Functions and Relationsphaas/INFO150/slides/s... · 2019. 10. 30. · Functions and Relations...
Transcript of Functions and Relationsphaas/INFO150/slides/s... · 2019. 10. 30. · Functions and Relations...
Functions and Relations
Reading: EC 4.1–4.5
Peter J. Haas
INFO 150Fall Semester 2019
Lecture 12 1/ 23
Functions and RelationsFunction Notation and TerminologyBinary RelationsInverse Relations and FunctionsComposition of FunctionsProperties of FunctionsOrdering RelationsEquivalence Relations
Lecture 12 2/ 23
Notation and Terminology of Functions 1
Definition
A function f : A ! B associates with each input from the domain A one and only oneoutput in the codomain B according to some rule
Terminology
I We say that “f is a function from A to B”
I If the rule associates to element a 2 A the element b 2 B, then we writef (a) = b and say that “f maps a to b” or “that value of f at a is b” or “f of aequals b”
Exercise: Define f : N ! N by the rule f (x) = 2x + 1
I Q: is every element of the codomain an output of one and only one input to f ?
Exercise: Define f : Z ! Z by the rule f (x) = x2
I Q: is every element of the codomain an output of one and only one input to f ?
Lecture 12 3/ 23
Notation and Terminology of Functions 1
Definition
A function f : A ! B associates with each input from the domain A one and only oneoutput in the codomain B according to some rule
Terminology
I We say that “f is a function from A to B”
I If the rule associates to element a 2 A the element b 2 B, then we writef (a) = b and say that “f maps a to b” or “that value of f at a is b” or “f of aequals b”
Exercise: Define f : N ! N by the rule f (x) = 2x + 1
I Q: is every element of the codomain an output of one and only one input to f ?
Exercise: Define f : Z ! Z by the rule f (x) = x2
I Q: is every element of the codomain an output of one and only one input to f ?
Lecture 12 3/ 23
Notation and Terminology of Functions 1
Definition
A function f : A ! B associates with each input from the domain A one and only oneoutput in the codomain B according to some rule
Terminology
I We say that “f is a function from A to B”
I If the rule associates to element a 2 A the element b 2 B, then we writef (a) = b and say that “f maps a to b” or “that value of f at a is b” or “f of aequals b”
Exercise: Define f : N ! N by the rule f (x) = 2x + 1
I Q: is every element of the codomain an output of one and only one input to f ?
Exercise: Define f : Z ! Z by the rule f (x) = x2
I Q: is every element of the codomain an output of one and only one input to f ?
Lecture 12 3/ 23
No .
fix )=O,
i - e; 2×41=0
,
impliesX= - Igel R
NO . ft - 2) = Plz ) = 4-
Notation and Terminology of Functions 2
Definition
A function f : A ! B associates with each input from the domain A one and only oneoutput in the codomain B according to some rule
Functions come in many guises
I Phone directories
I Word-processing software
I Addition: f (3, 4) = 7
I Truth tables: f : {T ,F}2 ! {T ,F}, e.g., f (p, q) = p ^ q
I Cutting the top card: (HCDS) = SHCD
Lecture 12 4/ 23
p q p ^ q
T T TT F FF T FF F F
Representing a Function
An example function
I Name: f
I Domain: {1, 2, 3, 4, 5}I Codomain: NI Rule: To each number in the domain, associate the square of the number
Representations of the rule
1. The above sentence
2. Algebraic formula: f (x) = x2
3. Set-based description: f = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25)}
4. Table:Input: 1 2 3 4 5
Output: 1 4 9 16 25
5. Arrow diagram:
Lecture 12 5/ 23
1 2 30
{} {a} {b} {c} {a,b} {a,c} {b,c} {a,b,c}
1 2 3 4 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 250
Another functionexample:
r inputoutput
exactlyone outgoing →
arrow
Binary Relations
Definition
A binary relation consists of a domain A, a codomain B, and a subset of A⇥ B calledthe rule for the relation.
Example: Relation E
I Domain: The set S of all UMass students this semester
I Codomain: The set C of classes o↵ered at UMass this semester
I Rule: (x , y) is in E if student x is enrolled in class y this semester
Example: Relation L
I Domain: A = {1, 2, 3, 4}I Codomain: B = {2, 3, 5}I Rule: L = {(1, 2), (1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}I Succinct representation: L = {(x , y) 2 A⇥ B : x < y}I Infix notation: 1 L 2, 1 L 3, 2 L 5, 4 L 5, . . .
Observation
A function F : A ! B is a special case of a relation such that for every x 2 A, thereexists exactly one element y 2 B for which (x , y) 2 F (exactly one outgoing arrow)
Lecture 12 6/ 23
Relation L
2 3 5
1 2 3 4
ILL , 1<3,225,. .
.
Inverse Relations 1
Definition
Given a relation R with domain A and codomain B, the inverse R�1 of R is the
relation with domain B and codomain A such that
(x , y) 2 R if and only if (y , x) 2 R�1.
ExampleI Relation R: domain N and codomain Z with rule R = {(x , y) 2 N⇥ Z : x = y
2}or equivalently R = {(y2, y) : y 2 Z}
I Relation S : domain Z and codomain N with rule S = {(x , y) 2 Z⇥ N : y = x2}
or equivalently S = {(x , x2) : x 2 Z}
I Claim: R and S are inverses of each other1. If (x , y) 2 R, then x = y
2, which means that (y , x) = (y , y2) 2 S X2. If (x , y) 2 S , then x
2 = y , which means that (y , x) = (x2, x) 2 R X
Lecture 12 7/ 23
ex i ( 4,4 , ( 4,
-2 )
ex : 12,4 ),
C - 44 )
ed in verse of a L b C i.e.,
a ab ) from last slide
is G such that Ca,b) G G if a > b
ca,b)c. L iff a ab
,so ( b
,
a ) G G,
since b > a
Inverse Relations 2
Example: Relation E
I Domain is A = {1, 2, 3} and codomain is P(A)
I (x , y) 2 E (or equivalently x E y) if and only if x 2 y
I (y , x) 2 E�1 (or equivalently y E
�1x) if and only if x 2 y (also written y 3 x)
b ca
{} {a} {b} {c} {a,b} {a,c} {b,c} {a,b,c}b ca
{} {a} {b} {c} {a,b} {a,c} {b,c} {a,b,c}
Lecture 12 8/ 23
E E�1
" ycontains
x"
. I
Inverse Relations 3
Example: Arrow diagram when domain and codomain are the same
R = {(A, A), (A, B), (A, C), (A, E), (C , B), (C ,D), (E , A), (E , B), (E , C), (E ,D)}
R�1 = {(A, A), (B, A), (C , A), (E , A), (B, C), (D, C), (A, E), (B, E), (C , E), (D, E)}
A
B
D
E
C
A
B
D
E
C
Lecture 12 9/ 23
R R�1
Inverse Functions 1
Definition
Functions f : A ! B and g : B ! A are inverses of each other if
f (a) = b if and only if g(b) = a
for all a 2 A and b 2 B. We often write f�1 for the inverse of f .
Example: Prove that f : Z ! Z with rule f (x) = x + 3 and g : Z ! Z with ruleg(y) = y � 3 are inverses of each other
I Claim 1: For all a 2 A and b 2 B, if f (a) = b then g(b) = a
I Let a, b 2 Z be given such that f (a) = b, i.e., a+ 3 = b
I Then a = b � 3, i.e., g(b) = a. X
I Claim 2: For all a 2 A and b 2 B, if g(b) = a then f (a) = b
I Let a, b 2 Z be given such that g(b) = a, i.e., b � 3 = a
I Then a+ 3 = b, i.e., f (a) = b. X
Lecture 12 10/ 23
Inverse Functions 1
Definition
Functions f : A ! B and g : B ! A are inverses of each other if
f (a) = b if and only if g(b) = a
for all a 2 A and b 2 B. We often write f�1 for the inverse of f .
Example: Prove that f : Z ! Z with rule f (x) = x + 3 and g : Z ! Z with ruleg(y) = y � 3 are inverses of each other
I Claim 1: For all a 2 A and b 2 B, if f (a) = b then g(b) = a
I Let a, b 2 Z be given such that f (a) = b, i.e., a+ 3 = b
I Then a = b � 3, i.e., g(b) = a. X
I Claim 2: For all a 2 A and b 2 B, if g(b) = a then f (a) = b
I Let a, b 2 Z be given such that g(b) = a, i.e., b � 3 = a
I Then a+ 3 = b, i.e., f (a) = b. X
Lecture 12 10/ 23
-
→
←
Inverse Functions 2: Computing an Inverse
Example: for f : Q ! Q with rule f (x) = 25 x � 2, find f
�1
I Let a, b 2 Q be given such that f (a) = b, i.e., 25a� 2 = b
I Solving for a, we have a = 52b + 5
I So take f�1(y) = g(y) = 5
2 y + 5
Lecture 12 11/ 23
Inverses and Arrow Diagrams
An inverse is obtained by reversing the arrows
2
3
4
a
b
c
d
1
b
c
d
1
2
3
4
a
Example: Why is there is no function whose inverse g : {a, b, c, d} ! {1, 2, 3, 4} isgiven below?
2
3
4
a
b
c
d
1
b
c
d
1
2
3
4
a
Lecture 12 12/ 23
g-' ?
g. yrs ) = ?
g-'
air ?
Composition of Functions
Definition
Given f : A ! B and g : B ! C , the composition g � f of g and f has domain A,codomain C and rule (g � f )(x) = g(f (x)).
Example:
I f : R�0 ! R with rule f (x) =px
I g : R ! R with rule g(x) = 2x
I Then (g � f )(x) = g(f (x)) = g(px) = 2
px
I Then (f � g)(x) = f (g(x)) = f (2x) =p2x (what is the problem here?)
Composition via arrow diagrams
Lecture 12 13/ 23
a
b
c
d
x
y
z
a
b
c
d
2
1
2
1x
y
z
a
b
c
d2
1x
y
z
yes::b.
ate
,evaluate
first
domain of fog = domain of g= IR
,but - I ER and
I fog)th -
- Fn 4 IR
if we define 5 : Pio → 11230 with rule gun-
- H
Ithen fog is well defined
Inverse Functions Revisited
Definition
For a given set A, the identity function on A is the function ◆A : A ! A with the rule◆A(x) = x for all x 2 A. We’ll often simply write ◆ when A is clear from context. Wecan also write ◆A = {(x , x) : x 2 A} when we wish to view ◆A as a binary relation.
Theorem
Functions f : A ! B and g : B ! A are inverses of each other if and only iff � g = ◆B and g � f = ◆A. In other words, f �1
�f (x)
�= x and f
�f�1(y)
�= y
Example 1:I Let f : Q ! Q be the function with rule f (x) = 2
5 x � 2
I Let g : Q ! Q be the function with rule g(x) = 52 x + 5
I Then (g � f )(x) = g(f (x)) = g( 25 x � 2) = 52 (
25 x � 2) + 5 = (x � 5) + 5 = x
I Also, (f � g)(x) = f (g(x)) = f ( 52 x + 5) = 25 (
52 x + 5)� 2 = (x + 2)� 2 = x
Example 2: f : A ! A⇥ A with f (a) = (a, a) and g : A⇥ A ! A with g(x , y) = x
I Given a 2 A: (g � f )(a) = g(f (a)) = g(a, a) = a, so g � f = ◆AI Given (1, 2) 2 A⇥ A: (f � g)(1, 2) = f (1) = (1, 1) 6= (1, 2), so f � g 6= ◆A⇥A
Lecture 12 14/ 23
Inverse Functions Revisited
Definition
For a given set A, the identity function on A is the function ◆A : A ! A with the rule◆A(x) = x for all x 2 A. We’ll often simply write ◆ when A is clear from context. Wecan also write ◆A = {(x , x) : x 2 A} when we wish to view ◆A as a binary relation.
Theorem
Functions f : A ! B and g : B ! A are inverses of each other if and only iff � g = ◆B and g � f = ◆A. In other words, f �1
�f (x)
�= x and f
�f�1(y)
�= y
Example 1:I Let f : Q ! Q be the function with rule f (x) = 2
5 x � 2
I Let g : Q ! Q be the function with rule g(x) = 52 x + 5
I Then (g � f )(x) = g(f (x)) = g( 25 x � 2) = 52 (
25 x � 2) + 5 = (x � 5) + 5 = x
I Also, (f � g)(x) = f (g(x)) = f ( 52 x + 5) = 25 (
52 x + 5)� 2 = (x + 2)� 2 = x
Example 2: f : A ! A⇥ A with f (a) = (a, a) and g : A⇥ A ! A with g(x , y) = x
I Given a 2 A: (g � f )(a) = g(f (a)) = g(a, a) = a, so g � f = ◆AI Given (1, 2) 2 A⇥ A: (f � g)(1, 2) = f (1) = (1, 1) 6= (1, 2), so f � g 6= ◆A⇥A
Lecture 12 14/ 23
Inverse Functions Revisited
Definition
For a given set A, the identity function on A is the function ◆A : A ! A with the rule◆A(x) = x for all x 2 A. We’ll often simply write ◆ when A is clear from context. Wecan also write ◆A = {(x , x) : x 2 A} when we wish to view ◆A as a binary relation.
Theorem
Functions f : A ! B and g : B ! A are inverses of each other if and only iff � g = ◆B and g � f = ◆A. In other words, f �1
�f (x)
�= x and f
�f�1(y)
�= y
Example 1:I Let f : Q ! Q be the function with rule f (x) = 2
5 x � 2
I Let g : Q ! Q be the function with rule g(x) = 52 x + 5
I Then (g � f )(x) = g(f (x)) = g( 25 x � 2) = 52 (
25 x � 2) + 5 = (x � 5) + 5 = x
I Also, (f � g)(x) = f (g(x)) = f ( 52 x + 5) = 25 (
52 x + 5)� 2 = (x + 2)� 2 = x
Example 2: f : A ! A⇥ A with f (a) = (a, a) and g : A⇥ A ! A with g(x , y) = x
I Given a 2 A: (g � f )(a) = g(f (a)) = g(a, a) = a, so g � f = ◆AI Given (1, 2) 2 A⇥ A: (f � g)(1, 2) = f (1) = (1, 1) 6= (1, 2), so f � g 6= ◆A⇥A
Lecture 12 14/ 23
fawnby
notinverses
Properties of Functions 1
Definition
The function f : A ! B is invertible if there is a function f�1 : B ! A such that
f (x) = y if and only if f �1(y) = x . By symmetry of the definition, (f �1)�1 = f .
Example
I With A = B = R�0, if f has rule f (x) = x2, then f
�1(x) =px
I Arrow diagram:
A non-invertible function g
I Problem 1: no arrow points to 4 (g is not onto)
I Problem 2: two arrows point to 3 (g is not one-to-one)
Lecture 12 15/ 23
2
3
4
a
b
c
d
1
b
c
d
1
2
3
4
a
(f �1�f )(x) = x
Properties of Functions 1
Definition
The function f : A ! B is invertible if there is a function f�1 : B ! A such that
f (x) = y if and only if f �1(y) = x . By symmetry of the definition, (f �1)�1 = f .
Example
I With A = B = R�0, if f has rule f (x) = x2, then f
�1(x) =px
I Arrow diagram:
A non-invertible function g
I Problem 1: no arrow points to 4 (g is not onto)
I Problem 2: two arrows point to 3 (g is not one-to-one)
Lecture 12 15/ 23
2
3
4
a
b
c
d
1
b
c
d
1
2
3
4
a
2
3
4
a
b
c
d
1
b
c
d
1
2
3
4
a
(f �1�f )(x) = x
-
Properties of Functions 2
Definition
The function f : A ! B is onto if 8y 2 B, 9x 2 A, f (x) = y . The function f isone-to-one if (x 6= y) !
�f (x) 6= f (y)
�.
Theorem
A function f : A ! B is invertible if and only if it is both one-to-one and onto.
Exercise: Which functions are invertible?
I f : Z ! Z with f (x) = 2x + 3
I g : Z ! N with g(x) =
(�2z if z 0
2z � 1 if z > 0
I h : N ! N with h(n) = sum of digits in the numeral n
Lecture 12 16/ 23
2
3
4
a
b
c
d
1
X
not onto.
flex ) -
-O implies LXt3=o ,
so HI -3g ¢ #✓ gt - 11--2
, go-25-4,91-31--6, - -
-
c. to . I and onto 90=9 guk '
9123=3,433--5,
. --
,
"
Anot I - tot : htt 3) = honor ) -
- 4
(Direct) Proofs About Functions
Proposition 1
If f : A ! B is one-to-one and g : B ! C is one-to-one, then (g � f ) : A ! C isone-to-one.
Proof
1. Write h = g � f and let x1, x2 2 A be given such that h(x1) = h(x2)
2. This means that g(f (x1)) = g(f (x2))
3. Since g is one-to-one, this means that f (x1) = f (x2)
4. Since f is one-to-one, this means that x1 = x2 X
Example: f : N ! N wih f (x) = 5x + 7 and g : N ! Q with g(n) = 5n+2
1. Write h = g � f and let x1, x2 2 A be given such that h(x1) = h(x2)
2. This means that g(f (x1)) = g(f (x2)) or 5f (x1)+2 = 5
f (x2)+2 or f (x1)+25 = f (x2)+2
5
3. Multiply by 5 and subtract 2 on both sides: f (x1) = f (x2) or 5x1 + 7 = 5x2 + 7
4. Subtract 7 and divide by 5 on both sides to get x1 = x2 X
Lecture 12 17/ 23
I - to -I : ( x
, + 4) → Lf exit ftp.D
Xcx,k f Hr ) ) → cry ,
= XD
(Direct) Proofs About Functions
Proposition 1
If f : A ! B is one-to-one and g : B ! C is one-to-one, then (g � f ) : A ! C isone-to-one.
Proof
1. Write h = g � f and let x1, x2 2 A be given such that h(x1) = h(x2)
2. This means that g(f (x1)) = g(f (x2))
3. Since g is one-to-one, this means that f (x1) = f (x2)
4. Since f is one-to-one, this means that x1 = x2 X
Example: f : N ! N wih f (x) = 5x + 7 and g : N ! Q with g(n) = 5n+2
1. Write h = g � f and let x1, x2 2 A be given such that h(x1) = h(x2)
2. This means that g(f (x1)) = g(f (x2)) or 5f (x1)+2 = 5
f (x2)+2 or f (x1)+25 = f (x2)+2
5
3. Multiply by 5 and subtract 2 on both sides: f (x1) = f (x2) or 5x1 + 7 = 5x2 + 7
4. Subtract 7 and divide by 5 on both sides to get x1 = x2 X
Lecture 12 17/ 23
of contrapositive I flexi -
-flair ) ) → µ,- uhh )
(Direct) Proofs About Functions
Proposition 1
If f : A ! B is one-to-one and g : B ! C is one-to-one, then (g � f ) : A ! C isone-to-one.
Proof
1. Write h = g � f and let x1, x2 2 A be given such that h(x1) = h(x2)
2. This means that g(f (x1)) = g(f (x2))
3. Since g is one-to-one, this means that f (x1) = f (x2)
4. Since f is one-to-one, this means that x1 = x2 X
Example: f : N ! N wih f (x) = 5x + 7 and g : N ! Q with g(n) = 5n+2
1. Write h = g � f and let x1, x2 2 A be given such that h(x1) = h(x2)
2. This means that g(f (x1)) = g(f (x2)) or 5f (x1)+2 = 5
f (x2)+2 or f (x1)+25 = f (x2)+2
5
3. Multiply by 5 and subtract 2 on both sides: f (x1) = f (x2) or 5x1 + 7 = 5x2 + 7
4. Subtract 7 and divide by 5 on both sides to get x1 = x2 X
Lecture 12 17/ 23
,Tracing the proof
Properties of Relations
Definition
Let R be a binary relation on a set A
1. R is reflexive if (a, a) 2 R for all a 2 A (must have loops)
2. R is antisymmetric if (a, b) 2 R and a 6= b implies (b, a) 62 R (no double arrows)
3. R is transitive if (a, b), (b, c) 2 R implies (a, c) 2 R (if you can get from a to c
following two arrows, you can also get there following one arrow)
Definition
A relation R on a set A is partial order if it is antisymmetric, transitive, and reflexive
Examples
1. A = {1, 2, 3, 4}: a R1 b means a b
2. A = P({1, 2, 3}): a R2 b means a ✓ b
3. A = {1, 2, 3, 6}: a R3 b means a divides b
Lecture 12 18/ 23
L-
- { C 1,27 ,( 3,4)
, ④7) I 122,324 ,"
LE s fl :D ,134 ,
.. . I
)
Properties of Relations
Definition
Let R be a binary relation on a set A
1. R is reflexive if (a, a) 2 R for all a 2 A (must have loops)
2. R is antisymmetric if (a, b) 2 R and a 6= b implies (b, a) 62 R (no double arrows)
3. R is transitive if (a, b), (b, c) 2 R implies (a, c) 2 R (if you can get from a to c
following two arrows, you can also get there following one arrow)
Definition
A relation R on a set A is partial order if it is antisymmetric, transitive, and reflexive
Examples
1. A = {1, 2, 3, 4}: a R1 b means a b
2. A = P({1, 2, 3}): a R2 b means a ✓ b
3. A = {1, 2, 3, 6}: a R3 b means a divides b
Lecture 12 18/ 23
Properties of Relations
Definition
Let R be a binary relation on a set A
1. R is reflexive if (a, a) 2 R for all a 2 A (must have loops)
2. R is antisymmetric if (a, b) 2 R and a 6= b implies (b, a) 62 R (no double arrows)
3. R is transitive if (a, b), (b, c) 2 R implies (a, c) 2 R (if you can get from a to c
following two arrows, you can also get there following one arrow)
Definition
A relation R on a set A is partial order if it is antisymmetric, transitive, and reflexive
Examples
1. A = {1, 2, 3, 4}: a R1 b means a b
2. A = P({1, 2, 3}): a R2 b means a ✓ b
3. A = {1, 2, 3, 6}: a R3 b means a divides b
Lecture 12 18/ 23
3
2
1
1
2 3
6
1
4{1,2,3}
{1,3}{2,3}
{1}{2}
{3}
{}
{1,2}
R1 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)} Hasse diagrams=p
antisymmetric : ( atb )s7[ca ,HeRrcb,aIER ]
Proofs About Properties
Example: Prove the reflexive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let a 2 Z be given
2. Since a� a = 0, which is even, we have that (a, a) 2 R X
Example: Prove the transitive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let (a, b), (b, c) 2 R be given [We’ll prove that (a, c) 2 R]
2. Since a� b and b � c are even, we can write a� b = 2K and b � c = 2L forsome integers K and L
3. Therefore a� c = (a� b) + (b � c) = 2K + 2L = 2(K + L)
4. Thus (a� c) is even and hence (a, c) 2 R X
Example: Prove the antisymmetric property for R = {(s, t) 2 P({1, 2, 3, 4})2 : s ✓ t}I Show: for all s, t 2 P({1, 2, 3, 4}), if (s, t) 2 R and (t, s) 2 R, then s = t
1. Let s, t 2 P({1, 2, 3, 4}) be given2. Since (s, t) 2 R and (t, s) 2 R, we have that s ✓ t and t ✓ s
3. It follows that s = t by definition of set equality X
Lecture 12 19/ 23
Proofs About Properties
Example: Prove the reflexive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let a 2 Z be given
2. Since a� a = 0, which is even, we have that (a, a) 2 R X
Example: Prove the transitive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let (a, b), (b, c) 2 R be given [We’ll prove that (a, c) 2 R]
2. Since a� b and b � c are even, we can write a� b = 2K and b � c = 2L forsome integers K and L
3. Therefore a� c = (a� b) + (b � c) = 2K + 2L = 2(K + L)
4. Thus (a� c) is even and hence (a, c) 2 R X
Example: Prove the antisymmetric property for R = {(s, t) 2 P({1, 2, 3, 4})2 : s ✓ t}I Show: for all s, t 2 P({1, 2, 3, 4}), if (s, t) 2 R and (t, s) 2 R, then s = t
1. Let s, t 2 P({1, 2, 3, 4}) be given2. Since (s, t) 2 R and (t, s) 2 R, we have that s ✓ t and t ✓ s
3. It follows that s = t by definition of set equality X
Lecture 12 19/ 23
Proofs About Properties
Example: Prove the reflexive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let a 2 Z be given
2. Since a� a = 0, which is even, we have that (a, a) 2 R X
Example: Prove the transitive property for R = {(a, b) 2 Z⇥ Z : a� b is even}1. Let (a, b), (b, c) 2 R be given [We’ll prove that (a, c) 2 R]
2. Since a� b and b � c are even, we can write a� b = 2K and b � c = 2L forsome integers K and L
3. Therefore a� c = (a� b) + (b � c) = 2K + 2L = 2(K + L)
4. Thus (a� c) is even and hence (a, c) 2 R X
Example: Prove the antisymmetric property for R = {(s, t) 2 P({1, 2, 3, 4})2 : s ✓ t}I Show: for all s, t 2 P({1, 2, 3, 4}), if (s, t) 2 R and (t, s) 2 R, then s = t
1. Let s, t 2 P({1, 2, 3, 4}) be given2. Since (s, t) 2 R and (t, s) 2 R, we have that s ✓ t and t ✓ s
3. It follows that s = t by definition of set equality X
Lecture 12 19/ 23
( at b) → Tla,
b) GR n lb,at ER ]
contrapositive : ca , b) f R A lb, a) ER# (a =D
Other Types of Orders
Definition
A relation R over a set A is irreflexive if (a, a) 62 R for all a 2 A. A strict partialordering on A is a relation R on A that is transitive, antisymmetric, and irreflexive.
Notes
I Irreflexive means no loops in an arrow diagram
I A relation R can be neither reflexive or irreflexive if some (but not all) nodes inthe arrow diagram have loops
Example:
I Strict subset relation: write A ⇢ B if A ✓ B and B � A 6= ;I Then R = {(A,B) 2 P({1, 2, 3, 4})2 : A ⇢ B} is a strict partial ordering
Definition
A relation R on A is a total ordering if it is a partial ordering and also satisfies theproperty:For all a, b 2 A, if a 6= b, then either (a, b) 2 R or (b, a) 2 R. A strict total orderinghas the same properties except that it is irreflexive.
Lecture 12 20/ 23
c :p no Boo
-
i.e. ,B has at least one element x
set . x ¢ A
Types of Orderings: Examples
Notation:
I A = {1, 2, 4, 8}, B = P({1, 2, 3}), C = {0, 1}4
I V (↵) = value of binary numeral ↵, e.g., V (0101) = 5
Example 1: R1 = {(x , y) 2 A2 : x y} total ordering
Example 2: R2 = {(x , y) 2 B2 : x ✓ y} partial ordering (incomparable subsets)
Example 3: R2 = {(S ,T ) 2 B2 : every element in S is every element in T} ?
Example 4: R2 = {(S ,T ) 2 B2 : n(S) < n(T )} ?
Example 5: R2 = {(S ,T ) 2 B2 : sum of elements in S is sum of elements in T} ?
Example 6: R2 = {(↵,�) 2 C2 : ↵ has fewer 1’s than � has} ?
Example 7: R2 = {(↵,�) 2 C2 : V (↵) V (�)} ?
Lecture 12 21/ 23
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= s
e.9 .1011
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Equivalence Relations
Definition
A partition of a set A is a set S = {S1, S2, S3, . . .} such that
1. For all i , Si 6= 0
2. For all i , j : if Si 6= Sj , then Si \ Sj = ;3. S1 [ S2 [ S3 [ · · · = A
Definition
A relation R on A is an equivalence relation if there is a partition S of A such that(x , y) 2 R if and only if x and y are in the same part of S. We call S the partition ofA induced by R.
Example 1: A = {1, 2, 3, 4, 5, 6}I R = {(a, b) 2 A⇥ A : a� b is even}I S = {{1, 3, 5}, {2, 4, 6}}
Example 2:
I R = {(a, b) 2 Z⇥ Z : a� b is divisible by 4}I S = {P0,P1,P2,P3} where Pi = {a 2 Z : a = 4k + i for some k 2 Z}
Lecture 12 22/ 23
disjoint
Equivalence Relations
Definition
A partition of a set A is a set S = {S1, S2, S3, . . .} such that
1. For all i , Si 6= 0
2. For all i , j : if Si 6= Sj , then Si \ Sj = ;3. S1 [ S2 [ S3 [ · · · = A
Definition
A relation R on A is an equivalence relation if there is a partition S of A such that(x , y) 2 R if and only if x and y are in the same part of S. We call S the partition ofA induced by R.
Example 1: A = {1, 2, 3, 4, 5, 6}I R = {(a, b) 2 A⇥ A : a� b is even}I S = {{1, 3, 5}, {2, 4, 6}}
Example 2:
I R = {(a, b) 2 Z⇥ Z : a� b is divisible by 4}I S = {P0,P1,P2,P3} where Pi = {a 2 Z : a = 4k + i for some k 2 Z}
Lecture 12 22/ 23
disjoint
Equivalence Relations
Definition
A partition of a set A is a set S = {S1, S2, S3, . . .} such that
1. For all i , Si 6= 0
2. For all i , j : if Si 6= Sj , then Si \ Sj = ;3. S1 [ S2 [ S3 [ · · · = A
Definition
A relation R on A is an equivalence relation if there is a partition S of A such that(x , y) 2 R if and only if x and y are in the same part of S. We call S the partition ofA induced by R.
Example 1: A = {1, 2, 3, 4, 5, 6}I R = {(a, b) 2 A⇥ A : a� b is even}I S = {{1, 3, 5}, {2, 4, 6}}
Example 2:
I R = {(a, b) 2 Z⇥ Z : a� b is divisible by 4}I S = {P0,P1,P2,P3} where Pi = {a 2 Z : a = 4k + i for some k 2 Z}
Lecture 12 22/ 23
disjoint
part ion induced by R
odd - Odd is even
even- even is even
even- odd is even ,
odd - even is even
Equivalence Relations
Definition
A partition of a set A is a set S = {S1, S2, S3, . . .} such that
1. For all i , Si 6= 0
2. For all i , j : if Si 6= Sj , then Si \ Sj = ;3. S1 [ S2 [ S3 [ · · · = A
Definition
A relation R on A is an equivalence relation if there is a partition S of A such that(x , y) 2 R if and only if x and y are in the same part of S. We call S the partition ofA induced by R.
Example 1: A = {1, 2, 3, 4, 5, 6}I R = {(a, b) 2 A⇥ A : a� b is even}I S = {{1, 3, 5}, {2, 4, 6}}
Example 2:
I R = {(a, b) 2 Z⇥ Z : a� b is divisible by 4}I S = {P0,P1,P2,P3} where Pi = {a 2 Z : a = 4k + i for some k 2 Z}
Lecture 12 22/ 23
if a,be Pi
,
then a- 4kt is bintifor integers k and L
So a- b -
- 14kt i ) - Ltthti ) -
- 41k - L ),which is div . by 4
-
✓
Formal Properties of an Equivalence Relation
Definition
A relation R on A is symmetric if for all a, b 2 A, if (a, b) 2 A then (b, a) 2 A.
Theorem
A relation R on A is an equivalence relation if and only if it is reflexive, symmetric,and transitive.
Example: For each relation, determine whether it is an equivalence relation
I T1 = {(a, b) 2 Z2 : b � a is divisible by 5}
I T2 = {(a, b) 2 Z2 : a2 � b2 is divisible by 5}
I T3 = {(a, b) 2 Z2 : |a� b| 2}
Lecture 12 23/ 23
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Transitive
( 54 ) ¢T ,not transitive ie . II Land 2%4 but not 1154