Relations and Functions-II PDF

Relations and Functions-II MODULE - VII 23 Relation and...

Relations and Functions-II MODULE - VII 23 Relation and Function RELATIONS AND FUNCTIONS-II Notes We have learnt about the basic concept of Relations and Functions. We know about the ordered pair, the cartesian product of sets, relation, functions, their domain, Co-doman and range. Now we will extend our knowledge to types of relations and functions, composition of functions, invertible functions and binary operations. OBJECTIVES After studying this lesson, you will be able to :  verify the equivalence relation in a set  verify that the given function is one-one, many one, onto/ into or one one onto  find the inverse of a given function  determine whether a given operation is binary or not.  check the commutativity and associativity of a binary operation.  find the inverse of an element and identity element in a set with respest to a binary operation. EXPECTED BACKGROUND KNOWLEDGE Before studying this lesson, you should know :  Concept of set, types of sets, operations on sets  Concept of ordered pair and cartesian product of set.  Domain, co-domain and range of a relation and a function 23.1 RELATION 23.1.1 Relation : Let A and B be two sets. Then a relation R from Set A into Set B is a subset of A × B. Thus, R is a relation from A to B R  A × B  If (a, b) R then we write aRb which is read as ‘a’ is related to b by the relation R, if (a, b) R, then we write a R b and we say that a is not related to b by the relation R.  If n(A) = m and n(B) = n, then A × B has mn ordered pairs, therefore, total number of relations form A to B is 2mn. MATHEMATICS 105  Relations and Functions-II MODULE - VII 23.1.2 Types of Relations Relation and Function (i) Reflexive Relation : A relation R on a set A is said to be reflexive if every element of A is related to itself. Thus, R is reflexive  (a, a) R for all a A Notes A relation R is not reflexive if there exists an element a A such that (a, a) R. Let A = {1, 2, 3} be a set. Then R = {(1, 1), (2, 2), (3, 3), (1, 3), (2, 1)} is a reflexive relation on A. but R1 = {(1, 1), (3, 3) (2, 1) (3, 2)} is not a reflexive relation on A, because 2 A but (2,2) R. (ii) Symmetric Relation A relation R on a set A is said to be symmetric relation if (a, b) R (b, a) R for all (a, b) A i.e. aRb bRa for all a, b A. Let A = {1, 2, 3, 4} and R1 and R2 be relations on A given by R1 = {(1, 3), (1, 4), (3, 1), (2, 2), (4, 1) and R2 = {(1, 1), (2, 2), (3, 3), (1, 3)}  R1 is symmetric relation on A because (a, b) R1 (b, a) R1 or aR1b bR1 a for all a,b A but R2 is not symmetric because (1, 3) R2 but (3, 1) R2. A reflexive relation on a set A is not necessarily symmetric. For example, the relation R ={(1, 1), (2, 2), (3, 3), (1, 3)} is a reflexive relation on set A = {1, 2, 3} but it is not symmetric. (iii) Transitive Relation: Let A be any set. A relation R on A is said to be transitive relation if (a, b) R and (b, c) R (a, c) R for all a, b, c A i.e. aRb and bRc aRc for all a, b, c A For example : On the set N of natural numbers, the relation R defined by xRy  ‘x is less than y’, is transitive, because for any x, y, z N x < y and y < z  x < z i.e. xRy and yRz  xRz Take another example Let A be the set of all straight lines in a plane. Then the relation ‘is parallel to’ on A is a transitive relation, because for any l1, l2, l3 A l, ||l2 and l2|| l3 l1|| l3 106 MATHEMATICS  Relations and Functions-II MODULE - VII Example 23.1 Check the relation R for reflexivity, symmetry and transitivity, where R is Relation and defined as l1Rl2 iff l1 l2 for all l1, l2 A Function Solution : Let A be the set of all lines in a plane. Given that l1 Rl2  l1  l2 for all l1, l2 A Reflexivity : R is not reflexive because a line cannot be perpendicular to itself i.e. l  Notes l is not true. Symmetry : Let l1, l2 A such that l1Rl2 Then l1 Rl2  l1  l2  l2  l1  l2 Rl1 So, R is symmetric on A Transitive R is not transitive, because l1  l2 and l2  l3 does not impty that l1 l3 23.2 EQUIVALENCE RELATION A relation R on a set A is said to be an equivalence relation on A iff (i) it is reflexive i.e. (a, a) R for all a A (ii) it is symmetric i.e. (a, b) R (b, a) R for all a, b A (iii) it is transitive i.e. (a, b) R and (b, c) R (a, c) R for all a, b, c A For example the relation ‘is congruent to’ is an equivalence relation because (i) it is reflexive as    (,)R for all S where S is a set of triangles. (ii) it is symmetric as  1R 2  1   2    2  1   2 R1 (iii) it is transitive as 1   2 and  2   3  1   3 it means (1, 2) R and (2, 3) R (1, 3)R Example 23.2 Show that the relation R defined on the set A of all triangles in a plane as R = {(T1,T2) : T1 is similar to T2) is an equivalence relation. Solution : We observe the following properties of relation R; Reflexivity we know that every triangle is similar to itself. Therefore, (T, T) R for all T A R is reflexive. Symmetricity Let (T1, T2) R, then (T1, T2) R  T1 is similar to T2  T2 is similar to T1  (T2, T1) R, So, R is symmetric. MATHEMATICS 107  Relations and Functions-II MODULE - VII Transitivity : Let T1, T2, T3 A such that (T1, T2) R and (T2, T3) R. Relation and Then (T1, T2) R and (T2, T3) R Function T1 is similar to T2 and T2 is similar to T3 T1 is similar to T3 (T1, T3) R Notes Hence, R is an equivalence relation. CHECK YOUR PROGRESS 23.1 1. Let R be a relation on the set of all lines in a plane defined by (l1, l2) R line l1 is parallel to l2. Show that R is an equivalence relation. 2. Show that the relation R on the set A of points in a plane, given by R = {(P, Q) : Distance of the point P from the origin is same as the distance of the point Q from the origin} is an equivalence relation. 3. Show that each of the relation R in the set A   x  z : 0  x  12 , given by (i) R   a , b  : a  b is multiple of 4 (ii) R  { a, b  : a  b } is an equivalence relation 4. Prove that the relation 'is a factor of' from R to R is reflexive and transitive but not symmetric. 5. If R and S are two equivalence relations on a set Athen R  S is also an equivalence relation. 6. Prove that the relation R on set N  N defined by (a,b) R (c,d)  a+d = b + c for all (a,b), (c,d)  N  N is an equivalence relation. 23.3 CLASSIFICATION OF FUNCTIONS Let f be a function from A to B. If every element of the set B is the image of at least one element of the set A i.e. if there is no unpaired element in the set B then we say that the function f maps the set A onto the set B. Otherwise we say that the function maps the set A into the set B. Functions for which each element of the set A is mapped to a different element of the set B are said to be one-to-one. One-to-one function Fig.23.27 108 MATHEMATICS  Relations and Functions-II MODULE - VII The domain is  A, B, C  Relation and The co-domain is 1, 2, 3, 4  Function The range is 1, 2, 3  A function can map more than one element of the set A to the same element of the set B. Such a type of function is said to be many-to-one. Notes Many-to-one function Fig. 23.2 The domain is  A, B, C  The co-domain is 1, 2, 3, 4  The range is 1, 4  A function which is both one-to-one and onto is said to be a bijective function. Fig. 23.3 Fig. 23.4 Fig. 23.5 Fig. 23.6 Fig. 23.3 shows a one-to-one function mapping  A, B, C  into 1, 2, 3, 4 . Fig. 23.4 shows a one-to-one function mapping  A, B, C  onto 1, 2, 3 . Fig. 23.5shows a many-to-one function mapping  A, B, C  into 1, 2, 3, 4 . Fig. 23.6 shows a many-to-one function mapping  A, B, C  onto 1, 2 . Function shown in Fig. 23.4 is also a bijective Function. MATHEMATICS 109  Relations and Functions-II MODULE - VII Note : Relations which are one-to-many can occur, but they are not functions. The following Relation and figure illustrates this fact. Function Notes Fig. 23.7 Example 23.3 Without using graph prove that the function f : R R defiend by f  x   4  3x is one-to-one. Solution : For a function to be one-one function f  x1   f  x 2   x1  x 2  x1, x 2  domain  Now f  x1   f  x 2  gives 4  3x1  4  3x 2 or x1  x 2  f is a one-one function. Example 23.4 Prove that f : R R defined by f  x   4x 3  5 is a bijection Solution : Now f  x1   f  x 2   x1, x 2 Domain  4x13  5  4x 23  5  x13  x 23  x13  x 23  0   x 2  x1   x12  x1x 2  x 22   0  x1  x 2 or x12  x1x 2  x 2 2  0 (rejected). It has no real value of x1 and x 2.  f is a one-one function. Again let y   x  where y  codomain, x  domain. 1/ 3  y5 We have y 4x 3  5 or x    4   For each y  codomain  x  domain such that f  x   y. Thus f is onto function.  f is a bijection. 110 MATHEMATICS  Relations and Functions-II Example 23.5 Prove that f : R R defined by f  x   x 2  3 is neither one-one nor MODULE - VII Relation and onto function. Function Solution : We have f  x1   f  x 2   x1, x 2  domain giving x12  3  x 22  3  x12  x 2 2 Notes or x12  x 2 2  0  x1  x 2 or x1   x 2 or f is not one-one function. Again let y  f  x  where y  codomain x  domain.  y  x2  3  x  y3   y  3  no real value of x in the domain.  f is not an onto finction. 23.4 GRAPHICAL REPRESENTATION OF FUNCTIONS Since any function can be represented by ordered pairs, therefore, a graphical representation of the function is always possible. For example, consider y  x 2. y  x2 x 0 1 1 2 2 3 3 4 4 y 0 1 1 4 4 9 9 16 16 Fig. 23.8 Does this represent a function? Yes, this represent a function because corresponding to each value of x  a unique value of y.. Now consider the equation x 2  y 2  25 x 2  y 2  25 MATHEMATICS 111  Relations and Functions-II MODULE - VII Relation and x 0 0 3 3 4 4 5 5 3 3 4 4 Function y 5 5 4 4 3 3 0 0 4 4 3 3 Notes Fig. 23.9 This graph represents a circle. Does it represent a function ? No, this does not represent a function because corresponding to the same value of x, there does not exist a unique value of y. CHECK YOUR PROGRESS 23.2 1. (i) Does the graph represent a function? Fig. 23.10 (ii) Does the graph represent a function ? Fig. 23.11 112 MATHEMATICS  Relations and Functions-II 2. Which of the following functions are into function ? MODULE - VII Relation and (a) Function Notes Fig.23.12 (b) f : N N, defined as f  x   x 2 Here N represents the set of natural numbers. (c) f : N N, defined as f  x   x 3. Which of the following functions are onto function if f : R R (a) f  x   115x  49 (b) f x  x 4. Which of the following functions are one-to-one functions ? (a) f :  20, 21, 22   40, 42, 44  defined as f  x   2x (b) f :  7, 8, 9  10  defined as f  x   10 (c) f : I R defined as f  x   x 3 (d) f : R R defined as f  x   2  x 4 (d) f : N N defined as f  x   x 2  2x 5. Which of the following functions are many-to-one functions ? (a) f :  2, 1,1, 2   2, 5  defined as f  x   x 2  1 (b) f :  0,1, 2  1 defined as f  x   1 (c) Fig.23.13 (d) f : N N defined as f  x   5x  7 MATHEMATICS 113  Relations and Functions-II MODULE - VII 23.5 COMPOSITION OF FUNCTIONS Relation and Function Consider the two functions given below: y  2x  1, x  1, 2, 3  z  y  1, y   3, 5, 7  Notes Then z is the composition of two functions x and y because z is defined in terms of y and y in terms of x. Graphically one can represent this as given below : Fig. 23.18 The composition, say, gof of function g and f is defined as function g of function f. If f : A B and g : B C then g o f : A to C Let f  x   3x  1 and g  x   x2  2 Then fog  x   f  g  x    f  x 2  2   3  x 2  2   1  3x 2  7 (i) and  gof   x   g  f  x    g  3x  1    3x  1 2  2  9x 2  6x  3 (ii) Check from (i) and (ii), if fog = gof Evidently, fog  gof Similarly,  fof   x   f  f  x    f  3x  1  [Read as function of function f ].  3  3x  1   1  9x  3  1  9x  4  gog   x   g  g  x    g  x 2  2  [ Read as function of function g ] 2   x 2  2   2  x 4  4x 2  4  2  x 4  4x 2  6 114 MATHEMATICS  Relations and Functions-II MODULE - VII Example 23. 6 If f  x   x  1 and g  x   x 2  2 , calculate fog and gof. Relation and Solution : fog  x   f  g  x   Function  f  x2  2   x2  2  1  x2  3  gof   x   g  f  x   Notes 2  g  x  1    x  1   2  x  1  2 = x + 3. Here again, we see that  fog   gof 1 Example 23. 7 If f  x   x 3 , f : R R and g  x   , g : R  0 R  0 x Find fog and gof. 3 1 1 1 Solution :  fog   x   f  g  x    f       x x x3 1  gof   x   g  f  x    g  x 3   x 3 Here we see that fog = gof CHECK YOUR PROGRESS 23.3 1. Find fog, gof, fof and gog for the following functions : 1 f  x   x 2  2, g  x   1  , x  1. 1 x 2. For each of the following functions write fog, gof, fof and gog. (a) f  x   x 2  4 , g  x   2x  5 (b) f  x   x2 , g  x   3 2 (c) f  x   3x  7 , g  x   ,x 0 x 3. Let f  x   | x |, g  x    x . Verify that fog  gof. 4. Let f  x   x 2  3, g  x   x  2 3 3 Prove that fog  gof and f  f     g  f      2    2 5. If f  x   x 2 , g  x   x. Show that fog  gof. 1 1 6. Let f  x   | x |, g  x    x  3 , h  x   ; x  0. x Find (a) fog (b) goh (c) foh (d) hog (e) fogoh MATHEMATICS 115  Relations and Functions-II MODULE - VII 23.6 INVERSE OF A FUNCTION Relation and Function (A) Consider the relation Notes Fig. 23.19 This is a many-to-one function. Now let us find the inverse of this relation. Pictorially, it can be represented as Fig 23.20 Clearly this relation does not represent a function. (Why ?) (B) Now take another relation Fig.23.21 It represents one-to-one onto function. Now let us find the inverse of this relation, which is represented pictorially as Fig. 23.22 116 MATHEMATICS  Relations and Functions-II This represents a function. (C) Consider the relation MODULE - VII Relation and Function Notes Fig. 23.23 Ir represents many-to-one function. Now find the inverse of the relation. Pictorially it is represented as Fig. 23.24 This does not represent a function, because element 6 of set B is not associated with any element of A. Also note that the elements of B does not have a unique image. (D) Let us take the following relation Fig. 23.25 It represent one-to-one into function. Find the inverse of the relation. Fig. 23.26 MATHEMATICS 117  Relations and Functions-II MODULE - VII It does not represent a function because the element 7 of B is not associated with any element of Relation and A. From the above relations we see that we may or may not get a relation as a function when we Function find the inverse of a relation (function). We see that the inverse of a function exists only if the function is one-to-one onto function i.e. only if it is a bijective function. Notes CHECK YOUR PROGRESS 23.4 1 (i) Show that the inverse of the function y  4x  7 exists. (ii) Let f be a one-to-one and onto function with domain A and range B. Write the domain and range of its inverse function. 2. Find the inverse of each of the following functions (if it exists) : (a) f x  x  3  xR (b) f  x   1  3x  xR (c) f  x   x2  xR x 1 (d) f x  , x 0 xR x 23.7 BINARY OPERATIONS : Let A, B be two non-empty sets, then a function from A × A to A is called a binary operation on A. If a binary operation onA is denoted by ‘*’, the unique element ofAassociated with the ordered pair (a, b) of A × A is denoted by a * b. The order of the elements is taken into consideration, i.e. the elements associated with the pairs (a, b) and (b, a) may be different i.e. a * b may not be equal to b * a. Let A be a non-empty set and ‘*’ be an operation on A, then 1. A is said to be closed under the operation * iff for all a, b A implies a * b A. 2. The operation is said to be commutative iff a * b = b * a for all a, b A. 3. The operation is said to be associative iff (a * b) * c = a * (b * c) for all a, b, c A. 4. An element e A is said to be an identity element iff e * a = a = a * e 5. An element a A is called invertible iff these exists some b A such that a * b = e = b * a, b is called inverse of a. 118 MATHEMATICS  Relations and Functions-II Note : If a non empty set A is closed under the operation *, then operation * is called a binary MODULE - VII operation on A. Relation and Function For example, let A be the set of all positive real numbers and ‘*’ be an operation on A defined ab by a * b = for all a, b A 3 For all a, b, c A, we have Notes ab (i) a*b= is a positive real number A is closed under the given operation. 3  * is a binary operation on A. ab ba (ii) a *b    b * a  the operation * is commutative. 3 3 ab.c bc a bc abc ab abc (iii) (a * b) * c  *c  3  and a *(b * c)  a * .  – 3 3 9 3 3 3 9  ( a * b) * c  a * (b * c )  the operation * is associative. a a (iv) There exists 3 A such that 3 * a = 3.  a .3  a *3 3 3 3 is an identity element. 9 9 a. 9 (v) For every a A, there exists  A such that a *  a  3 and a a 3 9.a 9 *a  a  3 a 3 9 9 9  a *  3  * a  every element of A is invertible, and inverse of a is a a a CHECK YOUR PROGRESS 23.5 1. Determine whether or not each of operation * defined below is a binary operation. ab (i) ab  , a, b  Z 2 (ii) a  b  a b , a, b  Z (iii) a  b  a 2  3b 2 , a, b  R 2. If A  1, 2 find total number of binary operations on A. MATHEMATICS 119  Relations and Functions-II MODULE - VII 3. Let a binary operation ‘*’ on Q (set of all rational numbers) be defined as Relation and a * b = a + 2b for all a, b Q. Function Prove that (i) The given operation is not commutative. (ii) The given operation is not associative. Notes  ab 4. Let * be the binary operation difined on Q by a * b  for all a, b Q  then find the 3 inwrse of 4*6. 5. Let A  N  N and * be the binary operation on A defined by (a,b)*(c,d)=(a+c, b+d). Show that * is commutative and associative. Find the identity element of on A if any 6. A binary operation * on Q - {-1} is defined by a * b = a+b+ab; for all a, b  Q  1. Find identity element on Q. Also find the inverse of an element in Q-{-1}. C 1A + % LET US SUM UP  Reflexive relation R in X is a relation with (a, a) R a X.  Symmetric relation R in X is a relation satisfying (a, b) R implies (b, a) R.  Transitive relation R in X is a relation satisfying (a, b) R and (b, c) R implies that (a, c)R.  Equivalence relation R in X is a relation which is reflexive, symmetric and transitive.  If range is a subset of co-domain that function is called on into function.  If f: A B , and f (x) = f (y) x = y that function is called one-one function.  Any function is inuertible if it is one-one-onto or bijective.  If more than one element of A has only one image in to than function is called many one function.  A binary operation * on a set A is a function * from A × A to A.  If a * b = b * a for all a, b A, then the operation is said to be commutative.  If (a * b) * c = a * (b * c) for all a, b, A, then the operation is said to be associative.  If e * a = a = a * e for all a A, then element e A is said to be an identity element.  If a * b = e = b * a then a and b are inverse of each other  A pair of elements grouped together in a particular order is called an a ordered pair.  If n(A) = p, n(B) = q then n(A × B) = pq  R × R = {(x, y) : x, y  R} and R × R × R = {(x, y, z) : x, y, z R} 120 MATHEMATICS  Relations and Functions-II  In a function f : A B, B is the codomain of f. MODULE - VII  f, g : X R and X  R, then Relation and Function (f + g)(x) = f(x) + g(x), (f – g)(x) = f(x) – g(x)  f f ( x) (f. g)x = f(x). g(x),   ( x)   g g ( x) Notes  A real function has the set of real number or one of its subsets both as its domain and as its range. SUPPORTIVE WEBSITES http://www.bbc.co.uk/education/asguru/maths/13pure/02functions/06composite/ index.shtml http://mathworld.wolfram.com/Composition.html http://www.cut-the-knot.org/Curriculum/Algebra/BinaryColorDevice.shtml http://mathworld.wolfram.com/BinaryOperation.html TERMINAL EXERCISE 1. Write for each of the following functions fog, gof, fof, gog. (a) f  x   x 3 g  x   4x  1 1 (b) f  x   ,x  0 g  x   x 2  2x  3 x2 (c) f  x   x  4,x  4 gx  x  4 (d) f  x   x 2  1 g  x   x2  1 1 1 2. (a) Let f  x   x , g  x   , x  0, h  x   x 3. Find fogoh x (b) f  x   x 2  3, g  x   2x 2  1 Find fog (3) and gof (3). 3. Which of the following equations describe a function whose inverse exists : (a) f  x   x (b) f  x   x, x  0 3x  5 3x  1 (c) f  x   x 2  1, x  0 (d) f  x   (e) f  x   x  1. 4 x 1 2 4.   If gof  x   sin x and gof  x   sin x then find f (x) and g (x) MATHEMATICS 121  Relations and Functions-II MODULE - VII a b Relation and 5. Let  be a binary operation on Q defined by a  b  for all a,b  Q , prove that 3 Function  is commutative on Q. ab 6. Let  be a binary operation on on the set Q of rational numbers define by a  b  5 Notes for all a,b  Q , show that  is associative on Q. 7. Show t hat the relation R in the set o f real numbers, defined as R = {(a, b)} : a  b2} is neither reflexive, nor symmetric nor transitive. 8. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric and transitive. 9. Show that the relation R in the set A defined as R = {(a, b)  : a = b} a, b A, is equivalence relation. 10. Let A = N × N, N being the set of natural numbers. Let * : A × A A be defined as (a, b) * (c, d) = {ad + bc, bd) for all (a, b), (c, d) A. Show that (i) * is commutative (ii) * is associative (iii) identity element w.r.t * does not exist. 11. Let * be a binary operation on the set N of natural numbers defined by the rule a * b = ab for all a, b N (i) Is * commutative? (ii) Is * associative? 122 MATHEMATICS  Relations and Functions-II MODULE - VII Relation and ANSWERS Function CHECK YOUR PROGRESS 23.2 1. (i) No (ii) Yes Notes 2. (a), (b) 3. (a), 4. (a), (c),(e) 5. (a), (b) CHECK YOUR PROGRESS 23.3 x2 x2  2 1. fog   2 , gof   1  x 2 x2  1 fof  x 4  4x 2  6 , gog  x 2. (a) fog  4x 2  20x  21 , gof  2x 2  3 fof  x 4  8x 2  12 , gog  4x  15 (b) fog  9, gof  3, fof  x 4 , gog  3 6  7x 2 (c) fog  , gof  , fof  9x  28, gog  x x 3x  7 1 1 1 6. (a) fog  x 3 (b) goh  1 (c) foh  x x3 1 (d) hog  (e) fogoh(1)  1 1 x3 CHECK YOUR PROGRESS 23.4 1. (ii) Domain is B. Range is A. 1 x 2. (a) f 1 (x)  x  3 (b) f 1 (x)  3 1 (c) Inverse does not exist. (d) f 1 (x)  x 1 MATHEMATICS 123  Relations and Functions-II MODULE - VII CHECK YOUR PROGRESS 23.5 Relation and Function 1. (i) No (ii) Yes (iii) Yes 2. 16 9 Notes 4. 8 5. (0,0) 1 a 6. identity = 0, a  a 1 TERMINAL EXERCISE 3 1. (a) fog   4 x  1 , gof  4 x3  1, fog  x 9 , gog  16 x  5 1 fog  , 3x 4  2 x 2  1 (b) 2 gof  , fof  x 4 , gog  x 4  4 x3  4 x 2 x 2  2 x  3 x4 (c) fog  x  8, gof  x  4  4 , fof  x  4  4 gog  x  8 (d) fog  x 4  2 x 2 , gof  x 4  2 x 2  2, fof  x 4  2 x 2 , gog  x 4  2 x 2  2, 1 2. (a) 1 , (b)  fog  3  364 ,  gof  3  289 x3 3. (c), (d), (e), 4. f  x   sin 2 x, g  x   x 8. Neither reflexive, nor symmetric, nor transitive 9. Yes, R is an equivalence relation 11. (i) Not commutative 124 MATHEMATICS