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LECTURE NOTES
on
DISCRETE MATHEMATICS
Eusebius Doedel

LOGIC
Introduction. First we introduce some basic concepts needed in our
discussion of logic. These will be covered in more detail later.
A set is a collection of “objects” (or “elements”).
EXAMPLES :
• the infinite set of all integers : Z ≡ {· · · , −2, −1, 0, 1, 2, 3, · · ·}.
• the infinite set of all positive integers : Z+
≡ {1, 2, 3, · · ·}.
• the infinite set R of all real numbers.
• the finite set {T , F }, where T denotes “True ” and F “False ”.
• the finite set of alphabetic characters : {a, b, c, · · · , z}.
1

A function (or “map”, or “operator”) is a rule that associates to every
element of a set one element in another set.
EXAMPLES :
• If S1 = {a, b, c} and S2 = {1, 2} then the associations
a 7→ 2, b 7→ 1, c 7→ 2,
define a function f from S1 to S2 . We write
f : S1 −→ S2 .
• Similarly f(n) = n2
defines a function f : Z+
−→ Z+
.
• f(n) = n2
can also be viewed as a function f : Z −→ Z.
2

EXAMPLE :
Let Pn denoet the infinite set of all polynomial functions p(x) of degree
n or less with integer coefficients.
• The derivative operator D restricted to elements of Pn can be
viewed as a function from Pn to Pn−1,
D : Pn −→ Pn−1, D : p 7→
dp
dx
.
For example, if p(x) = x3
+ 2x + 1, then
D : x3
+ 2x + 1 7→ 3x2
+ 2 ,
i.e.,
D(x3
+ 2x + 1) = 3x2
+ 2 .
3

EXAMPLE :
• We can also define functions of several variables, e.g.,
f(x, y) = x + y ,
can be viewed as a function “from Z+
cross Z+
into Z+
”.
We write
f : Z+
× Z+
−→ Z+
.
4

Basic logical operators.
The basic logical operators
∧ (“and” , “conjunction”)
∨ (“or” , “disjunction”)
¬ (“not” , “negation”)
are defined in the tables below :
p ¬p
T F
F T
p q p ∨ q
T T T
T F T
F T T
F F F
p q p ∧ q
T T T
T F F
F T F
F F F
5

Let B ≡ {T , F }. Then we can view ¬, ∨, and ∧, as functions
¬ : B −→ B, ∨ : B × B −→ B , ∧ : B × B −→ B .
We can also view the arithmetic operators −, +, and ×, as functions
− : Z −→ Z, + : Z × Z −→ Z, ∗ : Z × Z −→ Z,
defined by value tables, for example,
x −x
· ·
-2 2
-1 1
0 0
1 -1
2 -2
· ·
6

Logical expressions.
A logical expression (or “proposition”) P(p, q, · · ·) is a function
P : B × B × · · · × B −→ B .
For example,
P1(p, q) ≡ p ∨ ¬q and P2(p, q, r) ≡ p ∧ (q ∨ r)
are logical expressions.
Here
P1 : B × B −→ B ,
and
P2 : B × B × B −→ B .
7

The values of a logical expression can be listed in a truth table .
EXAMPLE :
p q ¬q p ∨ (¬q)
T T F T
T F T T
F T F F
F F T T
8

Analogously, arithmetic expressions such as
A1(x, y) ≡ x + (−y) and A2(x, y, z) ≡ x × (y + z)
can be considered as functions
A1 : R × R −→ R, and A2 : R × R × R −→ R ,
or, equivalently,
A1 : R2
−→ R, and A2 : R3
−→ R .
9

Two propositions are equivalent if they always have the same values.
EXAMPLE :
¬(p ∨ q) is equivalent to ¬p ∧ ¬q ,
(one of de Morgan’s laws), as can be seen in the table below :
p q p ∨ q ¬(p ∨ q) ¬p ¬q ¬p ∧ ¬q
T T T F F F F
T F T F F T F
F T T F T F F
F F F T T T T
10

NOTE :
In arithmetic
−(x + y) is equivalent to (−x) + (−y) ,
i.e., we do not have that
−(x + y) is equivalent to (−x) × (−y) .
Thus the analogy between logic and arithmetic is limited.
11

The three basic logical operators ¬ , ∨ , and ∧ , are all we need.
However, it is very convenient to introduce some additional operators,
much like in arithmetic, where we write x3
to denote x × (x × x).
Three such additional operators are
⊕ “exclusive or”
→ “conditional” , “if then”
↔ “biconditional” , “if and only if” , “iff”
defined as :
p q p ⊕ q p → q p ↔ q
T T F T T
T F T F F
F T T T F
F F F T T
12

EXAMPLE :
Suppose two persons, P and Q, are suspected of committing a crime.
• Let P denote the statement by P that
“Q did it, or we did it together”.
• Let Q denote the statement by Q that
“P did it, or if I did it then we did it together”.
• Suppose we know P always tells the truth and Q always lies.
Who committed the crime ?
NOTE : By “did it” we mean “was involved”.
13

Let p denote “P did it” and let q denote “Q did it”.
Then p and q are logical variables.
We are given that the value of the logical expression
q ∨ (p ∧ q) is True ,
and that
p ∨

q → (p ∧ q)

is False .
Equivalently we have that the value of the logical expression
¬

p ∨

q → (p ∧ q)

∧

q ∨ (p ∧ q)

is True .
Our problem is to find for what values of p and q this is the case.
14

As an analogy from arithmetic, consider the problem of finding the values
of x and y in Z so that
the value of the arithmetic expression x2
+ y is 5 ,
and such that
the value of x + y is 3 ,
i.e., we want to find all solutions of the the simultaneous equations
x2
+ y = 5 , x + y = 3 .
(How many solutions are there ?)
15

For the “crime problem” we have the truth table
p q ¬

p ∨

q → (p ∧q)

∧

q ∨ (p ∧ q)

T T F T T T F T T
T F F T T F F F F
F T T F F F T T F
F F F T T F F F F
(1) (2) (6) (5) (4) (3) (9) (8) (7)
The order of evaluation has been indicated at the bottom of the table.
The values of the entire expression are in column (9).
We observe that the expression is True only if p = F and q = T .
Therefore Q was involved in the crime, but P was not.
16

EXERCISE :
Consider the logical expression in the preceding “crime” example.
Find a much simpler, equivalent logical expression.
(It must have precisely the same values as listed in column “(9)”.)
17

EXERCISE :
Suppose three persons, P, Q, and R, are suspects in a crime.
• P states that “Q or R, or both, were involved”.
• Q states that “P or R, or both, were involved”.
• R states that “P or Q, but not both, were involved”.
• Suppose P and Q always tell the truth, while R always lies.
Who were involved in the crime ?
NOTE : there may be more than one solution · · ·
18

EXERCISE :
Construct a truth table for the logical expression
(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) .
Based on the truth table find a simpler, equivalent logical expression.
19

A contradiction is a logical expression whose value is always False .
For example p ∧ ¬p is a contradiction :
p ¬p p ∧ ¬p
T F F
F T F
A tautology is a logical expression that is always True .
For example p ∨ ¬p is a tautology :
p ¬p p ∨ ¬p
T F T
F T T
A logical expression that is neither a tautology nor a contradiction
is called a contingency.
20

EXERCISE :
Verify by truth table that the following expressions are tautologies :

(p → q) ∧ p

→ q ,

(p → q) ∧ ¬q

→ ¬p .
21

NOTATION :
We use the symbol “⇒” to indicate that a conditional statement is a
tautology.
For example, from the preceding exercise we have

(p → q) ∧ p

⇒ q , (”modus ponens”) ,

(p → q) ∧ ¬q

⇒ ¬p , (”modus tollens”) .
22

As another example we show that

(p → q) ∧ (¬p → r) ∧ (q → r)

→ r
is a tautology.
p q r

(p → q) ∧ (¬p → r) ∧ (q → r)

→ r
T T T T T F T T T T T
T T F T T F T F F T F
T F T F F F T F T T T
T F F F F F T F T T F
F T T T T T T T T T T
F T F T F T F F F T F
F F T T T T T T T T T
F F F T F T F F T T F
(1) (2) (3) (5) (9) (6) (7) (10) (8) (11) (4)
The last column (11) consist of True values only. Therefore we can write

(p → q) ∧ (¬p → r) ∧ (q → r)

⇒ r
QUESTION : Does it matter which of the two ∧’s is evaluated first?
23

EXAMPLE :
Here we illustrate another technique that can sometimes be used to show
that a conditional statement is a tautology.
Consider again the logical expression
P(p, q, r) ⇐⇒ (p → q) ∧ (¬p → r) ∧ (q → r) .
We want to show that
P(p, q, r) ⇒ r ,
i.e., that
P(p, q, r) → r always has value True .
NOTE : We need only show that:
We can’t have that P(p, q, r) is True , while r is False .
24

So suppose that P(p, q, r) is True , while r is False .
(We must show that this cannot happen ! )
Thus all three of
(a) p → q (b) ¬p → F and (c) q → F
have value True .
(c) can only have value True if q = F .
(b) can only have value True if ¬p = F , i.e., if p = T .
But then (a) becomes T → F which has value F .
So indeed, not all three, (a), (b), and (c) can have value True .
QED ! (“quod erat demonstrandum”: “which was to be shown”)
25

NOTE :
This was an example of a proof by contradiction.
(We will give more examples later · · · )
26

EXERCISE :
Use a proof “by contradiction” to prove the following:

(p → q) ∧ p

⇒ q ,

(p → q) ∧ ¬q

⇒ ¬p .
(Already done by truth table.)
27

EXERCISE :
Also use a ”direct proof” to prove that

(p → q) ∧ p

⇒ q ,

(p → q) ∧ ¬q

⇒ ¬p .
(In a direct proof one assumes that the LHS is True and then one shows
that the RHS must be True also.)
28

NOTATION :
From the definition of the ↔ operator we see that logical expressions
P1(p, q, · · ·) and P2(p, q, · · ·) ,
are equivalent if and only if
P1(p, q, · · ·) ↔ P2(p, q, · · ·)
is a tautology.
In this case we write
P1(p, q, · · ·) ⇐⇒ P2(p, q, · · ·) .
29

EXAMPLE :
¬(p ∧ q) is equivalent to ¬p ∨ ¬q ,
i.e.,
¬(p ∧ q) ⇐⇒ (¬p ∨ ¬q) ,
as seen in the truth table
p q ¬ (p ∧ q) ↔ (¬p ∨ ¬q)
T T F T T F F F
T F T F T F T T
F T T F T T T F
F F T F T T T T
30

The operators
⊕ , → , and ↔ ,
can be expressed in terms of the basic operators
¬ , ∧ , and ∨ ,
as verified below :
p q (p ⊕ q) ↔

(p ∨ q) ∧ ¬ (p ∧ q)

T T F T T F F T
T F T T T T T F
F T T T T T T F
F F F T F F T F
31

p q (p → q) ↔ (q ∨ ¬p)
T T T T T F
T F F T F F
F T T T T T
F F T T T T
p q (p ↔ q) ↔

(q ∨ ¬p) ∧(p ∨ ¬q)

T T T T T T T F
T F F T F F T T
F T F T T F F F
F F T T T T T T
32

Thus we can write
p ⊕ q ⇐⇒ (p ∨ q) ∧ ¬(p ∧ q) ,
p → q ⇐⇒ q ∨ ¬p ,
p ↔ q ⇐⇒ (q ∨ ¬p) ∧ (p ∨ ¬q) .
It also follows that
p ↔ q ⇐⇒ (p → q) ∧ (q → p) .
33

Basic logical equivalences.
The fundamental logical equivalences (“laws”) are :
p ∨ q ⇐⇒ q ∨ p p ∧ q ⇐⇒ q ∧ p commutative law
p ∨ (q ∧ r) ⇐⇒ p ∧ (q ∨ r) ⇐⇒
(p ∨ q) ∧ (p ∨ r) (p ∧ q) ∨ (p ∧ r) distributive law
p ∨ F ⇐⇒ p p ∧ T ⇐⇒ p identity law
p ∨ ¬p ⇐⇒ T p ∧ ¬p ⇐⇒ F complement law
34

Some useful additional laws are :
¬T ⇐⇒ F ¬F ⇐⇒ T negation law
p ∨ p ⇐⇒ p p ∧ p ⇐⇒ p idempotent law
p ∨ T ⇐⇒ T p ∧ F ⇐⇒ F domination law
p ∨ (p ∧ q) ⇐⇒ p p ∧ (p ∨ q) ⇐⇒ p absorption law
NOTE : Remember the absorption laws : they can be very useful !
35

Some more laws are :
(p ∨ q) ∨ r ⇐⇒ p ∨ (q ∨ r) (p ∧ q) ∧ r ⇐⇒ p ∧ (q ∧ r) associative law
¬(p ∨ q) ⇐⇒ ¬p ∧ ¬q ¬(p ∧ q) ⇐⇒ ¬p ∨ ¬q de Morgan
¬(¬p) ⇐⇒ p double negation
p → q ⇐⇒ ¬q → ¬p contrapositive
36

• All laws of logic can in principle be proved using truth tables.
• To illustrate the axiomatic nature of the fundamental laws,
we prove an additional law using only the fundamental laws.
• At every step the fundamental law used will be indicated.
• Once proved, additional laws may be used in further proofs.
37

EXAMPLE : Proof of the idempotent law
p ∨ p ⇐⇒ p
p ⇐⇒ p ∨ F identity law
⇐⇒ p ∨ (p ∧ ¬p) complement law
⇐⇒ (p ∨ p) ∧ (p ∨ ¬p) distributive law
⇐⇒ (p ∨ p) ∧ T complement law
⇐⇒ p ∨ p identity law
38

NOTE :
• Proving additional laws, using only the fundamental laws, is not as
easy as one might expect, because we have very few tools available.
• However after proving some of these additional equivalences we
have a more powerful set of laws.
39

Simplification of logical expressions.
It is useful to simplify logical expressions as much as possible.
This is much like in arithmetic where, for example, the expression
x3
+ 3x2
y + 3xy2
+ y3
is equivalent to
(x + y)3
.
40

EXAMPLE : Reconsider the logical expression
¬

p ∨

q → (p ∧ q)

∧

q ∨ (p ∧ q)

from the “crime example”.
It is equivalent to the much simpler logical expression
¬p ∧ q ,
because it has the same truth table values :
p q ¬p ¬p ∧ q
T T F F
T F F F
F T T T
F F T F
41

One way to simplify a logical expression is by using
• the fundamental laws of logic,
• known additional laws,
• the definitions of the additional logical operators.
For the “crime example” this can be done as follows :
42

¬

p ∨

q → (p ∧ q)

∧

q ∨ (p ∧ q)

⇐⇒ ¬

p ∨

q → (p ∧ q)

∧

q ∨ (q ∧ p)

commutative law
⇐⇒ ¬

p ∨

q → (p ∧ q)

∧ q absorption law
⇐⇒ ¬

p ∨

(p ∧ q) ∨ ¬q

∧ q equivalence of →
⇐⇒ ¬

p ∨ (p ∧ q)

∨ ¬q

∧ q associative law
⇐⇒ ¬

p ∨ ¬q

∧ q absorption law
⇐⇒ · · ·
43

¬

p ∨ ¬q

∧ q
⇐⇒

¬p ∧ ¬¬q

∧ q de Morgan
⇐⇒

¬p ∧ q

∧ q double negation
⇐⇒ ¬p ∧

q ∧ q

associative law
⇐⇒ ¬p ∧ q idempotent law
44

EXERCISE :
Use logical equivalences to simplify the logical expression
(p ∧ (¬(¬p ∨ q))) ∨ (p ∧ q) .
(This example was already considered before, using a truth table.)
EXERCISE :
Use logical equivalences to verify the following equivalence:
(¬p ∧ q) ∨ (¬q ∧ p) ⇐⇒ (p ∨ q) ∧ ¬(p ∧ q) .
(This was considered before in connection with the ⊕ operator.)
45

EXERCISE :
Use logical equivalences to show that the logical expression

(p → q) ∧ (¬p → r) ∧ (q → r)

→ r ,
is a tautology, i.e., show that

(p → q) ∧ (¬p → r) ∧ (q → r)

⇒ r .
NOTE : Earlier we proved this by truth table and by contradiction.
46

Predicate Calculus.
Let S be a set.
A predicate P is a function from S to {T , F } :
P : S −→ {T , F } ,
or
P : S × S × · · · × S −→ {T , F } .
or
P : S1 × S2 × · · · × Sn −→ {T , F } .
47

EXAMPLE : Let Z+
denote the set of all positive integers.
Define
P : Z+
−→ {T , F }
by
P(x) = T if x ∈ Z+
is even ,
P(x) = F if x ∈ Z+
is odd .
We can think of P(x) as the statement
“x is an even integer”,
which can be either True or False .
• What are the values of P(12), P(37), P(-3) ?
48

EXAMPLE :
Let U be a set and S a subset of U.
Let P(x) denote the statement “x ∈ S”, i.e.,
P(x) ⇐⇒ x ∈ S .
Then
P : U −→ {T , F } .
49

EXAMPLE :
Let P(x, y) denote the statement “x + y = 5”, i.e.,
P(x, y) ⇐⇒ x + y = 5 .
Then we can think of P as a function
P : Z+
× Z+
−→ {T , F } .
50

Quantifiers.
For a more compact notation we introduce the quantifiers
∀ , ∃ , and ∃!
DEFINITIONS : Let S be a set and P a predicate, P : S −→ {T , F } .
Then we define :
∀x ∈ S P(x) means “ P(x) = T for all x ∈ S ”.
∃x ∈ S P(x) means “ there exists an x ∈ S for which P(x) = T ”.
∃!x ∈ S P(x) means “ there is a unique x ∈ S for which P(x) = T ”.
51

If it is clear from the context what S is, then one often simply writes
∀x P(x) , ∃x P(x) , ∃!x P(x) .
If S has a finite number of elements then
∀x P(x) ⇐⇒ P(x1) ∧ P(x2) ∧ · · · ∧ P(xn)
∃x P(x) ⇐⇒ P(x1) ∨ P(x2) ∨ · · · ∨ P(xn)
and
∃!x P(x) ⇐⇒

P(x1) ∧ ¬P(x2) ∧ ¬P(x3) ∧ ¬ · · · ∧ ¬P(xn)

∨

¬P(x1) ∧ P(x2) ∧ ¬P(x3) ∧ ¬ · · · ∧ ¬P(xn)

∨ · · ·
∨

¬P(x1) ∧ ¬P(x2) ∧ ¬P(x3) ∧ ¬ · · · ∧ P(xn)

52

Let S be a set and
P : S × S −→ {T , F } .
Then
∀x, y P(x, y) and ∀x∀y P(x, y)
both mean
∀x

∀yP(x, y)

.
Thus ∀x, y P(x, y) means
“P(x, y) is True for any choice of x and y ”.
53

Similarly,
∃x, y P(x, y) and ∃x∃y P(x, y)
both mean
∃x

∃yP(x, y)

.
Thus ∃x, y P(x, y) means
“There exist an x and y for which P(x, y) is True ”.
54

EXAMPLE :
Let
P : S × S −→ {T , F } ,
where
S = {1, 2} .
Then
∀x, y P(x, y) ⇐⇒ P(1, 1) ∧ P(1, 2) ∧ P(2, 1) ∧ P(2, 2) ,
while
∃x, y P(x, y) ⇐⇒ P(1, 1) ∨ P(1, 2) ∨ P(2, 1) ∨ P(2, 2) .
55

EXERCISE : Let
P : Z × Z −→ {T , F } ,
where P(x, y) denotes
“ x + y = 5 ”.
What are the values of the following propositions ?
∀x, y P(x, y)
∃x, y P(x, y)
∀x ∃!y P(x, y)
∃x ∀y P(x, y)
56

EXERCISE : Let
P : Z+
× Z+
× Z+
−→ {T , F } ,
where P(x, y, z) denotes the statement
“ x2
+ y2
= z ”.
What are the values of the following propositions?
∃x, y, z P(x, y, z)
∀x, y, z P(x, y, z)
∀x, y ∃z P(x, y, z)
∀x, z ∃y P(x, y, z)
∀z ∃x, y P(x, y, z)
57

EXAMPLE :
Let
P, Q : S −→ {T , F } ,
where
S = {1, 2} .
Then
∀x

P(x) ∨ Q(x)

⇐⇒

P(1) ∨ Q(1)

∧

P(2) ∨ Q(2)

,
while
∀x, y

P(x) ∨ Q(y)

⇐⇒

P(1)∨Q(1)

∧

P(1)∨Q(2)

∧

P(2)∨Q(1)

∧

P(2)∨Q(2)

.
58

EXERCISE : ( see preceding example · · · ):
Show that
∀x

P(x) ∨ Q(x)

is not equivalent to
∀x, y

P(x) ∨ Q(y)

Hint: Take S = {1, 2} , and find predicates P and Q so that one of the
propositions is True and the other one False .
59

EXAMPLE : If
P, Q : S −→ {T , F } ,
where
S = {1, 2} ,
then
∃x

P(x) ∧ Q(x)

⇐⇒

P(1) ∧ Q(1)

∨

P(2) ∧ Q(2)

.
EXAMPLE : If again
P, Q : S −→ {T , F } ,
and
S = {1, 2} ,
then
∀x

P(x) → Q(x)

⇐⇒

P(1) → Q(1)

∧

P(2) → Q(2)

⇐⇒

¬P(1) ∨ Q(1)

∧

¬P(2) ∨ Q(2)

.
60

EXAMPLE :

∀xP(x)

∨

∀xQ(x)

6
⇐⇒ ∀x

P(x) ∨ Q(x)

i.e., there are predicates P and Q for which the equivalence not valid.
As a counterexample take
P, Q : Z+
−→ {T , F } ,
where
P(x) ⇐⇒ “x is even” , and Q(x) ⇐⇒ “x is odd” .
Then the RHS is True but the LHS is False .
61

EXERCISE :
Show that

∃xP(x)

∧

∃xQ(x)

6
⇐⇒ ∃x

P(x) ∧ Q(x)

by giving an example where LHS and RHS have a different logical value.
62

Some equivalences (Valid for any propositions P and Q) :
(1) ¬

∃xP(x)

⇐⇒ ∀x

¬P(x)

(2)

∀xP(x)

∧

∀xQ(x)

⇐⇒ ∀x

P(x) ∧ Q(x)

(3)

∀xP(x)

∧

∀xQ(x)

⇐⇒ ∀x∀y

P(x) ∧ Q(y)

(4)

∀xP(x)

∨

∀xQ(x)

⇐⇒ ∀x∀y

P(x) ∨ Q(y)

63

EXAMPLE : Proof of Equivalence (1) when the set S is finite .
¬

∃xP(x)

⇐⇒ ¬

P(x1) ∨ P(x2) ∨ · · · ∨ P(xn)

⇐⇒ ¬

P(x1) ∨

P(x2) ∨ · · · ∨ P(xn)

⇐⇒ ¬P(x1) ∧ ¬

P(x2) ∨ · · · ∨ P(xn)

⇐⇒ · · ·
⇐⇒ ¬P(x1) ∧ ¬P(x2) ∧ ¬ · · · ∧ ¬P(xn)
⇐⇒ ∀x¬P(x)
64

EXERCISES :
These equivalences are easily seen to be valid:
• Prove Equivalence 2 :

∀xP(x)

∧

∀xQ(x)

⇐⇒ ∀x

P(x) ∧ Q(x)

• Prove Equivalence 3 :

∀xP(x)

∧

∀xQ(x)

⇐⇒ ∀x∀y

P(x) ∧ Q(y)

65

EXAMPLE : Proof of Equivalence (4) .

∀xP(x)

∨

∀xQ(x)

⇐⇒ ∀x∀y

P(x) ∨ Q(y)

NOTE : A correct proof consist of verifying that
LHS ⇒ RHS and RHS ⇒ LHS
or equivalently
LHS ⇒ RHS and ¬ LHS ⇒ ¬ RHS
66

PROOF :
(i)

∀xP(x)

∨

∀xQ(x)

⇒ ∀x∀y

P(x) ∨ Q(y)

is easily seen to be True by a direct proof (with 2 cases).
(ii) ¬
h
∀xP(x)

∨

∀xQ(x)
i
⇒ ¬∀x∀y

P(x) ∨ Q(y)

can be rewritten as

∃x¬P(x)

∧

∃x¬Q(x)

⇒ ∃x∃y

¬P(x) ∧ ¬Q(y)

P and Q being arbitrary, we may replace them by ¬P and ¬Q :

∃xP(x)

∧

∃xQ(x)

⇒ ∃x∃y

P(x) ∧ Q(y)

which is easily seen to be True by a direct proof.
QED !
67

EXERCISE :
Prove that the equivalence

∀xP(x)

∧

∃xQ(x)

⇐⇒ ∀x∃y

P(x) ∧ Q(y)

is valid for all propositions P and Q.
Hint : This proof can be done along the lines of the preceding proof.
68

More equivalences (Valid for any propositions P and Q) :
(5) ¬

∀xP(x)

⇐⇒ ∃x¬P(x)
(6)

∃xP(x)

∨

∃xQ(x)

⇐⇒ ∃x

P(x) ∨ Q(x)

(7)

∃xP(x)

∨

∃xQ(x)

⇐⇒ ∃x∃y

P(x) ∨ Q(y)

(8)

∃xP(x)

∧

∃xQ(x)

⇐⇒ ∃x∃y

P(x) ∧ Q(y)

69

Equivalences (5)-(8) follow from
• negating equivalences (1)-(4), and
• replacing P by ¬P and Q by ¬Q ..
70

EXAMPLE : Proof of Equivalence (6), using Equivalence (2)
(2)

∀xP(x)

∧

∀xQ(x)

⇐⇒ ∀x

P(x) ∧ Q(x)

¬

∀xP(x)

∧

∀xQ(x)

⇐⇒ ¬∀x

P(x) ∧ Q(x)

∃x¬P(x)

∨

∃x¬Q(x)

⇐⇒ ∃x

¬P(x) ∨ ¬Q(x)

(6)

∃xP(x)

∨

∃xQ(x)

⇐⇒ ∃x

P(x) ∨ Q(x)

71

EXERCISE :
• Prove Equivalence 7 using Equivalence 3.
• Prove Equivalence 8 using Equivalence 4.
72

EXAMPLE (of negating a logical expression) :
¬∃x∀y∀zP(x, y, z) ⇐⇒ ∀x¬

∀y

∀zP(x, y, z)

⇐⇒ ∀x∃y¬

∀zP(x, y, z)

⇐⇒ ∀x∃y∃z¬P(x, y, z)
73

EXAMPLE (of transforming a logical expression) :
∃x

P(x) → Q(x)

⇐⇒ ∃x

¬P(x) ∨ Q(x)

⇐⇒

∃x¬P(x)

∨

∃xQ(x)

⋆
⇐⇒

¬∀xP(x)

∨

∃xQ(x)

⇐⇒

∀xP(x)

→

∃xQ(x)

⋆ This step follows from the earlier Equivalence 6.
74

EXAMPLE : (from Rosen’s book: in detail)
Let
D(x) denote the statement “x is a duck”
P(x) ,, “x is one of my poultry”
O(x) ,, “x is an officer”
W(x) ,, “x is willing to waltz”
75

Statement logic equivalent
Ducks never waltz ¬∃x

D(x) ∧ W(x)

∀x

(D(x) → ¬W(x)

Officers always waltz ¬∃x

(O(x) ∧ ¬W(x)

∀x

(O(x) → W(x)

All my poultry are ducks ∀x

(D(x) ∨ ¬P(x)

∀x

(P(x) → D(x)

My poultry are not officers ∀x¬

O(x) ∧ P(x)

∀x

(P(x) → ¬O(x)

QUESTION : Do the first three statements imply the last one?
76

Do the first three statements imply the last one, i.e., is
h
∀x(D(x) → ¬W(x))

∧

∀x(O(x) → W(x))

∧

∀x(P(x) → D(x))
i
→

∀x(P(x) → ¬O(x))

a tautology, i.e., do we have
h
∀x(D(x) → ¬W(x))

∧

∀x(O(x) → W(x))

∧

∀x(P(x) → D(x))
i
⇒

∀x(P(x) → ¬O(x))

The answer is YES .
In fact, it is a tautology for any predicates D, O, P, W.
77

h
∀x(D(x) → ¬W(x))

∧

∀x(O(x) → W(x))

∧

∀x(P(x) → D(x))
i
(1) (2) (3)
⇒

∀x(P(x) → ¬O(x))

(4)
To prove this, we must show that :
if (1) , (2) , and (3) are True then (4) is True .
To show (4) is True , we must show:
If, for arbitrary z, P(z) is True then, using (1,2,3), ¬O(z) is True .
78

h
∀x(D(x) → ¬W(x))

∧

∀x(O(x) → W(x))

∧

∀x(P(x) → D(x))
i
(1) (2) (3)
⇒

∀x(P(x) → ¬O(x))

(4)
PROOF : Let z be an arbitrary element from our set of objects.
Assume P(z) is True .
We must show that ¬O(z) is True , i.e., O(z) is False .
From (3) it follows D(z) is True .
From (1) follows ¬W(z) is True . i.e., W(z) is False .
From (2), since W(z) is False , it follows O(z) is False . QED !
79

EXERCISE (from Rosen):
Express each of the following using predicates and quantifiers:
(1) All clear explanations are satisfactory.
(2) Some excuses are unsatisfactory.
(3) Some excuses are not clear explanations.
Question : Does (3) follow from (1) and (2) ?
Hint : See preceding example.
80

REVIEW EXERCISES.
Problem 1. By truth table check if the ⊕ operator is associative:
(p ⊕ q) ⊕ r ⇐⇒ p ⊕ (q ⊕ r) .
Problem 2. Use logical equivalences to prove that
p → (q → r) ⇐⇒ (p ∧ q) → r .
Problem 3. By contradiction show that the following is a tautology:
[(p ∨ t) ∧ (p → q) ∧ (q → r) ∧ (r → s) ∧ (s → t)] → t .
Problem 4. Use logical equivalences to simplify
([(p ∧ q) ↔ p] → p) → (p → q) .
81

Problem 5. Verify the following basic tautologies, which are known as
“laws of inference”, and which are useful in proofs:
Tautology Name
(p ∧ (p → q)) ⇒ q modus ponens
(¬q ∧ (p → q)) ⇒ ¬p modus tollens
((p → q) ∧ (q → r)) ⇒ (p → r) hypothetical syllogism
((p ∨ q) ∧ ¬p) ⇒ q disjunctive syllogism
((p ∨ q) ∧ (¬p ∨ r)) ⇒ (q ∨ r) resolution
82

Problem 6. Express the following statements in predicate logic:
(a) “There is a unique x for which P(x) is True .”
(b) “There is no greatest integer.”
(c) “x0 is the smallest integer for which P(x) is True .”
(d) “Every person has exactly two parents.”
NOTE :
- Let P(x, y) denote “y is a parent of x ” .
- You may use the predicates x ≤ y , x y , and x 6= y .
83

Problem 7.
Let P(x, y, z) denote the statement
x2
y = z ,
where the universe of discourse of all three variables is the set Z.
What is the truth value of each of the following?
P(1, 1, 1) ∀y, z∃xP(x, y, z)
P(0, 7, 0) ∃!y, z∀xP(x, y, z)
∀x, y, zP(x, y, z) ∀x, y∃zP(x, y, z)
∃x, y, zP(x, y, z) ∀x∃y, zP(x, y, z)
84

Problem 8.
Let P(x, y, z, n) denote the statement
xn
+ yn
= zn
,
where x, y, z, n ∈ Z+
.
What is the truth value of each of the following:
P(1, 1, 2, 1) ∀x, y∃!zP(x, y, z, 1)
P(3, 4, 5, 2) ∀z∃x, yP(x, y, z, 1)
P(7, 24, 25, 2) ∃x, y∀zP(x, y, z, 2)
∃!x, y, zP(x, y, z, 2) ∃x, y, zP(x, y, z, 3)
NOTE : One of the above is very difficult!
85

Problem 9.
Give an example that shows that
∀x∃yP(x, y) 6
⇐⇒ ∃y∀xP(x, y) .
Problem 10.
Prove that
∀x[P(x) → Q(x)] ⇒ [∀xP(x) → ∀xQ(x)].
Problem 11.
Prove that
∀x∃y(P(x) ∨ Q(y)) ⇐⇒ ∀xP(x) ∨ ∃xQ(x) .
86

MATHEMATICAL PROOFS.
• We will illustrate some often used basic proof techniques .
(Some of these techniques we have already seen · · · )
• Several examples will be taken from elementary Number Theory.
87

DEFINITIONS : Let n, m ∈ Z+
.
• We call n odd if ∃k ∈ Z : k ≥ 0, n = 2k + 1.
• We call n even if n is not odd. (Then n = 2k for some k ∈ Z+
.)
• We say “m divides n”, and write m|n, if n = qm for some q ∈ Z+
.
• In this case we call m a divisor of n.
• If m|n then we also say that “n is divisible by m”.
• n (n ≥ 2) is a prime number if its only positive divisors are 1 and n.
• n (n ≥ 2) composite if it is not prime.
• n and m are relatively prime if 1 is their only common divisor.
88

Direct proofs.
Many mathematical statements have the form
“ if P then Q ”
i.e.,
P ⇒ Q ,
or, more often,
∀x

P(x) → Q(x)

,
where P and Q represent specific predicates .
RECALL : a direct proof consists of
• assuming that, for arbitrary x, P(x) is True
• demonstrating that Q(x) is then necessarily True also,
89

PROPOSITION : If n ∈ Z+
is odd then n2
is odd.
REMARK :
This proposition is of the form P ⇒ Q or, more specifically,
∀n ∈ Z+
: P(n) → Q(n) ,
where P and Q are predicates (functions)
P, Q : Z+
−→ {T , F } ,
namely,
P(n) ⇐⇒ ”n is odd ” , Q(n) ⇐⇒ ”n2
is odd ” .
NOTE : Actually Q(n) ⇐⇒ P(n2
) here !
90

If n ∈ Z+
is odd then n2
is odd.
PROOF :
Assume n ∈ Z+
is odd (i.e., assume P(n) = T ).
Then, by definition, n = 2k + 1 for some k ∈ Z, k ≥ 0.
By computation we find
n2
= (2k + 1)2
= 2(2k2
+ 2k) + 1 = 2m + 1 ,
where we have defined m ≡ 2k2
+ 2k.
Thus, by definition, n2
is odd, i.e., Q(n) = T . QED !
91

PROPOSITION : If n ∈ Z+
is odd then 8 | (n − 1)(n + 1) .
PROOF : If n ∈ Z+
is odd then we can write
n = 2k + 1 for some integer k, k ≥ 0 .
By computation we find
(n − 1)(n + 1) = n2
− 1 = 4k2
+ 4k = 4k(k + 1) .
Clearly
4 | 4k(k + 1) .
Note, however, that either k is even or k + 1 is even, i.e.,
2 | k or 2 | (k + 1) .
Thus
8 | 4k(k + 1) . QED !
92

LEMMA (Needed in the following example · · ·) For x ∈ R , x 6= 0, 1 :
n
X
k=0
xk
=
1 − xn+1
1 − x
, ∀n ≥ 0 , ( Geometric sum ) .
PROOF ( a ”constructive proof” ) :
Let
Sn =
n
X
k=0
xk
.
Then
Sn = 1 + x + x2
+ · · · + xn−1
+ xn
,
x · Sn = x + x2
+ · · · + xn−1
+ xn
+ xn+1
,
so that
Sn − x · Sn = (1 − x) · Sn = 1 − xn+1
,
from which the formula follows. QED !
93

DEFINITION :
A perfect number is a number that equals the sum of all of its divisors,
except the number itself.
EXAMPLES :
6 is perfect :
6 = 3 + 2 + 1 ,
and 28 is perfect :
28 = 14 + 7 + 4 + 2 + 1 ,
and so is 496 :
496 = 248 + 124 + 62 + 31 + 16 + 8 + 4 + 2 + 1 .
94

PROPOSITION : Let m ∈ Z+
, m 1.
If 2m
− 1 is prime, then n ≡ 2m−1
(2m
− 1) is perfect,
or, using quantifiers,
∀m ∈ Z+
: 2m
− 1 is prime → 2m−1
(2m
− 1) is perfect .
PROOF : Assume 2m
− 1 is prime.
Then the divisors of n = 2m−1
(2m
− 1) are
1, 2, 22
, 23
, · · · , 2m−1
,
and
(2m
− 1), 2(2m
− 1), 22
(2m
− 1), · · · , 2m−2
(2m
− 1), 2m−1
(2m
− 1) .
The last divisor is equal to n , so we do not include it in the sum.
95

The sum is then
m−1
X
k=0
2k
+ (2m
− 1)
m−2
X
k=0
2k
=
1 − 2m
1 − 2
+ (2m
− 1)
1 − 2m−1
1 − 2
= (2m
− 1) + (2m
− 1)(2m−1
− 1)
= (2m
− 1) (1 + 2m−1
− 1)
= (2m
− 1) 2m−1
= n. QED !
NOTE : We used the formula
m
X
k=0
xk
=
1 − xm+1
1 − x
, (the geometric sum) ,
(valid for x 6= 0, 1).
96

Proving the contrapositive.
It is easy to see (by Truth Table) that
p → q ⇐⇒ ¬q → ¬p .
EXAMPLE :
The statement
“n2
even ⇒ n even”,
proved earlier is equivalent to
“¬(n even) ⇒ ¬(n2
even)”,
i.e., it is equivalent to
n odd ⇒ n2
odd .
97

This equivalence justifies the following :
If we must prove
P ⇒ Q ,
then we may equivalently prove the contrapositive
¬Q ⇒ ¬P .
(Proving the contrapositive is sometimes easier .)
98

PROPOSITION : Let n ∈ Z+
, with n ≥ 2.
If the sum of the divisors of n is equal to n + 1 then n is prime.
PROOF : We prove the contrapositive :
If n is not prime then the sum of the divisors can not equal n + 1.
So suppose that n is not prime.
Then n has divisors
1, n, and m, for some m ∈ Z+
, m 6= 1, m 6= n ,
and possibly more.
Thus the sum of the divisors is greater than n + 1. QED !
99

Some specific contrapositives.
(p ∧ q) → r ⇐⇒ ¬r → (¬p ∨ ¬q)
(p ∨ q) → r ⇐⇒ ¬r → (¬p ∧ ¬q)

∀xP(x)

→ q ⇐⇒ ¬q →

∃x¬P(x)

∃xP(x)

→ q ⇐⇒ ¬q →

∀x¬P(x)

p → ∀xQ(x) ⇐⇒

∃x¬Q(x)

→ ¬p
p → ∃xQ(x) ⇐⇒

∀x¬Q(x)

→ ¬p
100

, with n ≥ 2.
∀a, b ∈ Z+
( n|a ∨ n|b ∨ n 6 |ab ) ⇒ n is prime .
PROOF : The contrapositive is
If n is not prime then ∃a, b ( n 6 |a ∧ n 6 |b ∧ n|ab ) .
Here the contrapositive is easier to understand and quite easy to prove :
Note that if n is not prime then
n = a b ,
for certain integers a and b, both greater than 1 and less than n.
Clearly n 6 |a , n 6 |b , and n|ab . QED !
101

. Then
5|n2
⇒ 5|n ,
PROOF : We prove the contrapositive , i.e.,
5 6 |n ⇒ 5 6 |n2
.
So suppose 5 6 |n.
Then we have the following cases :
n = 5k + 1 ⇒ n2
= 25k2
+ 10k + 1 = 5(5k2
+ 2k) + 1 ,
n = 5k + 2 ⇒ n2
= 25k2
+ 20k + 4 = 5(5k2
+ 4k) + 4 ,
n = 5k + 3 ⇒ n2
= 25k2
+ 30k + 9 = 5(5k2
+ 6k + 1) + 4 ,
n = 5k + 4 ⇒ n2
= 25k2
+ 40k + 16 = 5(5k2
+ 8k + 3) + 1 ,
for k ∈ Z, k ≥ 0.
This shows that 5 6 |n2
. QED !
102

Proof by contradiction.
To prove a statement P ⇒ Q by contradiction :
• assume P = T and Q = F ,
• show that these assumptions lead to an impossible conclusion
(a “contradiction”).
(We have already seen some proofs by contradiction.)
103

PROPOSITION :
If a prime number is the sum of two prime numbers
then one of these equals 2.
PROOF :
Let p1, p2, and p be prime numbers, with p1 + p2 = p.
Suppose that neither p1 nor p2 is equal to 2.
Then both p1 and p2 must be odd ( and greater than 2 ) .
Hence p = p1 + p2 is even, and greater than 2.
This contradicts that p is prime. QED !
104

PROPOSITION :
√
2 is irrational, i.e., if m, n ∈ Z+
then m
n
6=
√
2.
PROOF : Suppose m, n ∈ Z+
and m
n
=
√
2.
We may assume m and n are relatively prime (cancel common factors).
Then m =
√
2 n ⇒ m2
= 2n2
*
⇒ m2
even
⇒ m even (proved earlier)
⇒ ∃k ∈ Z+
(m = 2k)
⇒ 2n2
= m2
= (2k)2
= 4k2
(using * above)
⇒ n2
= 2k2
⇒ n2
even
⇒ n even
Thus both n and m are even and therefore both are divisible by two.
This contradicts that they are relatively prime. QED !
NOTATION : The “⇒” means that the immediately following state-
ment is implied by the preceding statement(s).
105

EXERCISE :
Use a proof by contradiction to show the following:
(p ∨ q) ∧ (p → r) ∧ (q → r) ⇒ r .
(p → q) ∧ (q → r) ⇒ p → r .
Hint : See a similar example earlier in the Lecture Notes.
106

PROPOSITION :
If the integers
1, 2, 3, · · · , 10,
are placed around a circle, in any order, then there exist three integers
in consecutive locations around the circle that have a sum greater than
or equal to 18.
107

PROOF : (by contradiction)
Suppose any three integers in consecutive locations around the circle
have sum less than 18, that is, less then or equal to 17.
Excluding the number 1, which must be placed somewhere, there remain
exactly three groups of three integers in consecutive locations.
The total sum is then less than or equal to 1 + 17 + 17 + 17 = 52.
However, we know that this sum must equal
1 + 2 + 3 + · · · + 10 = 55 .
Hence we have a contradiction. QED !
108

FACT : Any real number x can be uniquely written as
x = n + r ,
where
n ∈ Z and r ∈ R , with 0 ≤ r 1 .
DEFINITION : We then define the floor of x as
⌊x⌋ ≡ n .
EXAMPLES : ⌊7⌋ = 7 , ⌊−7⌋ = −7 , ⌊π⌋ = 3 , ⌊−π⌋ = −4 .
EXAMPLE : Use a proof by cases to show that
⌊2x⌋ = ⌊x⌋ + ⌊x +
1
2
⌋ .
110

⌊2x⌋ = ⌊x⌋ + ⌊x + 1
2
⌋ , x = n + r , 0 ≤ r 1
PROOF :
Case 1 : 0 ≤ r 1
2
: Then
0 ≤ 2r 1 and
1
2
≤ r +
1
2
1 .
LHS : ⌊2x⌋ = ⌊2n + 2r⌋ = 2n
RHS : ⌊x⌋ = ⌊n + r⌋ = n
⌊x +
1
2
⌋ = ⌊n + r +
1
2
⌋ = n
so that the identity is satisfied.
111

⌊2x⌋ = ⌊x⌋ + ⌊x + 1
2
⌋ , x = n + r , 0 ≤ r 1
PROOF : (continued · · · )
Case 2 : 1
2
≤ r 1 : Then
1 ≤ 2r 2 and 1 ≤ r +
1
2
1 +
1
2
.
LHS : ⌊2x⌋ = ⌊2n + 2r⌋ = 2n + 1
RHS : ⌊x⌋ = ⌊n + r⌋ = n
⌊x +
1
2
⌋ = ⌊n + r +
1
2
⌋ = n + 1
so that the identity is satisfied. QED !
112

Existence proofs.
• Mathematical problems often concern the existence, or non-existence,
of certain objects.
• Such problems may arise from the mathematical formulation of
problems that arise in many scientific areas.
• A proof that establishes the existence of a certain object is called
an existence proof.
113

EXAMPLE : For any positive integer n there exists a sequence
of n consecutive composite integers , i.e.,
∀n ∈ Z+
∃m ∈ Z+
: m + i is composite , i = 1, · · · , n .
For example,
n = 2 : (8,9) are 2 consecutive composite integers (m = 7),
n = 3 : (8,9,10) are 3 three consecutive composite integers (m = 7),
n = 4 : (24,25,26,27) are 4 consecutive composite integers (m = 23),
n = 5 : (32,33,34,35,36) are 5 consecutive composite integers (m = 31).
114

∀n ∈ Z+
∃m ∈ Z+
: m + i is composite , i = 1, · · · , n .
PROOF :
Let m = (n + 1)! + 1.
Then, clearly, the n consecutive integers
(n + 1)! + 1 + 1, (n + 1)! + 1 + 2, · · · , (n + 1)! + 1 + n ,
are composite. (Why ?) QED !
NOTE : This is a constructive existence proof : We demonstrated
the existence of m by showing its value (as a function of n).
115

PROPOSITION : There are infinitely many prime numbers.
The idea of the proof (by contradiction) :
• Assume there is only a finite number of prime numbers.
• Then we’ll show ∃N 1 ∈ Z+
that is neither prime nor composite .
• But this is impossible !
• Thus there must be infinitely many prime numbers !
116

PROOF :
Suppose the total number of primes is finite , say,
p1, p2, · · · , pn .
Let
N = p1p2 · · · pn + 1
• Then N cannot be prime. (Why not ?)
• Also, none of the p1, p2, · · · , pn divide N,
since N divided by pi gives a remainder of 1 , (i = 1, · · · , n) .
• Thus N cannot be composite either ! QED !
117

NOTE :
• This proof is a non-constructive existence proof.
• We proved the existence of an infinite number of primes
without actually showing them !
118

NOTE :
• There is no general recipe for proving a mathematical statement
and often there is more than one correct proof.
• One generally tries to make a proof as simple as possible,
so that others may understand it more easily.
• Nevertheless, proofs can be very difficult, even for relatively simple
statements such as Fermat’s Last Theorem :
¬∃x, y, z, n ∈ Z+
, n ≥ 3 : xn
+ yn
= zn
.
119

NOTE :
• There are many mathematical statements that are thought to be
correct, but that have not yet been proved (“open problems ”),
e.g., the “Goldbach Conjecture ” :
“Every even integer greater than 2 is the sum of two prime numbers”.
• Indeed, proving mathematical results is as much an art as it is
a science, requiring creativity as much as clarity of thought.
• An essential first step is always to fully understand the problem.
• Where possible, experimentation with simple examples may help
build intuition and perhaps suggest a possible method of proof.
120

REVIEW EXERCISES.
Problem 1. Use a direct proof to show the following:
(p ∨ q) ∧ (q → r) ∧

(p ∧ s) → t

∧

¬q → (u ∧ s)

∧ ¬r ⇒ t .
(Assuming the left-hand-side is True , you must show that t is True .)
Problem 2. Let n be a positive integer.
Prove the following statement by proving its contrapositive:
”If n3
+ 2n + 1 is odd then n is even ”.
121

Problem 3. Let n be an integer. Show that
3|n2
⇒ 3|n ,
by proving its contrapositive.
Hint : There are two cases to consider.
Problem 4. Give a direct proof to show the following:
The sum of the squares of any two rational numbers is a rational number.
122

Problem 5.
Show that for all positive real x
if x is irrational then
√
x is irrational .
by proving the contrapositive .
Problem 6. Use a proof by contradiction to prove the following:
If the integers 1, 2, 3, · · · , 7, are placed around a circle, in any order,
then there exist two adjacent integers that have a sum greater than or
equal to 9 .
(Can you also give a direct Proof ?)
123

FACT : Any real number x can be uniquely written as
x = n − r ,
where
n ∈ Z and r ∈ R , with 0 ≤ r 1 .
DEFINITION : We then define the ceiling of x as
⌈x⌉ ≡ n .
EXAMPLES : ⌈7⌉ = 7 , ⌈−7⌉ = −7 , ⌈π⌉ = 4 , ⌈−π⌉ = −3 .
Problem 7.
Is the following equality valid for all positive integers n and m ?
⌊
n + m
2
⌋ = ⌈
n
2
⌉ + ⌊
m
2
⌋ .
If Yes then give a proof. If No then give a counterexample.
124

SET THEORY
Basic definitions.
• Let U be the collection of all objects under consideration.
(U is also called the “universe” of objects under consideration.)
• A set is a collection of objects from U.
125

Let A , B , and C be sets.
x ∈ A ⇐⇒ “x is an element (a member) of A”.
x /
∈ A ⇐⇒ ¬(x ∈ A)
A ⊆ B ⇐⇒ ∀x ∈ U : x ∈ A ⇒ x ∈ B subset
A = B ⇐⇒ (A ⊆ B) ∧ (B ⊆ A) equality
The above take values in {T , F }.
126

The following are set-valued :
A ∪ B ≡ { x ∈ U : (x ∈ A) ∨ (x ∈ B) } union
A ∩ B ≡ { x ∈ U : (x ∈ A) ∧ (x ∈ B) } intersection
Ā ≡ { x ∈ U : x /
∈ A } complement
A − B ≡ { x ∈ U : (x ∈ A) ∧ (x /
∈ B) } difference
127

Venn diagram.
This is a useful visual aid for proving set theoretic identities.
EXAMPLE : For the two sides of the identity
(A ∩ B) − C = A ∩ (B − C)
we have the following Venn diagrams :
B B
C
C
A
A
128

The actual proof of the identity, using the above definitions and the laws
of logic, is as follows :
x ∈

(A ∩ B) − C

⇐⇒ x ∈ (A ∩ B) ∧ x /
∈ C
⇐⇒ (x ∈ A ∧ x ∈ B) ∧ x /
∈ C
⇐⇒ x ∈ A ∧ (x ∈ B ∧ x /
∈ C) associative law
⇐⇒ x ∈ A ∧ x ∈ (B − C)
⇐⇒ x ∈ A ∩ (B − C)
129

EXAMPLE :
For the two sides of the identity
A − (B ∪ C) = (A − B) ∩ (A − C)
we have the following Venn diagrams :
B B
C
C
A
A
130

The actual proof of the identity is as follows :
x ∈

A − (B ∪ C)

⇐⇒ x ∈ A ∧ x /
∈ (B ∪ C)
⇐⇒ x ∈ A ∧ ¬(x ∈ (B ∪ C))
⇐⇒ x ∈ A ∧ ¬(x ∈ B ∨ x ∈ C)
⇐⇒ x ∈ A ∧ x /
∈ B ∧ x /
∈ C de Morgan
⇐⇒ x ∈ A ∧ x ∈ A ∧ x /
∈ B ∧ x /
∈ C idempotent law
⇐⇒ x ∈ A ∧ x /
∈ B ∧ x ∈ A ∧ x /
∈ C commut.+assoc.
⇐⇒ x ∈ (A − B) ∧ x ∈ (A − C)
⇐⇒ x ∈ (A − B) ∩ (A − C)
131

EXAMPLE : A ∩ B = A ∪ B ⇒ A = B
PROOF : ( a direct proof · · · )
Assume (A ∩ B) = (A ∪ B). We must show that A = B.
This is done in two stages :
(i) show A ⊆ B and (ii) show B ⊆ A .
To show (i) :
Let x ∈ A. We must show that x ∈ B.
Since x ∈ A it follows that x ∈ A ∪ B.
Since (A ∩ B) = (A ∪ B) it follows that x ∈ (A ∩ B).
Thus x ∈ B also.
The proof of (ii) proceeds along the same steps.
132

Subsets.
S ⊆ U means S is a subset of a universal set U .
The set of all subsets of U is denoted by 2U
or P(U) , the power set .
This name is suggested by the following fact :
If U has n elements then P(U) has 2n
elements (sets) .
133

EXAMPLE :
Let
U = {1 , 2 , 3} .
Then
P(U) =
n
{}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
o
.
We see that P(U) has 23
= 8 elements .
NOTE : The empty set ∅ = {} and U itself are included in P(U).
134

Basic set theoretic identities :
A ∪ B = B ∪ A A ∩ B = B ∩ A commutative laws
A ∪ (B ∩ C) = A ∩ (B ∪ C) =
(A ∪ B) ∩ (A ∪ C) (A ∩ B) ∪ (A ∩ C) distributive laws
A ∪ ∅ = A A ∩ U = A identity laws
A ∪ Ā = U A ∩ Ā = ∅ complement laws
135

Some additional identities :
Ū = ∅ ¯
∅ = U
A ∪ A = A A ∩ A = A idempotent laws
A ∪ U = U A ∩ ∅ = ∅ domination laws
A ∪ (A ∩ B) = A A ∩ (A ∪ B) = A absorption laws
136

Some more identities :
(A ∪ B) ∪ C = (A ∩ B) ∩ C =
A ∪ (B ∪ C) A ∩ (B ∩ C) associative law
A ∪ B = Ā ∩ B̄ A ∩ B = Ā ∪ B̄ de Morgan’s laws
¯
Ā = A involution law
All the preceding identities can be proved using the definitions of set
theory and the laws of logic.
Note the close correspondence of these identities to the laws of logic.
137

NOTE : One can also proceed axiomatically by only assuming :
• the existence of a power set P(U) , where U is a universal set ,
• special elements U and ∅ ,
• a unary operator ¯ , and two binary operators ∪ and ∩ ,
• the basic set theoretic identities .
Given this setup one can derive all other set theoretic identities.
Note the close correspondence between the above axiomatic setup and
the axiomatic setup of logic !
138

EXAMPLE :
Prove the idempotent law
A ∪ A = A
using only the basic set theoretic identities :
A = A ∪ ∅ identity law
A ∪ (A ∩ Ā) complement law
(A ∪ A) ∩ (A ∪ Ā) distributive law
(A ∪ A) ∩ U complement law
A ∪ A identity law
139

EXAMPLE :
( to illustrate the close relation between Set Theory and Logic · · · )
Using another approach we prove the absorption law :
A ∪ (A ∩ B) = A
Thus we must prove
∀x ∈ U : x ∈ A ∪ (A ∩ B) ⇐⇒ x ∈ A
∀x ∈ U : x ∈ A ∨ x ∈ A ∩ B ⇐⇒ x ∈ A
∀x ∈ U : x ∈ A ∨ (x ∈ A ∧ x ∈ B) ⇐⇒ x ∈ A
140

∀x ∈ U : x ∈ A ∨ (x ∈ A ∧ x ∈ B) ⇐⇒ x ∈ A
Define logical predicates a(x) and b(x) :
a(x) ⇐⇒ x ∈ A , b(x) ⇐⇒ x ∈ B .
Then we must prove
∀x ∈ U : a(x) ∨ (a(x) ∧ b(x)) ⇐⇒ a(x) .
It suffices to prove that, for arbitrary logical variables a and b ,
a ∨ (a ∧ b) ⇐⇒ a .
But this is the absorption law from logic !
141

REVIEW EXERCISES.
For each of the following, determine whether it is valid or invalid.
If valid then give a proof. If invalid then give a counterexample.
(1) A ∩ (B ∪ A) = A
(2) A ∪ (B ∩ C) = (A ∪ B) ∩ C
(3) (A ∩ B) ∪ (C ∩ D) = (A ∩ D) ∪ (C ∩ B)
(4) (A ∩ B) ∪ (A ∩ B̄) = A
(5) A ∪ ((B ∪ C) ∩ A) = A
(6) A − (B ∪ C) = (A − B) ∩ (A − C)
(7) B ∩ C ⊆ A ⇒ (B − A) ∩ (C − A) = ∅
(8) (A ∪ B) − (A ∩ B) = A ⇒ B = ∅
142

FUNCTIONS
DEFINITIONS : Let A and B be sets.
Then f is called
a function from A to B
if to each element of A it associates exactly one element of B .
We write
f : A −→ B
and we call A the domain of f and B the codomain of f .
We also define the range of f to be
f(A) ≡ {b ∈ B : b = f(a) for some a ∈ A} .
143

We say that f is :
one-to-one (or injective) iff ∀a1, a2 ∈ A : a1 6= a2 ⇒ f(a1) 6= f(a2)
iff ∀a1, a2 ∈ A : f(a1) = f(a2) ⇒ a1 = a2
onto (or surjective) iff ∀b ∈ B ∃a ∈ A : f(a) = b
iff f(A) = B
bijective iff f is one-to-one and onto
144

EXAMPLE :
Let
A = {a, b, c} , B = {1, 2} ,
and let f : A −→ B be defined by
f : a 7→ 1 , f : b 7→ 2 , f : c 7→ 1 .
Then f not one-to-one, but f is onto.
a
b
c
1
2
A B
145

EXAMPLE :
Let
A = {a, b} , B = {1, 2, 3} ,
and let f : A −→ B be defined by
f : a 7→ 1 , f : b 7→ 3 .
Then f is one-to-one but not onto.
a
1
b
2
3
A B
146

EXAMPLE :
Let A = B = Z+
, and let
f : Z+
−→ Z+
be defined by
f : n 7→
n(n + 1)
2
,
i.e.,
f(n) =
n(n + 1)
2
.
Then f is one-to-one but not onto.
147

f(n) ≡ n(n + 1)/2
PROOF : (1) f is not onto :
Here
f(Z+
) = { 1 , 3 , 6 , 10 , 15 , 21 , · · · } ,
so it seems that f is not onto.
To be precise, we show that f(n) can never be equal to 2 :
f(n) = 2 ⇐⇒ n(n + 1)/2 = 2 ⇐⇒ n2
+ n − 4 = 0 .
But this quadratic equation has no integer roots.
148

(2) f is one-to-one : Assume that f(n1) = f(n2).
We must show that n1 = n2 :
f(n1) = f(n2) ⇐⇒ n1(n1 + 1)/2 = n2(n2 + 1)/2
⇐⇒ n2
1 + n1 = n2
2 + n2
⇐⇒ n2
1 − n2
2 = − (n1 − n2)
⇐⇒ (n1 + n2)(n1 − n2) = − (n1 − n2)
⇐⇒ n1 = n2 or n1 + n2 = − 1
However n1, n2 ∈ Z+
. Thus n1 + n2 cannot be negative.
It follows that n1 = n2. QED !
149

EXAMPLE :
Define
f : Z × Z −→ Z × Z
or equivalently
f : Z2
−→ Z2
by
f(m, n) = (m + n , m − n) ,
or equivalently, in matrix multiplication notation
f :

m
n

7→

1 1
1 −1

m
n

.
Then f is one-to-one, but not onto.
150

PROOF :
(i) One-to-one :
Suppose
f(m1, n1) = f(m2, n2) .
Then
(m1 + n1 , m1 − n1) = (m2 + n2 , m2 − n2) ,
i.e.,
m1 + n1 = m2 + n2 ,
and
m1 − n1 = m2 − n2 .
Add and subtract the equations, and divide by 2 to find
m1 = m2 and n1 = n2 ,
that is,
(m1, n1) = (m2, n2) .
Thus f is one-to-one.
151

(ii) Not onto :
Let (s, d) ∈ Z2
be arbitrary. Can we solve
f(m, n) = (s, d) ,
i.e., can we solve
m + n = s ,
m − n = d ,
for m, n ∈ Z ?
Add and subtract the two equations, and divide by 2 to get
m =
s + d
2
and n =
s − d
2
.
However, m and n need not be integers, e.g., take s = 1, d = 2.
Thus f is not onto. QED !
152

Given two functions
f : A −→ B and g : B −→ C ,
we can compose them :
(g ◦ f)(a) ≡ g

f(a)

.
A B C
f g
f(a) g(f(a))
a
o
g f
153

EXAMPLE :
Let A = B = C = Z (all integers), and define f, g : Z −→ Z by
f(n) ≡ n2
+ 2n − 1 , g(n) ≡ 2n − 1 .
• Let h1(n) ≡ f

g(n)

. Then
h1(n) = f(2n − 1) = (2n − 1)2
+ 2(2n − 1) − 1 = 4n2
− 2 .
• Let h2(n) ≡ g

f(n)

. Then
h2(n) = g(n2
+ 2n − 1) = 2(n2
+ 2n − 1) − 1 = 2n2
+ 4n − 3 .
• Let h3(n) ≡ g

g(n)

. Then
h3(n) = g(2n − 1) = 2(2n − 1) − 1 = 4n − 3 .
154

Inverses. Let
f : A −→ B ,
and
g : B −→ A .
Then g is called the inverse of f if
∀a ∈ A : g

f(a)

= a ,
and
∀b ∈ B : f

g(b)

= b .
If f has an inverse g then we say
f is invertible ,
and we write f−1
for g .
155

EXAMPLE :
Let f : Z −→ Z be defined by f : n 7→ n − 1 , i.e.,
f(n) = n − 1 (”shift operator”) .
• f is one-to-one :
If f(n1) = f(n2) then n1 − 1 = n2 − 1 , i.e., n1 = n2 .
• f is onto :
Given any m ∈ Z , can we find n such that f(n) = m ?
That is, can we find n such that n − 1 = m ?
Easy: n = m + 1 !
156

f(n) = n − 1 , f : Z −→ Z
It follows that f is invertible, with inverse
f−1
(m) = m + 1 .
Check :
f

f−1
(m)

= f(m + 1) = (m + 1) − 1 = m ,
f−1

f(n)

= f−1
(n − 1) = (n − 1) + 1 = n .
157

EXAMPLE :
Let f : R −→ R be defined by f : x 7→ 1 − 2x , i.e.,
f(x) = 1 − 2x .
• f is one-to-one :
If f(x1) = f(x2) then 1 − 2x1 = 1 − 2x2 , i.e., x1 = x2 .
• f is onto :
Given any y ∈ R , can we find x such that f(x) = y ?
That is, can we find x such that 1 − 2x = y ?
Easy: x = (1 − y)/2 !
( We actually constructed the inverse in this step : f−1
(y) = 1−y
2
. )
158

f(x) = 1 − 2x , f : R −→ R
We found that f is invertible, with inverse
f−1
(y) =
1 − y
2
.
Check ( not really necessary · · · ) :
f

f−1
(y)

= f((1 − y)/2) = 1 − 2

(1 − y)/2

= y ,
f−1

f(x)

= f−1
(1 − 2x) =
1 − (1 − 2x)
2
= x .
NOTE : We constructed f−1
(y) by solving f(x) = y for x .
159

THEOREM :
f : A −→ B is invertible if and only if f is 1 − 1 and onto .
REMARK :
• It is not difficult to see that this theorem holds for finite sets.
• However, the proof also applies to infinite sets.
160

PROOF :
(1a) First we show that if f is invertible then f is 1 − 1 .
By contradiction: Suppose f is invertible but not 1 − 1.
Since f is not 1 − 1 there exist a1, a2 ∈ A, a1 6= a2, such that
f(a1) = f(a2) ≡ b0 .
Since f is invertible there is a function g : B −→ A such that
g

(f(a)

) = a, ∀a ∈ A .
In particular
g

f(a1)

= a1 , and g

f(a2)

= a2 ,
i.e.,
g(b0) = a1 , and g(b0) = a2 .
Thus g is not single-valued (not a function). Contradiction !
161

(1b) Now we show that if f is invertible then f is onto.
By contradiction: Suppose f is invertible but not onto.
Since f is not onto there exists b0 ∈ B such that
f(a) 6= b0, ∀a ∈ A .
Since f is invertible there is a function g : B −→ A such that
f

g(b)

= b, ∀b ∈ B .
In particular
f

g(b0)

= b0 , where g(b0) ∈ A .
But this contradicts that f(a) 6= b0, ∀a ∈ A .
162

(2a) Next we show that if f is 1 − 1 and onto then f is invertible.
Define a function g : B −→ A as follows :
Since f is 1 − 1 and onto we have that for any b ∈ B
b = f(a) for some unique a ∈ A .
For each such a ∈ A set
g(b) = a .
Then g : B −→ A , and by construction
f

g(b)

= f(a) = b .
163

(2b) We still must show that ∀a ∈ A : g

f(a)

= a .
By contradiction : Suppose g

f(a0)

6= a0 for some a0 ∈ A .
Define b0 = f(a0) . Then b0 ∈ B and
g(b0) 6= a0 ,
where both g(b0) and a0 lie in A .
Since f is one-to-one it follows that
f

g(b0)

6= f(a0) ,
i.e.,
f

g(b0)

6= b0 .
But this contradicts the result of (2a) ! QED !
164

EXAMPLE :
• Define f : Z+
−→ Z+
by
f(n) = n(n − 2)(n − 4) + 4 .
Then f is not one-to-one ; for example, f(2) = f(4) = 4 :
n 1 2 3 4 5 6 7 · · ·
f(n) 7 4 1 4 19 52 109 · · ·
Using calculus one can show that f(n) is increasing for n ≥ 3.
Thus f is not onto ; for example,
∀n ∈ Z+
: f(n) 6= 2 .
165

• Now let
S = f(Z+
) = {1, 4, 7, 19, 52, 109, · · ·} ,
and consider f as a function
f : Z+
−→ S .
Then f is onto, but still not one-to-one, since f(2) = f(4) = 4.
• Finally let
D = Z+
− {2} ,
and consider f as a function
f : D −→ S .
Now f is one-to-one and onto, and hence invertible.
166

EXAMPLE : The floor and ceiling functions.
FACT :
∀x ∈ R ∃ ! n ∈ Z and ∃ ! r ∈ R with 0 ≤ r 1 such that
x = n + r .
We already defined the floor function, ⌊·⌋ , as
⌊x⌋ = n .
EXAMPLES :
⌊π⌋ = 3 , ⌊e⌋ = 2 , ⌊3⌋ = 3 , ⌊−7/2⌋ = − 4 ,
where e = 2.71828 · · · .
167

FACT :
∀x ∈ R ∃ ! n ∈ Z and ∃ ! r ∈ R with 0 ≤ r 1 such that
x = n − r .
We already defined the ceiling function, ⌈·⌉ , as
⌈x⌉ = n .
EXAMPLES :
⌈π⌉ = 4 , ⌈e⌉ = 3 , ⌈3⌉ = 3 , ⌈−7/2⌉ = − 3 .
168

We see that
⌊·⌋ : R −→ Z ,
and
⌈·⌉ : R −→ Z .
EXERCISE :
• Is ⌊·⌋ one-to-one? onto? invertible?
• Is ⌈·⌉ one-to-one? onto? invertible?
• Draw the graphs of ⌊·⌋ and ⌈·⌉ .
169

EXAMPLE : Let p , k ∈ Z+
.
Then
⌈
p
k
⌉
p + k
k
.
PROOF :
By definition of the ceiling function we can write
p
k
= ⌈
p
k
⌉ − r ,
where 0 ≤ r 1 .
Hence
⌈
p
k
⌉ =
p
k
+ r
p
k
+ 1 =
p + k
k
. QED !
170

EXAMPLE : Show that the linear function
f : R2
−→ R2
defined by
f(x, y) = (x + y , x − y) ,
or, in matrix form,
f :

x
y

7→

1 1
1 −1

x
y

,
is one-to-one and onto.
• One-to-one : Exercise!
Hint : See the earlier example where this function was considered as
f : Z2
−→ Z2
.
171

• Onto :
f(x, y) = (x + y , x − y) or f :

x
y

7→

1 1
1 −1

x
y

We can construct the inverse by solving
f(x, y) = (s, d) ,
that is, by solving
x + y = s , x − y = d ,
for x, y ∈ R :
x =
s + d
2
, y =
s − d
2
.
Thus the inverse is
g(s, d) = (
s + d
2
,
s − d
2
) or g :

s
d

7→
1
2
1
2
1
2 −1
2

s
d

.
172

f(x, y) = (x + y , x − y) , g(s, d) = (s+d
2
, s−d
2
)
Check ( not really necessary · · · ) :
f(g(s, d)) = f(
s + d
2
,
s − d
2
)
= (
s + d
2
+
s − d
2
,
s + d
2
−
s − d
2
) = (s, d) ,
and
g(f(x, y)) = g(x + y , x − y)
= (
(x + y) + (x − y)
2
,
(x + y) − (x − y)
2
) = (x, y) .
173

EXAMPLE :
More generally, a function
f : R2
−→ R2
is linear if it can be written as
f(x, y) = ( a11 x + a12 y , a21 x + a22 y ) ,
or equivalently, as matrix-vector multiplication ,
f :

x
y

7→

a11 a12
a21 a22

x
y

,
where aij ∈ R , (i, j = 1, 2) .
174

f : R2
−→ R2
, f :

x
y

7→

a11 a12
a21 a22

x
y

This function is invertible if the determinant
D ≡ a11a22 − a12a21 6= 0 .
In this case the inverse is given by
f−1
:

s
d

7→
1
D

a22 −a12
−a21 a11

s
d

.
EXERCISE : Check that
f−1
(f(x, y)) = (x, y) and f(f−1
(s, d)) = (s, d) .
175

Now consider the same linear function
f :

n
m

7→

a11 a12
a21 a22

n
m

,
but with aij ∈ Z , (i, j = 1, 2) , and as a function
f : Z2
−→ Z2
.
Is
f−1
:

s
d

7→
1
D

a22 −a12
−a21 a11

s
d

.
still the inverse?
176

ANSWER : Not in general !
f is now invertible only if the determinant
• D = a11a22 − a12a21 6= 0,
and
• ∀i, j : D | aij .
In this case f−1
is still given by
f−1
:

s
d

7→
1
D

a22 −a12
−a21 a11

s
d

.
177

EXERCISE : Show that
f :

n
m

7→

3 2
4 3

n
m

,
is invertible as a function f : Z2
−→ Z2
.
What is the inverse?
EXERCISE : Show that
f :

n
m

7→

3 1
4 3

n
m

,
is not invertible as a function f : Z2
−→ Z2
.
178

REVIEW EXERCISES.
Problem 1.
Define
f : R −→ R
by
f(x) ≡



1/x if x 6= 0 ,
0 if x = 0 .
• Draw the graph of f .
• Is f one-to-one?
• Is f onto?
• What is f−1
?
179

Problem 2. Consider a function
f : A −→ B .
For each of the following, can you find a function f that is
(i) one-to-one (ii) onto (iii) one-to-one and onto ?
• A = {1, 2, 3} , B = {1, 2}
• A = {1, 2} , B = {1, 2, 3}
• A = {all even positive integers} , B = {all odd positive integers}
180

Problem 3.
Can you find a function f : Z −→ Z+
that is one-to-one and onto ?
Can you find a function g : Z+
−→ Z that is one-to-one and onto ?
Problem 4. Let Sn be a finite set of n elements.
Show that a function
f : Sn −→ Sn
is one-to-one if and only if it is onto.
181

Problem 5. Let f : A −→ B and g : B −→ C be functions.
• Suppose f and g are one-to-one.
Is the composition g ◦ f necessarily one-to-one?
• Suppose the composition g ◦ f is one-to-one.
Are f and g necessarily one-to-one?
• Suppose f and g are onto.
Is the composition g ◦ f necessarily onto?
• Suppose the composition g ◦ f is onto.
Are f and g necessarily onto?
Justify your answers.
182

Problem 6.
Let P2 denote the set of all polynomials of degree 2 or less ,
i.e., polynomials of the form
p(x) = a x2
+ b x + c , a, b, c ∈ R , x ∈ R .
Let P1 denote the set of all polynomials of degree 1 or less ,
i.e., polynomials of the form
p(x) = d x + e , d, e ∈ R , x ∈ R .
Consider the derivative function (or derivative operator )
D : P2 −→ P1 .
183

For example,
D : 3x2
+ 7x − 4 7→ 6x + 7 ,
and
D : 5x − 2π 7→ 5 .
QUESTIONS :
• Is D indeed a function from P2 to P1 ?
• Is D one-to-one ?
• Is D onto ?
• Does D have an inverse ?
184

Problem 7. If A and B are sets, and if
f : A −→ B ,
then for any subset S of A we define the image of S as
f(S) ≡ {b ∈ B : b = f(a) for some a ∈ S} .
Let S and T be subsets of A . Prove that
• f(S ∪ T) = f(S) ∪ f(T) ,
• f(S ∩ T) ⊆ f(S) ∩ f(T) .
• Also give an example that shows that in general
f(S ∩ T) 6= f(S) ∩ f(T) .
185

Problem 8. If A and B are sets, and if
f : A −→ B ,
then for any subset S of B we define the pre-image of S as
f−1
(S) ≡ {a ∈ A : f(a) ∈ S} .
NOTE : f−1
(S) is defined even if f does not have an inverse!
Let S and T be subsets of B . Prove that
• f−1
(S ∪ T) = f−1
(S) ∪ f−1
(T) ,
• f−1
(S ∩ T) = f−1
(S) ∩ f−1
(T) .
186

THE DIVISION THEOREM :
∀n ∈ Z, ∀d ∈ Z+
, ∃! q, r ∈ Z : ( 0 ≤ r d , n = qd + r ) .
EXAMPLE : If n = 21 and d = 8 then
n = 2 · d + 5 .
Thus, here q = 2 and r = 5 .
• d is called the divisor,
• q is called the quotient; we write q = n div d ,
• r is called the remainder; we write r = n mod d .
187

EXAMPLES :
n = 14 , d = 5 : 14 = 2 · d + 4 , so
14 div 5 = 2 and 14 mod 5 = 4 .
n = −14 , d = 5 : −14 = (−3) · d + 1 , so
−14 div 5 = − 3 and − 14 mod 5 = 1 .
188

Let d ∈ Z+
and n, q, r ∈ Z .
From the definitions of “div” and “mod” it follows that :
PROPERTY 1 : n = (n div d) d + n mod d
Example : 23 = (23 div 7) · 7 + 23 mod 7
PROPERTY 2 : If 0 ≤ r d then (qd + r) mod d = r
Example : (5 · 7 + 3) mod 7 = 3
PROPERTY 3 : (qd + n mod d) mod d = n mod d
Example : (5 · 7 + 23 mod 7) mod 7 = 23 mod 7
189

PROPERTY 4 : Let a, b ∈ Z and d ∈ Z+
. Then
(ad + b) mod d = b mod d .
PROOF :
By the Division Theorem
b = qd + r, where r = b mod d , with 0 ≤ r d .
Thus, using Property 2
(ad + b) mod d =

(a + q)d + r

mod d = r = b mod d .
QED !
EXAMPLE : (57 · 7 + 13) mod 7 = 13 mod 7 .
190

PROPERTY 5 : Let a ∈ Z and d ∈ Z+
. Then
(a mod d) mod d = a mod d .
PROOF : Using Property 3,
(a mod d) mod d = (0 · d + a mod d) mod d = a mod d .
EXAMPLE :
(59 mod 7) mod 7 = 59 mod 7 .
191

PROPERTY 6 :
Let a, b ∈ Z and d ∈ Z+
. Then
(a + b) mod d = (a mod d + b mod d) mod d .
EXAMPLE :
(5 + 8) mod 3 = 1 = (5 mod 3 + 8 mod 3) mod 3 .
192

PROOF :
a = qad + ra , where ra = a mod d , with 0 ≤ ra d ,
b = qbd + rb , where rb = b mod d , with 0 ≤ rb d .
Thus
(a + b) mod d = (qad + ra + qbd + rb) mod d
=

(qa + qb)d + ra + rb

mod d
= (ra + rb) mod d (using Property 4)
= (a mod d + b mod d) mod d . QED !
193

DEFINITION :
If a, b ∈ Z , d ∈ Z+
, and if
a mod d = b mod d ,
then we also write
a ≡ b (mod d) ,
and we say
“a is congruent to b modulo d”.
EXAMPLE :
83 ≡ 31 (mod 26) .
Note that 83 − 31 = 52 , which is divisible by 26 , i.e.,
26 | (83 − 31) .
194

PROPOSITION : Let a, b ∈ Z , and d ∈ Z+
. Then
a ≡ b (mod d) if and only if d | (a − b) .
PROOF :
(⇒) First, if a ≡ b (mod d) then, by definition,
a mod d = b mod d .
Hence there exist qa, qb, r ∈ Z , with 0 ≤ r d , such that
a = qad + r and b = qbd + r (same remainder).
It follows that
a − b = (qa − qb) d ,
so that d | (a − b).
195

a ≡ b (mod d) if and only if d | (a − b)
(⇐) Conversely, if d | (a − b) then
a − b = qd ,
i.e. ,
a = b + qd ,
for some q ∈ Z .
It follows that
a mod d = (b + qd) mod d = b mod d .
196

PROPOSITION : If a, b ∈ Z , and c, d ∈ Z+
, then
a ≡ b (mod d) ⇒ ac ≡ bc (mod dc) .
PROOF :
a ≡ b (mod d) if and only if d | (a − b) ,
i.e.,
a − b = qd , for some q ∈ Z .
Then ac − bc = qdc , so that (dc) | (ac − bc) ,
i.e.,
ac ≡ bc (mod dc) .
NOTE : We also have that ac ≡ bc (mod d) .
197

PROPOSITION : Let a, b ∈ Z and d ∈ Z+
. Then
a ≡ b (mod 2d) ⇒ a2
≡ b2
(mod 4d) .
EXAMPLE : Let a = 13 , b = 7 d = 3 . Then
13 ≡ 7 (mod 2 · 3) ,
i.e.,
13 (mod 6) = 7 (mod 6) ,
and
132
≡ 72
(mod 4 · 3) ,
i.e.,
169 (mod 12) = 49 (mod 12) (Check!) .
198

a ≡ b (mod 2d) ⇒ a2
≡ b2
(mod 4d)
PROOF : Suppose a ≡ b (mod 2d) .
Then 2d|(a − b) , i.e., a − b = q2d , for some q ∈ Z .
Thus a and b differ by an even number.
It follows that a and b must be both even or both odd.
Hence a + b must be even, i.e., a + b = 2c for some c ∈ Z .
Then a2
− b2
= (a + b) (a − b) = (2c) (q2d) = cq 4d .
It follows that 4d|(a2
− b2
) , i.e., a2
≡ b2
(mod 4d) . QED !
NOTE : Also a2
≡ b2
(mod 2d) and a2
≡ b2
(mod d) .
199

PROPOSITION : If n 3 then not all of
n , n + 2 , n + 4 ,
can be primes.
Idea of the proof : Always one of these three numbers is divisible by 3 .
PROOF. By contradiction : Assume that n 3 and that
n , n + 2 and n + 4 are primes .
Since n is prime and n 3 we have
n mod 3 = 1 or n mod 3 = 2 . (Why ?)
Case 1 : If n mod 3 = 1 then n = 3k + 1 and
n + 2 = 3k + 3 , i.e., 3|(n + 2) .
Case 2 : If n mod 3 = 2 then n = 3k + 2 and
n + 4 = 3k + 6 , i.e., 3|(n + 4) .
Contradiction !
200

THE FACTORIZATION THEOREM :
∀n ∈ (Z+
− {1}) , ∃!

m, {pi}m
i=1, {ni}m
i=1

:
m ∈ Z+
,
∀i (i = 1, · · · , m) : pi, ni ∈ Z+
,
1 p1 p2 · · · pm ,
∀i (i = 1, · · · , m) : pi is a prime number ,
n = pn1
1 pn2
2 · · · pnm
m .
EXAMPLE : 252 = 22
32
71
.
201

PROPOSITION : log23 is irrational.
PROOF : By contradiction:
Suppose log23 is rational, i.e., ∃p, q ∈ Z+
, such that
log23 = p/q .
By definition of the log function it follows that
2p/q
= 3 ,
from which
2p
= 3q
.
Let n = 2p
. Then n ∈ Z+
, with n ≥ 2 .
Then n has two different prime factorizations, namely
n = 2p
and n = 3q
.
This contradicts the Factorization Theorem. QED !
202

REMARK :
The fact that
2p
6= 3q
,
also follows from the facts that 2p
is even and 3q
is odd.
203

DEFINITION :
We call n ∈ Z+
a perfect square if
∃k ∈ Z+
: n = k2
.
FACT :
The factorization of a perfect square has only even powers :
If
k = pn1
1 pn2
2 · · · pnm
m ,
then
n = k2
= p2n1
1 p2n2
2 · · · p2nm
m .
204

PROPOSITION :
If n ∈ Z+
is not a perfect square then
√
n is irrational.
PROOF :
By contradiction :
Suppose n is not a perfect square, but
√
n is rational.
Thus
√
n =
p
q
,
i.e.,
p2
= n q2
,
for some p, q ∈ Z+
.
205

p2
= n q2
• The prime factorization of p2
has even powers only.
• The prime factorization of q2
has even powers only.
• The prime factorization of n must have an odd power,
(otherwise n would be a perfect square).
• Thus the factorization of nq2
must have an odd power.
• Thus p2
has two distinct factorizations:
one with even powers and one with at least one odd power.
This contradicts the uniqueness in the Factorization Theorem.
QED !
206

DEFINITION :
k ∈ Z+
is the greatest common divisor of n, m ∈ Z+
,
k = gcd(n, m) ,
if
• k|n and k|m ,
• no positive integer greater than k divides both n and m .
207

REMARK :
One can determine
gcd( n , m )
from the minimum exponents in the prime factorizations of n and m.
EXAMPLE : If
n = 168 , m = 900 ,
then
n = 23
31
71
, m = 22
32
52
,
and
gcd(168, 900) = 22
31
= 12 .
208

THE EUCLIDEAN THEOREM :
Let n , d ∈ Z+
, and let
r = n mod d , (the remainder)
Then
gcd(n, d) =







gcd(d , r) if r 0 ,
d if r = 0 .
209

EXAMPLE :
gcd(93 , 36) = gcd(36 , 93 mod 36) = gcd(36 , 21)
= gcd(21 , 36 mod 21) = gcd(21 , 15)
= gcd(15 , 21 mod 15) = gcd(15 , 6)
= gcd(6 , 15 mod 6) = gcd(6 , 3)
= 3 .
210

EXAMPLE :
gcd(2008 , 1947) = gcd(1947 , 2008 mod 1947) = gcd(1947 , 61)
= gcd(61 , 1947 mod 61) = gcd(61 , 56)
= gcd(56 , 61 mod 56) = gcd(56 , 5)
= gcd(5 , 56 mod 5) = gcd(5 , 1)
= 1 .
Thus 2008 and 1947 are relatively prime .
211

LEMMA :
Let a, b, c ∈ Z , and d ∈ Z+
.
Then
(1) ( a = b + c , d|a , d|b ) ⇒ d|c ,
(2) ( a = b + c , d|b , d|c ) ⇒ d|a ,
(3) ( a = bc , d|c ) ⇒ d|a .
212

gcd(n, d) =

gcd(d, r) if r 0
d if r = 0
PROOF OF THE EUCLIDEAN THEOREM :
n = q · d + r ,
where
q = n div d and r = n mod d .
Case 1 : r = 0 .
Then clearly d|n .
Also d|d and no greater number than d divides d .
Hence d = gcd(n, d) .
214

gcd(n, d) =

gcd(d, r) if r 0
d if r = 0
Case 2 : r 0 :
Let k = gcd(n, d) .
Then k|n and k|d .
n = q · d + r ,
By Lemma (3) k|qd .
By Lemma (1) k|r .
Thus k divides both d and r .
215

gcd(n, d) =

gcd(d, r) if r 0
d if r = 0
k = gcd(n, d) , n = q · d + r .
Show k is the greatest common divisor of d and r :
By contradiction :
Suppose k1 k and k1 = gcd(d, r) .
Thus k1|d and k1|r
By Lemma (3) k1|qd .
By Lemma (2) k1|n .
Thus k1 divides both n and d .
Since k1 k this contradicts that k = gcd(n, d) . QED !
216

REVIEW EXERCISES.
Problem 1.
Prove that a composite number n has a factor k ≤
√
n .
Thus to check if a number n is prime one needs only check whether
n mod k = 0 , k = 2, 3, · · · , ⌊
√
n⌋ .
Problem 2. Use the above fact to check whether 143 is prime.
217

Problem 3. Find all integer solutions of
2x ≡ 7(mod 17) .
Problem 4. Find all integer solutions of
4x ≡ 5(mod 9) .
Problem 5.
Does there exists an integer x that simultaneously satisfies
x ≡ 2(mod 6) and x ≡ 3(mod 9) ?
218

Problem 6. Let
S = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } ,
and define
f : S −→ S ,
by
f(k) = (5k + 3) mod 10 .
Is f invertible?
Problem 7. with S as above, also consider
f(k) = (6k + 3) mod 10 ,
and
f(k) = (7k + 3) mod 10 .
219

Problem 8. Let n ≥ 2 ,
Sn = { 0 , 1 , 2 , 3 , · · · , n − 1 } ,
and define
f : Sn −→ Sn ,
by
f(k) = (pk + s) mod n ,
where p is prime, with p n , and s ∈ Sn .
Prove that f is one-to-one (and hence onto and invertible ).
220

THE PRINCIPLE OF INDUCTION
Let
S = { s1 , s2 , s3 , · · · }
be a countably infinite set .
Suppose P is a predicate,
P : S −→ { T , F } ,
such that :
(i) P(s1) = T ,
(ii) P(sn) = T ⇒ P(sn+1) = T .
Then
P(s) = T , for all s ∈ S .
221

EXAMPLE :
n
X
k=1
k =
n(n + 1)
2
, ∀n ∈ Z+
.
Here S = Z+
and
P(n) = T if
n
X
k=1
k =
n(n + 1)
2
,
and
P(n) = F if
n
X
k=1
k 6=
n(n + 1)
2
.
We must show that P(n) = T for all n .
222

PROOF :
(i) “By inspection” the formula holds if n = 1 , i.e., P(1) = T .
(ii) Suppose P(n) = T for some arbitrary n ∈ Z+
, i.e.,
n
X
k=1
k =
n(n + 1)
2
.
We must show that P(n + 1) = T , i.e.,
n+1
X
k=1
k =
(n + 1)

(n + 1) + 1

2
.
This is done as follows:
n+1
X
k=1
k =
n
X
k=1
k

+ (n + 1) =
n(n + 1)
2
+ (n + 1)
=
(n + 1)

(n + 1) + 1

2
. QED !
223

EXAMPLE :
n
X
k=1
k2
=
n(n + 1)(2n + 1)
6
, ∀n ∈ Z+
.
PROOF :
(i) Again the formula is valid if n = 1 .
(ii) Suppose
n
X
k=1
k2
=
n(n + 1)(2n + 1)
6
,
for some arbitrary n ∈ Z+
.
We must show that
n+1
X
k=1
k2
=
(n + 1)

(n + 1) + 1

2(n + 1) + 1

6
.
224

Pn
k=1 k2
= n(n+1)(2n+1)
6
⇒
Pn+1
k=1 k2
= (n+1) ((n+1)+1) (2(n+1)+1)
6
To do this :
n+1
X
k=1
k2
=
n
X
k=1
k2

+ (n + 1)2
= n(n + 1)(2n + 1)/6 + (n + 1)2
= (n + 1)

n(2n + 1) + 6(n + 1)

/6
= (n + 1) (2n2
+ 7n + 6)/6
= (n + 1) (n + 2) (2n + 3)/6
= (n + 1)

(n + 1) + 1

2(n + 1) + 1

/6 . QED !
225

EXAMPLE :
(n3
− n) mod 3 = 0 , ∀n ∈ Z+
.
PROOF :
(i) By inspection P(1) = T .
(ii) Suppose P(n) = T , i.e.,
(n3
− n) mod 3 = 0 , for some arbitrary n ∈ Z+
.

(n + 1)3
− (n + 1)

mod 3 = 0 .
226

(n3
− n) mod 3 = 0 ⇒ ( (n + 1)3
− (n + 1) ) mod 3 = 0
To do this :

(n + 1)3
− (n + 1)

mod 3 = (n3
+ 3n2
+ 3n − n) mod 3
=

3(n2
+ n) + n3
− n

mod 3
= (n3
− n) mod 3 = 0 . QED !
227

EXAMPLE :
Let P(n) denote the statement
“A set Sn of n elements has 2n
subsets”.
CLAIM : P(n) = T for all n ≥ 0 .
PROOF :
(i) P(0) = T because the empty set has one subset, namely itself.
(ii) Suppose that P(n) = T for some arbitrary n , ( n ≥ 0 ) ,
i.e., Sn has 2n
subsets.
We must show that P(n+1) = T , i.e., Sn+1 has 2n+1
subsets.
228

To do this write
Sn+1 = { s1 , s2 , · · · , sn , sn+1 } = Sn ∪ {sn+1} .
Now count the subsets of Sn+1 :
(a) By inductive hypothesis Sn has 2n
subsets.
These are also subsets of Sn+1 .
(b) All other subsets of Sn+1 have the form
T ∪ {sn+1} ,
where T is any subset of Sn .
Thus there are 2n
such additional subsets.
The total number of subsets of Sn+1 is therefore
2n
+ 2n
= 2n+1
. QED !
229

EXAMPLE :
3n
n!
CLAIM :
P(n) = T for all integers n with n 6 .
REMARK : P(n) is False for n ≤ 6 . (Check!)
230

PROOF :
(i) P(7) = T , because
37
= 2187 5040 = 7!
(ii) Assume P(n) = T for some arbitrary n , (n ≥ 7) ,
i.e.,
3n
n! (n ≥ 7) .
3n+1
(n + 1)!
231

3n
n! ⇒ 3n+1
(n + 1)!
To do this :
3n+1
= 3 · 3n
3 · n! (by inductive assumption)
(n + 1) n! (since n ≥ 7)
= (n + 1)!
232

EXERCISE : For which nonnegative integers is
n2
≤ n! ?
Prove your answer by induction.
EXERCISE : For which positive integers n is
n2
≤ 2n
?
Prove your answer by induction.
233

EXAMPLE :
Let
Hm ≡
m
X
k=1
1
k
. (”Harmonic numbers.”)
H2n ≥ 1 +
n
2
.
CLAIM : P(n) = T for all n ∈ Z+
.
PROOF :
(i) It is clear that P(1) = T , because
H21 =
2
X
k=1
1
k
= 1 +
1
2
.
234

(ii) Assume that
P(n) = T for some arbitrary n ∈ Z+
,
i.e.,
H2n ≥ 1 +
n
2
.
We must show that
P(n + 1) = T ,
i.e.,
H2n+1 ≥ 1 +
n + 1
2
.
235

To do this :
H2n+1 =
2n+1
X
k=1
1
k
=
2n
X
k=1
1
k
+
2n+1
X
k=2n+1
1
k
=
2n
X
k=1
1
k
+
2n+2n
X
k=2n+1
1
k
≥ (1 +
n
2
) + 2n 1
2n + 2n
= (1 +
n
2
) +
1
2
= 1 +
n + 1
2
. QED !
236

REMARK :
It follows that
∞
X
k=1
1
k
diverges ,
i.e.,
n
X
k=1
1
k
−→ ∞ as n −→ ∞ .
237

EXAMPLE : (The Binomial Formula.)
For n ≥ 0 , a, b nonzero,
(a + b)n
=
n
X
k=0

n
k

an−k
bk
,
where

n
k

≡
n!
k! (n − k)!
.
REMARK : Thus we can write
(a+b)n
= an
+

n
1

an−1
b+

n
2

an−2
b2
+· · ·+

n
n − 1

abn−1
+bn
.
238

PROOF :
The formula holds if n = 0 . (Check!)
Assume that for some arbitrary n, ( n ≥ 0 )
(a + b)n
=
n
X
k=0

n
k

an−k
bk
.
We must show that the formula is also valid for n + 1 , i.e., that
(a + b)n+1
=
n+1
X
k=0

n + 1
k

an+1−k
bk
.
239

This can be done as follows :
(a + b)n+1
= (a + b)(a + b)n
= (a + b)
n
X
k=0

n
k

an−k
bk
= an+1
+
n
X
k=1

n
k

an−k+1
bk
+
n−1
X
k=0

n
k

an−k
bk+1
+ bn+1
= an+1
+
n
X
k=1

n
k

an−k+1
bk
+
n
X
k=1

n
k − 1

an−k+1
bk
+ bn+1
= an+1
+
n
X
k=1
n
n
k

+

n
k − 1
o
an−k+1
bk
+ bn+1
= an+1
+
n
X
k=1

n + 1
k

an−k+1
bk
+ bn+1
=
n+1
X
k=0

n + 1
k

an+1−k
bk
. QED !
240

REMARK :
In the proof we used the fact that

n
k

+

n
k − 1

=
n!
k! (n − k)!
+
n!
(k − 1)! (n − k + 1)!
=
n! (n − k + 1) + n! k
k! (n − k + 1)!
=
(n + 1)!
k! (n − k + 1)!
=

n + 1
k

.
241

REMARK :
One can order the binomial coefficients in Pascal’s triangle as follows :
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
. . . . . . . . . .
242

Observe that every entry can be obtained by summing the closest entries
in the preceding row.
This is so because the (n + 1)st and (n + 2)nd rows look like :
1 · · · · · · · · ·

n
k − 1

n
k

· · · · · · · · · 1
1 · · · · · · · · ·

n + 1
k

· · · · · · · · · 1
and we have shown above that

n
k

+

n
k − 1

=

n + 1
k

.
243

REMARKS :
• The advantage of a proof by induction is that it is systematic.
• A disadvantage is that the result (e.g., a formula) must be known in
advance from a heuristic argument or by trial and error.
• In contrast, a constructive proof actually derives the result.
244

EXAMPLE : For x ∈ R , x 6= 0, 1 :
n
X
k=0
xk
=
1 − xn+1
1 − x
, ∀n ≥ 0 , ( Geometric sum ) .
PROOF ( a constructive proof : already done earlier ) :
Let
Sn =
n
X
k=0
xk
.
Then
Sn = 1 + x + x2
+ · · · + xn−1
+ xn
,
x · Sn = x + x2
+ · · · + xn−1
+ xn
+ xn+1
,
so that
Sn − x · Sn = (1 − x) · Sn = 1 − xn+1
,
from which the formula follows. QED !
245

ALTERNATE PROOF ( by induction ) :
(i) “By inspection” we find that the formula holds if n = 0 .
(ii) Suppose that Sn = 1−xn+1
1−x
, for some arbitary n, ( n ≥ 0 ) .
Show Sn+1 = 1−x(n+1)+1
1−x
:
Sn+1 = Sn + xn+1
=
1 − xn+1
1 − x
+ xn+1
=
(1 − xn+1
) + xn+1
(1 − x)
1 − x
=
1 − xn+1
+ xn+1
− xn+2
1 − x
=
1 − x(n+1)+1
1 − x
. QED !
246

EXERCISE :
Use mathematical induction to prove that
21 | (4n+1
+ 52n−1
) ,
whenever n is a positive integer.
EXERCISE :
The Fibonacci numbers are defined as: f1 = 1 , f2 = 1 , and
fn = fn−1 + fn−2 , for n ≥ 3 .
Use a proof by induction to show that
3 | f4n ,
for all n ≥ 1 .
247

Variations on the Principle of Induction.
Let S = { s1, s2, s3, · · · } be a countably infinite set and P a predicate :
P : S −→ { T , F } .
VARIATION 1 : ( as used so far · · · )

P(s1) ∧
h
∀n ≥ 1 : P(sn) ⇒ P(sn+1)
i
⇒ ∀n : P(sn) .
VARIATION 2 :

P(s1) ∧ P(s2) ∧
h
∀n ≥ 2 : P(sn−1) ∧ P(sn) ⇒ P(sn+1)
i
⇒ ∀n : P(sn) .
VARIATION · · ·
STRONG INDUCTION :

P(s1) ∧ ∀n ≥ 1 :
h
P(s1) ∧ P(s2) ∧ · · · ∧ P(sn) ⇒ P(sn+1)
i
⇒ ∀n : P(sn) .
248

EXAMPLE : The Fibonacci Numbers.
The Fibonacci numbers are defined recursively as
f1 = 1 ,
f2 = 1 ,
fn = fn−1 + fn−2 , for n ≥ 3 .
249

f1 = 1 f11 = 89 f21 = 10946
f2 = 1 f12 = 144 f22 = 17711
f3 = 2 f13 = 233 f23 = 28657
f4 = 3 f14 = 377 f24 = 46368
f5 = 5 f15 = 610 f25 = 75025
f6 = 8 f16 = 987 f26 = 121393
f7 = 13 f17 = 1597 f27 = 196418
f8 = 21 f18 = 2584 f28 = 317811
f9 = 34 f19 = 4181 f29 = 514229
f10 = 55 f20 = 6765 f30 = 832040
250

PROPERTY :
There is an explicit formula for fn , namely
fn =
1
√
5
h 1 +
√
5
2
n
−
1 −
√
5
2
n i
.
251

We can also write
fn =
1
√
5
h 1 +
√
5
2
n
−
1 −
√
5
2
n i
=
1
√
5
1 +
√
5
2
n h
1 −
1 −
√
5
1 +
√
5
n i
≈
1
√
5
1 +
√
5
2
n h
1 −
1 − 2.236
1 + 2.236
n i
=
1
√
5
1 +
√
5
2
n h
1 + (−1)n+1

0.3819
n i
≈
1
√
5
1 +
√
5
2
n
.
252

f1 = 1 ≈ 0.72361 f11 = 89 ≈ 88.99775
f2 = 1 ≈ 1.17082 f12 = 144 ≈ 144.00139
f3 = 2 ≈ 1.89443 f13 = 233 ≈ 232.99914
f4 = 3 ≈ 3.06525 f14 = 377 ≈ 377.00053
f5 = 5 ≈ 4.95967 f15 = 610 ≈ 609.99967
f6 = 8 ≈ 8.02492 f16 = 987 ≈ 987.00020
f7 = 13 ≈ 12.98460 f17 = 1597 ≈ 1596.99987
f8 = 21 ≈ 21.00952 f18 = 2584 ≈ 2584.00008
f9 = 34 ≈ 33.99412 f19 = 4181 ≈ 4180.99995
f10 = 55 ≈ 55.00364 f20 = 6765 ≈ 6765.00003
253

fn = 1
√
5
h
1+
√
5
2
n
−

1−
√
5
2
n i
PROOF (By Induction, using Variation 2) :
The formula is valid when n = 1 :
f1 =
1
√
5
h 1 +
√
5
2
−
1 −
√
5
2
i
= 1 .
The formula is also valid when n = 2 :
f2 =
1
√
5
h 1 +
√
5
2
2
−
1 −
√
5
2
2 i
=
1
√
5
√
5 = 1 .
(Check!)
254

Inductively, assume that we have
fn−1 =
1
√
5
h 1 +
√
5
2
n−1
−
1 −
√
5
2
n−1 i
,
and
fn =
1
√
5
h 1 +
√
5
2
n
−
1 −
√
5
2
n i
.
We must show that
fn+1 =
1
√
5
h 1 +
√
5
2
n+1
−
1 −
√
5
2
n+1 i
.
255

Using the inductive hypothesis for n and n − 1 we have
fn+1 = fn−1 + fn (by definition)
=
1
√
5
h 1 +
√
5
2
n−1
−
1 −
√
5
2
n−1
+
1 +
√
5
2
n
−
1 −
√
5
2
n i
=
1
√
5
h 1 +
√
5
2
n−1
1 +
1 +
√
5
2

−
1 −
√
5
2
n−1
1 +
1 −
√
5
2
i
=
1
√
5
h 1 +
√
5
2
n−1 3 +
√
5
2

−
1 −
√
5
2
n−1 3 −
√
5
2
i
=
1
√
5
h 1 +
√
5
2
n−1 1 +
√
5
2
2
−
1 −
√
5
2
n−1 1 −
√
5
2
2 i
=
1
√
5
h 1 +
√
5
2
n+1
−
1 −
√
5
2
n+1 i
. QED !
256

Direct solution of the Fibonacci recurrence relation.
f1 = 1 , f2 = 1 ,
fk = fk−1 + fk−2 , for k ≥ 3 .
Try solutions of the form
fn = c zn
,
This gives
c zn
= c zn−1
+ c zn−2
, or zn
− zn−1
− zn−2
= 0 ,
from which we obtain the characteristic equation
z2
− z − 1 = 0 .
257

z2
− z − 1 = 0
The characteristic equation has solutions (“roots”):
z =
1 ±
√
1 + 4
2
,
that is,
z1 =
1 +
√
5
2
, z2 =
1 −
√
5
2
.
The general solution of the recurrence relation is then
fn = c1 zn
1 + c2 zn
2 .
258

fn = c1 zn
1 + c2 zn
2
The constants c1 and c2 are determined by the initial conditions :
f1 = 1 ⇒ c1 z1 + c2 z2 = 1 ,
and
f2 = 1 ⇒ c1 z2
1 + c2 z2
2 = 1 ,
that is,
c1
1 +
√
5
2
+ c2
1 −
√
5
2
= 1 ,
and
c1
1 +
√
5
2
2
+ c2
1 −
√
5
2
2
= 1 ,
from which we find
c1 =
1
√
5
and c2 = −
1
√
5
. (Check!)
259

fn = c1 zn
1 + c2 zn
2
We found that
z1 =
1 +
√
5
2
, z2 =
1 −
√
5
2
.
and
c1 =
1
√
5
and c2 = −
1
√
5
. (Check!)
from which
fn =
1
√
5
1 +
√
5
2
n
−
1
√
5
1 −
√
5
2
n
.
260

REVIEW EXERCISES.
Problem 1.
Prove that the Fibonacci numbers satisfy the following relations:
•
Pn
k=1 f2k−1 = f2n , for n ∈ Z+
.
• fn−1 fn+1 − f2
n = (−1)n
, for n ∈ Z+
, (n ≥ 2) .
• −f1 + f2 − f3 + · · · − f2n−1 + f2n = f2n−1 − 1 , for n ∈ Z+
.
261

Problem 2. The recurrence relation
xn+1 = c xn (1 − xn) , n = 1, 2, 3, · · · ,
known as the logistic equation, models population growth when there are
limited resources.
Write a small computer program (using real arithmetic) to see what
happens to the sequence xn, n = 1, 2, 3, · · · , with 0 x1 1 , for
each of the following values of c :
(a) 0.5 , (b) 1.5 , (c) 3.2 , (d) 3.5 , (e) 3.9
Problem 3. Find an explicit solution to the recurrence relation
xn+1 = 3 xn − 2 xn−1 , n = 1, 2, 3, · · · ,
with x1 = 1 and x2 = 3 .
262

RELATIONS
A binary relation relates elements of a set to elements of another set.
EXAMPLE :
The operator “≤” relates elements of Z to elements of Z. e.g.,
2 ≤ 5 , and 3 ≤ 3 , but 7 6≤ 2 .
We can also view this relation as a function
≤ : Z × Z −→ {T , F } .
263

RECALL :
Let
A = {a1, a2, · · · , anA
} and B = {b1, b2, · · · , bnB
} .
The product set A × B is the set of all ordered pairs from A and B.
More precisely,
A × B ≡ {(a, b) : a ∈ A, b ∈ B} .
264

NOTE :
• The product set A × B has nA · nB elements.
• If A and B are distinct and nonempty then A × B 6= B × A.
EXAMPLE :
If
A = { ! , ? } and B = { • , ◦ , } ,
then
A × B = { (!, •) , (!, ◦) , (!, ) , (?, •) , (?, ◦) , (?, ) } .
265

We can now equivalently define :
DEFINITION :
A binary relation R from A to B is a subset of A × B.
NOTATION :
If R ⊆ A × B, and
(a, b) ∈ R ,
then we say that
“a is R-related to b”,
and we also write
a R b .
266

EXAMPLE :
Let A = {1, 3} and B = {3, 5, 9}.
Let R denote the relation “divides” from A to B, i.e.,
aRb if and only if a|b .
Then
1R3 , 1R5 , 1R9 , 3R3 , and 3R9 .
Thus
R = { (1, 3) , (1, 5) , (1, 9) , (3, 3) , (3, 9) } .
267

We can represent R by the following diagram
3
A B
1
3
5
9
R
This representation is an example of a directed bipartite graph.
Note that R is not a function, since it is multi-valued.
268

REMARK :
We see that a relation generalizes the notion of a function.
Unlike functions from a set A to a set B :
• A relation does not have to be defined for all a ∈ A .
• A relation does not have to be single-valued.
269

A finite relation from a set A into itself can be represented
by an ordinary directed graph.
EXAMPLE :
Let
A = {1 , 2 , 3 , 4 , 5 , 6} ,
and let R denote the relation “divides” from A to A .
We say that R is a relation “on A” .
This relation has the following directed graph representation :
270

1
3
2
5
6
4
The “divides” relation on the set A .
271

We can compose relations as follows :
Let R be a relation from A to B , and S a relation from B to C.
B C
R
S
S
A
R
o
Then S ◦ R is the relation from A to C defined by
a(S ◦ R)c if and only if ∃b ∈ B : aRb ∧ bSc .
272

EXAMPLE :
Let
A = {1, 2, 3}, B = {2, 6}, and C = {1, 9, 15},
and define the relations R and S by
aRb if and only if a|b ,
and
bSc if and only if b + c is prime .
Define
T = S ◦ R .
273

Then from the diagram below we see that
T = { (1, 1) , (1, 9) , (1, 15) , (2, 1) , (2, 9) , (2, 15) , (3, 1) } .
C
R
S
S
A
R
o
1
2
3
2
6
1
9
15
B
= T
274

Let A, B, C, and D be sets, and let R S, and T be relations :
R : A −→ B , S : B −→ C , T : C −→ D .
PROPOSITION :
The composition of relations is associative, i.e.,
(T ◦ S) ◦ R = T ◦ (S ◦ R) .
A B C
R S T
R
T
S
S R
o
o o
o
T S
D
275

A B C
R S T
R
T
S
S R
o
o o
o
T S
D
(T ◦ S) ◦ R = T ◦ (S ◦ R) .
PROOF : Let a ∈ A and d ∈ D. Then
a(T ◦ S) ◦ Rd ⇐⇒ ∃b ∈ B : (aRb ∧ bT ◦ Sd)
⇐⇒ ∃b ∈ B, ∃c ∈ C : (aRb ∧ bSc ∧ cTd)
⇐⇒ ∃c ∈ C, ∃b ∈ B : (aRb ∧ bSc ∧ cTd)
⇐⇒ ∃c ∈ C : (aS ◦ Rc ∧ cTd) ⇐⇒ aT ◦ (S ◦ R)d .
QED !
276

EXAMPLE : Let the relation R on
A = { 2, 3, 4, 8, 9, 12 } ,
be defined by
(a, b) ∈ R if and only if (a|b ∧ a 6= b) .
Then
R = { (2, 4) , (2, 8) , (2, 12) , (3, 9) , (3, 12) , (4, 8) , (4, 12) } ,
and
R2
≡ R ◦ R = { (2, 8) , (2, 12) } ,
R3
≡ R2
◦ R = R ◦ R ◦ R = { } .
277

0
0
1
1 0
0
1
1 0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
00
00
11
11
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
00
00
00
11
11
11
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1 0
0
1
1
R R R
R
R
2
3
12
3
4
9
3
4
8
9
3
4
8
9
8
4
3
2
9
2
8
12
2
12
2
12
278

EXAMPLE :
Let R be the relation on the set of all real numbers defined by
xRy if and only if xy = 1
Then
xR2
z ⇐⇒ ∃y : xRy and yRz
⇐⇒ ∃y : xy = 1 and yz = 1
⇐⇒ x = z and x 6= 0 . (Why ?)
279

The last equivalence in detail:
∃y : xy = 1 and yz = 1 if and only if x = z and x 6= 0 .
PROOF : (⇒) Let x and z be real numbers, and assume that
∃y : xy = 1 and yz = 1 .
Then x and z cannot equal zero.
Thus we can write
y =
1
x
and y =
1
z
.
Hence 1/x = 1/z, i.e., x = z.
Thus x = z and x 6= 0 .
280

∃y : xy = 1 and yz = 1 if and only if x = z and x 6= 0 .
(⇐)
Conversely, suppose x and z are real numbers with
x = z and x 6= 0 .
Let y = 1/x.
Then xy = 1 and yz = 1.
QED !
281

Similarly
xR3
z ⇐⇒ ∃y : xR2
y and yRz
⇐⇒ ∃y : x = y and x 6= 0 and yz = 1
⇐⇒ xz = 1 (Why ?)
Thus
R3
= R ,
R4
= R3
◦ R = R ◦ R = R2
,
R5
= R4
◦ R = R2
◦ R = R3
= R ,
and so on · · ·
Thus we see that
Rn
= R2
if n is even ,
and
Rn
= R if n is odd .
282

EXAMPLE :
Let R be the relation on the real numbers defined by
xRy if and only if x2
+ y2
≤ 1 .
Then
xR2
⇐⇒ ∃y : x2
+ y2
≤ 1 and y2
+ z2
≤ 1
⇐⇒ x2
≤ 1 and z2
≤ 1 (Why ?)
⇐⇒ | x | ≤ 1 and | z | ≤ 1 .
283

y
x x
z
R R2
1 −1
1
1
−1
The relations R and R2
as subsets of R2
.
284

Similarly
xR3
z ⇐⇒ ∃y : xR2
y and yRz
⇐⇒ ∃y : | x | ≤ 1 and | y | ≤ 1 and y2
+ z2
≤ 1
⇐⇒ ∃y : | x | ≤ 1 and y2
+ z2
≤ 1 (Why ?)
⇐⇒ | x | ≤ 1 and | z | ≤ 1 (Why ?)
Thus R3
= R2
.
285

Similarly
R4
= R3
◦ R = R2
◦ R = R3
= R2
,
R5
= R4
◦ R = R2
◦ R = R3
= R2
,
and so on · · ·
Thus we see that
Rn
= R2
for all n ≥ 2 .
286

The relation matrix.
A relation between finite sets can be represented by a relation matrix.
(Also known as the transition matrix).
For a relation R from A to B the relation matrix R has entries
Rij =



0 if (ai, bj) 6∈ R ,
1 if (ai, bj) ∈ R .
287

EXAMPLE :
Reconsider the example where
A = {1, 2, 3}, B = {2, 6}, and C = {1, 9, 15} ,
aRb if and only if a|b, and bSc if and only if b + c is prime .
C
R
S
S
A
R
o
1
2
3
2
6
1
9
15
B
= T
288

C
R
S
S
A
R
o
1
2
3
2
6
1
9
15
B
= T
The relation matrices of R and S are
2 6 1 9 15
R =
1
2
3


1 1
1 1
0 1

 and S =
2
6

1 1 1
1 0 0

.
289

We found that
T = S ◦R = { (1, 1) , (1, 9) , (1, 15) , (2, 1) , (2, 9) , (2, 15) , (3, 1) } .
C
R
S
S
A
R
o
1
2
3
2
6
1
9
15
B
= T
290

C
R
S
S
A
R
o
1
2
3
2
6
1
9
15
B
= T
The relation matrices of R and S were found to be
R =


1 1
1 1
0 1

 , S =

1 1 1
1 0 0

.
The relation matrix of T = S ◦ R is
T =


1 1 1
1 1 1
1 0 0

 .
291

PROPOSITION :
Let A, B, and C be finite sets.
Let R be a relation from A to B.
Let S be a relation from B to C.
Let T = S ◦ R.
Then the relation matrix of T has the same zero-structure
as the matrix product RS.
292

PROOF :
Tij = 1 ⇐⇒ aiTcj
⇐⇒ aiRbk and bkScj, for some bk ∈ B
⇐⇒ Rik = 1 and Skj = 1 for some k
⇐⇒
PnB
ℓ=1 Riℓ Sℓj 6= 0
⇐⇒ [RS]ij 6= 0 . QED !
293

REMARK :
If we use Boolean arithmetic, then T = RS .
EXAMPLE :
The matrix product RS in the preceding example is


1 1
1 1
0 1



1 1 1
1 0 0

=


2 1 1
2 1 1
1 0 0

 .
Using Boolean arithmetic the matrix product is
T =


1 1 1
1 1 1
1 0 0

 .
294

The inverse of a relation.
Let R be a relation from A to B.
Then the inverse relation R−1
is the relation from B to A defined by
b R−1
a if and only if a R b ,
or, in equivalent notation,
(b, a) ∈ R−1
if and only if (a, b) ∈ R .
Thus, unlike functions, relations are always invertible.
295

EXAMPLE :
1
2
3
a
b
A
B
Here
R = {(1, a), (1, b), (2, a), (3, b)} ,
and
R−1
= {(a, 1), (a, 2), (b, 1), (b, 3)} .
The relation matrices are
R =


1 1
1 0
0 1

 , R−1
=

1 1 0
1 0 1

.
296

Note that
R−1
= RT
(transpose) .
This holds in general, because if
A = {a1, a2, · · · , anA
} , B = {b1, b2, · · · , bnB
} ,
then, by definition of R−1
we have for any ai, bj that
bjR−1
ai ⇐⇒ aiRbj .
Hence [R−1
]ji = Rij .
297

EXERCISE : Let R be the relation on the set
A = { 1 , 2 , 3 , 4 } ,
defined by
a1Ra2 if and only if a1 a2 .
• Write down R as a subset of A × A .
• Show the relation matrix of R .
• Do the same for R2
, R3
, · · ·
EXERCISE : Do the same for a1Ra2 if and only if a1 ≤ a2 .
EXERCISE : Do the same for a1Ra2 if and only if a1 + a2 = 5 .
EXERCISE : Do the same for the set Z instead of A .
298

DEFINITION : Let R be a relation on A (i.e., from A to A).
• R is called reflexive if
∀a ∈ A : (a, a) ∈ R, i.e., ∀a ∈ A : aRa ,
i.e., if the relation matrix R (for finite A) satisfies
∀i : Rii = 1 .
EXAMPLES :
The “divides” relation on Z+
is reflexive.
The “≤” relation on Z is reflexive.
The “⊆” relation on a power set 2A
is reflexive.
The “” relation on Z is not reflexive.
299

• R is symmetric if
∀a, b ∈ A : aRb → bRa ,
or, equivalently,
aRb ⇒ bRa ,
i.e., if the relation matrix (for finite A) is symmetric :
∀i, j : Rij = Rji .
EXAMPLES :
The relation on the real numbers defined by
+ y2
≤ 1 ,
is symmetric.
is not symmetric.
300

• R is antisymmetric if for all a, b ∈ A we have
aRb ∧ bRa ⇒ a = b ,
or equivalently,
a 6= b ⇒ ¬(aRb) ∨ ¬(bRa) ,
i.e., if the relation matrix (for finite A) satisfies
∀i, j with i 6= j : RijRji 6= 1 .
EXAMPLES :
The “≤” relation on Z is antisymmetric.
is antisymmetric.
The “⊆” relation on a power set 2A
is antisymmetric.
301

• R is transitive if
aRb ∧ bRc ⇒ aRc .
We’ll show later that R is transitive if and only if
n
X
k=1
Rk
= R
in Boolean arithmetic
EXAMPLES :
is transitive.
The “≤” relation on Z is transitive.
The “⊆” relation on a power set 2S
is transitive.
The relation aRb ⇐⇒ “a + b is prime” on Z+
is not transitive.
302

EXERCISE : Let A be a set of n elements.
• How many relations are there on A ?
How many relations are there on A that are :
• symmetric ?
• antisymmetric ?
• symmetric and antisymmetric ?
• reflexive ?
• reflexive and symmetric ?
• transitive (∗)
?
(∗)
Hint : Search the web for “the number of transitive relations” !
303

• An equivalence relation is a relation that is
- reflexive
- symmetric
- transitive.
EXAMPLE :
The following relation on Z is an equivalence relation :
aRb if and only if a mod m = b mod m .
(Here m ≥ 2 is fixed.)
304

• A partial order is a relation that is
- reflexive
- antisymmetric
- transitive.
EXAMPLES :
.
- The “≤” relation on Z+
.
- The “⊆” relation on the power set 2S
.
- The operator “” on Z is not a partial order:
(It is antisymmetric (!) and transitive, but not reflexive.)
305

• A relation R on a set A is called a total order if
- R is a partial order, and
- ∀a, b ∈ A we have aRb or bRa .
EXAMPLES :
- The partial order “≤” is also a total order on Z+
.
- The partial order m|n on Z+
is not a total order.
(For example 5 6 |7 and 7 6 |5 .)
- The partial order “⊆” on 2S
is not a total order.
306

Equivalence classes.
Let A be a set and let R be an equivalence relation on A .
Let a1 ∈ A.
Define
[a1] = {a ∈ A : aRa1} ,
that is
[a1] = all elements of A that “are equivalent ” to a1 .
Then [a1] is called the equivalence class generated by a1 .
307

EXAMPLE :
Let R be the relation “congruence modulo 3” on Z+
, i.e.,
aRb if and only if a mod 3 = b mod 3 .
For example
1R1 , 1R4 , 1R7 , 1R10 , · · · ,
2R2 , 2R5 , 2R8 , 2R11 , · · · ,
3R3 , 3R6 , 3R9 , 3R12 , · · · .
308

Thus
[1] = { 1 , 4 , 7 , 10 , 13 , · · · } ,
[2] = { 2 , 5 , 8 , 11 , 14 , · · · } ,
[3] = { 3 , 6 , 9 , 12 , 15 , · · · } .
We see that
Z+
= [1] ∪ [2] ∪ [3] .
• The relation R has partitioned Z+
into the subsets [1] , [2] , [3] .
• Any member of a subset can represent the subset, e.g.,
[11] = [2] .
309

EXAMPLE :
Consider Z × Z , the set of all ordered pairs of integers.
Define a relation R on Z × Z by
(a1, b1)R(a2, b2) if and only if a1 − b1 = a2 − b2 .
Note that R can be viewed as subset of
(Z × Z) × (Z × Z) .
310

(a1, b1)R(a2, b2) if and only if a1 − b1 = a2 − b2
R is an equivalence relation :
• R is reflexive : (a, b)R(a, b) ,
• R is symmetric : (a1, b1)R(a2, b2) ⇒ (a2, b2)R(a1, b1) ,
• R is transitive:
(a1, b1)R(a2, b2) ∧ (a2, b2)R(a3, b3) ⇒ (a1, b1)R(a3, b3) .
.
311

(a1, b1)R(a2, b2) if and only if a1 − b1 = a2 − b2
The equivalence classes are
Ak = { (a, b) : a − b = k } .
For example,
A1 = [(2, 1)] = {· · · , (−1, −2) , (0, −1) , (1, 0) , (2, 1) , · · ·} .
The sets Ak partition the set Z × Z , namely,
Z × Z = ∪∞
k=−∞ Ak .
312

A
A
A
A
A A
A
0
1
2
−1
−2
3
−3
b
a
313

DEFINITION :
• The reflexive closure of R is the smallest relation containing R that
is reflexive.
• The symmetric closure of R is the smallest relation containing R
that is symmetric.
• The transitive closure of R is the smallest relation containing R
that is transitive.
314

EXAMPLE :
Let A = {1, 2, 3, 4} , and let R be the relation on A defined by
R = { (1, 4) , (2, 1) , (2, 2) , (3, 2) , (4, 1) } .
Then R is not reflexive, not symmetric, and not transitive :
00
00
00
11
11
11
00
00
00
11
11
11
00
00
00
11
11
11
00
00
00
11
11
11
2
4 3
1
315

The reflexive closure of R is :
0
0
1
1
0
0
0
1
1
1
0
0
1
1 0
0
1
1
2
4 3
1
316

The symmetric closure of R is :
00
00
00
11
11
11
00
00
00
11
11
11
00
00
11
11 00
00
11
11
2
4 3
1
317

To get the transitive closure of R :
00
00
00
11
11
11
00
00
00
11
11
11
00
00
11
11 00
00
11
11
2
4 3
1
318

00
00
00
11
11
11
00
00
00
11
11
11
00
00
11
11 00
00
11
11
2
4 3
1
319

00
00
00
11
11
11
00
00
00
11
11
11
00
00
11
11 00
00
11
11
2
4 3
1
320

The transitive closure of R is :
00
00
11
11
0
0
1
1
00
00
00
11
11
11
0
0
0
1
1
1
2
4 3
1
321

PROPERTY : The transitive closure R∗
of a relation R is given by
R∗
= ∪∞
k=1 Rk
.
PROOF : Later · · ·
PROPERTY : For a finite set of n elements, the relation matrix of
the transitive closure is
R∗
=
n
X
k=1
Rk
(using Boolean arithmetic) .
NOTE : It suffices to sum only the first n powers of the matrix R !
322

EXAMPLE : In the preceding example,
R = { (1, 4) , (2, 1) , (2, 2) , (3, 2) , (4, 1) } ,
0
0
1
1
0
0
0
1
1
1
0
0
1
1
0
0
1
1
2
4 3
1
we have the relation matrix :
R =




0 0 0 1
1 1 0 0
0 1 0 0
1 0 0 0



 .
323

Thus, using Boolean arithmetic,
R2
= R · R =




0 0 0 1
1 1 0 0
0 1 0 0
1 0 0 0








0 0 0 1
1 1 0 0
0 1 0 0
1 0 0 0



 =




1 0 0 0
1 1 0 1
1 1 0 0
0 0 0 1



 ,
R3
= R · R2
=




0 0 0 1
1 1 0 0
0 1 0 0
1 0 0 0








1 0 0 0
1 1 0 1
1 1 0 0
0 0 0 1



 =




0 0 0 1
1 1 0 1
1 1 0 1
1 0 0 0



 ,
R4
= R · R3
=




0 0 0 1
1 1 0 0
0 1 0 0
1 0 0 0








0 0 0 1
1 1 0 1
1 1 0 1
1 0 0 0



 =




1 0 0 0
1 1 0 1
1 1 0 1
0 0 0 1



 .
324

Therefore
R∗
=
4
X
k=1
Rk
= R + R2
+ R3
+ R4
=




0 0 0 1
1 1 0 0
0 1 0 0
1 0 0 0



 +




1 0 0 0
1 1 0 1
1 1 0 0
0 0 0 1



 +




0 0 0 1
1 1 0 1
1 1 0 1
1 0 0 0



 +




1 0 0 0
1 1 0 1
1 1 0 1
0 0 0 1




=




1 0 0 1
1 1 0 1
1 1 0 1
1 0 0 1



 .
325

The graph of R∗
(shown before) is :
00
00
11
11
0
0
1
1
00
00
11
11
0
0
1
1
2
4 3
1
which indeed has the relation matrix
R∗
=
4
X
k=1
Rk
=




1 0 0 1
1 1 0 1
1 1 0 1
1 0 0 1



 .
326

RECALL : The transitive closure R∗
R∗
= ∪∞
k=1 Rk
(to be proved later · · · )
EXAMPLE : Consider the relation R on the real numbers
xRy if and only if xy = 1 .
Earlier we found that
xR2
y if and only if x = y ∧ x 6= 0 ,
and
R = R3
= R5
= · · · ,
R2
= R4
= R6
= · · · .
Thus the transitive closure is
xR∗
y if and only if xy = 1 ∨ (x = y ∧ x 6= 0) .
327

EXAMPLE : Again consider the relation R on the real numbers
+ y2
≤ 1 .
Earlier we found that
xR2
y if and only if | x | ≤ 1 ∧ | y | ≤ 1 ,
and
Rn
= R2
, for n ≥ 2 .
Thus the transitive closure is
xR∗
y if and only if x2
+ y2
≤ 1 ∨ ( | x |≤ 1 ∧ | y |≤ 1 ) ,
that is,
xR∗
y if and only if | x | ≤ 1 ∧ | y | ≤ 1 . (Why ?)
328

EXERCISE : Let R be the relation on the real numbers given by
+ y2
= 1 .
• Draw R as a subset of the real plane.
• Is R reflexive?
• Is R symmetric?
• Is R antisymmetric?
• Is R transitive?
• What is R2
? (Be careful!)
• What is R3
?
• What is the transitive closure of R ?
329

xRy if and only if xy ≤ 1 .
• Is R reflexive?
• Is R symmetric?
• What is R2
?
• What is R3
?
330

≤ y .
• Is R reflexive?
• Is R symmetric?
• What is R2
?
• What is R3
?
• What is Rn
? (Prove your formula for Rn
by induction.)
331

≤ y
Then
xR2
⇐⇒ ∃y : x2
≤ y and y2
≤ z
⇐⇒ x4
≤ z .
Similarly
xR3
z ⇐⇒ ∃y : xR2
y and yRz
⇐⇒ ∃y : x4
≤ y and y2
≤ z
⇐⇒ x8
≤ z .
By induction one can prove that
xRn
z ⇐⇒ x2n
≤ z .
332

We see that
• The relation R corresponds to the area of the x, y-plane that
lies on or above the curve y = x2
.
• The relation R2
corresponds to the area of the x, y-plane that
lies on or above the curve y = x4
.
• The relation Rn
corresponds to the area of the x, y-plane that
lies on or above the curve y = x2n
.
333

The transitive closure is
R∗
= ∪∞
k=1 Rk
.
(to be proved later · · · )
We find that R∗
is the union of the following two regions:
• The area of the x, y-plane that lies on or above the curve
y = x2
(i.e., the area that corresponds to the relation R ).
• The area inside the rectangle whose corners are located at
(x, y) = (−1, 0) , (1, 0) , (1, 1) , (−1, 1) ,
(excluding the border of this rectangle).
(Check!)
334

Let R be a relation on a set A .
Recursively define
R1
= R , Rn+1
= Rn
◦ R , n = 1, 2, 3, · · · .
Then for all n ∈ Z+
we have:
PROPERTY 1 : Rn
◦ R = R ◦ Rn
PROPERTY 2 : R symmetric ⇒ Rn
is symmetric
EXERCISE : Use induction to prove these properties.
335

PROPERTY 1 : Rn
◦ R = R ◦ Rn
, for all n ∈ Z+
PROOF : Clearly, the equality holds if n = 1.
Inductive assumption :
Rn
◦ R = R ◦ Rn
, for some arbitrary n ∈ Z+
.
We must show that
Rn+1
◦ R = R ◦ Rn+1
.
To do this :
Rn+1
◦ R = (Rn
◦ R) ◦ R (by definition)
= (R ◦ Rn
) ◦ R (by inductive assumption)
= R ◦ (Rn
◦ R) (by associativity)
= R ◦ Rn+1
(by definition). QED !
336

PROPERTY 2 : R symmetric ⇒ Rn
is symmetric
PROOF : Clearly True if n = 1 .
Inductively assume that Rn
is symmetric.
We must show that Rn+1
is symmetric:
aRn+1b ⇐⇒ aRn ◦ Rb (by definition of power)
⇐⇒ ∃p(aRp ∧ pRnb) (by definition of composition)
⇐⇒ ∃p(pRa ∧ bRnp) (since R and Rn are symmetric)
⇐⇒ ∃p(bRnp ∧ pRa) (commutative law of logic)
⇐⇒ bR ◦ Rna (by definition of composition)
⇐⇒ bRn ◦ Ra (by Property 1)
⇐⇒ bRn+1a (by definition of power) . QED !
337

Let R be a relation on a set A and let n ∈ Z+
.
PROPERTY 3 : R transitive ⇒ Rn
is transitive
PROOF :
Let R be transitive.
Obviously R1
is transitive.
By induction assume that Rn
is transitive for some n ∈ Z+
.
We must show that Rn+1
is transitive, i.e., we must show that
aRn+1
b ∧ bRn+1
c ⇒ aRn+1
c .
338

Given R and Rn
are transitive. Show Rn+1
is transitive
· · · continuation of proof · · ·
aRn+1
b ∧ bRn+1
c
⇒ aRn
◦ Rb ∧ bRn
◦ Rc (power)
⇒ aRn
◦ Rb ∧ bR ◦ Rn
c (by Property 1)
⇒ aRp ∧ pRn
b ∧ bRn
q ∧ qRc (∃p, q: composition)
⇒ aRp ∧ pRn
q ∧ qRc (inductive assumption)
⇒ aRn
◦ Rq ∧ qRc (composition)
⇒ aR ◦ Rn
q ∧ qRc (by Property 1)
339

Given R and Rn
are transitive. Show Rn+1
is transitive.
· · · continuation of proof · · ·
aR ◦ Rn
q ∧ qRc
⇒ aRn
s ∧ sRq ∧ qRc (∃s: composition)
⇒ aRn
s ∧ sRc (since R is transitive)
⇒ aR ◦ Rn
c (composition)
⇒ aRn
◦ Rc (by Property 1)
⇒ aRn+1
c (power) QED !
340

Let R and S be relations on a set A .
Recall that we can also think of R as a subset of A × A .
We have:
PROPERTY 4 : R ⊆ S ⇒ Rn
⊆ Sn
PROOF : The statement clearly holds when n = 1 .
Inductive step:
Given R ⊆ S and Rn
⊆ Sn
. Show Rn+1
⊆ Sn+1
To do this :
Suppose that (x, z) ∈ Rn+1
.
Then ∃y : (x, y) ∈ R and (y, z) ∈ Rn
.
By the assumptions (x, y) ∈ S and (y, z) ∈ Sn
.
Hence (x, z) ∈ Sn+1
. QED !
341

Let S be a relation on a set A .
PROPERTY 5 : S is transitive if and only if ∀n ∈ Z+
: Sn
⊆ S
PROOF :
(⇐) (∀n ∈ Z+
: Sn
⊆ S) ⇒ S is transitive
Let (x, y) ∈ S and (y, z) ∈ S .
Then, by definition of composition, (x, z) ∈ S2
.
Since, in particular, S2
⊆ S it follows that (x, z) ∈ S .
Hence S is transitive.
342

(⇒) S is transitive ⇒ ∀n ∈ Z+
: Sn
⊆ S
By induction :
Clearly Sn
⊆ S if n = 1 .
Suppose that for some n we have Sn
⊆ S .
We must show that Sn+1
⊆ S .
343

Given (1): S is transitive, and (2): Sn
⊆ S . Show Sn+1
⊆ S
To do this, suppose that (x, z) ∈ Sn+1
.
We must show that (x, z) ∈ S .
By definition of composition, (x, z) ∈ Sn
◦ S , and hence
∃y : (x, y) ∈ S and (y, z) ∈ Sn
.
By inductive hypothesis (2) (y, z) ∈ S .
Thus (x, y) ∈ S and (y, z) ∈ S .
By assumption (1) S is transitive, so that (x, z) ∈ S . QED !
344

THEOREM :
The transitive closure R∗
R∗
= ∪∞
k=1 Rk
.
PROOF : Let U = ∪∞
k=1 Rk
.
We must show that
(1) U is transitive.
(2) U is the smallest transitive relation containing R .
If so, then R∗
= U .
345

U = ∪∞
k=1 Rk
⋆
(1) We first show that U is transitive :
Suppose (x, y) ∈ U and (y, z) ∈ U .
We must show that (x, z) ∈ U .
From ⋆ it follows that
(x, y) ∈ Rm
and (y, z) ∈ Rn
, for some m, n ∈ Z+
.
By definition of composition
(x, z) ∈ Rn+m
.
Thus, using ⋆ again, it follows that
(x, z) ∈ U .
346

U = ∪∞
k=1 Rk
⋆
(2) Show U is the smallest transitive relation containing R :
To do this it suffices to show that :
( S transitive and R ⊆ S ) ⇒ U ⊆ S .
R
U
S
347

U = ∪∞
k=1 Rk
⋆
R ⊆ S ⇒ Rn
⊆ Sn
Property 4
S is transitive ⇐⇒ ∀n ∈ Z+
: Sn
⊆ S Property 5
To do: Given S transitive and R ⊆ S . Show U ⊆ S
Let (x, y) ∈ U . Then, by ⋆ we have
(x, y) ∈ Rn
for some n ∈ Z+
.
By Property 4 : (x, y) ∈ Sn
.
By Property 5 : (x, y) ∈ S . QED !
348

REVIEW PROBLEMS
and
REVIEW CLICKER QUESTIONS
349

Review Problem 1.
Prove that for every integer n we have
n5
− n ≡ 0 (mod 30)
350

Review Problem 2.
Suppose m and n are relatively prime integers; m ≥ 2 , n ≥ 2 .
Prove that logmn is an irrational number.
351

Review Problem 3. If A and B are sets, and if
f : A −→ B ,
then for any subset S of B we define the pre-image of S as
f−1
(S) ≡ {a ∈ A : f(a) ∈ S} .
NOTE : f−1
(S) is defined even if f does not have an inverse!
Let S and T be subsets of B .
Prove that
f−1
(S ∩ T) = f−1
(S) ∩ f−1
(T)
352

Review Problem 4.
Prove that if a , b , and c are integers such that
m ≥ 2 and a ≡ b(mod m)
then
gcd(a, m) = gcd(b, m) .
353

Review Problem 5.
Use mathematical induction to prove that
21 | (4n+1
+ 52n−1
) ,
whenever n is a positive integer.
354

Review Problem 6.
The Fibonacci numbers are defined as: f1 = 1 , f2 = 1 , and
fn = fn−1 + fn−2 , for n ≥ 3 .
Use a proof by induction to show that
fn−1 fn+1 − f2
n = (−1)n
for all n ≥ 2 .
355

Review Problem 7.
Let A and B be non-empty sets.
Let f be a 1 − 1 function from A to B .
Suppose S is an partial order on B .
Define a relation R on A as follows:
∀a1, a2 ∈ R : a1Ra2 ⇐⇒ f(a1)Sf(a2) .
Prove that R is an partial order on A .
356

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