CSci 2011 Discrete

MathematicsLecture 9, 10

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RecapPropositional operation summary

Check translationDefinition

Tautology, Contradiction, logical equicalence

not not and or conditional Bi-conditionalp q p q pq pq pq pqT T F F T T T TT F F T F T F FF T T F F T T FF F T T F F T T

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Recapp T pp F p Identity Laws (p q) r p (q r)

(p q) r p (q r) Associative laws

p T Tp F F Domination Law p (q r) (p q) (p r)

p (q r) (p q) (p r) Distributive laws

p p pp p p

Idempotent Laws

(p q) p q (p q) p q De Morgan’s laws

( p) p Double negation law

p (p q) pp (p q) p Absorption laws

p q q pp q q p

Commutative Laws

p p Tp p F Negation lows

pq pq Definition of Implication p q (p q) (q p) Definition of

Biconditional

Recap Quantifiers

Universal quantifier: x P(x) Negating quantifiers

¬x P(x) = x ¬P(x) ¬x P(x) = x ¬P(x) xy P(x, y)

Nested quantifiers xy P(x, y): “For all x, there exists a y such that P(x,y)” xy P(x,y): There exists an x such that for all y P(x,y) is true” ¬x P(x) = x ¬P(x), ¬x P(x) = x ¬P(x)

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RecapModus ponens

p p q q

Modus tollens q p q p

Hypothetical syllogism

p q q r p r

Disjunctive syllogism

p q p q

Addition p p q Simplification p q

p

Conjunction p q p q

Resolution p q p r q r

Recap p→q

Direct Proof: Assume p is true. Show that q is also true. Indirect Proof: Assume ¬q is true. Show that p is true.

Proof by contradiction Proving p: Assume p is not true. Find a contradiction. Proving p→q

¬(p→q) (p q) p ¬q Assume p is tue and q is not true. Find a contradiction.

Proof by Cases: [(p1p2…pn)q] [(p1q)(p2q)…(pnq)] If and only if proof: pq (p→q)(q→p) Existence Proof: Constructive vs. Non-constructive Proof Uniqueness: 1) Show existence 2) find contradiction if not unique Forward and backward reasoning Counterexample

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What is a set? A set is a unordered collection of “objects”

Order does not matter Sets do not have duplicate elements set-builder notation: D = {x | x is prime and x > 2}

a S if an element a is an element of a set S. a S if not. U is the universal set. empty (or null) set = { }, if a set has zero elements. Sets can contain other sets

S = {{1},{2},{3}} V = {{{1},{{2}}},{{{3}}},{{1},{{2}},{{{3}}}}}

Venn Diagram S T if x (x S x T) S = T if S T and if S T.

S S S, S

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Set Equality, Subsets Two sets are equal if they have the same elements

{1, 2, 3, 4, 5} = {5, 4, 3, 2, 1} {1, 2, 3, 2, 4, 3, 2, 1} = {4, 3, 2, 1} Two sets are not equal if they do not have the same

elements {1, 2, 3, 4, 5} ≠ {1, 2, 3, 4}

If all the elements of a set S are also elements of a set T, then S is a subset of T If S = {2, 4, 6}, T = {1, 2, 3, 4, 5, 6, 7}, S is a subset of T This is specified by S T meaning that x (x S x T)

For any set S, S S (S S S) For any set S, S (S S)

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If S is a subset of T, and S is not equal to T, then S is a proper subset of T Can be written as: R T and R T Let T = {0, 1, 2, 3, 4, 5} If S = {1, 2, 3}, S is not equal to T, and S is a subset of T A proper subset is written as S T Let Q = {4, 5, 6}. Q is neither a subset or T nor a proper

subset of T

The difference between “subset” and “proper subset” is like the difference between “less than or equal to” and “less than” for numbers

Proper Subsets

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Set cardinality The cardinality of a set is the number of elements in

a set, written as |A|

Examples Let R = {1, 2, 3, 4, 5}. Then |R| = 5 || = 0 Let S = {, {a}, {b}, {a, b}}. Then |S| = 4

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Power Sets Given S = {0, 1}. All the possible subsets of S?

(as it is a subset of all sets), {0}, {1}, and {0, 1} The power set of S (written as P(S)) is the set of all the

subsets of S P(S) = { , {0}, {1}, {0,1} }

Note that |S| = 2 and |P(S)| = 4 Let T = {0, 1, 2}. The P(T) = { , {0}, {1}, {2},

{0,1}, {0,2}, {1,2}, {0,1,2} } Note that |T| = 3 and |P(T)| = 8

P() = { } Note that || = 0 and |P()| = 1

If a set has n elements, then the power set will have 2n elements

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Tuples In 2-dimensional space, it is a (x, y) pair of numbers

to specify a location In 3-dimensional (1,2,3) is not the same as (3,2,1) –

space, it is a (x, y, z) triple of numbers In n-dimensional space, it is a

n-tuple of numbers Two-dimensional space uses

pairs, or 2-tuples Three-dimensional space uses

triples, or 3-tuples Note that these tuples are

ordered, unlike sets the x value has to come first +x

+y

(2,3)

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Cartesian products A Cartesian product is a set of all ordered 2-tuples

where each “part” is from a given set Denoted by A x B, and uses parenthesis (not curly brackets) For example, 2-D Cartesian coordinates are the set of all

ordered pairs Z x Z Recall Z is the set of all integers This is all the possible coordinates in 2-D space

Example: Given A = { a, b } and B = { 0, 1 }, what is their Cartiesian product?

C = A x B = { (a,0), (a,1), (b,0), (b,1) } Formal definition of a Cartesian product:

A x B = { (a,b) | a A and b B }

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Cartesian Products 2 All the possible grades in this class will be a

Cartesian product of the set S of all the students in this class and the set G of all possible grades Let S = { Alice, Bob, Chris } and G = { A, B, C } D = { (Alice, A), (Alice, B), (Alice, C), (Bob, A), (Bob, B),

(Bob, C), (Chris, A), (Chris, B), (Chris, C) } The final grades will be a subset of this: { (Alice, C), (Bob,

B), (Chris, A) } Such a subset of a Cartesian product is called a relation (more

on this later in the course)

ch2.2 Set Operations

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Set operations: Union Formal definition for the union of two sets:

A U B = { x | x A or x B }

Further examples {1, 2, 3} U {3, 4, 5} = {1, 2, 3, 4, 5} {a, b} U {3, 4} = {a, b, 3, 4} {1, 2} U = {1, 2}

Properties of the union operation A U = A Identity law A U U = U Domination law A U A = A Idempotent law A U B = B U A Commutative law A U (B U C) = (A U B) U C Associative law

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Set operations: Intersection Formal definition for the intersection of two sets: A ∩

B = { x | x A and x B }

Examples {1, 2, 3} ∩ {3, 4, 5} = {3} {a, b} ∩ {3, 4} = {1, 2} ∩ =

Properties of the intersection operation A ∩ U = A Identity law A ∩ = Domination law A ∩ A = A Idempotent law A ∩ B = B ∩ A Commutative law A ∩ (B ∩ C) = (A ∩ B) ∩ C Associative law

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Disjoint setsFormal definition for disjoint sets: two sets

are disjoint if their intersection is the empty set

Further examples{1, 2, 3} and {3, 4, 5} are not disjoint{a, b} and {3, 4} are disjoint{1, 2} and are disjoint

Their intersection is the empty set and are disjoint!

Their intersection is the empty set

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Formal definition for the difference of two sets:A - B = { x | x A and x B }

Further examples{1, 2, 3} - {3, 4, 5} = {1, 2}{a, b} - {3, 4} = {a, b}{1, 2} - = {1, 2}

The difference of any set S with the empty set will be the set S

Set operations: Difference

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Complement setsFormal definition for the complement of a

set: A = { x | x A } = Ac

Or U – A, where U is the universal set

Further examples (assuming U = Z){1, 2, 3}c = { …, -2, -1, 0, 4, 5, 6, … }

Properties of complement sets (Ac)c = A Complementation lawA U Ac = U Complement lawA ∩ Ac = Complement law

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Set identitiesA = AAU = A Identity Law AU = U

A = Domination law

AA = AAA = A

Idempotent Law (Ac)c = A Complement

LawAB = BAAB = BA

Commutative Law

(AB)c = AcBc

(AB)c = AcBcDe Morgan’s

LawA(BC)

= (AB)CA(BC)

= (AB)C

Associative Law

A(BC) = (AB)(AC)A(BC) =

(AB)(AC)

Distributive Law

A(AB) = AA(AB) = A

Absorption Law

A Ac = UA Ac =

Complement Law

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How to prove a set identityFor example: A∩B=B-(B-A)Four methods:

Use the basic set identitiesUse membership tablesProve each set is a subset of each otherUse set builder notation and logical equivalences

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What we are going to prove…A∩B=B-(B-A)

A B

A∩B B-AB-(B-A)

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Proof by Set IdentitiesA B = A - (A - B)Proof) A - (A - B) = A - (A Bc)

= A (A Bc)c

= A (Ac B) = (A Ac) (A B) = (A B) = A B

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Showing each is a subset of the others

(A B)c = Ac Bc

Proof) Want to prove that (A B)c Ac Bc and Ac Bc (A B)c

x (A B)c

x (A B) (x A B) (x A x B) (x A) (x B) x A x B x Ac x Bc

x Ac Bc

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Examples Let A, B, and C be sets. Show that:a) (AUB) (AUBUC)b) (A∩B∩C) (A∩B)c) (A-B)-C A-Cd) (A-C) ∩ (C-B) =

ch2.3 Functions

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Definition of a functionA function takes an element from a set and

maps it to a UNIQUE element in another set

R Zf

4.3 4

Domain Co-domain

Pre-image of 4 Image of 4.3

f maps R to Z

f(4.3)

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More functions

1

2

3

4

5

“a”

“bb“

“cccc”

“dd”

“e”

A string length function

A

B

C

D

F

Alice

Bob

Chris

Dave

Emma

A class grade function

Domain Co-domainA pre-image

of 1The image

of “a”

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Even more functions

1

2

3

4

5

“a”

“bb“

“cccc”

“dd”

“e”

Not a valid function!Also not a valid function!

1

2

3

4

5

a

e

i

o

u

Some function…

Range

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Function arithmeticLet f1(x) = 2xLet f2(x) = x2

f1+f2 = (f1+f2)(x) = f1(x)+f2(x) = 2x+x2

f1*f2 = (f1*f2)(x) = f1(x)*f2(x) = 2x*x2 = 2x3

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One-to-one functions

1

2

3

4

5

a

e

i

o

A one-to-one function

1

2

3

4

5

a

e

i

o

A function that is not one-to-one

A function is one-to-one if each element in the co-domain has a unique pre-image

Formal definition: A function f is one-to-one if f(x) = f(y) implies x = y.

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More on one-to-oneInjective is synonymous with one-to-one

“A function is injective”A function is an injection if it is one-to-one

Note that there can be un-used elements in the co-domain

1

2

3

4

5

a

e

i

o

A one-to-one function

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Onto functions

1

2

3

4

5

a

e

i

o

A function that is not onto

A function is onto if each element in the co-domain is an image of some pre-image

Formal definition: A function f is onto if for all y C, there exists x D such that f(x)=y.

1

2

3

4

a

e

i

o

u

An onto function

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1

2

3

4

a

e

i

o

u

An onto function

More on ontoSurjective is synonymous with onto

“A function is surjective”A function is an surjection if it is onto

Note that there can be multiply used elements in the co-domain

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Onto vs. one-to-oneAre the following functions onto, one-to-

one, both, or neither?1

2

3

4

a

b

c

1

2

3

a

b

c

d

1

2

3

4

a

b

c

d

1

2

3

4

a

b

c

d

1

2

3

4

a

b

c

1-to-1, not onto

Onto, not 1-to-1

Both 1-to-1 and onto Not a valid function

Neither 1-to-1 nor onto

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BijectionsConsider a function that is

both one-to-one and onto:

Such a function is a one-to-one correspondence, or a bijection

1

2

3

4

a

b

c

d

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Identity functionsA function such that the image and the pre-

image are ALWAYS equal

f(x) = 1*xf(x) = x + 0

The domain and the co-domain must be the same set

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Inverse functions

R Rf

4.3 8.6

Let f(x) = 2*x

f-1

f(4.3)

f-1(8.6)

Then f-1(x) = x/2

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More on inverse functions Can we define the inverse of the following functions?

An inverse function can ONLY be done defined on a bijection

1

2

3

4

a

b

c

1

2

3

a

b

c

d

What is f-1(2)?Not onto!

What is f-1(2)?Not 1-to-1!

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Compositions of functions

g f

f ○ g

g(a) f(b)

(f ○ g)(a)

b = g(a)f(g(a))a

A B C

(f ○ g)(x) = f(g(x))

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Compositions of functions

g f

f ○ g

g(1) f(5)

(f ○ g)(1)

g(1)=5

f(g(1))=131

R R R

Let f(x) = 2x+3 Let g(x) = 3x+2

f(g(x)) = 2(3x+2)+3 = 6x+7

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Compositions of functionsDoes f(g(x)) = g(f(x))?

Let f(x) = 2x+3 Let g(x) = 3x+2

f(g(x)) = 2(3x+2)+3 = 6x+7g(f(x)) = 3(2x+3)+2 = 6x+11

Function composition is not commutative!

Not equal!

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Useful functionsFloor: x means take the greatest integer

less than or equal to the number

Ceiling: x means take the lowest integer greater than or equal to the number

round(x) = x+0.5

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Ceiling and floor propertiesLet n be an integer(1a) x = n if and only if n ≤ x < n+1(1b) x = n if and only if n-1 < x ≤ n(1c) x = n if and only if x-1 < n ≤ x(1d) x = n if and only if x ≤ n < x+1(2) x-1 < x ≤ x ≤ x < x+1 (3a) -x = - x(3b) -x = - x(4a) x+n = x+n (4b) x+n = x+n

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Ceiling property proofProve rule 4a: x+n = x+n

Where n is an integerWill use rule 1a: x = n if and only if n ≤ x < n+1

Direct proof!Let m = xThus, m ≤ x < m+1 (by rule 1a)Add n to both sides: m+n ≤ x+n < m+n+1By rule 1a, m+n = x+nSince m = x, m+n also equals x+nThus, x+n = m+n = x+n

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FactorialFactorial is denoted by n!

n! = n * (n-1) * (n-2) * … * 2 * 1

Thus, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720

Note that 0! is defined to equal 1

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Proving Function problemsLet f be an invertible function from Y to ZLet g be an invertible function from X to YShow that the inverse of f○g is:

(f○g)-1 = g-1 ○ f-1

(Pf) Thus, we want to show, for all zZ and xX ((f g) (g-1 f-1)) (x) = x and ((f-1 g-1) (g f)) (z) = z

((f g) (g-1 f-1)) (x) = (f g) (g-1 f-1)) (x)) = (f g) g-1 f-1(x))) = (f g g-1 f-1(x))))) = (f f-1(x)) = x

The second equality is similar