1 / 49

Mathematical Preliminaries

Mathematical Preliminaries. Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques. SETS. A set is a collection of elements. We write. Set Representations C = { a, b, c, d, e, f, g, h, i, j, k } C = { a, b, …, k } S = { 2, 4, 6, … }

ceana
Download Presentation

Mathematical Preliminaries

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mathematical Preliminaries Prof. Busch - LSU

  2. Mathematical Preliminaries • Sets • Functions • Relations • Graphs • Proof Techniques Prof. Busch - LSU

  3. SETS A set is a collection of elements We write Prof. Busch - LSU

  4. Set Representations • C = { a, b, c, d, e, f, g, h, i, j, k } • C = { a, b, …, k } • S = { 2, 4, 6, … } • S = { j : j > 0, and j = 2k for some k>0 } • S = { j : j is nonnegative and even } finite set infinite set Prof. Busch - LSU

  5. U A 6 8 2 3 1 7 4 5 9 10 A = { 1, 2, 3, 4, 5 } • Universal Set: all possible elements • U = { 1 , … , 10 } Prof. Busch - LSU

  6. B A • Set Operations • A = { 1, 2, 3 } B = { 2, 3, 4, 5} • Union • A U B = { 1, 2, 3, 4, 5 } • Intersection • A B = { 2, 3 } • Difference • A - B = { 1 } • B - A = { 4, 5 } 2 4 1 3 5 U 2 3 1 Venn diagrams Prof. Busch - LSU

  7. Complement • Universal set = {1, …, 7} • A = { 1, 2, 3 } A = { 4, 5, 6, 7} 4 A A 6 3 1 2 5 7 A = A Prof. Busch - LSU

  8. { even integers } = { odd integers } Integers 1 odd 0 5 even 6 2 4 3 7 Prof. Busch - LSU

  9. DeMorgan’s Laws A U B = A B U A B = A U B U Proof it! Prof. Busch - LSU

  10. Empty, Null Set: = { } S U = S S = S - = S - S = U = Universal Set Prof. Busch - LSU

  11. U A B U A B Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } Proper Subset: B A Prof. Busch - LSU

  12. A B = U Disjoint Sets A = { 1, 2, 3 } B = { 5, 6} A B Prof. Busch - LSU

  13. Set Cardinality • For finite sets A = { 2, 5, 7 } |A| = 3 (set size) Prof. Busch - LSU

  14. Powersets A powerset is a set of sets S = { a, b, c } Powerset of S = the set of all the subsets of S 2S = { , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } Observation: | 2S | = 2|S| ( 8 = 23 ) Prof. Busch - LSU

  15. Cartesian Product A = { 2, 4 } B = { 2, 3, 5 } A X B = { (2, 2), (2, 3), (2, 5), ( 4, 2), (4, 3), (4, 5) } |A X B| = |A| |B| Generalizes to more than two sets A X B X … X Z Prof. Busch - LSU

  16. FUNCTIONS domain range B 4 A f(1) = a a 1 2 b c 3 5 f : A -> B If A = domain then f is a total function otherwise f is a partial function Prof. Busch - LSU

  17. RELATIONS R = {(x1, y1), (x2, y2), (x3, y3), …} xi R yi e. g. if R = ‘>’: 2 > 1, 3 > 2, 3 > 1 Prof. Busch - LSU

  18. Equivalence Relations • Reflexive: x R x • Symmetric: x R y y R x • Transitive: x R y and y R z x R z • Example: R = ‘=‘ • x = x • x = y y = x • x = y and y = z x = z Prof. Busch - LSU

  19. Equivalence Classes For equivalence relation R equivalence class of x = {y : x R y} Example: R = { (1, 1), (2, 2), (1, 2), (2, 1), (3, 3), (4, 4), (3, 4), (4, 3) } Equivalence class of 1 = {1, 2} Equivalence class of 3 = {3, 4} Prof. Busch - LSU

  20. GRAPHS A directed graph e b node d a edge c • Nodes (Vertices) • V = { a, b, c, d, e } • Edges • E = { (a,b), (b,c), (b,e),(c,a), (c,e), (d,c), (e,b), (e,d) } Prof. Busch - LSU

  21. Labeled Graph 2 6 e 2 b 1 3 d a 6 5 c Prof. Busch - LSU

  22. e b d a c Walk Walk is a sequence of adjacent edges (e, d), (d, c), (c, a) Prof. Busch - LSU

  23. e b d a c Path Path is a walk where no edge is repeated Simple path: no node is repeated Prof. Busch - LSU

  24. Cycle e base b 3 1 d a 2 c Cycle: a walk from a node (base) to itself Simple cycle: only the base node is repeated Prof. Busch - LSU

  25. Euler Tour 8 base e 7 1 b 4 6 5 d a 2 3 c A cycle that contains each edge once Prof. Busch - LSU

  26. Hamiltonian Cycle 5 base e 1 b 4 d a 2 3 c A simple cycle that contains all nodes Prof. Busch - LSU

  27. Finding All Simple Paths e b d a c origin Prof. Busch - LSU

  28. Step 1 e b d a c origin (c, a) (c, e) Prof. Busch - LSU

  29. Step 2 e b d a (c, a) (c, a), (a, b) (c, e) (c, e), (e, b) (c, e), (e, d) c origin Prof. Busch - LSU

  30. Step 3 e b d a c (c, a) (c, a), (a, b) (c, a), (a, b), (b, e) (c, e) (c, e), (e, b) (c, e), (e, d) origin Prof. Busch - LSU

  31. Step 4 e b d a (c, a) (c, a), (a, b) (c, a), (a, b), (b, e) (c, a), (a, b), (b, e), (e,d) (c, e) (c, e), (e, b) (c, e), (e, d) c origin Prof. Busch - LSU

  32. Trees root parent leaf child Trees have no cycles Prof. Busch - LSU

  33. root Level 0 Level 1 Height 3 leaf Level 2 Level 3 Prof. Busch - LSU

  34. Binary Trees Prof. Busch - LSU

  35. PROOF TECHNIQUES • Proof by induction • Proof by contradiction Prof. Busch - LSU

  36. Induction We have statements P1, P2, P3, … • If we know • for some b that P1, P2, …, Pb are true • for any k >= b that • P1, P2, …, Pk imply Pk+1 • Then • Every Pi is true Prof. Busch - LSU

  37. Proof by Induction • Inductive basis • Find P1, P2, …, Pb which are true • Inductive hypothesis • Let’s assume P1, P2, …, Pk are true, • for any k >= b • Inductive step • Show that Pk+1 is true Prof. Busch - LSU

  38. Example Theorem:A binary tree of height n has at most 2n leaves. Proof by induction: let L(i) be the maximum number of leaves of any subtree at height i Prof. Busch - LSU

  39. We want to show: L(i) <= 2i • Inductive basis • L(0) = 1 (the root node) • Inductive hypothesis • Let’s assume L(i) <= 2i for all i = 0, 1, …, k • Induction step • we need to show that L(k + 1) <= 2k+1 Prof. Busch - LSU

  40. Induction Step height k k+1 From Inductive hypothesis: L(k) <= 2k Prof. Busch - LSU

  41. Induction Step height L(k) <= 2k k k+1 L(k+1) <= 2 * L(k) <= 2 * 2k = 2k+1 (we add at most two nodes for every leaf of level k) Prof. Busch - LSU

  42. Proof by Contradiction • We want to prove that a statement P is true • we assume that P is false • then we arrive at an incorrect conclusion • therefore, statement P must be true Prof. Busch - LSU

  43. Example • Theorem: Jika n2 genap maka n genap • Proof: • Dengan contradiction kita asumsikan • jika n2 genap maka n ganjil • We will show that this is impossible

  44. Bukti: Diketahui n2 genap (kita asumsikan benar) Andaikan  n ganjil (negasinya) Karena  n ganjil maka nbisa ditulis dengan bentuk n = 2m + 1 dengan m bilangan bulat

  45. Diperoleh n2 = (2m + 1)2 n2 = 4m2 + 4m +1 n2 = 2(2m2 + 2m) + 1 Itu berarti n2 = 2(2m2 + 2m) + 1 ganjil. Kontradiksi dengan asumsi “n2 genap” karena terjadi kontradiksi maka dapat disimpulkan jika n2 genap maka n genap

  46. Example • Theorem: is not rational • Proof: • Assume by contradiction that it is rational • = n/m • n and m have no common factors • We will show that this is impossible Prof. Busch - LSU

  47. = n/m 2 m2 = n2 n is even n = 2 k Therefore, n2 is even m is even m = 2 p 2 m2 = 4k2 m2 = 2k2 Thus, m and n have common factor 2 Contradiction! Prof. Busch - LSU

  48. Latihan

  49. Latihan • Buatlah notasi bagi set Z yang mempunyai anggota {0,1,2,3,4} • N mewakili bilangan natural (bilangan asli) • Buatlah notasi bagi set Y yang merupakan bilangan ganjil • Buatlah notasi bagi set yang terdiri dari 2 pasang (2-tuples) bilangan natural dengan syarat nilai elemen pertama lebih kecil dari elemen kedua pada tuples

More Related