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# Relations & functions

## by indu thakur, teacher at delhi on May 05, 2011

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assignment and some important points.

assignment and some important points.

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• Putcha Narasimham, Founder Professor & Proprietor--Coach & Consultant at Knowledge Enaber Systems I have read a part of this PPT and I found an example that supports my view that relation defined as 'any subset of Cartesian Product' is WRONG.

I claim that the 'Set of principles and criteria according to which ordered pairs are formed' is a 'relation' but NOT the 'sets of ordered pairs formed by themselves'. This applies to definition and examples given in slide 2 Equivalence Relation on A.

Please see my posts and comment. Thanks

26JUL13
4 months ago
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## Relations & functionsPresentation Transcript

• TYPES OF RELATIONS , TYPES OFFUNCTIONS & BINARY OPERATIONS BY:--INDU THAKUR
• EQUIVALENCE RELATION ON A: Let R be a relation on a non empty set A, then R is called an equivalence relation iff it is: (i) Reflexive , (ii) Symmetric & (iii) Transitive. Note: If a Relation does not satisfy any of the above properties it is not considered as an equivalence Relation.Identity Relation: Let A be a set. Then the relationIA = {(a, a) for all a є A} is called the identity relation on A. In otherwords a relation IA on A is called Identity Relation if every element of Ais related to itself only. Example: Let A={1,2,3} be a set. The relation IA ={(1,1),(2,2),(3,3)} is the identity relation on set A. But the relation R1 ={(1,1),(2,2)}and R2={(1,1),(2,2),(3,3),(1,3)} are notidentity relations on A , because (3, 3) ∉ R1 and in R2 element 1 is related to elements 1 and 3as well.
• Let A={1,2,3} be a set. Then R= {(1,1),(1,2),(2,1),(3,2)} is not a ReflexiveRelation on A. But R1= {(1,1),(2,2),(2,1),(3,2)} is not a ReflexiveRelation on A, because 2ЄA but(2,2) ∉ R. Note: Identity Relation is always reflexive on a non-void Set.However a Reflexive is not necessarily the Identity Relation on thatset. EMPTY RELATION : A relation in a set A is called empty relation , if no element of A is related to any element of A , R = φ ⊆ AXA. For example, consider a relation R in the set A = {1,2,3} given by R = {(a,b) : |a – b|= 8} . UNIVERSAL RELATION : A relation in a set A is called universal relation , if each element of A is related to every element of A , i.e., R = AXA. For example, consider a relation R in the set A = {1,2,3} given by R={(a,b) : |a – b|≥ 0} . NOTE : Both the empty relation and the universal relation are some times called trivial relations.
• EXAMPLE : Let A ={1,2,3} , find the number of relations on Acontaining (1,2) and (1,3) which are reflexive and symmetric but nottransitive. SOL. The smallest relation R1 on A containing (1, 2) and (1,3) which is reflexiveand symmetric but not transitive is {(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)}. We areleft with two pairs (2,3) and (3,2). Now to get another relation R2 , if we add anypair , say (2,3) to R1 then we must add the remaining pair (3,2) in order tomaintain symmetric of R2 and then R2 becomes transitive also. Hence , the numberof relation is one.EXAMPLE : Let A ={1,2,3} , find the number of equivalencerelations on A containing (1,2) . SOL. The smallest relation R1 on A containing (1, 2) is{(1,1),(2,2),(3,3),(1,2),(2,1)}. We are left with two pairs (1,3),(3,1),(2,3) and (3,2).Now to get another relation R2 , if we add any pair , say (2,3) to R1 then we mustadd the remaining pair (3,2) in order to maintain symmetric of R2. Also tomaintain transitivity we are forced to add (1,3) and (3,1) thus the only bigger thatR1 is the universal relation. Hence the total number of equivalence relations on Acontaining (1,2) is two.
• EXAMPLE : Let Q be the set all rational numbers and relation on Q defined by R={(X,Y) : 1 + XY > 0}. Prove that R is reflexive and symmetric but not transitive. SOL. Consider any x,y Є Q , since 1+x.x = 1+x2 ≥ 1(x,x) ЄR reflexiveLet (x,y) Є R 1+xy > 0 1+yx > 0 (y,x) ЄR symmetric.But not transitive . Since (-1,0) and (0,2) Є R , because 1 > 0 by putting values. But (-1,2) ∉ R because -1<0 Greatest Integer Function (Floor or step Function) f : R → R defined as f(x) = [x] ∀ x εR , {x} = x – [x] where {x} is Fractional part or decimal part of x. For Example, {3.45} = 0.45, {-2.75} = 0.25 From the definition of [x] , greatest integer less than or equal to x we can see that [x] = -1 for -1≤ x <0 = 0 for 0 ≤ x <1 = 1 for 1 ≤ x <2 and so on. For smallest Integer Function (ceiling Function) ,we take smallest integer greater than or equal to x. For example , ⌈4.7⌉ =5 , ⌈-7.2⌉ = -7, ⌈0.75⌉ = 1 and so on. It is neither 1-1 and onto.
• Example : Show that the relation R defined by (a,b) R (c,d) ⇨ a+d = b+c on the set N XN is an equivalence relation. SOL. Let (a,b) , (c,d) ,(e,f) Є NXN. (i) reflexive : since a+b = b+a a,b Є N ⇨ (a,b) R (b,a) (ii) symmetric : let (a,b)R(c,d) ⇨ a+d = b+c ⇨ d+a = c+b ⇨ (c,d) R (a,b) [by commutative law in N] (iii) transitive : let (a,b) R(c,d) , (c,d) R (e,f) ⇨ a+d = b+c and c+f = d+e by adding we will get a+d+c+f = b+c+d+e ⇨ (a,b) R (e,f) therefore R is an equivalence relation. EXAMPLE: Let R be a relation on N × N, defined by (a,b) R (c,d) ad=bc ∀ (a,b) ,(c,d) Є N × N. Show that R is an equivalence relation on N × N.SOL. Let (a,b) be an arbitrary element of N × N. Then (a,b) Є N × N. ⇨ ab = ba (by commutative on N) ⇨ R is reflexive on N × N. Let (a,b),(c,d) Є N × N such that (a,b) R (c,d) ⇨ ad=bc ⇨ cb=da (by commutative) Let (a,b),(c,d),(e,f) ЄN × N such that (a,b) R (c,d) and (c,d) R (e,f) then ad=bc and cf=de ⇨(ad)(cf)=(bc)(de) ⇨ af = be ⇨ (a,b) R (e,f)
• Example:Find the domain ofSolution:We set up the inequality x2 - x - 6 > 0and use our steps of quadratic inequalities to solve. Factoring we get (x - 3) (x + 2) > 0Putting -2 and 3 on a number line, gives three regions. The table shows x+2 x-3 Total Left (-3) - - + Hence the solution is Middle(0) + - - ( -∞ ,-2] U [3 , ∞) Right(4) + + +
• WHICH ARE RELATIONS BUT NOT FUNCTIONS ?
• The Vertical Line TestWHAT IS THE RANGE OF ABOVE GRAPH?If any vertical line passes through the graph at most once thenthe graph is the graph of a function.
• Example The Circle is not the graph of a function since there are vertical lines that cross it twice. Horizontal line TestIf any horizontal line passesthrough the graph at more thantwo points then graph is not 1-1or many one .
• Types of functions: Let f be a function from X to Y then f iscalled (i) One-one function (or injection) iff different elements of Xhave different images in Y. i.e. f(x1) =f(x2) ⇨ x1 = x2 for all x1, x2ЄX or ∀ x1≠x2 ⇨ f(x1) ≠ f(x2) .(ii) Many-one iff two or more elements of X have same image in Y,i.e., f is not one-one. (iii) Onto (or surjection) iff each element of Y isthe image of at least one element of X, i.e., iff range of f = Y.(iv) Into iff there exists at least one element in Y which is not theimage of any element of X, i.e., iff range of f is a proper subset of Y.One-one correspondence (or bijection) iff f is both one-one andonto.
• "One-to-One" NOT "One-to-One"A function f from A to B is called one-to-one (or 1-1) if wheneverf (a) = f (b) then a = b. No element of B is the image of more than oneelement in A. The function f (x) = x³ is One-to-One.
• "Onto" NOT "Onto"(all elements in B are used) (the 8 and 1 in Set B are not used) A function f from A to B is called onto if for all b in B there is an a in A such that f (a) = b. All elements in B are used. The function g (x) = | x - 2 | NOT One-to-One.
• Example (i): Let f: N→N be defined by f(n) = (n + 1)/2, if n is odd Or =n/2, ifn is even for all n Є N. Examine the function is onto, one-one or bijective. Solution: The function f is onto. Reason: Consider any nεN ( Codomain of f).Certainly 2nεN Also 2n is even. Thus, ∀ nЄ N (co- domain of f), ⁆ 2n Є N(domain of f) such that f(2n) = 2n/2 = n f is onto. But f is not 1-1Reason : As 1, 2Є N (domain of f) and f (1) = (1+1)/2=1, f (2) = 2/2 = 1, so thatthe different elements 1and 2 of the domain of f have the same image 1. Hence,f is not bijective. Example (ii): Let A=R-{3}and B=R-{1}. Consider the function f: A→Bdefined by f(x) =(x-2)/(x-3).Is f one-one and onto? Justify your answer.Solution: The function is one-one. Let x1, x2εA be such that f(x1)= f(x2)( x1-2)/(x1-3) = (x2-2)/(x2-3) ⇨ x1x2 - 2x1 - 3x2 +6= x1x2 - 3x1 - 2x2 +6 then -2x1-3x2 = -3x1-2x2 ⇨ x1= x2 , Hence f is one-one.We shall find the range of the function, Let y=f(x), then y= (x-2)/(x-3) xy-3y=x-2 then x= (3y-2)/(y-1) but xε R.(3y-2)/(y-1)εR it implies y≠1. Range of f= R-{1}=B. Thus the range ofthe function f= Codomain of f. Therefore f is onto. We can say it bijective.
• The composition of two functions is also called the resultant of two functionsor function of a function. The function on the right acts first and the function on theleft acts second, reversing English reading order. We remember the order byreading the notation as "g of ƒ". The order is important, because rarely do we getthe same result both ways. Let f : A → B and g : B → C be two functions . Then the composition of f and g , denoted by gof, is defined as the function gof : A → C given by gof(x) = g(f(x)) , ∀ x Є A . h = gof Some properties of composition of functions: 1. The composition of functions is associative i.e., if f : A → B , g : B→ C ,h : C →D are the functions then ho(gof) = (hog)of 2. Let f : A → B , g : B→ C be two functions. (i) If f , g are both 1-1 , then gof is also 1-1. (ii) If f , g are both onto , then gof is also onto.
• Remark: If A and B are (non-empty) finite sets containing m and n elements respectively, then (i) the number of functions from A to B =nm or n(B)n(A) .Remark : If f and g are functions such that gof is defined and (i) If gof is 1-1 , then f and g both necessarily 1-1. (ii) If gof is onto , then f and g both need not be necessarily onto. It can be verified in general that gof is 1-1 implies that f is 1-1.Similarly, gof is onto implies that g is onto.Examples : 1. f : {1,2,3,4} → {1,2,3,4,5,6} defined as f(x) = x ∀ x andg : {1,2,3,4,5,6} → {1,2,3,4,5,6} as g(x) = x , for x=1,2,3,4 and g(5)= g(6) = 5. then gof(x)= x ∀ x , which shows that gof is 1-1 butg is not 1-1. 2. f : {1,2,3,4} → {1,2,3,4} and g : {1,2,3,4} → {1,2,3} defined asf(1) =1, f(2) = 2, f(3)=f(4) =3 , g(1)=1,g(2)=2 and g(3)=g(4)=3. It showsthat gof is onto but f is not onto.
• INVERTIBLE FUNCTION: A function f : X →Y is definedto be invertible, if there exists a function g: Y →X suchthat gof = IX and fog = IY. The function g is called theinverse of ‘f ‘ and is denoted by f –1. Thus, if f isinvertible, then f must be one-one and onto andconversely, If f is one-one and onto, then f must beinvertible. We state some properties of invertible functions : 1. Inverse of a bijective function is unique. 2. Inverse of a bijective function is also bijective and (f-1)-1 =f. 3. If f: A → B is bijective then (i) f-1of = IA (ii) fof-1 = IB 4. If f: A → B and g : B → C are both bijectives , then gof : A → C and (gof)-1 = f-1og-1. 5. Let A be a non-empty set and f : A → A , g : A → A be two functions such that gof = IA = fog then f & g are both bijectives and g = f-1
• Example : If f : R → R and g : R → R are given byf(x) =|x| and g(x) = |2x – 5| , show that gof ≠ fog. SOL. gof : R → R is given by gof(x) = g(f(x)) =g(|x|)= |2|x| - 5| and fog : R → R is given byfog(x) = f(g(x))= f(|2x-5|) = ||2x-5||=|2x-5| . Also(gof) (-1) = |2|-1|-5|= |2-5|= |-3|=3 and (fog)(-1)= |2(-1)-5| =|-7| =7 ⇨ fog ≠ gof.Example: Show that if f: A →B and g : B →C are one-one, then gof :A →C is also one-one.Solution : Suppose gof (x1) = gof (x2) then g (f (x1)) = g (f (x2)) f(x1) = f (x2),as g is one-one x1 = x2, as f is one-one Hence, gof is one-one.Example: Show that if f: A →B and g: B →C are onto, then gof : A→C is also onto.Solution: an arbitrary element z Є C, ⁆ a pre-image y of z under gsuch that g (y) = z, since g is onto. Further, for y Є B, ⁆ an element x in Awith f (x) = y, since f is onto. Therefore, gof (x) = g (f (x)) = g (y) = z, showing that gofis onto.
• BINARY OPERATIONS : Let S be a non void set. A function Ffrom SxS to S is called a binary operation on S. i.e. F: SxS →S isa binary operation on set S. Generally binary operations arerepresented by the symbols * , instead of letters f, g etc. Thus aBinary operation * on a set S associates each ordered pair (a, b)of elements of S. (or any two elements of S) to a uniqueelement a*b of S.
• TYPES OF BINARY OPERATIONS:COMMUTATIVE BINARY OPERATION :A Binary Operation * on a set S is said to be commutative if (a*b) = (b*a) ∀ a, b є S. ASSOCIATIVE BINARY OPERATION:A Binary Operation * on a set S is said to be associative if (a*b)*c = a*(b*c) ∀ a, b, c єS. DISTRIBUTIVE BINARY OPERATION:Let * and 0 be two Binary operations on a set S. Then * is said to be (i) Left distributive over 0 if a*(b0c) =(a*b)0(a*c) for all a, b, c є S.(ii) Right distributive over 0 if (b0c)*a= (b*a) 0(c*a) for all a, b, c є S.IDENTITY BINARY OPERATION:Let * be a Binary Operation on a set S. An element e є S is said to be anidentity element for the binary operation * if a*e=a=e*a, for all a є S.INVERSE BINARY OPERATION : Let * be a Binary Operation on a set S. An element a є S is said to beinvertible with respect to the operation * , if ⁆ an element b in S such thata*b= =b*a = e and b is called inverse of a. NUMBER OF BINARY OPERATIONS: Let A be a finite set containing m elements then the number of binary operations on A is mmxm e.g. If A= {a, b} then the no. of binary operations on A = 24 = 16 is equal to number of functions from A × A to A.
• Is the operation * commutative?(Does the property x + y = y + x hold for ALL possible arrangements of values?)Start testing values:a * b = b * a is true since both sides equal b.c * d = d * c is true since both sides equal b.WOW! Having to test ALL possible arrangements could take forever! There mustbe an easier way......... The operation * is commutative. Testing for Commutativity Shortcut: It is easy to check whether an operation defined by a table is commutative. Simply draw a diagonal line from upper left to lower right, and see if the table is symmetric about this line. Reading the table: Read the first value from This table shows the operation * the left hand column and the second value ("star"). The operation is working on from the top row. The answer is in the cell the finite set where the row and column intersect. A = { a, b, c, d }. The table shows the For example, a * b = b, b * b = c, c * d = 16 possible calculations using the b, d * b = a and so on. elements of set A.
• What is the identity element for the operation * ?(What single element will always return the original value?)The identity element is a.a * a = a, b * a = b, c * a = c, d * a = dChecking for the Identity Element:You will know the identity element when you see it, because all of the values in itsrow or column are the same as the row or column headings.What is the inverse element for b ?(What element, when paired with b, will return the identity element a?)The inverse element of b is d. b*d=a Remark : Zero is identity operation on R but it is not identity for the addition operation on N , as 0 ∉N. In fact the addition operation on N does not have any identity. One further notices that for the addition operation + : R × R → R , given any aЄR, ⁆ -a in R such that a+(-a) = 0 (identity for ‘+’) =(-a)+a . Similarly for multiplication operation on R , given that a ≠0 in R ,⁆ 1/a in R such that a×(1/a) = 1(identity for ‘×’) = (1/a)×a.
• EXAMPLE : Define a binary operation * on the set {0,1,2,3,4,5} as a*b = Show that 0 is the identity for this operation and each element a of the set is invertible with 6 – a being the inverse of a. Ans. First of all we write down the operation table. * 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4 From the second row of operation table it is clear that 0 is the identity element. It is also clear that Inverse of 1 is 5 ; inverse of 2 is 4 Inverse of 3 is 3 ; inverse of 4 is 2 Inverse of 5 is 1 ; inverse of 6 is 0 , Thus inverse of a is 6 – a.
• Q.1 Show that the number of binary operations on {1,2}having 1 as identity and having 2 as inverse of 2 is exactly one.Ans.1 Binary operation on set S is a function from SXS →S i.e.,a function from {(1,1),(1,2),(2,1),(2,2)} to {1,2}. 1*1= 1, 1*2 = 2 = 2*1, and 2*2 = 1. DIAGRAM SXS * S (1,1) (1,2) 1 (2,1) 2 (2,2) BY:--INDU THAKUR
• Q.2 Determine the number of binary operations on {1,2} having 1 as identity element. Ans.2 Binary operation on set S is a function from SXS →S i.e., afunction from {(1,1),(1,2),(2,1),(2,2)} to {1,2}. 1*1 = 1, 1*2 = 2 = 2*1, and 2*2 = 1 or 2. DIAGRAM (there aretwo desired binary operations on S). SXS * S SXS * S (1,1) (1,1) (1,2) 1 (1,2) 1 (2,1) 2 (2,1) 2 (2,2) (2,2)
• EXAMPLE : Let A = NU{0} X NU{0} and let * be a binary operation on A defined by(a,b)*(c,d) = (a+c,b+d) ∀ (a,b) , (c,d) ЄA . Show that (i) * is commutative on A. (ii) * is associative on A. Also ,find the identity element, if any , in A.SOL. (i) Let (a,b) ,(c,d) ЄA Then (a,b)*(c,d) = (a+c,b+d) and (c,d)*(a,b) = (c+a,d+b) a+c= c+a and b+d= d+b ∀ a,b,c,d Є NU{0} Therefore (a,b)*(c,d) = (c,d)*(a,b) ∀ (a,b) ,(c,d) ЄA ‘*’ is commutative on A . (ii) For any (a,b) ,(c,d) ,(e,f) ЄA { ((a,b) *(c,d)}*(e,f) = (a+c,b+d) *(e,f) = ((a+c)+e, (b+d)+f) = (a+(c+e) , b+(d+f)) [by associative law on N] = (a,b)*(c+e , d+f) = (a,b) * { (c,d)*(e,f)} ⇨ ‘*’ is associative on A. L et (x,y) be the identity element in A . Then (a,b)* (x,y) = (a,b) ∀ (a,b) ЄA (a+x , b+y) = (a,b) a+x =a ,b+y = b , it gives x=0,y=0, therefore (0,0) ЄA (identity) Also (0,0) *(a,b) = (0+a , 0+b) = (a,b) ∀ (a,b) ЄA. Thus (0,0) is the identity element in A. BY:-- INDU THAKUR