discrete mathematics

1. UNIT-I: Relation
1.0 Introduction
1.1 Objective
1.2 Relation
1.3 Type of relation
1.4 composition of relations
1.5 Pictorial representation of relations
1.6 Closures of relations
1.7 Equivalence relations
1.8 Partial ordering relation.
1.9 Function
1.10 Various types of functions
1.11 Composition of function
1.12 Recursively defined function
1.13 Mathematical Induction: Piano’s axioms
1.14 Mathematical Induction
1.15 Discrete numeric functions
1.16 Generating functions
1.17 Simple recurrence relation with constant coefficients
1.18 Linear recurrence relation with constant coefficients.
1.19 Asymptotic behavior of functions
1.0 INTRODUCTION (Relation)
We start by considering a simple example.
Let S denote the set of all students at UPTec University,
Lucknow and
Let T denote the set of all teaching staff there.
For every student s∈S and every teaching staff t∈T,
exactly one of the following is true:
• s has attended a lecture given by t, or
• s has not attended a lecture given by t.
We now define a relation R as follows.
Let s∈S and t∈T.
We say that sRt if s has attended a lecture given by t. If
we now look at all possible pairs (s,t) of students and
teaching staff, then some of these pairs will satisfy the
relation while other pairs may not.
1

2. To put it in a slightly different way, we can say that the
relation R can be represented by the collection of all
pairs (s,t) where sRt. This is a sub collection of the set of
all possible pairs (s,t).
Formally, we define a relation in terms of these “ordered
pairs”.
Relations, as noted above, will be defined in terms of
ordered pairs (a, b) of elements, where a is designated as
the first element and b as the second element.
There are three kinds of relations which play a major role
in our theory:
(i) Equivalence relations,
(ii) Partial order relations,
(iii) Functions.
All these relations will be discussed here.
1.1 OBJECTIVE
After going through this unit-I you will be able to:
• Define a Relation and various types of relations
• Discuss a pictorial representation of relations.
• Explain the closure of reflexive, symmetric and transitive
relations.
• Define and explain the Equivalence relations and partial
order relation (POSet).
• Define and explain the difference between a relation and a
function.
• Discuss the various types of functions such as One-One,
into, onto and Inverse functions.
• Discuss the composition of functions
• Discuss the Recursively defined functions.
2

3. • Discuss various proof methods such as proof by
Counterexample, by Contra positive and by Contradictions
• Define and explain Piano’s axioms and mathematical
induction
1.1Some definitions required to define relation
Definition1: Ordered Pair:
Let A and B are two sets and let a∈A and b∈B then a set of two
elements whose elements have been listed in a specific order is
called an ordered pair. It is denoted by (a,b). Particularly:
 For different a and b: (a,b)≠(b,a) and
 If (a1,b1)=(a2,b2) ⇔ a1=a2 and b1=b2
Thus in case of relation (a,b)≠(b,a) unless a=b, whereas in case
of Sets, the order of elements is irrelevant; for example
{2,3}={3,2}.
Definition2: (Cartesian product of two sets):
Let A and B be two nonempty sets. The set A×B = {(a,b) :
a∈A and b∈B} is called the Cartesian product of the sets A
and B. In other words, A×B is the set of all ordered pairs
(a,b), where a∈A and b∈B. In short this product A×B is
read as “A cross B”.
Example1.1: Let A = {1, 2} and B= {a, b, c}. Then
A×B = {(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)}
BXA = {(a, 1), (a, 2) (b, 1) (b, 2), (c, 1), (c, 2)}
And AXA = {(1, 1), (1,2),(2,1),(2,2)}
Clearly, from this example, we can note down the
following points:
3

4.  A×B≠B×A
 If A has n elements and B has m elements than A×B
has m.n elements.
 If A=φ and (or) B=φ then A×B=φ
 If either A or B has infinite set then A×B is also an
infinite set.
 If A×B=B×A ⇔ A=B
1.2RELATION
Let A and B are two nonempty sets. A binary relation or,
simply, relation from A to B is a subset of A X B i.e.
R is a relation from A to B ⇔ R⊆ (A×B)
Example1.2: Let A = {1, 2,3} and B= {a, b, c}
Then A×B={(1,a),(1,b),(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c)}
R1={(1,a),(1,c)}
R2={(1,a),(2,a),(2,c)}
R3={(3,c)} are all examples of relations from A to B.
Suppose R is a relation from A to B (i.e. R⊆ (A×B)). Then R
is a set of ordered pairs where each first element comes
from A and each second element comes from B. That is,
for each pair a∈A and b∈B, exactly one of the following is
true:
(i) (a,b)∈R, we then say “a is R-related to b”. We write
aRb.
(ii) (a, b)∉R, we then say “a is not R-related to b”. We
write a Rb .
/
4

5. There are many instances when A = B. In this case, R is a
relation from a set A to itself i.e. R is a subset of A2=A X A.
Then we say that R is a relation on A.
Definition1: Domain and Range of a relation:
If R⊆ (A×B) is a relation from A×B, then
Domain(R)={a: (a,b)∈R} and
Range(R)={b: (a,b)∈R}.
The domain of a relation R is the set of all first elements of
the ordered pairs which belong to R, and the range of R is
the set of second elements.
Example2.3: In the example1.2 above, for relation
R2={(1,a),(2,a),(2,c)}
Domain(R)={1,2}
Range(R)={a,c}.
Example2.4: Let A={1,2,3,4}. Define a relation R on A by
writing (x,y)∈R if x < y. Then
R = {(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}.
Example2.5: Let A={1,2,3}. Define a relation R on A as
R={{a,b}: a is divisible by b. We have R = {(1,1),(2,1),(3,1),
(2,2),(3,3)}.
5

6. Example2.6:Let Abe the power set of the set {1,2} in
other words, A= {φ,{1},{2},{1,2}} is the set of subsets of the
set {1,2}. Write a relation on A, where (P,Q)∈R, if P⊂Q.
In this case we have:
R= {(φ,{1}), (φ,{2}), (φ,{1,2}), ({1},{1,2}), ({2},{1,2})}.
Example2.7 :If set A has n elements and B has m
elements, how many relations are there on the set A×B.
Let|A|=nand|B|=m.WeknowthatR⊆ (A×B) and |
A×B|=m.n, Also set of all possible subsets of A×B is power
set of A×B i.e. P(A×B).
Thus if | A×B|=m.n, then |P(A×B)|=2mn.
Hence If set A has n elements and B has m elements, then
there are 2mn relations on it.
Definition2: (Inverse Relation):
If R⊆ (A×B) is a relation from A×B then the inverse
relation of R (denoted by R-1), is a relation from B to A. It
is defined as R-1={(b,a): (a,b)∈R}.
Also the domain and range of R-1 are equal to the range
and domain of R. Clearly for any relation R, (R-1)-1=R.
Example2.8: Let A={1,2,3} and B={a,b,c}, if R is a relation
from A to B such that R={(1,b),(2,a),(2,b)};
Dom(R)={1,2}andRange(R)={a,b},then
R-1={(b,1),(a,2),(b,2)} and Dom(R-1)={a,b} and
range(R-1)={1,2}.
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7. Check your progress-1:
Q.1: Define the following: a) Identity relation b) Universal relation
c) Void relation
Q.2: Let A={1,2,3,4,5,6} and let R be a relation on A defined by “x
divides y”. Write R as a set of ordered pairs.
Q.3: Find the inverse relation on the relation R, above, i.e. “x is
multiple of y”.
Q.4: Let S be the relation on the set N of +ve integers, defined by
the equation x+3y=13 i.e.
S={(x,y): x+3y=13}. Find the relation S?
Q.5: Find the Domain and Range of the above relation?
Q.5: Find the inverse of the following relations
a) is shorter than b) “is younger than” c) “is child of” d) “is
a sibling of”
e) “is parallel to” f) “lies above” g) “is perpendicular to”
1.3 TYPES OF RELATIONS:
Let A be a given non empty set then a relation R⊆A×A is
called a binary relation on A. Binary relations that satisfy
certain special properties can be very useful in solving
computation problems. So let’s discuss some of these
properties:
We have following types of properties in a (Binary)
relation on a given set A.
1. Reflexive
2. Irreflexive
3. Symmetric
7

8. 4. Asymmetric
5. Anti-symmetric
6. Transitive
1. Reflexive Relations
A relation R on a set A is reflexive if for every a∈A, aRa. that is, a
relation R in a set A is said to be reflexive if every element of A is
related to itself i.e. aRa is true for every a∈A.
Definition2: (In terms of directed graph): R is reflexive if there
must be a loop at each node a∈A.
Example1: Let A be the set of all straight lines in a plane. The
relation R “x is parallel to y” is reflexive since every straight line
is parallel to itself.
Example2: Let A be the set of numbers and relation R in A is
defined by “x is equal to y” is reflexive” since each number is
equal to itself.
Example3: Let A={1,2,3} and the relation R in A is defined by
R={(1,1),(2,2),(2,3)} is not reflexive because (3,3) does not belongs
to R. The given relation R will be reflexive, if every ordered pair
(a,a)∈R for all a∈A.
Example4:
Consider the following five relations on the set A = {1, 2, 3, 4, 5}:
R1 = {(1, 1), (1, 2), (2,2),(2, 3), (1, 3), (3,3),(4,2),(4, 4),(5,5)}
R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (4, 4)}
R3 = {(1, 3), (2, 1)}
R4 = Ø, the empty relation
R5 = A X A, the universal relation
Determine which of the relations are reflexive.
Solution:
8

9. The only relations R1 and R5 are reflexive, since A contains the
five elements 1, 2, 3, 4 and 5, a relation R on A is reflexive if it
contains the five pairs (1, 1), (2, 2), (3, 3), (4, 4) and (5,5). Thus only
R1 and R5 are reflexive.
R2, R3, and R4 are not reflexive since, for example, (5, 5) does not
belong to any of them.
Check Your progress2:
Consider the following five relations:
1. Relation ≤ (Less than or equal ) on the set Z of integers
2. Set inclusion ⊆ on a collection C of sets
3. Relation ┴ (perpendicular) on the set L of lines in the plane.
4. Relation││ (parallel) on the set L of lines in the plane.
5. Relation │ of divisibility on the set N of positive integers.
(Recall x│y if there exists z such that xz = y.)
Determine which of the relations are reflexive.
The relation (3) is not reflexive since no line is perpendicular to itself.
Also (4) is not reflexive since no line is parallel to itself. The other
relations are reflexive; that is, x ≤ x for every integer x in Z, A⊆A for
any set A in C, and n│n for every positive integer n in N.
2. Irreflexive Relations
A relations R on a set is irreflexive if (a, a)∉R for every a є
A. Thus R is not irreflexive if there exist at least one a∈A
such that (a, a)∈R.
Definition2: (In terms of directed graph): R is Irreflexive if there
is no loop at any node a∈A.
Example1: Let A= {1,2,3} and let R= {(1, 1),(3,2)}.
9

10. Here R is not reflexive since (2,2) or (3,3)∉R. Also R is not
irreflexive, since (1, 1)∈R.
Example2: Let A={a,b,c}be a non empty set. Let R={(a,b),(b,c),(c,a)}
Here R is irreflexive since (a,a) )∉R for every a∈A. Also note that
there is no loop at any node.
3. Symmetric Relations:
A relation R on a set A is symmetric if ∀a,b in A, if aRb, then
bRa. In other words a relation R is symmetric if in R whenever
(a, b)∈R then (b, a)∈R.
Thus R is not symmetric if there exists a, b∈A such that (a,
b)∈R but (b, a)∉R.
Definition2: (In terms of directed graph): A relation (R) is
symmetric; if one node (x) is connected to node (y) then there must
be a return arc from node (y) to node (x).
• A relation R is said to be symmetric if R=R-1
Example1: Let A={set of all straight lines in a plane}. The Relation R
on A is defined by “ a is perpendicular to be” is a symmetric relation
because a⊥b ⇒ b⊥a.
Example2: Let N={set of Natural numbers}. The Relation R on N is
defined by “ a is equal to b” is symmetric since aRb⇒bRa.
Example3: Consider the following five relations on the set
A = {1, 2, 3, 4, 5}:
R1 = {(1, 1), (1, 2), (2,2),(2, 3), (1, 3), (3,3),(4,2),(4, 4),(5,5)}
R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (4, 4)}
R3 = {(1, 3), (2, 1)}
R4 = Ø, the empty relation
10

11. R5 = A X A, the universal relation.
Determine which of the above relations are symmetric.
Solution:
The relations R2, R4 and R5 are symmetric, since in R whenever (a,
b)∈R then (b, a)∈R .
The other relations R1 and R3 are not symmetric, since in R1, (1, 2)∈R1
but (2, 1)∉R1 and in R3; (1, 3)∈R3 but (3, 1)∉R3.
Example4: Each of the following defines a relation on set N of positive
integers:
R: x is greater than y
S: x+y=10
T: x+4y=10
Determine which of the relation are reflexive and which of them are
symmetric.
Solution: a) None are reflexive. For example (1,1) ∉ R , S or T.
b) Only S is symmetric, since whenever (a, b)∈S then (b, a)∈S.
The other relations R and T are not symmetric.
**************
4: Anti-symmetric:
A relation R on a set A is anti-symmetric if whenever aRb and
bRa then a = b. That is if (a,b),(b,a)∈R then there must be the
case that a=b. Thus R is not antisymmetric if there exists a, b∈A
such that (a, b) and (b, a) belong to R, but a ≠ b
• Definition2: A relation R is said to be Anti-symmetric if
(a,b)∈R⇒(b,a)∉R unless a=b.
• Definition3: (In terms of directed graph): R is anti-symmetric, if
whenever there is an edge from xy, with x≠y, then there is no
edge from yx.
Note: Symmetric and anti-symmetric relations are not negatives to each
other. For example, the relation R={(1,4),(4,1),(2,4)} is neither
symmetric nor anti-symmetric and the relation S={(1,1),(2,2)} is both
symmetric and anti symmetric.
11

12. Example1: Let A be a set of positive integers and R be a
relation on A such that
R={(a,b): a,b∈A and a≥b}. This relation R is an anti-symmetric
relation because if (a,b),(b,a)∈R ⇒a=b
Example2: Determine which of the following in above
example3 is antisymmetric.
Solution: R2 is not anti-symmetric since (1,2),(2,1)∈R2, but
1≠2. Similarly R5 is also not a anti-symmetric. All the other
relations are anti-symmetric.
5.Asymmetric Relation
A relation R on set A is asymmetric if (a, b)∈R then (b, a)∉R. It means
that R is not asymmetric if for some a and b from A both (a, b)∈R and
(b, a)∈R.
• Definition2: R is Asymmetric, if there is an edge from xy then
there must not be an edge from yx.
• Symmetric and Asymmetric relations are negatives to each other
i.e. if a relation is symmetric then it will not be asymmetric or
vice-versa.
Example1: Let A = {a, b, c, d} and let R = {(a, b), (b, b), (b, d),
(d, a)}. Determine whether the relation is symmetric,
asymmetric and anti-symmetric.
Solution: Here (a, b)∈R but (b, a)∉R, hence R is not symmetric.
Also since (b, b)∈R, R is not asymmetric and since if a ≠ b,
either (a, b)∉R or (b, a)∉R, ∀a,b∈A, the R is not anti-
symmetric.
Example 2
Let A be a set of integers and R be a relation on A such that
R = {(x, y)│ x ∈A, y∈A and x<y}.
Determine whether the relation is symmetric, asymmetric and
anti-symmetric.
Solution:
12

13. Here for every (x, y)∈R such that x<y, it is true that (y, x)∉R,
hence R is not symmetric. Also R is asymmetric. If x ≠ y, then
either (a, b) Є R or (b, a) Є R, hence R is antisymmetric.
6. Transitive Relations:
A relation R on a set A is said to be transitive if (a,b)∈R
and (b,c)∈R ⇒(a,c)∈R, for all a,b,c∈A.
In other words relation R is transitive if aRb and bRc
implies that aRc, for all a,b,c∈A
Thus R is not transitive if there exist a, b, c ∈A such that
(a, b), (b, c)∈R but (a, c)∉R.
Definition2: (In terms of directed graph): R is transitive, if
whenever there is an edge from xy and yz then there must also
be an edge from xz.
Example1: Determine which of the relations in above Example3
are transitive.
Solution: The relation R1 is not transitive since(4,2), (2,3)∈R1
but (4, 3)∉R1. All the other relations are transitive. Also the
relation R3 is not transitive since (2,1), (1,3)∈R3, but (2, 3)∉R3.
All the other relations are transitive relation.
Example2: Determine which of the relations are transitive.
a) Relation ≤ on the set Z of integers
b) Relation || on the set of line in the plane.
c) Relation set inclusion ⊆on the collection S of sets.
Solution: Relations (a) and (c) are transitive whereas (b) is not
transitive, since if a||b and b||a then a is not parallel to itself.
1.4: Composition of Relations
13

14. Let A, B and C is sets, and let R1 be a relation from A to B
and R2 be a relation from B to C.
That is, R1⊆ (A×B) and R2⊆ (B×C).
Then the composite of R1 and R2 is a relation from A to C,
denoted by R2οR1 and defined by
R1οR2={(a,c): there exists b∈B such that (a,b)∈R1 and (b,c)∈R2}
The relation R1οR2 is called the composition of R1 and R2.
Suppose R is a relation on a set A, that is R1⊆ (A×A). Then
RοR, the composition of R with itself is always defined. RοR is
sometimes denoted by R2.
Similarly, R3= RοRοR, and so on. Thus Rn is defined for all
positive n.
To find composition of relations using Matrix form:
We can also find the composition of relations R1 and R2 (i.e.
R1οR2) using matrices. Suppose MR1 and MR2 denotes the
matrices of the relations R1 and R2. Then by multiplying MR1
and MR2, we get the matrix MR1.MR2(=M). The nonzero entries of
this matrix M gives us the elements related to R1οR2.
Example 2.12:
Let A = {1, 2, 3}, B = {a, b, c}, C= {x, y, z}. Consider the following
relation R1 from A to B and R2 from B to C.
R1= {(1, b), (2, a), (3, c)} and
R2= {(a, z), (b, x), (c, y), (c, z)}
a) Find the composite relation R1οR2.
b) Find the matrix for MR1οR2 and compare the results obtained
in part (a).
Solution: (a) The arrow diagram of the relations R1 and R2 is
shown in figure-1.
A B C
1 a x
b 14
2 y
3 c z
Figure-1

15. Since 1 in A is connected to x in C by the path 1bx , so (1,x)∈ R1οR2..
Similarly 2 in A is connected to z in C by the path 2az , so (2,z)∈ R1οR2
3 in A is connected to y in C by the path 3cy , so (3,y)∈ R1οR2
3 in A is connected to z in C by the path 3cz , so (3,z)∈ R1οR2
So finally R1οR2={(1,x),(2,z),(3,y),(3,z)}
(b) The matrices for MR1 , MR2 and MR1oR2 can be obtained as
follows:
a b c x y z x y z
1 0 1 0 a 0 0 1 1 1 0 0
MR1= 2 1 0 0 MR2= b 1 0 0 MR1. MR2 = 2 0 0 1
3 0 0 1 c 0 1 0 3 0 1 1
The non zero entries in the matrix MR1.MR2 (=M) gives a
elements belongs to R1oR2. So
R1οR2={(1,x),(2,z),(3,y),(3,z)}
Theorem 2.1: (Show that the composition of relations is Associative).
Let A, B, C and D be sets. Suppose
R is a relation from A to B,
S is a relation from B to C, and
T is a relation from C to D. Then (RοS)οT = Rο(SοT)
Solution:
L.H.S.: Suppose (a,d)∈ (RοS)οT ⇒ (a,c)∈RοS and (c,d)∈T
Since (a,c)∈RοS ⇒ (a,b)∈R and (b,c)∈S
Now (SοT) ⇒ (b,d)∈SοT
Since (a,b)∈R and (b,d)∈SοT ⇒(a,d)∈ Rο(SοT)
Hence proved.
Note: 1) composition of relations is Associative i.e. (RοS)οT =
Rο(SοT).
15

16. 2) (RοS)≠(SοR)
Check your progress-3:
Q.1:
Let R = {(1, 2), (3, 4), (2, 2)}and S = {(4, 2), (2, 5), (3, 1), (1, 3)}.
Compute RοS, SοR, Rο(SοR), (RοS)οR, RοRοR.
Solution:
a) RοS = {(1,5), (3, 2), (2, 5)}
SοR = {(4, 2), (3, 2), (1, 4)}. Clearly RοS≠ SοR
b) Rο(SοR)= {(3, 2)}.
(RοS)οR = {(3, 2)}. Clearly Rο(SοR)= (RοS)οR.
c) RοRοR= {(1, 2), (2, 2)}
Q.2:
1.5: PICTORIAL REPRESENTATION OF RELATIONS
The diagrammatical representation (also called graph) of a
relation R is called a pictorial representation of the given relation
R.
Example 2.9
Consider the relations R defined by the equation: x2 + y2 = 36
That is, R consists of all ordered pairs (x, y) which satisfy the
given equation. The graph of the equation is a circle having its
center at the origin and radius 6. The pictorial representation
(graph) of this equation is shown in figure-1
6
-6 0 6
0
-6
16

17. Figure-1: Graphical representation of the equation: x2 + y2 = 36
Method to represent a given relations (on finite set) in pictorial
form:
If A and B are finite sets, then there are two ways of picturing a
relation R from A to B.
1. By using matrix of the relation
2. By using Arrow diagram
3. By using directed graph of relations
• Matrix of the relation (denoted by MR) is a 2D rectangular
array whose rows are labeled by the elements of A and
whose columns are labeled by the elements of B. Put a 1 or
0 in each position of the array according as follows:
1 if (a,b)∈R
MR= 0 if (a,b)∉R
Where a∈A and b∈B .
• In arrow diagram, we write down the elements of A and B in
two disjoint disks, and then draw as arrow from a∈A to b∈B
whenever (a,b)∈R.
• Directed graph method is used when R is a relation from a
finite set to itself. .First we write down the elements of the
set, and when we draw an arrow from each element a to
each element b whenever a is related to b.
Example2.10:
Suppose A={1,2,3} and B={a,b,c}, if R is a relation from A to B such
that R={(1,b),(2,a),(2,c),(3,c)}
The following figure-a and figure-b shows these two ways of
representation.
17

18. a b c
1 0 1 0 1 a
2 1 1 0 2 b
3 0 0 1 3 c
Figure-a Figure-b
Example2.11: Let A={1,2,3,4} and a relation R from A to itself i.e.
R⊆ (A×A) is defined as:
R= {(1, 2), (2, 2), (2, 4), (3, 2), (3,3), (3, 4), (4, 1)}
1 2
3 4
Check your progress-2:
Q.1: Given a relation R={(1,y),(2,z),(3,y),(4,x),(4,z)} on sets
A={1,2,3,4} and B={x,y,z}
a) Draw the arrow diagram of R
b) Find the Matrix of R
c) Find the Inverse relation (R-1) of R in matrix form.
d) Determine the Domain and Range of R.
Q.2: Draw the directed graph of the following relation on a set
A={1,2,3,4}
i. R={(1,2)(2,2),(2,4),(3,2),(3,4)(4,1),(4,3)}
ii. R={(1,1),(2,2),(2,3),(3,2),(4,2),(4,4)}
1.6: CLOSURE OF RELATIONS:
If R is a binary relation and p is some property, and then the p
closure of R is the smallest binary relation containing R that
18

19. satisfies property p. Our goal is to construct closures for the
reflexive, symmetric and transitive properties. These are
commonly known as:
• Reflexive closure (denoted by r(R))
• Symmetric closure (denoted by s(R))
• Transitive closure (denoted by t(R))
Closures of relations are used to make the given relation
Reflexive, Symmetric and Transitive, whenever the given relation
is not in proper form (i.e. reflexive, symmetric or transitive).
Reflexive, Symmetric & Transitive Closures
Let R be a relation on a set A. Then reflexive and symmetric closure of
R, is defined as:
1. r(R)=R ∪ DA , where DA={(a,a):a∈A} is the diagonal or
equality relation on A.
2. s(R)=R ∪ R-1, where R-1 is the inverse relation of R, i.e.
R-1={(y,x): (x,y)∈R}
3. t(R)=R∪R2∪R3∪…..
4. If A is finite with n elements, then t(R)= R∪R2∪R3∪….Rn.
In other words, If R is a binary relation on A, then the reflexive
closure r(R) can be constructed by including all pairs (a,a) that are
not already in R, and
To construct symmetric closure s(R), we must include all pairs for
(y,x) for which (x,y)∈R.
To construct Transitive closure t(R), if R contains the pairs (a,b)
and (b,c) then t(R) must contains the pair (a,b), for all a,b,c∈A.
19

20. Finding t(R) can take a lot of time when A has a large number of
elements. There exist an efficient way for computing t(R), known
as warshall’s algorithm (which may be discussed latter).
Example 1
Consider the following relation R on the set A = {1, 2, 3, 4}:
R = {(1, 1), (1, 3), (2, 4), (3, 1), (3, 3), (4, 3)}.
Then r( R ) = R ∪ DA =R ∪ {(2, 2),(4, 4)}={(1, 1), (1, 3), (2, 4), (3, 1),
(3, 3), (4, 3),(2,2),(4,4)}
and s( R) = R ∪ R-1= R∪{(4, 2), (3, 4)}= {(1, 1), (1, 3), (2, 4), (3, 1),
(3, 3), (4, 3),(4,2),(3,4)
t(R)= R∪R2∪R3∪R4={…..}
Example2: Determine which of the following is transitive relation:
a) Relation ≤ on the set Z of integers.
b) Relation || on the set of lines in the plane.
Solution: a) The relation ≤ is transitive, since a≤b and b≤c then a≤c.
b) The relation || is not transitive, since a||b and b||a then a is
not || to a itself.
1.7: EQUIVALENCE RELATIONS (or RST relation):
A relation R in a set A is said to be an equivalence relation if
1. R is Reflexive i.e. aRa ∀a∈A.
2. R is Symmetric i.e. aRa⇒bRa, ∀a,b∈A.
3. R is Transitive i.e. If aRb and bRc⇒ aRc ∀a,b,c∈A.
Example1: Relation R on a set A defined by “a is equal to b” is
an equivalence relation, since it is reflexive (a = a for every
a∈A), symmetric (If a = b, then b = a , ∀a,b∈A) and transitive
(If a = b and b = c, then a = c, ∀a,b,c∈A).
20

21. Example2: Relation R in a set A defined by “ x is parallel to y”
is also an equivalence relation since it it is reflexive, symmetric
and transitive.
Example3: The relation ⊆ (set inclusion) is not an equivalence
relation, since it is reflexive and transitive, but it is not
symmetric. Since A⊆B does not imply B⊆A.
Example4: A = {1, 2, 3, 4, 5}. Let R be relation on A such that
R = {(x, y) │ x + y = 5}
We get R as {(1, 4), (2, 3), (4, 1), (3, 2)}. We can say that, R is
not reflexive as for every a, (a, a)∈R. R is symmetric as if (a,
b)∈R then (b, a)∈R and R is antisymmetric relation also. R is
not transitive as (1, 4) ∈R and (4, 1)∈R but (1, 1)∉R. Hence R
is not equivalence relation.
PARTIAL ORDERING RELATIONS (POR):
1.8
A relation R on a set A is called a Partial ordering or a partial
order relation, if it is:
1. Reflexive, i.e. aRa ∀a∈A.
2. Anti-symmetric, i.e. If aRb and bRa⇒ a=b ∀a,b∈A.
3. Transitive i.e. If aRb and bRc⇒ aRc ∀a,b,c∈A.
The set over which a partial order is defined is called a partially ordered
set (or POSET). It is denoted by (A,R) where A is a given set and R is a
relation which satisfy the above three conditions. POSET is discussed in
detail, in the Lattice chapter (Unit-II).
Example1:The relation ⊆ of set inclusion is a partial ordering on any
collection of sets since set inclusion has the following three desired
properties.
Reflexive, i.e. a ⊆ a ∀a∈A.
Anti-symmetric, i.e. a ⊆ b and b⊆ a⇒ a=b, ∀a,b∈A
Transitive i.e. if a⊆b and b⊆c, then a⊆c ∀a,b,c∈A.
Example2: The relation ≤ (less than or equal to) on the set R of real
numbers is a partial order relation.
Since the relation (≤) is:
1. Reflexive i.e. a≤a, ∀a∈R
21

22. 2. Anti-symmetric i.e. a ≤b and b≤ a⇒a=b, ∀a,b∈R
3. Transitive i.e a≤b and b≤c, ⇒a≤c ∀a,b,c∈R
Example3: Let N be the set of all positive integers. The relation “a
divides b” is a partial ordering on N. However, “a divides b” is not a
partial ordering on the set Z of integers since a|b and b|a does not imply
a = b. For example, 2|-2 and - 2│2 but 2 ≠ -2.
Check your progress-3:
1.9: Functions:
A function is a special kind of relation. For example suppose
X= the set of students of UP Technical university, and
Y=the set of their enrolment numbers.
Now consider a relation R between A to B, i.e. R ={(a,b)∈AxB |
b is enrollment number of a }.
It is a ‘special’ relation, because to each a∈A ∃! b such that aRb.
We call such a relation a function from A to B.
Definition:
Suppose X and Y be two nonempty sets. A rule or a
correspondence which assign each element x∈X to a unique
element y∈Y is called a function or mapping from X to Y and
written as
f: XY (read as “ f is a function which maps X into Y)
• The main idea is that each element of X is associated with
exactly one element of B. In other words if x∈X is associated
with y∈Y, then x is not associated with any other elements of Y.
• The element y is called the image of x under f and is denoted by
f(x) i.e. y=f(x)
• The element x is called the pre-image of y.
22

23. • Domain: The set X is called the domain of the function f, and
• Co-domain: The set B is called the co-domain of the function f.
• Range of f: The range of f, denoted by Range (f), is the subset of
elements in the co-domain Y that are associated with some element
of X. In other words Range (f)={f(x): x∈X}. It is also denoted as
f(X).
• Given an element x∈X, the unique element of Y to which the
function f associates, is denoted by f (x) and is called the f-image
(or image) of x or the value of the function f for x. We also say
that f maps x to f(x). The element x is referred to as the pre-image
of f (x).
Example1: If A = {1,2,3,4,5}, B = {1,8,27,64,125}, and the rule f
assigns to each member in A its cube, then f is a function from A to B.
The domain of f is A, its co-domain is B and its range is {1,8,27,64}.
Example2: Find the domain and range for f : f(x) =x/1-x
Solution: We can see that 1–x = 0, if x = 1, in this case f(x) will be
undefined.
Domain of f can be taken as R~{1}and co-domain can be R.
(Necessary conditions to be a function f: XY):
• A single element in domain X cannot have more than one image in Y. However, two or more than
two elements in X may have the same image in Y (see figure-1).
• Every element in domain X must have its image in Y but every element in Y may not have its pre-
image in X (otherwise it is not a function, see figure-
a 1 a 1
b 2 b 2
c 3 c 3
Figure1-2: Two associations (mapping) that are not function
23

24. Frequently, a function can be expressed by means of a
mathematical formula. For example, consider the function f,
which maps each natural number N to its square. We may
describe this function by the formula:
f(x) = x2 or y = x2
In the first notation, x is called a variable and the letter f denotes
the function.
In the second notation, x is called the independent variable and
y is called the dependent variable since the value of y will
depend on the value of x.
Remark: Whenever a function is given by a formula in terms of
a variable x, we assume, unless it is otherwise stated, that the
domain of the function is R (or the largest subset of R for which
the formula has meaning) and the co-domain in R.
Example1: Suppose A ={1,2,3}, B= {1,4,9,11} and f assigns to
each member in A its square values. Then f is a function from A
to B.
But if A={1,2,3,4}, B={1,4,9,10} and f is the same rule, then f is
not a function from A to B since no member of B is assigned to
the element 4 in A.
Example1: Let X={1,2,3,4,5} and Y={a,b,c,d,e}. Determine
whether or not each relation below is a function from XY. If
they are functions, give the domain, co-domain and Range of
each, if they are not tell why?
a) f={(1,a),(2,b),(3,b),(5,e)}
b) g={(1,e),(5,d),(3,a),(2,b),(1,d),(4,a)}
c) h={(5,a),(1,e),(4,b),(3,c),(2,d)}
24

25. Solution:
a) f is not a function, since 4∈X is not associated with any element of
Y or f(4) don’t have any image inY.
b) g is not a function since 1∈X is associated with two different
elements, namely e and d.
c) h is a function from X to Y since each member of X appears as the
first coordinate in exactly one ordered pair in function (say f); here
f(1)=e, f(2)=d, f(3)=c, f(4)=b and f(5)=a. Domain={1,2,3,4,5}, Co-
domain={a,b,c,d,e} and Range(h)={a,b,c,d,e}
Remark: Every relation is not necessarily a function but
every function is a relation.
Example2: State whether or not each diagram in figure-1 defines a
function from X={a,b,c} into Y={1,2,3}.
a 1 a 1 a 1
b 2 b 2 b 2
c 3 c 3 c 3
Figure-a figure-b Figure-c
Solution:
a) No. There is nothing to assigned to the element b∈A
b) No. f(c) has two images x and z in Y.
c) Yes.
Difference between Relation and function:
• Every function f: XY gives rise a relation from X to Y.
• Every relation (R⊆X×Y) is not necessarily a function (f: XY)..
Example1: Let X={1,2,3,4} and Y={a,b,c}. Consider the following
relations R1 and R2 (i.e. R1,R2⊆X×Y:
25

26. R1={(1,a),(2,a),(3,b),(4,c)}
R2={(1,a),(2,b),(1,c),(3,a),(4,b)}
Determine whether or not each relation below is a function from X to Y.
Solution: Yes, R1 is a function from X to Y. Each element of X appears
as the first element in one and only one ordered pair in R1, i.e. every
element in X must have its image in Y. Obviously R1 is also a relation
from X to Y. But R2 is not a function from X to Y since 1 is associated
with two different elements a and c of Y.
Hence every relation is not necessarily a function.
Example 2: Consider the following relations on the set A = {1, 2, 3}:
f = {(1, 3), (2, 3), (3, 1)}
g= {(1, 2), (3, 1)}
h= {(1, 3), (2, 1), (1, 2), (3, 1)}
Determine whether or not each relation above is a function from A into
A.
Solution:
f is a function from A into A since each member of A appears as the first
coordinate in exactly one ordered pair in f; here f(1) = 3, f(2) = 3 and
f(3) =1. g is not a function from A into A since 2 Є A is not the first
coordinate of any pair in g and so g does not assign any image to 2. Also
h is not a function from A into A since 1 Є A appears as the first
coordinate of two distinct ordered pairs in h, (1, 3) and (1, 2). If h is to
be a function it cannot assign both 3 and 2 to the element 1 Є A.
Equality of two functions: Two functions f and g of AB are said to
be equal iff f(x)=g(x) for all x∈A, and we write f=g. Note that two equal
function f and g are defined on same domain A.
For two unequal mappings from A to B, there must exist at least one
element x∈A such that f(x)≠g(x).
Example1: If f(x)=x3+1 where x is any real number and g(x)= x3+1
where x is any complex number then f≠g because the domain of f and g
are different.
Example2: Let A={3,4} and B={2,4,9,16}. Let a function f be defined
from A to B by f(x)= x2 and g={(3,9),(4,16)}, then f=g because f and g
26

27. both have the same domain={3,4} and each of them assigns the same
image to each element in the domain.
1.10 TYPES OF FUNCTIONS (or mappings):
Let f: AB is a given function. We have following types of functions
(or mapping) between A to B.
1. One-to-one mapping
2. Onto mapping
3. Invertible function (i.e. one-one onto)
4. Into mapping
5. Many one
6. Many one onto
7. Many one into
1.0 One-one mapping:
A function f: A → B is said to be one-to-one or one-one or
injective, written as 1-1, if different elements in the domain A
have different images in B. Another way to say this is that f is
one-one if f(x) =f(y) implies x= y (x,y∈A). An injective
function is called an injection. The following figure-1 illustrates
an injection from AB.
1 a
2 b
3 c
d
Figure-1: An injection
For example f: RR be defined by f(x) = 2x+1, x∈R, then for
x1 ,x2∈R (x1≠x2) we have f(x1)≠f(x2). So, f is 1–1.
27

28. 2: Onto mapping
A function f: A → B is said to be an onto or surjective if each
element of B is the image of some element of A. In other words,
f: A → B is onto if the image of f is the entire co-domain, that is,
f(A)=B. [Equivalently, we say that f is onto if Range(f)=B]. If a
function is mapping of A onto B, we write:
onto
f: A B.
An surjective function is called an surjection. The following
figure-2 illustrates an injection from AB.
1 a
2 b
3 c
4
Figure-2: A surjection
For example, f: ZZ : f(x) = x+1, x∈Z, then every element y in
the co-domain Z has a pre-image y–1 in the domain Z.
Therefore, f(Z) = Z ,and f is an onto mapping.
3: Invertible (one-one and Onto) Mapping
A function f; A→ B is invertible (or bijective) if its inverse
relation f -1 is a function from B to A. In general, the inverse
relation f -1 may not be a function.
Another term for bijective is “one-to-one and onto”. A bijective
function is called a bijection or a “one-to-one correspondence”
The following figure-3 illustrates an bijection from AB.
1 a
2 b
3 c
4 d 28
Figure-3: A bijection

29. The following theorem gives a necessary condition for invertible (or
one-one and onto) function:
Theorem1: A function f: A → B is invertible if and only if f is both
one-to-one (i.e. 1-1) and onto.
• If f: A→ B is both one-to-one and onto, then f is called a one-to-
one correspondence between A and B. This is called so because
each element of A corresponds to a unique element of B and vice
versa.
• In this case, each element of A maps to a distinct element of B, and
vice-versa.
For example, f: ZZ : f(x) = x+2, x∈Z is both injective and surjective.
So, f is bijective.
Example1: Consider the functions e: AB, f: BC, g: CD and h;
DE defined by the following diagram:
Determine whether the above functions are one-one, onto and
invertible function.
Solution:
e: is one-to-one as no element in B is image of more than one element in A.
f: is one-to-one as no element in C is image of more than one element in B. but
g: is not one-to-one as g(r)=g(u)=v
h: is not 1-1 as h(v)=h(w)=z.
e: is not onto since 3∈B is not the image under e of any element of A.
f: is onto since every element of C is the image under f of some element of B, i.e.
f(B)=C
29

30. g: is onto since every element of D is the image under g of some element of C, i.e.
g(C)=D.
h: is not onto since x,y∈E are not the image under h of any element of D.
only f is invertible since f-1 is exist and it is a function from C to B.
Example2: Let A be set of employees of a company and let B be the set
of their telephone extensions. Assuming that all the employees are listed
and that every person has his/her own extension, then the mapping from
A to B is invertible.
Example3:Thefunctione: RR; e(x) = x2 is neither 1-1 nor
onto.
Example4: Thefunctionf: RR; f(x) = 2x is 1-1 and not onto.
Example5: the function h(x) = x3 is 1-1 and onto.
Note:increasingfunctionsare1-1.
Example6: corman p.163:
Example1:
Example2:
Example3:
4: Into mapping: A function f: A → B is said to be into
function if there exits at least one element in (co-domain set) B
which is not the f-image of any element in (domain set) A. We
say that f is a mapping of A ‘into’ B. In this case the range of f
is a proper subset of the co-domain of f i.e. f(A)⊂B.
1 a
2 b
3 c
Figure-2: An into mapping
5: Many one: A mapping f: A → B is said to be many-one if
two (or more than two) distinct elements in A has the same
image in B i.e. x,y∈A and x≠y ⇒f(x)=f(y).
-2 4
Figure-2: Many-one mapping 9 30
2
3
(Because two different elements in A (i.e. -2,2} have the same
f-image in B) 4 16

31. 6: many-one onto: A mapping f: A → B is said to be many-one
onto mapping if it is many-one and onto.
In such a mapping following two conditions are hold:
a) if x,y∈A and x≠y ⇒f(x)=f(y) and
b) the image of f is the entire co-domain, that is, f(A)=B
Example1: If A={2,3,4}, B={x,y}, then f={(2,x),(3,y),(4,y)} is a many-
one onto mapping.
7: Many one into:
A mapping f: A → B is said to be many-one into if it is many-one and
into.
In such a mapping following two conditions are hold:
a) if x,y∈A and x≠y ⇒f(x)=f(y) and
b) if there exits at least one element in B which is not the f-image of
any element of A.
Example1: If A={-1,1,-3,3,4} and B={1,9,16}. Show that f:
AB={(x,y): y=x2, x∈A, y∈B} is a many-one onto mapping.
Proof: i) As f(-1)=f(1)=1; f(-3)=f(3)=9; and f(4)=16
So every element of A has f-image in B. Thus it is a mapping.
ii) f(-1)=f(1)=1 and f(-3)=f(3)=9 so two elements of A have the
same image in B, thus it is many-one mapping
iii) Every element of B is f-image of some element of A. Thus it is
onto mapping. Therefore f is a many-one onto mapping.
Example1(P-68,69)-book
Example2: (p3.23)(SS)
Check your progress-4:
31

32. 1.11COMPOSITION OF FUNCTION
Let A,B,C be three sets and f and g be two functions such that f: AB
and g: BC; where the co-domain of f is the domain of g. Then we may
define a new function from A to C [(gof): AC], called the product or
composition of f and g and denoted by (gof), as follows:
(gof)(x)=g[f(x)], ∀x∈A
That is, to find (gof)(x), first we have to find the image of x under f and
then find the image of f(x) under g.
Note: if we think f and g as relation then the composition of function is
same as the composition of relation. Only the notation is different. We
use the gof for the composition of f and g, instead of fog, which is used
for relation.
Theorem: (Associativity of composites of functions):
Let A,B,C and D be four sets and f, g, h be three functions such that f:
AB, g: BC, h: CD then
(hog)of=ho(gof)
Proof: It can be easily seen that both (hog)f and ho(gof) are mapping of
AD. These two mappings will be equal if they assign the same image
to each element x in the domain A. i.e. if
[(hog)of](x)=[ho(gof)](x)
By the definition of composites of functions,
[(hog)of](x)=(hog)[f(x)]=h[g{f(x)}]=h[(gof)(x)]=[ho(gof)](x).
Hence (hog)of=ho(gof)
Example 1,2,3,4,5 (pundir)-p-32,33
Example3.8,3.9.3.10 (SS)-p-3.23,3.24,3.27.3.28.
1.12RECURSIVELY DEFINED FUNCTIONS:
A function is said to be recursively defined if the function definition
refers to itself or call itself repeatedly. The recursive definition must
have the following two properties:
32

33. A recursive definition has two parts:
1. Base step (B): Definition of the smallest argument (usually f (0) or
f (1)), for which the function does not call itself.
2. Recursive step (R): Definition of f (n), given f (n - 1), f (n - 2), etc.
In other words a rule (or rules) that show how to construct new
elements of f from old ones. The argument of f must be closure to
the Base value as defined in step-1.
Here is an example of a recursively defined function:
We can calculate the values of this function:
f (0)=5
f (1)= f (0) + 2 = 5 + 2 = 7
f (2)= f (1) + 2 = 7 + 2 = 9
f (3)= f (2) + 2 = 9 + 2 = 11
This recursively defined function is equivalent to the explicitly
defined (by a formula in terms of the variable) function f (n) = 2n + 5.
However, the recursive function is defined only for nonnegative
integers.
Here is another example of a recursively defined function:
The values of this function are:
f (0)=0
f (1)= f (0) + (2)(1) - 1 = 0 + 2 - 1 = 1
f (2)= f (1) + (2)(2) - 1 = 1 + 4 - 1 = 4
f (3)= f (2) + (2)(3) - 1 = 4 + 6 - 1 = 9
f (4)= f (3) + (2)(4) - 1 = 9 + 8 - 1 = 16
…….
This recursively defined function is equivalent to the explicitly defined
function f (n) = n2. Again, the recursive function is defined only for
nonnegative integers.
33

34. Here is one more example of a recursively defined function:
The values of this function are:
f (0)= 1
f (1)= 1⋅f (0) = 1⋅1 = 1
f (2)= 2⋅f (1) = 2⋅1 = 2
f (3)= 3⋅f (2) = 3⋅2 = 6
f (4)= 4⋅f (3) = 4⋅6 = 24
f (5)= 5⋅f (4) = 5⋅24 = 120
…….
This is the recursive definition of the factorial function, F(n) = n!. Thus
a factorial function may also be defined as:
a) If n=0, then f(n)=n!=1 {here 0 is a base value}
b) If n>0, then f(n)=n!=n.(n-1)! { Here the n! is defined in terms of
smaller value of n which is closure to the base value 0}.
Observe that this (and all the above) definition are recursive since all
these functions definition are refers to itself.
Note: Not all recursively defined functions have an explicit definition
(by a formula in terms of the variable).
Example1:
Let us calculate 3! using the recursive definitions. This calculation requires the
following seven Steps:
1) 3! = 3 · 2!
2) 2! = 2 · 1!
3) 1! = 1 · 0!
4) 0! = 1
5) 1! = 1 · 1 = 1
6) 2! = 2 · 1 = 2
7) 3! = 3 · 2 = 6
34

35. Here we first defines 3! In terms of 2!, so we must postpone evaluating
3! Until we evaluate 2!. In step2 we define 2! in terms of 1!, so we must
postpone evaluating 2! Until we evaluate 1! An so on.
step5 can explicitly evaluate 0!(=1), which is the base value (i.e. 0) of
the recursive definition.
In step5 to step7, we backtrack, using 0! to find 1!, using 1! to find 2!,
and finally using 2! to find 3!. The final answer is 6.
2. The Fibonacci Numbers:
One special recursively defined function, which has no simple explicit
definition, yields the Fibonacci numbers (usually denoted by fn, f1, f2,
….}. The main characteristic of Fibonacci number is that the sums of
two previous numbers give a next number. If f0=0 and f1=1(base value)
then the sequence of Fibonacci numbers is:
0, 1, 1, 2, 3, 5, 8, 13, 21,…….
Thus a Fibonacci number can be defined recursively as:
a) If n = 0 or n= 1, then Fn = n. {here 0 and 1 are base values}
b) If n > 1, then Fn = Fn – 2 + Fn – 1. {Here Fn is defined in terms of
smaller value of n which is closure to the base values 0 and 1}.
3. Ackermann Function:
The Ackermann function is a function with two arguments, each of
which can be assigned any non-negative integer, that is, 0, 1, 2 … The
Ackermann function is defined recursively for non-negative integer’s m
and n as follows:
35

36. n+1 if m=0
A(m,n)= A(m-1,1) if m>0 and n=0
A(m-1, A(m,n-1)) if m>0 and n>0
Observe that A (m, n) is explicitly given only when m = 0. The base
criteria are the pairs
(0, 0), (0, 1), (0, 2), (0, 3),… (0, n), …
Although it is not obvious from the definition, the value of any A(m, n)
may eventually be expressed in terms of the value of the function on one
or more of the base pairs.
Lets see some special values for integer m:
A(0,n)=n+1
A(1,y)=n+2
A(2,n)=2n+3
A(3,n)=2n+3-3
A(4,n)=
Expressions of the latter form are sometimes called
power towers.
A(0,n) follows trivially from the definition. A(1,n) can be derived as
follows:
A(1,n)=A(0,A(1,n-1))
=A(1,n-1)+1
=A(0,A(1,n-2))+1
=A(1,n-2)+2
=….
=A(1,0)+n
=A(0,1)+n=2+n
Similarly we can calculate A(2,n)=2n+3.
Table of values of A(m,n): The following tables can be used for calculating the value for A(m,n).
mn 0 1 2 3 4 n
0 1 2 3 4 5 n+1
1 2 3 4 5 6 n + 2 = 2 + (n + 3) − 3
2 3 5 7 9 11 2n + 3 = 2 * (n + 3) − 3
36

37. 3 5 13 29 61 125 2(n + 3) − 3
65536
4 13 6553 2 −3 22↑65536-3 A(3, A(4, 3)) 2 (2 (2 (... 2))) - 3 (n + 3 twos)
3 =2 (n + 3) - 3
5 6553 A(4, A(5, 1) A(4, A(5, 2) A(4, A(5, 3)) A(4, A(5, n-1))
3 ) )
*see Wikipedia.
Note that from the table:
A(1, n) = 2 + (n + 3) - 3
A(2, n) = 2 × (n + 3) - 3
A(3, n) = 2↑(n + 3) - 3
A(4, n) = 2 (2 (2 (... 2))) - 3 (n + 3 twos) = 2 (n + 3) - 3
A(5, n) = 2 (n + 3) - 3 etc.
Example1:
By using the definition of the Ackermann function, find the value of
A(1,3).
Solution:
1. A(1,3)=A(0,A(1,2))
2. A(1,2)=A(0,A(1,1))
3. A(1,1)=A(0,(1,0))
4. A(1,0)=A(0,1)
5. A(0,1)=1+1=2
6. A(1,0)=2
7. A(1,1)=A(0,2)
8. A(0,2)=2+1=3
9. A(1,1)=3
10. A(1,2)=A(0,3)
11. A(0,3)=3+1=4
12. A(1,2)=4
13. A(1,3)=A(0,4)
14. A(0,4)=4+1=5
15. A(1,3)=5
So here we can see that the value of A (1, 3) is requires 15 steps. In the
above calculation Forward movement means we are postponing an
evaluation and calling the definition. The backward movement means
we are backtracking. The result of A(1,3)=5 can also be verified from
the above table.
37

38. Check your progress-5:
Q.1 P-3.33, Q3.30,3.31(SS)
1.13: Piano’s Axiom
In mathematical logic, the Peano axioms, also known as the Dedekind-Peano axioms or the Peano
postulates, are a set of axioms for the natural numbers presented by the 19th century Italian
mathematician Giuseppe Peano. The Peano axioms define the properties of natural numbers,
usually represented as a set N. Peano defines natural numbers in terms of three postulates (called
Peano’s Pastulates in his honour).
N is a set with the following properties:
P1: 1 is a natural number i.e. 1∈N
P2: For each n∈N, there exist a unique successor n*∈N such that
i) m*=n* if and only if m=n {i.e. Two numbers of which the successors are equal are themselves
equal}.
ii) There is no element k∈N such that k*=1 {i.e. for every natural number n, n* ≠ 1. That is, there is
no natural number whose successor is 1}.
P3: (induction axiom):
Let K be a subset on N such that
• 1∈K and
• For every natural number n, if n is in K, then n* (successor of n) is in K,
then K contains every natural number i.e. K=N.
(i.e. If a set K of natural numbers contains 1 and also the successor of every natural number in K, then every
natural number is in K}.
The elements of N are called the natural numbers. The successor n* of n is nothing but n+1 in
the usual sense. Thus
2=1*
3=2*=(1*)*
4=3*=(2*)*=((1*)*)* etc.
In this way all the natural numbers can be obtained.
1.13: Mathematical Induction
P-105(DU book)
38

39. P-1.22 (SS)
Example-1,2,3,(P-105)
39