# Unit 2 Relations and Functions Introduction to Relation

- Slides: 109

Unit 2 Relations and Functions

Introduction to Relation • A relation is a type of association that exists between two or more elements. • Consider the eg. 1. x is father of y. 2. x was born in the city y in the year z. 3. The number x is greater than number y. • In general one can have relation with ‘n’ objects. (where n is positive integer) • In describing the relation, it is necessary not only to specify the objects but also the order in which they appear.

Necessary ordering of objects • An ordered n- tuple, for n>0, is a sequence of objects , denoted by (a 1 , a 2, -----an). • If n=2, the ordered n-tuple is called an ordered pair. • If n=3, the ordered n-tuple is called an ordered triple and so on.

Application of relation • We can describe languages (e. g. , compiler grammar, a universal Turing machine) using sets and set relations. • Graph traversal requires sets to track node visits. • Data structures are inherently set-based. • Relational databases are entirely premised on set theory insofar as table operations are concerned.

Product set (cartesian product) • Let A & B be non empty sets. we define the product set (AXB)as. AXB = {(a, b)| a ε A and b ε B} If A= Ø or B= Ø, then AXB= Ø Eg. Let A={a, b, c}, B={1, 2} then AXB=? Solution: AXB: {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)} • The product set is not commutative i. e. AXB ≠ BXA

• following theorem establishes certain important properties of the product operation Theorem: If A, B, C are sets then 1. Ax(B U C)= (AXB) U (AXC) 2. AX(B ᴒ C) = (AXB) ᴒ (AXC) 3. (A U B)XC=(AXC) U (BXC) 4. (A ᴒ B)XC = (AXC) ᴒ (BXC)

• Theorem: If A and B are finite sets with cardinalities m, n resp. then |Ax. B|=m. n Proof: since |A|=m Let A={a 1 , a 2, -----am}, Similarly , |B|=n Let B={b 1 , b 2, -----bn} Now AXB={(ai, bj)|1<=i<=m, 1<=j<=n} Since for each element ai in A there exist a corresponding element bj in B in ordered pair (ai , bj), the set AXB consists of exactly m. n elements hence |AXB|=m. n Is proved Eg. If A={n ∈ N|1<=n<=100} , B={n ∈N|1<=n<=50} Then AXB = 100 X 50= 5000

• Empty relation: if R= Ø, then that relation is called empty or void relation. • If R=A 1 XA 2 XA 3 X-----An, then R is called the universal relation. • If R=1, 2 or 3 then R is called unary, binary or ternary relation. • Eg. 1 Let A={1, 2, 5, 6} and let R be relation characterized by the property “x is less than y” then R={(1, 2), (1, 5), (1, 6), (2, 5), (2, 6), (5, 6)} where R is binary. Eg 2. Let A={1, 2, 3} and R be the relation characterized by the property “x+y is less than equal to z” then R is{(1, 2, 3)} is ternary.

Binary Relation • Let A and B are non empty sets. Then binary relation R from A to B is subset of AXB, i. e. R c AXB. • The Domain of R is denoted by D(R), is the set of elements in A that are related to some elements in B i. e. D(R)= {a ε A|for some b ε B, (a, b) ε R} • The range of R denoted by Rn (R) is the set of elements in B that are related to some element in A i. e. Rn (R)={b ε B|for some a ε A, (a, b) ε R} Clearly we can say D(R) c A and Rn (R) c B

Eg. Let A={2, 3, 4, 5} and let R be the relation on A defined as a. Rb iff a<b find D(R) & Rn (R) Solution: R={(2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)} Then D(R)={2, 3, 4} , Rn (R) ={3, 4, 5}

Complement of relation (R ) • A relation as a set has its complement, which is defined below • The complement of relation R, denoted by R , is defined as. R ={(a, b)|(a, b) ∉ R} i. e. a. Rb iff a. Rb Eg. Let A ={1, 2, 3, 4} and B={a, b, c} Let R ={(1, a)(1, b), (2, c), (3, a), (4, b)} & S={(1, b), (1, c), (2, a), (3, b), (4, b)} Find 1. R and S 2. verify Demorgan’s law for R and S

1. R and S =? Solution: AXB={(1, a), (1, b), (1, c), (2, a), (2, b), (2, c), (3, a), (3, b), (3, c), (4, a), (4, b), (4, c)} R = {(1, c), (2, a), (2, b), (3, c), (4, a), (4, c)} S = {(1, a), (2, b), (2, c), (3, a), (3, c), (4, a), (4, c)} 2. Demorgan’s law states that – RUS =RᴒS R U S ={(1, a), (1, b), (1, c), (2, a), (2, c), (3, a), (3, b), (4, b)} R U S ={(2, b), (3, c), (4, a), (4, c)} R ᴒ S ={(2, b), (3, c), (4, a), (4, c)} Therefore R U S = R ᴒ S

Eg. R ᴒ S = R U S

Converse of Relation (Rc) • Given a relation from A to B one may define a relation from B to A as follows. Let R be a relation from A to B then the converse of R, denoted by Rc is the relation form B to A defined as. Rc = {(b, a)|(a, b) ∈ R} Clearly we can say Rc c BXA. If A=N the set of natural number and R is the relation <, then Rc is the relation >. The converse relation is also called as the inverse relation and is denoted by R-1.

• Following theorem gives important properties of the converse relation • Let R , S be the relations from A to B then – i) (Rc )c =R ii) (RUS) c = R c U S c iii) (R ᴒ S) c = R c ᴒ S c Eg. Let A ={1, 2, 3, 4} and B={a, b, c} Let R={(1, a), (3, c)} Find i) R c ii) D(R c ) iii) Rn (R c)

i) R c ={(a, 1), (a, 3), (c, 3)} ii) D(R c )= {(a, c)} = Rn (R) iii) Rn (R c)={(1, 3)} = D(R)

Composition of binary relation (R 1. R 2) • Relations that are formed from an existing sequence of relation. They are called as composite relation. • The concept of composite relation plays an important role in the execution of program where a sequence of data conversion takes place from decimal to binary and from binary to floating point. • Definition- Let R 1 is a relation from A to B and R 2 be the relation from B to C. The composite relation from A to C denoted by R 1. R 2 or (R 1 R 2) is defined as-

R 1. R 2= {(a, c)|a ε A ᴒ c ε C ᴒ ⱻ b[b ε B ᴒ (a, b) ε R 1 ᴒ (b, c) ε R 2]} • If R 1 is a relation from A to B and R 2 is the relation from C to D, R 1. R 2 is not defined unless B=C. • The operation of composite relation is not commutative. i. e. R 1. R 2 ≠ R 2. R 1 Eg. Let A={a, b, c, d} where R 1={(a, a), (a, b), (b, d)} and R 2={(a, d), (b, c), (b, d), (c, b)} Find R 1. R 2=? , R 2. R 1=? , R 2 2 =?

1. R 2 = {(a, d), (a, c)} 2. R 1 = {(c, d)} 3. R 2 2 = {(b, b), (c, c), (c, d)} Eg 2. Let A={2, 3, 4, 5, 6} and let R 1. R 2 be the relations on A such that. R 1={(a, b)|a-b=2} and R 2={(a, b)|a+1=b or a=2 b} find composite relation i)R 1. R 2 ii)R 2. R 1 iii) R 1 2

Solution: R 1={(4, 2), (5, 3), (6, 4)} R 2={(2, 3), (3, 4), (4, 5), (5, 6), (4, 2), (6, 3)} i)R 1. R 2 ={(4, 3), (5, 4), (6, 5), (6, 2)} ii)R 2. R 1={(3, 2), (5, 4), (4, 3)} iii) 2 =? R 1

Matrix representation of a relation(MR ) • Let A={a 1, a 2, ----am } and B={b 1, b 2, ----- bn} be finite sets containing resp. m & n elements. • Let R be the relation from A to B by definition R c AXB , hence we can represent R by a m. Xn matrix MR=[mij], which is defined as follows mij = 1 if (ai, bj)ε R =0 if (ai, bj) ∉ R

eg. 1 Let A={a, b, c, d} and B={1, 2, 3} Let R={(a, 1), (a, 2), (b, 1), (c, 2), (d, 1)} find the relation matrix Solution: MR will have 4 rows and 3 column MR = 1 1 0 1 0 0 0 0 Eg. 2 Let A={1, 2, 3, 4, 8} , B={1, 4, 6, 9} let a. Rb iff b/a find the relation matrix

Solution: R={(1, 1), (1, 4), (1, 6), (1, 9), (2, 4), (2, 6), (3, 9), (4, 4)} MR = 1 0 0 1 1 1 0 0 Eg. Let A = {a, b, c, d} and let MR = Find R 1 0 0 1 1 1 0

Relation Matrix operation • A relation matrix has entries which are either one or zero, such a matrix is called Boolean matrix. • Let A= [aij] and B= [bij] be mxn Boolean matrix we define. A+B =[ cij] where cij = 1 if aij = 1 or bij = 1 = 0 if aij & bij are both zero similarly if A= [aij] is an mxn boolean matrix and B= [bij] is an nxr matrix then A. B = [dij] an mxr matrix. where dij = 1 if aij = bij =1 = 0 if aij = 0 or bij =0

eg A= 1 0 1 1 0 then A+ B= 1 1 1 0 and B= 1 0 0 1 A. B = 1 1 0 1 1 • Properties of Relation matrix: Let R 1 be a relation from A to B, R 2 from B to C. Then the relation metrices satisfy the following properties. 1. MR 1. R 2 = MR 1 x MR 2 2. MRc = transpose of MR

eg. 1 Let A= {1, 2, 3, 4} and let R 1={(1, 1), (1, 2), (2, 3), (2, 4), (3, 4), (4, 1), (4, 2)} R 2={(3, 1), (4, 4), (2, 3), (2, 4), (1, 1), (1, 4)} verify i) MR 1. R 2 = MR 1 x MR 2 solution: MR 1 = 1 1 0 0 0 1 1 0 MR 2 = 1 0 0 0 0 1 1 0 1 R 1. R 2 = {(1, 1), (1, 4), (1, 3), (2, 1), (2, 4), (3, 4), (4, 1), (4, 3)} MR 1. R 2 = 1 0 1 1 1 0 0 0 1 1

MR 1 x MR 2 = = c M ii) R 1 1 1 0 0 1 1 0 0 0 0 1 1 0 0 0 1 1 1 = transpose of MR 1 c = {(1, 1), (2, 1), (3, 2), (4, 3), (1, 4), (2, 4)} MR 1 c = 1 0 0 0 1 1 0 0

Graphical representation of a relation • If A is a finite set and R is a relation on A, it is possible to represent R pictorically by means of a graph • The element of A are represented by points or circles called as nodes or vertices. • if a. Rb , this is indicated by drawing an arc from a to b with an arrow head pointing in the direction a b • If a. Ra , this is shown by drawing a loop around a. • These arcs or loop are called as edges of the graph. • The resulting graph is called a directed graph of diagraph of R.

eg. Let A={2, 3, 4, 5} and let R={(2, 3), (3, 2), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5)} draw its diagraph Solution:

eg. Let A ={a, b, c, d} and MR = Solution : 1 0 0 1 1 1 0 0 0 1 1 0 1 0 draw its digraph

Special properties of binary relation • Let R be a relation on a set A 1. Reflexive relation(a. Ra) R is reflexive if for every element a ∈ A, a. Ra i. e. (a, a) ∈ R Eg. Let A={a, b} and let R={(a, a), (a, b), (b, b)} then R is reflexive 2. Irreflexive relation (a. Ra) R is said to be irreflexive if for every element a ∈ A, a. Ra i. e. (a, a) ∉ R eg. Let A={1, 2} and let R ={(1, 2), (2, 1)}then R is irreflexive since (1, 1), (2, 2) ∉ R

eg. Let A={1, 2} and let R ={(1, 2), (2, 2)} then R is not irreflexive since (2, 2) ∈ R note: R is not reflexive either since (1, 1) ∉ R • If R is reflexive, the corresponding relation matrix(MR) will have its diagonal entries as one. • If R is irreflexive, the diagonal elements will be zero. • Diagraph of reflexive relation:

3. Symmetric relation: (a. Rb then b. Ra) R is said to be symmetric, if whenever a. Rb , then b. Ra. eg 1 Let A be the set of people let a. Rb if a is a friend of b, then obviously b is related to a , hence the relation of being friend is symmetric relation. Diagraph of symmetric relation:

4. Asymmetric relation: R is said to be asymmetric if whenever a. Rb then b. Ra. Eg. 1 Let A=R , the set of real number and let R be the relation “<“ then a<b b<a hence < is asymmetric. 5. Antisymmetric relation: (a. Rb and b. Ra then a=b) R is antisymmetric if whenever , a. Rb and b. Ra then a=b. eg Let A=R and let R be the relation “<=“ then a<=b and b<=a a=b. hence “<=“ is an antisymmetric relation.

6. Transitive relation: (a to b to c ) R is said to be transitive , if whenever a. Rb and b. Rc, then a. Rc. eg. Let A = set of triangles and let R be the relation of being congruent, then for triangle a, b, c ∈ A and a. Rb and b. Rc a. Rc. Diagraph of transitive relation

Equivalence relation • A binary relation R on a set A is called as equivalence relation if it is reflexive, symmetric and transitive. eg. 1 Let A=R and R be equality of number eg. 2 A is a set of triangles and R is similarity of triangles • Diagraph of an equivalence relation will have the following characteristics • Every vertex will have loop. • if there is an arc from a to b there should be an arc from b to a. • If there is an arc from a to b and one from b to c there should be an arc from a to c.

eg 2. Let A={a, b, c, d} R={(a, a), (b, b), (c, c), (d, d), (d, c)} determine whether R is equivalence relation? solution: R is reflexive since (a, a), (b, b), (c, c) and (d, d) ∈ R But R is not symmetric since (b, a) ∈ R but (a, b) ∉ R hence R is not an equivalence relation. eg. 3 Let A ={a, b, c} and let MR = 1 0 0 0 1 1 determine whether R is an equivalence relation?

Solution: R={(a, a), (b, b), (b, c), (c, b), (c, c)} R is reflexive since (a, a), (b, b)and (c, c) ∈ R R is symmetric since (b, c) ∈ R (c, b) ∈ R R is transitive since (b, b)and (b, c) ∈ R (b, c) and (c, b) ∈ R (b, c) and (c, c) ∈ R (b, c) ∈ R (c, b) and(b, b) ∈ R (c, b) and(b, c) ∈ R (c, c) ∈ R hence R is an equivalence relation.

• If R 1 and R 2 are equivalence relation on a set A then R 1 ᴒ R 2 is an equivalence relation • If R 1 and R 2 are equivalence relation on a set A then R 1 U R 2 is an equivalence relation Equivalence class: Let R be the equivalence relation on a set A, for every a ∈ A, let [a]R denote the set { x ∈ A |x R a} then [a]R is called as the equivalence class of a with respect to R. The rank of R is number of distinct equivalence classes of R if the number of classes is finite otherwise the rank is said to be infinite We can simply denote equivalence class as [a].

eg. 1 Let A={a, b, c} and let R={(a, a), (b, b), (c, c), (a, b), (b, a)} where R is clearly an equivalence relation find equivalence class. solution: The equivalence classes of elements of A are(element [a] is related with (a, a), (a, b)) [a]={a, b} [b]={b, a} [c]={c} The rank of R is 2

eg. 2. Let A={1, 2, 3, 4} and let R={(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (2, 3), (3, 2), (3, 3), (4, 4)} show that R is an equivalence relation and determine equivalence classes and hence find rank of R.

Partition(Π) • A partition of a non empty set A is a collection of sets {A 1, A 2, A 3 ------An} such that – A= i=1 UA I Ai ᴒ Aj = Φ for i not equal to j • We denote partition by Π. • An element of partition is called a block. • The rank of partition is number of blocks of Π eg. Let A={1, 2, 3} then Π 1= {{1, 2}, {3}} Π 2={{1, 3}, {2}} Π 3= {{1}, {2}, {3}}

eg. Let A={a, b, c, d}, Π ={{a, b}, {c}, {d}} find the equivalence relation induced by Π and construct its diagraph solution: R={(a, a), (a, b), (b, a), (b, b), (c, c), (d, d)} the diagraph of R is

Eg. 2 Let A ={1, 2, 3, 4, 5} and Π={{1, 2}, {3}, {4, 5}}find equivalence relation determined by Π and draw its diagraph. Solution: R ={(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (4, 4), (4, 5), (5, 4), (5, 5)} diagraph of R

Compatible relation • A relation R on set A is said to be compatible if it is reflexive and symmetric eg. 1. All equivalence are compatible relations. eg. 2 The relation of “being friend of” is compatible relation.

Closure of relation • Depending upon nature of relations, there are 3 types of closure of relations. 1. Reflexive closure (R 1 = R U Δ) 2. Symmetric closure(R 1 = R U R-1) 3. Transitive closure(R* = R 1 U R 2 U R 3 U----- Rn) 1. Reflexive closure (R 1 = R U Δ): Let R be the relation on a set A which is not reflexive relation. A relation R 1 = R U Δ is called the reflexive closure of R if R U Δ is the smallest reflexive relation containing R. A={a, b, c, d} then Δ ={(a, a), (b, b), (c, c), (d, d)}

eg. Let A={1, 2, 3} R 1, R 2, R 3 are relations on set A. find reflexive closure of R 1, R 2, R 3. where R 1={(1, 1), (2, 1)} R 2={(1, 1), (2, 2), (3, 3)} R 3={(3, 1), (1, 3), (2, 3)} solution: we have A={1, 2, 3} therefore Δ ={(1, 1), (2, 2), (3, 3)} then i)reflexive closure of R 1 is. R=R 1 U Δ therefore R={(1, 1), (2, 2), (3, 3)} ii)The reflexive closure of R 2 is. R=R 2 U Δ therefore R={(1, 1), (2, 2), (3, 3)}

iii)The reflexive closure of R 3 is. R=R 3 U Δ therefore R={(3, 1), (1, 3), (2, 3), (1, 1), (2, 2), (3, 3)} 2. Symmetric closure(R 1 = R U R-1) Let R be the relation on a set A and R is not symmetric relation. A relation R 1= R U R-1 is called symmetric closure of R if R U R-1 is smallest symmetric relation containing R. eg. Find the symmetric closure of following relation on A={1, 2, 3} where R 1={(1, 1), (2, 1)} R 2={(1, 2), (2, 1), (3, 2)(2, 2)} R 3={(1, 1), (2, 2), (3, 3)} solution: given that A={1, 2, 3} i) R 1 -1 ={(1, 1), (1, 2)} R=R 1 U R 1 -1 R={(1, 1, ), (2, 1), (1, 2)} is a symmetric closure.

ii) R 2 -1 ={(2, 1), (1, 2), (2, 3), (2, 2)} R=R 2 U R 2 -1 R={(1, 2), (2, 1), (3, 2), (2, 3)} iii)R 3 is symmetric relation therefore R 3 itself is symmetric closure

Transitive closure(R* ) • Let R be the relation on set A which is not transitive relation. The transitive closure of a relation R is the smallest transitive relation containing R. It is denoted by R*. • Theorem: |A|=n , R be the relation on A then transitive closure (R* )= R 1 U R 2 U R 3 U----- Rn Eg. 1 If A={1, 2, 3, 4, 5} and R={(1, 2), (3, 4), (4, 5), (4, 1), (1, 1)} find its transitive closure Solution: Let R* be the transitive closure of given relation R. therefore R* = R 1 U R 2 U R 3 U R 4 U R 5 now R 2 = R. R ={(3, 5), (3, 1), (4, 2), (1, 1)} R 3 = R 2. R= {(3, 2), (3, 1), (4, 2), (4, 1), (1, 2), (1, 1)} R 4 = R 3. R= {(3, 2), (3, 1), (4, 2), (4, 1), (1, 2), (1, 1)= R 3 R 5 = R 4. R={(3, 2), (3, 1), (4, 2), (4, 1), (1, 2), (1, 1)}= R 4

therefore R* = R 1 U R 2 U R 3 U R 4 U R 5 R* ={(1, 2), (3, 4), (4, 5), (4, 1), (1, 1), (3, 5), (3, 1), (4, 2), (3, 2)} eg. 2 Let A={1, 2, 3, 4}and R ={(1, 2), (2, 3), (3, 4)} be a relation on Set A find R* and draw its diagraph.

Warshall’s Algorithm • Let S be the finite set {v 1, v 2, v 3 ------vn}, R is a relation on S. The adjacency matrix A of R is an nxn Boolean matrix defined by – Aij = 1 if the diagraph D has an edge from vi to vj. = 0 if the diagraph D has no edge from vi to vj. • Warshall’s Algorithm: It is an efficient method of finding the adjacency matrix of the transitive closure of relation R on a finite set S. It uses properties of diagraph D, in a particular walk of various lengths in D.

• Warshall’s algorithm to find transitive closure: eg 1. Find the transitive closure of the relation R on set A A={1, 2, 3, 4} defined by. R={(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (3, 4), (3, 2), (4, 3)} solution: Step 1 - we have |A|=4 thus we have to find warshall’s sets - w 0 w 1 w 2 w 3 w 4 The first set w 0 = MR = Step 2: To find w 1 from w 0 we consider the first column and first row

In C 1, 1 is present at R 2 In R 1 , 1 is present at c 2, c 3, c 4 Thus add new entries in w 1 at (R 2, C 2), (R 2, C 3), (R 2, C 4) which is given beloww 1= Step 3 - To find w 2 from w 1 we consider 2 nd column and 2 nd row In C 2 , 1 is present at R 1, R 2, R 3, R 4 In R 2 , 1 is present at C 1, C 2, C 3, C 4 thus we add new entries in w 2 at (R 1, C 1), (R 1, C 2), (R 1, C 3), (R 1, C 4), (R 2, C 1), (R 2, C 2), (R 2, C 3), (R 2, C 4), (R 3, C 1), (R 3, C 2), (R 3, C 3), (R 3, C 4), (R 4, C 1), (R 4, C 2), (R 4, C 3), (R 4, C 4) which is given below

w 2 = All entries in w 2 are 1 Hence w 2 is relation matrix of transitive closure of R and R*={(1, 1), (1, 2)(1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

eg 2. Use warshall’s algorithm to find transitive closure of R where MR = and A={1, 2, 3}. solution: step 1 -We have|A|=3 thus we have to find warshall’s set w 0 w 1 w 2 w 3. The first set is – w 0 = MR = Step 2 –To find w 1 from w 0 we consider the first column and the first row In C 1, 1 is present at R 1, R 3. In R 1, 1 is present at C 1, C 3. thus add new entries in w 1 at (R 1, C 1), (R 1, C 3), (R 3, C 1), (R 3, C 3).

w 1 = Step 3 : To find w 2 from w 1 we consider 2 nd column and 2 nd row In R 2, 1 is present at C 2 In C 2, 1 present at R 2, R 3 thus add new entries in w 2 at (C 2, R 2), (C 2, R 3) which is given below w 2 = = w 1

Step 4: To find w 3. from w 2 we consider 3 rd column and 3 rd row In C 3, 1 is present at R 1, R 3 In R 3, 1 is present at C 1, C 2, C 3 Thus add new entries in w 3 at (R 1, C 1), (R 1, C 2), (R 1, C 3), (R 3, C 1), (R 3, C 2), (R 3, C 3) which is given below w 3 = Hence w 3 is the relation matrix of R* and R* ={{1, 1), (1, 2), (1, 3), (2, 2), (3, 1), (3, 2), (3, 3)}

eg 3. Let R ={(a, d), (b, a), (b, d), (c, b), (c, d), (d, c)} use warshall’s algorithm to find the matrix of transitive closure where A={a, b, c, d}

Partial ordering relation(POSET) • A binary relation R on a non empty set A is a partial order if R is reflexive, antisymmetric and transitive. • An ordered relation is a transitive relation on a set by means of which we can compare elements of a set. • The ordered pair (A, R) is called a partially ordered set or poset. • If the two objects are always related in a poset it is called a total order or linear order or simple order. Eg. (Z, <=) is a poset. In this case either a<=b or b<=a so two things are always related hence <= is a total order and (Z, <=) is a chain.

Hasse diagram • The poset can be represented by diagraph however more econonical way to describe poset is hasse diagram. • It is useful tool which completely describes the associated partially ordered relation. • A diagram which is drawing by considering comparable and non comparable elements is called hasse diagram. Therefore while drawing hasse diagram following points must be followed. 1. The element of a relation R are called vertices and denoted by points. 2. All loops are omited as relation is reflexive on poset.

3. If a. Rb or a<=b then join a to b by a straight line called an edge, the vertex b appears above the level of vertex a. therefore the arrows may be omitted from the edges in hasse diagram. 4. If a<= b and b<=a i. e. a and b are non comparable elements, then they lie on same level and there is no edge between a and b. 5. If a<=b and b<=c then a<=c so there is a path a b c. therefore do not join a to c directly i. e delete all edges that are implied by transitive relation.

eg. Let A ={2, 3, 4, 6} and let a. Rb if a divides b. show that R is partial order and draw its hasse diagram. solution: R={(2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)} R is reflexive since (2, 2), (3, 3), (4, 4), (6, 6) ε R, R is antisymmetric since if a. Rb , b. Ra unless a=b. R is also transitive since a. Rb and b. Rc implies a. Rc hence R is a partial order. hasse diagram for R.

eg. 2. If A={1, 2, 3, 4} R={(1, 1), (1, 2), (2, 4), (1, 3), (3, 4), (1, 4), (4, 4)} then show that R is a partial order and draw its hasse diagraph. solution: R is reflexive since (1, 1), (2, 2), (3, 3), (4, 4)εR. R is antisymmetric since (1, 2)εR but (2, 1) ∉ R. (2, 4) εR but (4, 2) ∉ R. similaraly (1, 3), (1, 4), (3, 4) εR but (3, 1), (4, 3) ∉ R. one can also similarly check that R is transitive

eg 3. Draw hasse diagram of a poset (p(s), c) where S={a, b, c}. solution: P(s)={Ø, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} now find the comparable and non-comparable element Ø c {a} , Ø c {b}, Ø c {c} therefore {a}, {b}, {c} lie above the level Ø. {a} c {a, b} , {b} c {b, c} , {c} c {a, c}. therefore {a, b}, {b, c}, {a, c} lie above the level of {a}, {b}, {c}. {a, b} c S , {b, c} c S {a, c} c S therefore S lie above the level of {a, b}, {a, c}, {b, c} but {a}, {b}, {c} are non comparable therefore {a}, {b}, {c} lie on same level. {a, b}, {a, c}, {b, c} are non comparable therefore lie on same level by considering the above observations the hasse diagram is

Chain and antichains • Chain: Let (A, <=) be a poset. A subset of A is called chain if every pair of elements in the subset are related. • If A iteself is a chain the poset (A, <=) is called a totally ordered set or linearly ordered set. Eg. Let A ={1, 2, 3} and let the partial order <= mean “less than or equalto” then (A, <=) is chain and its hasse diagram is • Antichain: A subset of A is called antichain if no two distinct elements in a subset are related.

Eg. Let A={a, b} and consider its poset {p(A), c } P(A)={Ø , {a}, {b}, {a, b}} Then the following subsets are chains {Ø, {a}, {a, b}}, {Ø, {b}, {a, b}}, {Ø, {a}}, {Ø, {b}}, {{a}, {a, b}}, {{b}, {a, b}} The following subset is an antichain is {{a}, {b}} In the above eg length of longest chain is 3. And number of elements in antichain are 2.

In this eg chains are{ {Ø, {a}, {a, b, c}}, {{a}, {a, c}, {a, b, c}}, {{b, c}, {a, b, c}} Antichains are: {{a}, {b, }, {c}}

Maximal and minimal elements • Let (A, R)be a poset then a in A is a minimal element if there doesnot exist an element b in A such that b. Ra. similarly for a maximal elements • There can be more than one minimal and maximal element in a poset. • In this hasse diagram Ø is a minimal element and {a, b, c} is a maximal element

• maximal and minimal elements Eg. 1 Maximal elements are: {a, e} Minimal elements are : {c, f} Eg. 2 S={1, 2, 3} Maximal elements are: {2, 3} Minimal elements are: {1}

Eg 3. Find maximal elements for set S={a, b, c, d} Solution: a. Rb so a is not maximal element. a. Rc so a is not maximal element c. Rd so c is not maximal element b. Rd so b is not maximal element Therefore d is maximal element Eg. 4 maximal elements of set S={6, 7, 8, 9} solution: {9}.

• Minimal elements: Eg. 1 S={1, 2, 3} minimal elements are : {2, 3} there is no any element related to 2 and 3. Eg. 2 S={4, 5, 6} consider above hasse diagram Minimal elements are: {4, 5} Eg. 3 S={1, 2, 3} find minimal elements solution: {2, 3} S={4, 5, 6} solution: {4, 5}

Upper bound and lower bound • Let S be a subset of A in the poset (A, R). If there exist an element a in A such that s. Ra for all s in S, then a is called an upper bound similarly for lower bounds. Note: To be an upper bound you must be related to every element in the set similarly for lower bound. Eg. 1 upper bound for {a, b, c} therefore { g } is upper bound. Sometimes we might get more than 1 UB

• Lower bound: S={a, b} Lower bound: {a} a. Ra and a. Rb so a is only lower bound Eg. 1 upper bound for {c, e} is {d} lower bound for {c, e} is {b, a, f}

Least upper bound and greatest lower bound • Let a, b be elements in a poset (A, <=) an element c is said to be an upper bound of a and b if a<=b and b<=c. • An element c is said to be a least upper bound (lub) of a and b if c is an upper bound of a and b and if there is no other upper bound d of a and b such that d<=c. • Similarly an element e is said to be a lower bound of a and b if e<=a and e<=b and e is called greatest lower bound (glb) of a and b if there is no other lower bound f of a, b such that e<=f.

Eg. 1 least upper bound(lub) of {c, e} =d greatest lower bound (glb) of {c, e}=b Upper bound of {a, f} are elements {b, c, d, e} lub{a, f}= {b} Lower bound of {a, f} : do not exist Upper bound of {b, d} is d. lub{b, d}= d lower bound for {b, d} is {b, a, f} glb{b, d}=b

Lattice • A Lattice is a poset in which every pair of elements has a least upper bound (lub)and a gretest lower bound(glb). • Let (A, <=)be a poset and a, b ε A then • lub of a & b is denoted by a v b. It is called join of a & b. i. e. a vb =lub(a, b)=join(OR). • The greatest lower bound of a and b is called the meet of a and b and it is denoted by a۸ b. therefore a۸ b = glb(a, b)=and=meet. lattice is a mathematical structure with two binary oprations v (join) and ۸ (meet) It is denoted by {L, v, ۸ }

Eg. 1 Let A={1, 2, 3}, P(A)={Ø, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} show that (P(A), c ) is a lattice Solution: The hasse diagram of the poset (P(A), c ) is Here every pair of elements of a poset has lub and glb hence (P(A), c ) is a lattice.

Eg 2. Determine which of the following posets are lattice In figure every pair of elements has lub and glb. It is a lattice. Eg. 3 Every pair of element has lub and glb. It is a lattice

meet (glb)a۸ b does not exist therefore it is not a lattice join c v d (lub) does not exist. therefore it is not a lattice.

Properties of lattice • Let (L, ۸ , v) be a lattice and a, b, c ∉ L then L satisfies the following properties. 1. Commutative property: a۸ b=b ۸ a and avb =bva 2. Associative property: av(b۸ c)= (avb)۸ (avc) a۸ (bvc)=(a۸ b)v(a۸ c) 3. Absorption law: a۸ (a v b)=(a۸ a )v (a۸ b)=a a۸ a = a , ava =a a۸ b=a iff avb =b

Eg 1. Determine whether the poset represented by each of the hasse diagram are lattices. justify your answer. every pair of element has glb and lub therefore it is a lattice Eg. 2 Is it a lattice?

Eg 3 Let A be the set of positive factors of 15 and R be a relation on A. R={ x. Ry |x divides y, x, y ∈ A} Draw hasse diagram and give meet and join for lattice. Solution: we have A={1, 3, 5, 15} R={(1, 1), (1, 3), (1, 5), (1, 15), (3, 15), (5, 15), (15, 15)} Hasse diagram- Every pair of element has lub and glb. therefore it is a lattice.

Function • Let A and B be the non empty set. A function “f” from A to B, denoted as f : A B , is a relation from A to B such that for every a ε A, there exists a unique b ε B such that (a, b) ε f. • Normally if (a, b) ε f , we write f(a)=b. • f is a relation with following property If f(a)=b and f(a)=c then a=c f: A B, it is read as f is a function of A to B. • Set A is known as Domain of f. • Set B is known as co-domain of f. • In function one to many relation is not allowed.

eg. 1 Let A={1, 2, 3}, B={1, 4, 9}and f: A B : f(x)=x 2 here , f assigns to each member in A its square, whether f is a function ? Solution: f is a function

eg. 2 Let A ={a, b, c} and B={1, 2, 3} then f={(a, 1), (b, 2), (a, 3), (c, 3)}. Solution: It is not a function from A to B as two elements 1 and 3 ε B are assigned to the same element a ε A It is not a function as a is mapped with 1 as well as 3

eg 3. Let A ={1, 2, 3} and B={1, 4, 5} Let f={(1, 1), (2, 4)} then f is not a function A to B since no member of B is assigned to the element 3 ε A It is not a function because 5 is not mapped with any element from set A

eg 4. Let A={a, b, c} and B={a, b, c, d} then f={(a, b), (b, c), (c, c)}. whether it is function?

eg 4. Let A={a, b, c} and B={a, b, c, d} then f={(a, b), (b, c), (c, c)}. whether it is function? Solution: It is a function. since each element of A is mapped to a unique element of B The domain of f is A={a, b, c} The range or codomain of B={b, c}

Types of functions • One-one (Injective) • onto(surjective) • one-one and onto(bijective) 1. one-one(injective): A function f : X Y is defined to be one –one or injective , if the images of distinct elements of X under f are distinct, i. e. for every x 1, x 2 ε X, f(x 1)=f(x 2) implies x 1=x 2, otherwise many one.

If we give 1 as input then we get a as output. one i/p will have one distinct image or output.

eg 2. • function many –one (non injective): function that not one-one is called many one function

• outputs are not distinct , there are 2 values 1 &2 share the same o/p. • therefore function is many one or non injective

2. onto (surjective): A function f: X Y is said to be onto(surjective) if every element of Y is the image of some element of X under f, that is for every y ε Y, there exists an element x in Xsuch that f(x)=y Every co-domain element is mapped and every element in codomain has preimage so it is onto function

Every co-domain element is mapped and every element in codomain has preimage so it is onto function

3. one-one and onto(bijective): A function f: A B is said to be bijective if f is both injective and surjective. • In other words , a function f: A B is a bijection if 1. It is one-one i. e f(x)=f(y) x=y for all x, y ε A. 2. It is onto i. e. for all y ε B there exists x εA , such that f(x)=y • If function is 1: 1 and onto then it is having special meaning i. e invetible function.

Composite function or product of function • Let f: A B & g: B C be two functions. The composite function of f and g denoted by gof : A C is a function such that gof(a)=g[f(a)] =g[b] • domain of gof is set A and codomain of gof is set B eg. 1 Let f(x)=x+2, g(x)=x-2 find i) gof ii)fog iii)gog iv) fof [May 08, Dec 12 4 M] solution: f(x)=x+2 g(x)=x-2

i) gof solution: gof(x)=g[f(x)] =g[x+2] =x+2 -2 =x ii) fog solution: f[g(x)] =f[x-2] = x-2+2 =x

iii) gog solution: gog(x) =g[x-2] =x-2 -2 =x-4 iv) fof solution: fof(x) = f[x+2] =x+2+2 =x+4

2. f(x)=2 x+1 , g(x)= X 2 -2 find i) gof(4) & fog(4) ii)gof(a+2) & fog(a+2) iii)fog(5) iv)gof(a+3) solution: i) gof(4)=g[f(4)] = g[2(4)+1] = g[9] = 92 -2 =81 -2 =79

Q-1. Given a relation whether it is a function Q-2. How many function possible from set A to B Q 2. the number of function from set A to B where |A|=m, |B|=n? The number of tasks available T 1 is mapped with a 1. T 2 will mapped with a 2 and so on. a 1 can be mapped with any value of B same with a 2, a 3 --am. The task a 1 can be mapped with m ways.

therefore Ta 1 =n ways, Ta 2 =n ways -----nm times. Tf = Ta 1 Ta 2 Tam | Tf |=| Ta 1| |Ta 2 | | Tam | = nxnxn------m times = nm eg 1. f(x)= x 2 is it a function? solution: x is mapping with x 2 i. e. x x 2 random number we take it will mapped with random number. 2 4 3 9 4 16 every number has single image. therefore it is a function

eg 2. f(x) = is it a function? domain is real solution: x mapped with every element is mapped with exactly one element R mapped with where R is real number 4 mapped with +2 or -2 therefore element have multiple images it is not real number , it is imaginary number. therefore it is not a function

How to check it is one to one function or not • image of x 1 = image of x 2 therefore f(x 1)=f(x 2) then x 1=x 2 eg 1. f(x)= X 2 is it one to one or not? R implies R(real no) Solution: f(x 1)=f(x 2) x 12 =x 2 2 = ( x 12 - x 2 2 )= 0 ( x 1 - x 2) (x 1+ x 2)=0 x 1 = x 2 or x 1 = - x 2 but we have to prove x 1=x 2 , but here another case is present so it is not one to one.

eg f(x)=

Pigeonhole principle

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