Mathematical Preliminaries
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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, … }

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Mathematical Preliminaries

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Mathematical preliminaries

Mathematical Preliminaries

Prof. Busch - LSU


Mathematical preliminaries

  • Mathematical Preliminaries

  • Sets

  • Functions

  • Relations

  • Graphs

  • Proof Techniques

Prof. Busch - LSU


Mathematical preliminaries

SETS

A set is a collection of elements

We write

Prof. Busch - LSU


Mathematical preliminaries

  • 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


Mathematical preliminaries

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


Mathematical preliminaries

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


    Mathematical preliminaries

    • 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


    Mathematical preliminaries

    { even integers } = { odd integers }

    Integers

    1

    odd

    0

    5

    even

    6

    2

    4

    3

    7

    Prof. Busch - LSU


    Mathematical preliminaries

    DeMorgan’s Laws

    A U B = A B

    U

    A B = A U B

    U

    Proof it!

    Prof. Busch - LSU


    Mathematical preliminaries

    Empty, Null Set:

    = { }

    S U = S

    S =

    S - = S

    - S =

    U

    = Universal Set

    Prof. Busch - LSU


    Mathematical preliminaries

    U

    A B

    U

    A B

    Subset

    A = { 1, 2, 3} B = { 1, 2, 3, 4, 5 }

    Proper Subset:

    B

    A

    Prof. Busch - LSU


    Mathematical preliminaries

    A B =

    U

    Disjoint Sets

    A = { 1, 2, 3 } B = { 5, 6}

    A

    B

    Prof. Busch - LSU


    Mathematical preliminaries

    Set Cardinality

    • For finite sets

    A = { 2, 5, 7 }

    |A| = 3

    (set size)

    Prof. Busch - LSU


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    RELATIONS

    R = {(x1, y1), (x2, y2), (x3, y3), …}

    xi R yi

    e. g. if R = ‘>’: 2 > 1, 3 > 2, 3 > 1

    Prof. Busch - LSU


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    Labeled Graph

    2

    6

    e

    2

    b

    1

    3

    d

    a

    6

    5

    c

    Prof. Busch - LSU


    Mathematical preliminaries

    e

    b

    d

    a

    c

    Walk

    Walk is a sequence of adjacent edges

    (e, d), (d, c), (c, a)

    Prof. Busch - LSU


    Mathematical preliminaries

    e

    b

    d

    a

    c

    Path

    Path is a walk where no edge is repeated

    Simple path: no node is repeated

    Prof. Busch - LSU


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    Hamiltonian Cycle

    5

    base

    e

    1

    b

    4

    d

    a

    2

    3

    c

    A simple cycle that contains all nodes

    Prof. Busch - LSU


    Mathematical preliminaries

    Finding All Simple Paths

    e

    b

    d

    a

    c

    origin

    Prof. Busch - LSU


    Mathematical preliminaries

    Step 1

    e

    b

    d

    a

    c

    origin

    (c, a)

    (c, e)

    Prof. Busch - LSU


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    Trees

    root

    parent

    leaf

    child

    Trees have no cycles

    Prof. Busch - LSU


    Mathematical preliminaries

    root

    Level 0

    Level 1

    Height 3

    leaf

    Level 2

    Level 3

    Prof. Busch - LSU


    Mathematical preliminaries

    Binary Trees

    Prof. Busch - LSU


    Mathematical preliminaries

    PROOF TECHNIQUES

    • Proof by induction

    • Proof by contradiction

    Prof. Busch - LSU


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    • 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


    Mathematical preliminaries

    Induction Step

    height

    k

    k+1

    From Inductive hypothesis: L(k) <= 2k

    Prof. Busch - LSU


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    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


    Mathematical preliminaries

    = 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


    Latihan

    Latihan


    Latihan1

    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


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