1. CSci 2011 Discrete Mathematics Lecture 9, 10

2. Recap Propositional operation summary Check translation Definition
Tautology, Contradiction, logical equicalence
not
and or
conditional
Bi-conditional p q p q
pq
pq
pq
pq T F
CSci 2011

3. Recap CSci 2011 p  T  p p  F  p (p  q)  r  p  (q  r)
Identity Laws
(p  q)  r  p  (q  r)
(p  q)  r  p  (q  r)
Associative laws
p  T  T
p  F  F
Domination Law
p  (q  r)  (p  q)  (p  r)
p  (q  r)  (p  q)  (p  r)
Distributive laws
p  p  p
p  p  p
Idempotent Laws
 (p  q)   p   q
 (p  q)   p   q
De Morgan’s laws
( p)  p
Double negation law
p  (p  q)  p
p  (p  q)  p
Absorption laws
p  q  q  p
p  q  q  p
Commutative Laws
p   p  T
p   p  F
Negation lows
pq  pq
Definition of Implication
p  q  (p  q)  (q  p)
Definition of Biconditional
CSci 2011

4. Recap Quantifiers CSci 2011 Nested quantifiers
Universal quantifier: x P(x)
Negating quantifiers
¬x P(x) = x ¬P(x)
¬x P(x) = x ¬P(x) xy P(x, y)
Nested quantifiers
xy P(x, y): “For all x, there exists a y such that P(x,y)”
xy 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)
CSci 2011

5. Recap p p  q  q  q   p q  r  p  r p  q  p  p  q p  q  p
Modus ponens p
p  q
 q
Modus tollens
 q
  p
Hypothetical syllogism
q  r
 p  r
Disjunctive syllogism
p  q
 p
Addition
 p  q
Simplification
p  q
 p
Conjunction q
 p  q
Resolution
 p  r
 q  r
CSci 2011

6. Recap CSci 2011 p→q Proof by contradiction
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: [(p1p2…pn)q]  [(p1q)(p2q)…(pnq)]
If and only if proof: pq (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
CSci 2011

7. What is a set? CSci 2011 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
CSci 2011

8. 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)
CSci 2011

9. Proper Subsets
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
CSci 2011

10. 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
CSci 2011

11. 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
CSci 2011

12. 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)
CSci 2011

13. 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 }
CSci 2011

14. 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)
CSci 2011

15. ch2.2 Set Operations
CSci 2011

16. 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
CSci 2011

17. 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
CSci 2011

18. Disjoint sets
Formal 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!
CSci 2011

19. Set operations: Difference
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
CSci 2011

20. Complement sets
Formal 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 law
A U Ac = U Complement law
A ∩ Ac =  Complement law
CSci 2011

21. Set identities CSci 2011 A = A AU = A Identity Law AU = U A = 
Domination law
AA = A
AA = A
Idempotent Law
(Ac)c = A
Complement Law
AB = BA
AB = BA
Commutative Law
(AB)c = AcBc
(AB)c = AcBc
De Morgan’s Law
A(BC)
= (AB)C
A(BC)
= (AB)C
Associative Law
A(BC) =
(AB)(AC)
A(BC) =
(AB)(AC)
Distributive Law
A(AB) = A
A(AB) = A
Absorption Law
A  Ac = U
A  Ac = 
CSci 2011

22. How to prove a set identity
For example: A∩B=B-(B-A)
Four methods:
Use the basic set identities
Use membership tables
Prove each set is a subset of each other
Use set builder notation and logical equivalences
CSci 2011

23. What we are going to prove…
A∩B=B-(B-A) A B
B-(B-A)
A∩B
B-A
CSci 2011

24. Proof by Set Identities
A  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
CSci 2011

25. 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
CSci 2011

26. Examples Let A, B, and C be sets. Show that: (AUB)  (AUBUC)
(A∩B∩C)  (A∩B)
(A-B)-C  A-C
(A-C) ∩ (C-B) = 
CSci 2011

27. ch2.3 Functions
CSci 2011

28. Definition of a function
A function takes an element from a set and maps it to a UNIQUE element in another set
f maps R to Z R Z
Domain
Co-domain f
f(4.3)
4.3 4
Pre-image of 4
Image of 4.3
CSci 2011

29. A string length function
More functions
The image
of “a”
A pre-image
of 1
Domain
Co-domain A B C D F
Alice
Bob
Chris
Dave
Emma
A class grade function 1 2 3 4 5
“a”
“bb“
“cccc”
“dd”
“e”
A string length function
CSci 2011

30. Also not a valid function!
Even more functions
Range 1 2 3 4 5 a e i o u
Some function… 1 2 3 4 5
“a”
“bb“
“cccc”
“dd”
“e”
Not a valid function!
Also not a valid function!
CSci 2011

31. Function arithmetic Let f1(x) = 2x Let 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
CSci 2011

32. One-to-one functions
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. 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
CSci 2011

33. More on one-to-one Injective 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
CSci 2011

34. A function that is not onto
Onto functions
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 1 2 3 4 5 a e i o
A function that is not onto
CSci 2011

35. More on onto Surjective 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 1 2 3 4 a e i o u
An onto function
CSci 2011

36. Onto vs. one-to-one
Are the following functions onto, one-to-one, both, or neither? 1 2 3 4 a b c 1 2 3 4 a b c d 1 2 3 4 a b c
1-to-1, not onto
Both 1-to-1 and onto
Not a valid function 1 2 3 a b c d 1 2 3 4 a b c d
Onto, not 1-to-1
Neither 1-to-1 nor onto
CSci 2011

37. Bijections Consider 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
CSci 2011

38. Identity functions
A function such that the image and the pre-image are ALWAYS equal
f(x) = 1*x
f(x) = x + 0
The domain and the co-domain must be the same set
CSci 2011

39. Inverse functions Let f(x) = 2*x R f R f-1 f(4.3) 4.3 8.6 f-1(8.6)
Then f-1(x) = x/2
CSci 2011

40. 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!
CSci 2011

41. Compositions of functions
(f ○ g)(x) = f(g(x))
f ○ g A B C g f
g(a)
f(b) a
f(g(a))
b = g(a)
(f ○ g)(a)
CSci 2011

42. Compositions of functions
Let f(x) = 2x+3 Let g(x) = 3x+2
f ○ g R R R g f
g(1)
f(5)
f(g(1))=13 1
g(1)=5
(f ○ g)(1)
f(g(x)) = 2(3x+2)+3 = 6x+7
CSci 2011

43. Compositions of functions
Does f(g(x)) = g(f(x))?
Let f(x) = 2x+3 Let g(x) = 3x+2
f(g(x)) = 2(3x+2)+3 = 6x+7
g(f(x)) = 3(2x+3)+2 = 6x+11
Function composition is not commutative!
Not equal!
CSci 2011

44. Useful functions
Floor: 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 
CSci 2011

45. Ceiling and floor properties
Let 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
CSci 2011

46. Ceiling property proof
Prove rule 4a: x+n = x+n
Where n is an integer
Will use rule 1a: x = n if and only if n ≤ x < n+1
Direct proof!
Let m = x
Thus, m ≤ x < m+1 (by rule 1a)
Add n to both sides: m+n ≤ x+n < m+n+1
By rule 1a, m+n = x+n
Since m = x, m+n also equals x+n
Thus, x+n = m+n = x+n
CSci 2011

47. Factorial Factorial 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
CSci 2011

48. Proving Function problems
Let f be an invertible function from Y to Z
Let g be an invertible function from X to Y
Show that the inverse of f○g is:
(f○g)-1 = g-1 ○ f-1
(Pf) Thus, we want to show, for all zZ and xX ((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
CSci 2011