UNIT-I: Relation1.0 Introduction1.1 Objective1.2 Relation1.3 Type of relation1.4 composition of relations1.5 Pictorial representation of relations1.6 Closures of relations1.7 Equivalence relations1.8 Partial ordering relation.1.9 Function1.10 Various types of functions1.11 Composition of function1.12 Recursively defined function1.13 Mathematical Induction: Piano’s axioms1.14 Mathematical Induction1.15 Discrete numeric functions1.16 Generating functions1.17 Simple recurrence relation with constant coefficients1.18 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 sS and every teaching staff tT, 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 sS and tT. 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. To put it in a slightly different way, we can say that the relation R can be represented by the

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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. Discuss various proof methods such as proof by

Counterexample, by Contra positive and by Contradictions Define and explain Piano’s axioms and mathematical

induction

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1.1 Some definitions required to define relation

Definition1: Ordered Pair:Let A and B are two sets and let aA and bB 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) andIf (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 AB = {(a,b) : aA and bB} is called the Cartesian product of the sets A and B. In other words, AB is the set of all ordered pairs (a,b), where aA and bB. In short this product AB is read as “A cross B”.

Example1.1: Let A = {1, 2} and B= {a, b, c}. Then AB = {(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:

ABBA If A has n elements and B has m elements than

AB has m.n elements.3

If A= and (or) B= then AB= If either A or B has infinite set then AB is

also an infinite set. If AB=BA A=B

1.2 RELATION

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 (AB)

Example1.2: Let A = {1, 2,3} and B= {a, b, c}Then AB={(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 (AB)). 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 aA and bB, 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 .

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.

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Definition1: Domain and Range of a relation: If R (AB) is a relation from AB, 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. ThenR = {(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)}.

Example2.6:Let A be the power set of the set {1,2} in other words, A = {,{1},{2},{1,2}} is the set of subsets of the set {1,2}. Write a relation on A, where (P,Q)R, if PQ. In this case we have:R = {(,{1}), (,{2}), (,{1,2}), ({1},{1,2}), ({2},{1,2})}.

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Example2.7: If set A has n elements and B has m elements, how many relations are there on the set AB.Let |A|=n and |B|=m. We know that R (AB) and | AB|=m.n, Also set of all possible subsets of AB is power set of AB i.e. P(AB).Thus if | AB|=m.n, then |P(AB)|=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 (AB) is a relation from AB 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} and Range(R)={a,b}, then R-1={(b,1),(a,2),(b,2)} and Dom(R-1)={a,b} and range(R-1)={1,2}.

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.

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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 RAA 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. Reflexive2. Irreflexive3. Symmetric4. Asymmetric5. Anti-symmetric6. Transitive

1. Reflexive RelationsA relation R on a set A is reflexive if for every aA, 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 aA.

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Definition2: (In terms of directed graph): R is reflexive if there must be a loop at each node aA.

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 aA.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 relationR5 = A X A, the universal relationDetermine which of the relations are reflexive.

Solution: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 integers2. Set inclusion on a collection C of sets3. Relation ┴ (perpendicular) on the set L of lines in the plane.4. Relation││ (parallel) on the set L of lines in the plane.

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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, AA for any set A in C, and n│n for every positive integer n in N.

2. Irreflexive RelationsA 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 aA such that (a, a)R.

Example1: Let A= {1,2,3} and let R= {(1, 1),(3,2)}. 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 aA. 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, bA such that (a, b)R but (b, a)R.

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Definition2: (In terms of directed graph): R is Irreflexive if there is no loop at any node aA.

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 ab ba.Example2: Let N={set of Natural numbers}. The Relation R on N is defined by “ a is equal to b” is symmetric since aRbbRa.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 relationR5 = 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 yS: x+y=10T: x+4y=10Determine 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. **************

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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, bA such that (a, b) and (b, a) belong to R, but a ≠ b

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.

Example1: Let A be a set of positive integers and R be a relation on A such thatR={(a,b): a,bA and ab}. This relation R is an anti-symmetric relation because if (a,b),(b,a)R a=bExample2: Determine which of the following in above example3 is antisymmetric.Solution: R2 is not anti-symmetric since (1,2),(2,1)R2, but 12. 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.

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Definition2: R is Asymmetric, if there is an edge from xy then there must not be an edge from yx.

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.

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 xy, with xy, then there is no edge from yx.

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,bA, the R is not anti-symmetric.Example 2Let A be a set of integers and R be a relation on A such that

R = {(x, y)│ x A, yA and x<y}.Determine whether the relation is symmetric, asymmetric and anti-symmetric.Solution: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,cA.In other words relation R is transitive if aRb and bRc implies that aRc, for all a,b,cAThus R is not transitive if there exist a, b, c A such that (a, b), (b, c)R but (a, c)R.

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Definition2: (In terms of directed graph): R is transitive, if whenever there is an edge from xy and yz then there must also be an edge from xz.

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 integersb)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

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 (AB) and R2 (BC). Then the composite of R1 and R2 is a relation from A to C, denoted by R2R1 and defined by R1R2={(a,c): there exists bB such that (a,b)R1 and (b,c)R2}

The relation R1R2 is called the composition of R1 and R2.Suppose R is a relation on a set A, that is R1 (AA). Then RR, the composition of R with itself is always defined. RR is sometimes denoted by R2. Similarly, R3= RRR, 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. R1R2) using matrices. Suppose MR1 and MR2 denotes

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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 R1R2.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 R1R2.b)Find the matrix for MR1R2 and compare the results

obtained in part (a).Solution: (a) The arrow diagram of the relations R1 and R2 is shown in figure-1.

Since 1 in A is connected to x in C by the path 1bx , so (1,x) R1R2..Similarly 2 in A is connected to z in C by the path 2az , so (2,z) R1R2 3 in A is connected to y in C by the path 3cy , so (3,y) R1R2 3 in A is connected to z in C by the path 3cz , so (3,z) R1R2 So finally R1R2={(1,x),(2,z),(3,y),(3,z)} (b) The matrices for MR1 , MR2 and MR1oR2 can be obtained as follows:

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A B C

Figure-1

123

abc

xyz

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. SoR1R2={(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 (RS)T = R(ST)Solution: L.H.S.: Suppose (a,d) (RS)T (a,c)RS and (c,d)T Since (a,c)RS (a,b)R and (b,c)SNow (ST) (b,d)STSince (a,b)R and (b,d)ST (a,d) R(ST)Hence proved.Note: 1) composition of relations is Associative i.e. (RS)T = R(ST). 2) (RS)(SR)

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 RS, SR, R(SR), (RS)R, RRR. Solution:

a) RS = {(1,5), (3, 2), (2, 5)} SR = {(4, 2), (3, 2), (1, 4)}. Clearly RS≠ SR

b) R(SR)= {(3, 2)}. (RS)R = {(3, 2)}. Clearly R(SR)= (RS)R.c) RRR= {(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

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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

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 relation2.By using Arrow diagram3.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 aA and bB .

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6

-6 0 6

-6

0

In arrow diagram, we write down the elements of A and B in two disjoint disks, and then draw as arrow from aA to bB 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.

Example2.11: Let A={1,2,3,4} and a relation R from A to itself i.e. R (AA) is defined as: R= {(1, 2), (2, 2), (2, 4), (3, 2), (3,3), (3, 4), (4, 1)}

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 Rc) Find the Inverse relation (R-1) of R in matrix form.d) Determine the Domain and Range of R.

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a b c

1 0 1 0 2 1 1 0 3 0 0 1

Figure-a Figure-b

123

abc

1

3 4

2

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 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:

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

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1. r(R)=R DA , where DA={(a,a):aA} 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)=RR2R3…..4. If A is finite with n elements, then t(R)= RR2R3….Rn.

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,cA.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)= RR2R3R4={…..}

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 ab and bc then ac. 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 aA.2. R is Symmetric i.e. aRabRa, a,bA.3. R is Transitive i.e. If aRb and bRc aRc a,b,cA.

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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 aA), symmetric (If a = b, then b = a , a,bA) and transitive (If a = b and b = c, then a = c, a,b,cA).

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 AB does not imply BA.

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.1.8 PARTIAL ORDERING RELATIONS (POR):

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 aA.2. Anti-symmetric, i.e. If aRb and bRa a=b a,bA.3. Transitive i.e. If aRb and bRc aRc a,b,cA.

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 aA.Anti-symmetric, i.e. a b and b a a=b, a,bA

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Transitive i.e. if ab and bc, then ac a,b,cA.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. aa, aR2. Anti-symmetric i.e. a b and b aa=b, a,bR3. Transitive i.e ab and bc, ac a,b,cRExample3: 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 aA ! 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 xX to a unique element yY is called a function or mapping from X to Y and written as f: XY (read as “ f is a function which maps X into Y)

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The main idea is that each element of X is associated with exactly one element of B. In other words if xX is associated with yY, 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.

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): xX}. It is also denoted as f(X).

Given an element xX, 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-xSolution: 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.

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(Necessary conditions to be a function f: XY): 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-

Figure1-2: Two associations (mapping) that are not function

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.

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abc

123

abc

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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 XY. 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)}

Solution: a) f is not a function, since 4X is not associated with any element of

Y or f(4) don’t have any image inY. b) g is not a function since 1X 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}

Example2: State whether or not each diagram in figure-1 defines a function from X={a,b,c} into Y={1,2,3}.

Solution: a) No. There is nothing to assigned to the element bA

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Remark: Every relation is not necessarily a function but every function is a relation.

Figure-a figure-b Figure-c

abc

123

abc

123

abc

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b) No. f(c) has two images x and z in Y.c) Yes.

Difference between Relation and function :

Every function f: XY gives rise a relation from X to Y. Every relation (RX×Y) is not necessarily a function (f: XY)..

Example1: Let X={1,2,3,4} and Y={a,b,c}. Consider the following relations R1 and R2 (i.e. R1,R2X×Y: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.

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Equality of two functions: Two functions f and g of AB are said to be equal iff f(x)=g(x) for all xA, 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 xA 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 fg 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 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: AB is a given function. We have following types of functions (or mapping) between A to B.

1. One-to-one mapping2. Onto mapping3. Invertible function (i.e. one-one onto)4. Into mapping5. 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,yA). An injective function is called an injection. The following figure-1 illustrates an injection from AB.

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For example f: RR be defined by f(x) = 2x+1, xR, then for x1 ,x2R (x1x2) we have f(x1)f(x2). So, f is 1–1.

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: ontof: A B. An surjective function is called an surjection. The following figure-2 illustrates an injection from AB.

For example, f: ZZ : f(x) = x+1, xZ, 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

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Figure-1: An injection

123

abcd

Figure-2: A surjection

1234

a b c

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 AB.

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: ZZ : f(x) = x+2, xZ is both injective and surjective. So, f is bijective.

Example1: Consider the functions e: AB, f: BC, g: CD and h; DE defined by the following diagram:

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Figure-3: A bijection

1234

a b cd

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)=vh: is not 1-1 as h(v)=h(w)=z.e: is not onto since 3B 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)=Cg: 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,yE 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: The function e : RR; e(x) = x2 is neither 1-1 nor onto.Example4: The function f : RR; f(x) = 2x is 1-1 and not onto.Example5: the function h(x) = x3 is 1-1 and onto. Note: increasing functions are 1-1.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

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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.

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,yA and xy f(x)=f(y).

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,yA and xy 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,yA and xy f(x)=f(y) and

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Figure-2: An into mapping

123

a b c

Figure-2: Many-one mapping (Because two different elements in A (i.e. -2,2} have the same f-image in B)

-2234

4 9 16

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: AB={(x,y): y=x2, xA, yB} 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)-bookExample2: (p3.23)(SS)

Check your progress-4:

1.11 COMPOSITION OF FUNCTION

Let A,B,C be three sets and f and g be two functions such that f: AB and g: BC; where the co-domain of f is the domain of g. Then we may define a new function from A to C [(gof): AC], called the product or composition of f and g and denoted by (gof), as follows: (gof)(x)=g[f(x)], xAThat 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: AB, g: BC, h: CD then (hog)of=ho(gof)Proof: It can be easily seen that both (hog)f and ho(gof) are mapping of AD. 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)

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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:

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)=5f (1)= f (0) + 2 = 5 + 2 = 7f (2)= f (1) + 2 = 7 + 2 = 9f (3)= f (2) + 2 = 9 + 2 = 11This 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:

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The values of this function are:f (0)=0f (1)= f (0) + (2)(1) - 1 = 0 + 2 - 1 = 1f (2)= f (1) + (2)(2) - 1 = 1 + 4 - 1 = 4f (3)= f (2) + (2)(3) - 1 = 4 + 6 - 1 = 9f (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.

Here is one more example of a recursively defined function:

The values of this function are:f (0)= 1f (1)= 1⋅f (0) = 1⋅1 = 1f (2)= 2⋅f (1) = 2⋅1 = 2f (3)= 3⋅f (2) = 3⋅2 = 6f (4)= 4⋅f (3) = 4⋅6 = 24f (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.

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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

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}.

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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:

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) 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 =….

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A(0,n)=n+1A(1,y)=n+2A(2,n)=2n+3A(3,n)=2n+3-3

A(4,n)=

Expressions of the latter form are sometimes called power towers.

=A(1,0)+n =A(0,1)+n=2+nSimilarly 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).

m\n 0 1 2 3 4 n0 1 2 3 4 5 n+1 1 2 3 4 5 6 n + 2 = 2 + (n + 3) − 32 3 5 7 9 11 2n + 3 = 2 * (n + 3) − 33 5 13 29 61 125 2(n + 3) − 34 13 65533 265536 − 3 2265536-3 A(3, A(4, 3)) 2 (2 (2 (... 2))) - 3 (n + 3 twos)

= 2 (n + 3) - 3

5 65533 A(4, A(5, 1))

A(4, A(5, 2))

A(4, A(5, 3)) A(4, A(5, n-1))

*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=26. A(1,0)=27. A(1,1)=A(0,2)8. A(0,2)=2+1=39. A(1,1)=310. A(1,2)=A(0,3)11. A(0,3)=3+1=412. A(1,2)=413. A(1,3)=A(0,4)14. A(0,4)=4+1=5

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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.

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).

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N is a set with the following properties:

P1: 1 is a natural number i.e. 1NP2: For each nN, 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 kN 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

1K 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. Thus2=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)P-1.22 (SS)

Example-1,2,3,(P-105)

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