Prof. Busch - LSU1 Mathematical Preliminaries. Prof. Busch - LSU2 Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques.

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1 Prof. Busch - LSU1 Mathematical Preliminaries

2 Prof. Busch - LSU2 Mathematical Preliminaries Sets Functions Relations Graphs Proof Techniques

3 Prof. Busch - LSU3 A set is a collection of elements SETS We write

4 Prof. Busch - LSU4 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

5 Prof. Busch - LSU5 A = { 1, 2, 3, 4, 5 } Universal Set: all possible elements U = { 1, …, 10 } 1 2 3 4 5 A U 6 7 8 9 10

6 Prof. Busch - LSU6 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 } U A B 2 3 1 4 5 2 3 1 Venn diagrams

7 Prof. Busch - LSU7 A Complement Universal set = {1, …, 7} A = { 1, 2, 3 } A = { 4, 5, 6, 7} 1 2 3 4 5 6 7 A A = A

8 Prof. Busch - LSU8 0 2 4 6 1 3 5 7 even { even integers } = { odd integers } odd Integers

9 Prof. Busch - LSU9 DeMorgan’s Laws A U B = A B U A B = A U B U Proof it!

10 Prof. Busch - LSU10 Empty, Null Set: = { } S U = S S = S - = S - S = U = Universal Set

11 Prof. Busch - LSU11 Subset A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 } A B U Proper Subset:A B U A B

12 Prof. Busch - LSU12 Disjoint Sets A = { 1, 2, 3 } B = { 5, 6} A B = U AB

13 Prof. Busch - LSU13 Set Cardinality For finite sets A = { 2, 5, 7 } |A| = 3 (set size)

14 Prof. Busch - LSU14 Powersets A powerset is a set of sets Powerset of S = the set of all the subsets of S S = { a, b, c } 2 S = {, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} } Observation: | 2 S | = 2 |S| ( 8 = 2 3 )

15 Prof. Busch - LSU15 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

16 Prof. Busch - LSU16 FUNCTIONS domain 1 2 3 a b c range f : A -> B A B If A = domain then f is a total function otherwise f is a partial function f(1) = a 4 5

17 Prof. Busch - LSU17 RELATIONS R = {(x 1, y 1 ), (x 2, y 2 ), (x 3, y 3 ), …} x i R y i e. g. if R = ‘>’: 2 > 1, 3 > 2, 3 > 1

18 Prof. Busch - LSU18 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

19 Prof. Busch - LSU19 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}

20 Prof. Busch - LSU20 GRAPHS A directed graph 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) } node edge a b c d e

21 Prof. Busch - LSU21 Labeled Graph a b c d e 1 3 5 6 2 6 2

22 Prof. Busch - LSU22 Walk a b c d e Walk is a sequence of adjacent edges (e, d), (d, c), (c, a)

23 Prof. Busch - LSU23 Path a b c d e Path is a walk where no edge is repeated Simple path: no node is repeated

24 Prof. Busch - LSU24 Cycle a b c d e 1 2 3 Cycle: a walk from a node (base) to itself Simple cycle: only the base node is repeated base

25 Prof. Busch - LSU25 Euler Tour a b c d e 1 2 3 4 5 6 7 8 base A cycle that contains each edge once

26 Prof. Busch - LSU26 Hamiltonian Cycle a b c d e 1 2 3 4 5 base A simple cycle that contains all nodes

27 Prof. Busch - LSU27 Finding All Simple Paths a b c d e origin

28 Prof. Busch - LSU28 (c, a) (c, e) Step 1 a b c d e origin

29 Prof. Busch - LSU29 (c, a) (c, a), (a, b) (c, e) (c, e), (e, b) (c, e), (e, d) Step 2 a b c d e origin

30 Prof. Busch - LSU30 Step 3 a b c d e origin (c, a) (c, a), (a, b) (c, a), (a, b), (b, e) (c, e) (c, e), (e, b) (c, e), (e, d)

31 Prof. Busch - LSU31 Step 4 a b c d e origin (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)

32 Prof. Busch - LSU32 Trees root leaf parent child Trees have no cycles

33 Prof. Busch - LSU33 root leaf Level 0 Level 1 Level 2 Level 3 Height 3

34 Prof. Busch - LSU34 Binary Trees

35 Prof. Busch - LSU35 PROOF TECHNIQUES Proof by induction Proof by contradiction

36 Prof. Busch - LSU36 Induction We have statements P 1, P 2, P 3, … If we know for some b that P 1, P 2, …, P b are true for any k >= b that P 1, P 2, …, P k imply P k+1 Then Every P i is true

37 Prof. Busch - LSU37 Proof by Induction Inductive basis Find P 1, P 2, …, P b which are true Inductive hypothesis Let’s assume P 1, P 2, …, P k are true, for any k >= b Inductive step Show that P k+1 is true

38 Prof. Busch - LSU38 Example Theorem: A binary tree of height n has at most 2 n leaves. Proof by induction: let L(i) be the maximum number of leaves of any subtree at height i

39 Prof. Busch - LSU39 We want to show: L(i) <= 2 i Inductive basis L(0) = 1 (the root node) Inductive hypothesis Let’s assume L(i) <= 2 i for all i = 0, 1, …, k Induction step we need to show that L(k + 1) <= 2 k+1

40 Prof. Busch - LSU40 Induction Step From Inductive hypothesis: L(k) <= 2 k height k k+1

41 Prof. Busch - LSU41 L(k) <= 2 k L(k+1) <= 2 * L(k) <= 2 * 2 k = 2 k+1 Induction Step height k k+1 (we add at most two nodes for every leaf of level k)

42 Prof. Busch - LSU42 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

43 43 Example Theorem: Jika n 2 genap maka n genap Proof: Dengan contradiction kita asumsikan jika n 2 genap maka n ganjil We will show that this is impossible

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

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

46 Prof. Busch - LSU46 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

47 Prof. Busch - LSU47 = n/m 2 m 2 = n 2 Therefore, n 2 is even n is even n = 2 k 2 m 2 = 4k 2 m 2 = 2k 2 m is even m = 2 p Thus, m and n have common factor 2 Contradiction!

48 Latihan 48

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 49