapositive: If n is an even integer, then n2 is an even integer • Proof: n=2k for some integer k (definition of even numbers) • n2 = (2k)2 = 4k2 = 2(2k2) • Since n2 is 2 times an integer, it is even

  20. In 10th place (2/5) (13 votes)

  21. Proof by contradiction • Given a statement p, assume it is false • Assume ¬p • Prove that ¬p cannot occur • A contradiction exists • Given a statement of the form p→q • To assume it’s false, you only have to consider the case where p is true and q is false

  22. Proof by contradiction • Theorem (by Euclid): There are infinitely many prime numbers. • Proof. Assume there are a finite number of primes • List them as follows: p1, p2 …, pn. • Consider the number q = p1p2 … pn + 1 • This number is not divisible by any of the listed primes • If we divided pi into q, there would result a remainder of 1 • We must conclude that q is a prime number, not among the primes listed above • This contradicts our assumption that all primes are in the list p1, p2 …, pn.

  23. Proof by contradiction example 2 • Rosen, section 1.5, question 21 (b) • Prove that if n is an integer and n3+5 is odd, then n is even • Rephrased: If n3+5 is odd, then n is even • Thus, p is “n3+5” is odd, q is “n is even” • Assume p and q • Assume that n3+5 is odd, and n is odd • Since n is odd: • n=2k+1 for some integer k (definition of odd numbers) • n3+5 = (2k+1)3+5 = 8k3+12k2+6k+6 = 2(4k3+6k2+3k+3) • As n3+5 = 2(4k3+6k2+3k+3) is 2 times an integer, n must be even • Thus, we have concluded q • Contradiction! • We assumed q was false, and showed that this assumption implies that q must be true • As q cannot be both true and false, we have reached our contradiction

  24. In 10th place (3/5) (13 votes)

  25. Vacuous proof example • When the antecedent is false • Consider the statement: • All criminology majors in CS 202 are female • Rephrased: If you are a criminology major and you are in CS 202, then you are female • Could also use quantifiers! • Since there are no criminology majors in this class, the antecedent is false, and the implication is true

  26. Trivial proof example • When the consequence is true • Consider the statement: • If you are tall and are in CS 202 then you are a student • Since all people in CS 202 are students, the implication is true regardless

  27. Proof by cases example • Prove that • Note that b ≠ 0 • Cases: • Case 1: a ≥ 0 and b > 0 • Then |a| = a, |b| = b, and • Case 2: a ≥ 0 and b < 0 • Then |a| = a, |b| = -b, and • Case 3: a < 0 and b > 0 • Then |a| = -a, |b| = b, and • Case 4: a < 0 and b < 0 • Then |a| = -a, |b| = -b, and

  28. In 10th place (4/5) (13 votes)

  29. Proofs of equivalence example • Rosen, section 1.5, question 40 • Show that m2=n2 if and only if m=n or m=-n • Rephrased: (m2=n2) ↔ [(m=n)(m=-n)] • Need to prove two parts: • [(m=n)(m=-n)] → (m2=n2) • Proof by cases! • Case 1: (m=n)→ (m2=n2) • (m)2 = m2, and (n)2 = n2, so this case is proven • Case 2: (m=-n) → (m2=n2) • (m)2 = m2, and (-n)2 = n2, so this case is proven • (m2=n2) → [(m=n)(m=-n)] • Subtract n2 from both sides to get m2-n2=0 • Factor to get (m+n)(m-n) = 0 • Since that equals zero, one of the factors must be zero • Thus, either m+n=0 (which means m=n) • Or m-n=0 (which means m=-n)

  30. In 10th place (5/5) (13 votes)

  31. Constructive existence proof example • Show that a square exists that is the sum of two other squares • Proof: 32 + 42 = 52 • Show that a cube exists that is the sum of three other cubes • Proof: 33 + 43 + 53 = 63

  32. Non-constructive existence proof example • Rosen, section 1.5, question 50 • Prove that either 2*10500+15 or 2*10500+16 is not a perfect square • A perfect square is a square of an integer • Rephrased: Show that a non-perfect square exists in the set {2*10500+15, 2*10500+16} • Proof: The only two perfect squares that differ by 1 are 0 and 1 • Thus, any other numbers that differ by 1 cannot both be perfect squares • Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1 • Note that we didn’t specify which one it was!

  33. In 9th place (14 votes)

  34. Uniqueness proof example • If the real number equation 5x+3=a has a solution then it is unique • Existence • We can manipulate 5x+3=a to equal (a-3)/5 • Is this constructive or nonconstructive? • Uniqueness • If there are two such numbers, then they would fulfill the following: a = 5x+3 = 5y+3 • We can manipulate this to yield that x = y • Thus, the one solution is unique!

  35. In 8th place (15 votes)

  36. Counterexamples • Given a universally quantified statement, find a single example which it is not true • Note that this is DISPROVING a UNIVERSAL statement by a counterexample • x ¬R(x), where R(x) means “x has red hair” • Find one person (in the domain) who has red hair • Every positive integer is the square of another integer • The square root of 5 is 2.236, which is not an integer

  37. In 7th place (16 votes)

  38. Weak mathematical induction (§ 3.3) • Rosen, question 7: Show • Base case: n = 1 • Inductive hypothesis: assume

  39. Weak mathematical induction (§ 3.3) • Inductive step: show

  40. In 6th place (18 votes)

  41. Strong mathematical induction (§ 3.3) • Show that any number > 1 can be written as the product of primes • Base case: P(2) • 2 is the product of 2 (remember that 1 is not prime!) • Inductive hypothesis: P(1), P(2), P(3), …, P(k) are all true • Inductive step: Show that P(k+1) is true

  42. Strong mathematical induction (§ 3.3) • Inductive step: Show that P(k+1) is true • There are two cases: • k+1 is prime • It can then be written as the product of k+1 • k+1 is composite • It can be written as the product of two composites, a and b, where 2 ≤ a ≤ b < k+1 • By the inductive hypothesis, both P(a) and P(b) is true

  43. In 5th place (1/2) (21 votes)

  44. Structural induction (§ 3.4) • Show that n(T) ≥ 2h(T) + 1 • Inductive hypothesis: • Let T1 and T2 be full binary trees • Assume that n(T1) ≥ 2h(T1) + 1 for some tree T1 • Assume that n(T2) ≥ 2h(T2) + 1 for some tree T2 • Recursive step: • Let T = T1 ∙ T2 • Here the ∙ operator means creating a new tree with a root note r and subtrees T1 and T2 • New element is T • By the definition of height and size, we know: • n(T) = 1 + n(T1) + n(T2) • h(T) = 1 + max ( h(T1), h(T2) ) • Therefore: • n(T) = 1 + n(T1) + n(T2) • ≥ 1 + 2h(T1) + 1 + 2h(T2) + 1 • ≥ 1 + 2*max ( h(T1), h(T2) ) the sum of two non-neg #’s is at least as large as the larger of the two • = 1 + 2*h(T) • Thus, n(T) ≥ 2h(T) + 1

  45. In 5th place (2/2) (21 votes)

  46. Induction methods compared

  47. In 4th place (25 votes)

  48. Pigeonhole principle (§4.2) • Consider 5 distinct points (xi, yi) with integer values, where i = 1, 2, 3, 4, 5 • Show that the midpoint of at least one pair of these five points also has integer coordinates • Thus, we are looking for the midpoint of a segment from (a,b) to (c,d) • The midpoint is ( (a+c)/2, (b+d)/2 ) • Note that the midpoint will be integers if a and c have the same parity: are either both even or both odd • Same for b and d • There are four parity possibilities • (even, even), (even, odd), (odd, even), (odd, odd) • Since we have 5 points, by the pigeonhole principle, there must be two points that have the same parity possibility • Thus, the midpoint of those two points will have integer coordinates

  49. In 3rd place (1/2) (27 votes)

  50. Combinatorial proof (§4.3) • Let n be a non-negative integer. Then • Combinatorial proof • A set with n elements has 2n subsets • By definition of power set • Each subset has either 0 or 1 or 2 or … or n elements • There are subsets with 0 elements, subsets with 1 element, … and subsets with n elements • Thus, the total number of subsets is • Thus,

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