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- 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 functions1.0 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 “orderedpairs”.Relations, as noted above, will be defined in terms ofordered pairs (a, b) of elements, where a is designated asthe first element and b as the second element.There are three kinds of relations which play a major rolein our theory: (i) Equivalence relations, (ii) Partial order relations, (iii) Functions.All these relations will be discussed here.1.1 OBJECTIVEAfter 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 relationDefinition1: Ordered Pair:Let A and B are two sets and let a∈A and b∈B then a set of twoelements whose elements have been listed in a specific order iscalled 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=b2Thus in case of relation (a,b)≠(b,a) unless a=b, whereas in caseof 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 Aand 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 isread 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 thefollowing 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=B1.2RELATIONLet 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 Ris a set of ordered pairs where each first element comesfrom A and each second element comes from B. That is,for each pair a∈A and b∈B, exactly one of the following istrue: (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 arelation 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 ofthe ordered pairs which belong to R, and the range of R isthe set of second elements.Example2.3: In the example1.2 above, for relationR2={(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 bywriting (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 asR={{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} inother words, A= {φ,{1},{2},{1,2}} is the set of subsets of theset {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 melements, 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 powerset 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 inverserelation of R (denoted by R-1), is a relation from B to A. Itis 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},thenR-1={(b,1),(a,2),(b,2)} and Dom(R-1)={a,b} andrange(R-1)={1,2}. 6
- 7. Check your progress-1:Q.1: Define the following: a) Identity relation b) Universal relation c) Void relationQ.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 iscalled a binary relation on A. Binary relations that satisfycertain special properties can be very useful in solvingcomputation problems. So let’s discuss some of theseproperties: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. Transitive1. Reflexive RelationsA relation R on a set A is reflexive if for every a∈A, aRa. that is, arelation R in a set A is said to be reflexive if every element of A isrelated 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. Therelation R “x is parallel to y” is reflexive since every straight lineis parallel to itself.Example2: Let A be the set of numbers and relation R in A isdefined by “x is equal to y” is reflexive” since each number isequal to itself.Example3: Let A={1,2,3} and the relation R in A is defined byR={(1,1),(2,2),(2,3)} is not reflexive because (3,3) does not belongsto 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 relationR5 = A X A, the universal relationDetermine which of the relations are reflexive.Solution: 8
- 9. The only relations R1 and R5 are reflexive, since A contains thefive elements 1, 2, 3, 4 and 5, a relation R on A is reflexive if itcontains the five pairs (1, 1), (2, 2), (3, 3), (4, 4) and (5,5). Thus onlyR1 and R5 are reflexive.R2, R3, and R4 are not reflexive since, for example, (5, 5) does notbelong 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 otherrelations are reflexive; that is, x ≤ x for every integer x in Z, A⊆A forany 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 єA. Thus R is not irreflexive if there exist at least one a∈Asuch 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 notirreflexive, 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 thatthere 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, thenbRa. 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-1Example1: Let A={set of all straight lines in a plane}. The Relation Ron A is defined by “ a is perpendicular to be” is a symmetric relationbecause a⊥b ⇒ b⊥a.Example2: Let N={set of Natural numbers}. The Relation R on N isdefined by “ a is equal to b” is symmetric since aRb⇒bRa.Example3: Consider the following five relations on the setA = {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)∈R1but (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 positiveintegers:R: x is greater than yS: x+y=10T: x+4y=10Determine which of the relation are reflexive and which of them aresymmetric.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 andbRa then a = b. That is if (a,b),(b,a)∈R then there must be thecase that a=b. Thus R is not antisymmetric if there exists a, b∈Asuch 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 xy, with x≠y, then there is no edge from yx.Note: Symmetric and anti-symmetric relations are not negatives to eachother. For example, the relation R={(1,4),(4,1),(2,4)} is neithersymmetric nor anti-symmetric and the relation S={(1,1),(2,2)} is bothsymmetric and anti symmetric. 11
- 12. Example1: Let A be a set of positive integers and R be arelation on A such thatR={(a,b): a,b∈A and a≥b}. This relation R is an anti-symmetricrelation because if (a,b),(b,a)∈R ⇒a=bExample2: Determine which of the following in aboveexample3 is antisymmetric.Solution: R2 is not anti-symmetric since (1,2),(2,1)∈R2, but1≠2. Similarly R5 is also not a anti-symmetric. All the otherrelations are anti-symmetric.5.Asymmetric RelationA relation R on set A is asymmetric if (a, b)∈R then (b, a)∉R. It meansthat 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 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.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 2Let 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 andanti-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, theneither (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)∈Rand (b,c)∈R ⇒(a,c)∈R, for all a,b,c∈A.In other words relation R is transitive if aRb and bRcimplies that aRc, for all a,b,c∈AThus 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 xy and yz then there must also be an edge from xz.Example1: Determine which of the relations in above Example3are transitive.Solution: The relation R1 is not transitive since(4,2), (2,3)∈R1but (4, 3)∉R1. All the other relations are transitive. Also therelation 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 nottransitive, 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 Band 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 byR1οR2={(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). ThenRοR, the composition of R with itself is always defined. RοR issometimes denoted by R2.Similarly, R3= RοRοR, and so on. Thus Rn is defined for allpositive 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 thematrices of the relations R1 and R2. Then by multiplying MR1and MR2, we get the matrix MR1.MR2(=M). The nonzero entries ofthis 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 followingrelation R1 from A to B and R2 from B to C.R1= {(1, b), (2, a), (3, c)} andR2= {(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 isshown 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 1bx , so (1,x)∈ R1οR2..Similarly 2 in A is connected to z in C by the path 2az , so (2,z)∈ R1οR2 3 in A is connected to y in C by the path 3cy , so (3,y)∈ R1οR2 3 in A is connected to z in C by the path 3cz , so (3,z)∈ R1οR2So finally R1οR2={(1,x),(2,z),(3,y),(3,z)}(b) The matrices for MR1 , MR2 and MR1oR2 can be obtained asfollows: 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. SupposeR is a relation from A to B,S is a relation from B to C, andT 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)∈SNow (SοT) ⇒ (b,d)∈SοTSince (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 RELATIONSThe diagrammatical representation (also called graph) of arelation R is called a pictorial representation of the given relationR.Example 2.9Consider the relations R defined by the equation: x2 + y2 = 36That is, R consists of all ordered pairs (x, y) which satisfy thegiven equation. The graph of the equation is a circle having itscenter 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 = 36Method to represent a given relations (on finite set) in pictorialform:If A and B are finite sets, then there are two ways of picturing arelation 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 ofrepresentation. 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 4Check your progress-2:Q.1: Given a relation R={(1,y),(2,z),(3,y),(4,x),(4,z)} on setsA={1,2,3,4} and B={x,y,z}a) Draw the arrow diagram of Rb) Find the Matrix of Rc) 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 setA={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 pclosure of R is the smallest binary relation containing R that 18
- 19. satisfies property p. Our goal is to construct closures for thereflexive, symmetric and transitive properties. These arecommonly 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 relationReflexive, Symmetric and Transitive, whenever the given relationis not in proper form (i.e. reflexive, symmetric or transitive).Reflexive, Symmetric & Transitive ClosuresLet R be a relation on a set A. Then reflexive and symmetric closure ofR, 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 reflexiveclosure r(R) can be constructed by including all pairs (a,a) that arenot already in R, andTo 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 ofelements. There exist an efficient way for computing t(R), knownas 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” isan equivalence relation, since it is reflexive (a = a for everya∈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, symmetricand transitive.Example3: The relation ⊆ (set inclusion) is not an equivalencerelation, since it is reflexive and transitive, but it is notsymmetric. Since A⊆B does not imply B⊆A.Example4: A = {1, 2, 3, 4, 5}. Let R be relation on A such thatR = {(x, y) │ x + y = 5}We get R as {(1, 4), (2, 3), (4, 1), (3, 2)}. We can say that, R isnot 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 isnot transitive as (1, 4) ∈R and (4, 1)∈R but (1, 1)∉R. Hence Ris not equivalence relation. PARTIAL ORDERING RELATIONS (POR):1.8A relation R on a set A is called a Partial ordering or a partialorder 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 orderedset (or POSET). It is denoted by (A,R) where A is a given set and R is arelation which satisfy the above three conditions. POSET is discussed indetail, in the Lattice chapter (Unit-II).Example1:The relation ⊆ of set inclusion is a partial ordering on anycollection of sets since set inclusion has the following three desiredproperties.Reflexive, i.e. a ⊆ a ∀a∈A.Anti-symmetric, i.e. a ⊆ b and b⊆ a⇒ a=b, ∀a,b∈ATransitive 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 realnumbers 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∈R3. Transitive i.e a≤b and b≤c, ⇒a≤c ∀a,b,c∈RExample3: Let N be the set of all positive integers. The relation “adivides b” is a partial ordering on N. However, “a divides b” is not apartial ordering on the set Z of integers since a|b and b|a does not implya = 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 supposeX= the set of students of UP Technical university, andY=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 acorrespondence which assign each element x∈X to a uniqueelement y∈Y is called a function or mapping from X to Y andwritten asf: XY (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 fassigns 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 beundefined.Domain of f can be taken as R~{1}and co-domain can be R. (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- 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 amathematical formula. For example, consider the function f,which maps each natural number N to its square. We maydescribe this function by the formula: f(x) = x2 or y = x2In the first notation, x is called a variable and the letter f denotesthe function.In the second notation, x is called the independent variable andy is called the dependent variable since the value of y willdepend on the value of x.Remark: Whenever a function is given by a formula in terms ofa variable x, we assume, unless it is otherwise stated, that thedomain of the function is R (or the largest subset of R for whichthe formula has meaning) and the co-domain in R.Example1: Suppose A ={1,2,3}, B= {1,4,9,11} and f assigns toeach member in A its square values. Then f is a function from Ato B.But if A={1,2,3,4}, B={1,4,9,10} and f is the same rule, then f isnot a function from A to B since no member of B is assigned tothe element 4 in A.Example1: Let X={1,2,3,4,5} and Y={a,b,c,d,e}. Determinewhether or not each relation below is a function from XY. Ifthey are functions, give the domain, co-domain and Range ofeach, 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-cSolution: 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: XY gives rise a relation from X to Y. • Every relation (R⊆X×Y) is not necessarily a function (f: XY).. 25
- 26. Example1: Let X={1,2,3,4} and Y={a,b,c}. Consider the followingrelations R1 and R2 (i.e. R1,R2⊆X×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 appearsas the first element in one and only one ordered pair in R1, i.e. everyelement in X must have its image in Y. Obviously R1 is also a relationfrom X to Y. But R2 is not a function from X to Y since 1 is associatedwith 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 intoA.Solution:f is a function from A into A since each member of A appears as the firstcoordinate in exactly one ordered pair in f; here f(1) = 3, f(2) = 3 andf(3) =1. g is not a function from A into A since 2 Є A is not the firstcoordinate of any pair in g and so g does not assign any image to 2. Alsoh is not a function from A into A since 1 Є A appears as the firstcoordinate of two distinct ordered pairs in h, (1, 3) and (1, 2). If h is tobe 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 AB are said tobe equal iff f(x)=g(x) for all x∈A, and we write f=g. Note that two equalfunction f and g are defined on same domain A. For two unequal mappings from A to B, there must exist at least oneelement x∈A such that f(x)≠g(x).Example1: If f(x)=x3+1 where x is any real number and g(x)= x3+1where x is any complex number then f≠g because the domain of f and gare different. 26
- 27. Example2: Let A={3,4} and B={2,4,9,16}. Let a function f be definedfrom A to B by f(x)= x2 and g={(3,9),(4,16)}, then f=g because f and gboth have the same domain={3,4} and each of them assigns the sameimage 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 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 into1.0 One-one mapping:A function f: A → B is said to be one-to-one or one-one orinjective, written as 1-1, if different elements in the domain Ahave different images in B. Another way to say this is that f isone-one if f(x) =f(y) implies x= y (x,y∈A). An injectivefunction is called an injection. The following figure-1 illustratesan injection from AB. 1 a 2 b 3 c d Figure-1: An injection 27
- 28. For example f: RR be defined by f(x) = 2x+1, x∈R, then forx1 ,x2∈R (x1≠x2) we have f(x1)≠f(x2). So, f is 1–1.2: Onto mappingA function f: A → B is said to be an onto or surjective if eachelement 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 afunction is mapping of A onto B, we write: ontof: A B.An surjective function is called an surjection. The followingfigure-2 illustrates an injection from AB. 1 a 2 b 3 c 4 Figure-2: A surjectionFor example, f: ZZ : f(x) = x+1, x∈Z, then every element y inthe 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) MappingA function f; A→ B is invertible (or bijective) if its inverserelation f -1 is a function from B to A. In general, the inverserelation f -1 may not be a function.Another term for bijective is “one-to-one and onto”. A bijectivefunction is called a bijection or a “one-to-one correspondence”The following figure-3 illustrates an bijection from AB. 28
- 29. 1 a 2 b 3 c 4 d Figure-3: A bijectionThe following theorem gives a necessary condition for invertible (orone-one and onto) function:Theorem1: A function f: A → B is invertible if and only if f is bothone-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, x∈Z 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:Determine whether the above functions are one-one, onto andinvertible 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. butg: is not one-to-one as g(r)=g(u)=vh: is not 1-1 as h(v)=h(w)=z. 29
- 30. 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)=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,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 setof their telephone extensions. Assuming that all the employees are listedand that every person has his/her own extension, then the mapping fromA to B is invertible.Example3:Thefunctione: RR; e(x) = x2 is neither 1-1 noronto.Example4: Thefunctionf: RR; 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 intofunction if there exits at least one element in (co-domain set) Bwhich is not the f-image of any element in (domain set) A. Wesay that f is a mapping of A ‘into’ B. In this case the range of fis 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 30
- 31. 5: Many one: A mapping f: A → B is said to be many-one iftwo (or more than two) distinct elements in A has the sameimage in B i.e. x,y∈A and x≠y ⇒f(x)=f(y). -2 4 2 9 3 4 16 Figure-2: Many-one mapping (Because two different elements in A (i.e. -2,2} have the same f-image in B)6: many-one onto: A mapping f: A → B is said to be many-oneonto 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)=BExample1: 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 andinto.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:AB={(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 thesame image in B, thus it is many-one mapping iii) Every element of B is f-image of some element of A. Thus it isonto mapping. Therefore f is a many-one onto mapping. 31
- 32. Example1(P-68,69)-bookExample2: (p3.23)(SS)Check your progress-4:1.11COMPOSITION OF FUNCTIONLet A,B,C be three sets and f and g be two functions such that f: ABand g: BC; where the co-domain of f is the domain of g. Then we maydefine a new function from A to C [(gof): AC], called the product orcomposition of f and g and denoted by (gof), as follows: (gof)(x)=g[f(x)], ∀x∈AThat is, to find (gof)(x), first we have to find the image of x under f andthen find the image of f(x) under g.Note: if we think f and g as relation then the composition of function issame as the composition of relation. Only the notation is different. Weuse the gof for the composition of f and g, instead of fog, which is usedfor 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 ofAD. These two mappings will be equal if they assign the same imageto 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-32,33Example3.8,3.9.3.10 (SS)-p-3.23,3.24,3.27.3.28.1.12RECURSIVELY DEFINED FUNCTIONS: 32
- 33. A function is said to be recursively defined if the function definitionrefers to itself or call itself repeatedly. The recursive definition musthave 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)=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)=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……. 33
- 34. This recursively defined function is equivalent to the explicitly definedfunction f (n) = n2. Again, the recursive function is defined only fornonnegative 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!. Thusa 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 allthese 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 thefollowing seven Steps: 1) 3! = 3 · 2! 2) 2! = 2 · 1! 3) 1! = 1 · 0! 4) 0! = 1 5) 1! = 1 · 1 = 1 34
- 35. 6) 2! = 2 · 1 = 2 7) 3! = 3 · 2 = 6Here we first defines 3! In terms of 2!, so we must postpone evaluating3! Until we evaluate 2!. In step2 we define 2! in terms of 1!, so we mustpostpone evaluating 2! Until we evaluate 1! An so on.step5 can explicitly evaluate 0!(=1), which is the base value (i.e. 0) ofthe 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 explicitdefinition, yields the Fibonacci numbers (usually denoted by fn, f1, f2,….}. The main characteristic of Fibonacci number is that the sums oftwo 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 ofwhich can be assigned any non-negative integer, that is, 0, 1, 2 … The 35
- 36. Ackermann function is defined recursively for non-negative integer’s mand 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>0Observe that A (m, n) is explicitly given only when m = 0. The basecriteria 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 oneor 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 asfollows: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+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).mn 0 1 2 3 4 n0 1 2 3 4 5 n+1 36
- 37. 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) − 3 655364 13 6553 2 −3 22↑65536-3 A(3, A(4, 3)) 2 (2 (2 (... 2))) - 3 (n + 3 twos) 3 =2 (n + 3) - 35 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 ofA(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=515. A(1,3)=5So here we can see that the value of A (1, 3) is requires 15 steps. In theabove calculation Forward movement means we are postponing anevaluation and calling the definition. The backward movement meanswe are backtracking. The result of A(1,3)=5 can also be verified fromthe above table. 37
- 38. Check your progress-5:Q.1 P-3.33, Q3.30,3.31(SS)1.13: Piano’s AxiomIn mathematical logic, the Peano axioms, also known as the Dedekind-Peano axioms or the Peanopostulates, are a set of axioms for the natural numbers presented by the 19th century Italianmathematician 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 (calledPeano’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 inthe 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 38
- 39. P-105(DU book)P-1.22 (SS)Example-1,2,3,(P-105) 39

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