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define composition and inverse relation with example in discrete mathematics

R = {(a, b) / a, b ∈ A} Then, the inverse relation R-1 on A is given by R-1 = {(b, a) / (a, b) ∈ R} That is, in the given relation, if "a" is related to "b", then "b" will be related to "a" in the inverse relation . Instructor: Is l Dillig, CS311H: Discrete Mathematics Recurrence Relations 2/23 Recurrence Relations I Recurively de ned sequences are often referred to as recurrence relations I The base cases in the recursive de nition are calledinitial valuesof the recurrence relation I Example:Write recurrence relation representing number of Given the bijections \(f\) and \(g\), find \(f\circ g\), \((f\circ g)^{-1}\) and \(g^{-1}\circ f^{-1}\). Let \(I_A\) and \(I_B\) denote the identity function on \(A\) and \(B\), respectively. To find the inverse function of \(f :{\mathbb{R}}\to{\mathbb{R}}\) defined by \(f(x)=2x+1\), we start with the equation \(y=2x+1\). \cr}\], \[{g\circ f}:{A}\to{C}, \qquad (g\circ f)(x) = g(f(x)).\], \[(g\circ f)(x) = g(f(x)) = 5f(x)-7 = \cases{ 5(3x+1)-7 & if $x < 0$, \cr 5(2x+5)-7 & if $x\geq0$. Therefore, \[(f^{-1}\circ f)(a) = f^{-1}(f(a)) = f^{-1}(b) = a,\]. Example 2: Give an example of an Equivalence relation. In the morning assembly at schools, students are supposed to stand in a queue in ascending order of the heights of all the students. 2 converse inverse? \(f :{\mathbb{Q}-\{2\}}\to{\mathbb{Q}^*}\), \(f(x)=1/(x-2)\); \(g :{\mathbb{Q}^*}\to{\mathbb{Q}^*}\), \(g(x)=1/x\). If  \(g\circ f\) is bijective, then \((g\circ f)^{-1}= f^{-1}\circ g^{-1}\). After simplification, we find \(g \circ f: \mathbb{R} \to \mathbb{R}\), by: \[(g\circ f)(x) = \cases{ 15x-2 & if $x < 0$, \cr 10x+18 & if $x\geq0$. Hence, \(|A|=|B|\). Read Inverse Functions for more. Definition Of Matrix • A matrix is a rectangular array of numbers. Example \(\PageIndex{3}\label{eg:invfcn-03}\). \(f :{\mathbb{Q}}\to{\mathbb{Q}}\), \(f(x)=5x\); \(g :{\mathbb{Q}}\to{\mathbb{Q}}\), \(g(x)=\frac{x-2}{5}\). In the mathematics of binary relations, the composition relations is a concept of forming a new relation R ; S from two given relations R and S.The composition of relations is called relative multiplication in the calculus of relations.The composition is then the relative product: 40 of the factor relations. \[f^{-1}(x) = \cases{ \textstyle\frac{1}{3}\,x & if $x\leq 3$, \cr \textstyle\frac{1}{2} (x-1) & if $x > 3$. To compute \(f\circ g\), we start with \(g\), whose domain is \(\mathbb{R}\). This video contains 1. IntroductionIntroduction Relationships … Exercise caution with the notation. In this article, we will learn about the relations and the different types of relation in the discrete mathematics. If \(p,q:\mathbb{R} \to \mathbb{R}\) are defined as \(p(x)=2x+5\), and \(q(x)=x^2+1\), determine \(p\circ q\) and \(q\circ p\). In brief, an inverse function reverses the assignment rule of \(f\). \cr}\], \[f(x) = 3x+2, \qquad\mbox{and}\qquad g(x) = \cases{ x^2 & if $x\leq5$, \cr 2x-1 & if $x > 5$. Let us start to learn the composition of functions and invertible function… Exercise \(\PageIndex{11}\label{ex:invfcn-11}\). Composite Functions. If \(n=2m\), then \(n\) is even, and \(m=\frac{n}{2}\). Then, because \(f^{-1}\) is the inverse function of \(f\), we know that \(f^{-1}(b)=a\). The function \(f :{\mathbb{Z}}\to{\mathbb{N}}\) is defined as \[f(n) = \cases{ -2n & if $n < 0$, \cr 2n+1 & if $n\geq0$. We find, \[\displaylines{ (g\circ f)(x)=g(f(x))=3[f(x)]+1=3x^2+1, \cr (f\circ g)(x)=f(g(x))=[g(x)]^2=(3x+1)^2. Do not forget to describe the domain and the codomain, Define \(f,g :{\mathbb{R}}\to{\mathbb{R}}\) as, \[f(x) = \cases{ 3x+1 & if $x < 0$, \cr 2x+5 & if $x\geq0$, \cr}\], Since \(f\) is a piecewise-defined function, we expect the composite function \(g\circ f\) is also a piecewise-defined function. discrete-mathematics elementary-set-theory relations function-and-relation-composition. Discrete MathematicsDiscrete Mathematics and Itsand Its ApplicationsApplications Seventh EditionSeventh Edition Chapter 9Chapter 9 RelationsRelations Lecture Slides By Adil AslamLecture Slides By Adil Aslam mailto:adilaslam5959@gmail.commailto:adilaslam5959@gmail.com 2. Find the inverse of each of the following bijections. More than 1,700 students from 120 countries! R is symmetric x R y implies y R x, for all x,y∈A The relation is reversable. Welcome to this course on Discrete Mathematics. Examples: If f(x) = x + 5 and g(x) = 3x 2 find (a) (f ∘ g)(x) (b) (f ∘ g)(2) (c) g(f(x)) share | cite | improve this question | follow | edited Jun 12 '20 at 10:38. A study guide for discrete mathematics, including course notes, worked exercises, and a mock exam. \(f(a_1) \in B\) and \(f(a_2) \in B.\)  Let \(b_1=f(a_1)\) and \(b_2=f(a_2).\) Substituting into equation 5.5.3, \[g(b_1)=g(b_2).\] Since \(f\) is a piecewise-defined function, we expect its inverse function to be piecewise-defined as well. Assume \(f(a)=b\). Composition of functions is a special case of composition of relations. & if $x > 3$. The image is computed according to \(f(g(x)) = 1/g(x) = 1/(3x^2+11)\). Why is \(f^{-1}:B \to A\) a well-defined function? A bijection is a function that is both one-to-one and onto. Here, the function \(f\) can be any function. (Beware: some authors do not use the term codomain(range), and use the term range inst… CS340-Discrete Structures Section 4.1 Page 6 Properties of Binary Relations: R is reflexive x R x for all x∈A Every element is related to itself. You'll meet many others as you learn more! Featured on Meta “Question closed” notifications experiment results and graduation Now, since \(f\) is one-to-one, we know \(a_1=a_2\) by definition of one-to-one. Definition: Inverse Function. If \(n=-2m-1\), then \(n\) is odd, and \(m=-\frac{n+1}{2}\). Be sure to specify their domains and codomains. Community ♦ 1. asked Nov 5 '12 at 14:10. In mathematics, relations and functions are the most important concepts. Evaluate \(f(g(f(0)))\). & if $x > 3$. The relation R S is known the composition of R and S; it is sometimes denoted simply by RS. Suppose, there is a relation $R = \lbrace (1, 1), (1,2), (3, 2) \rbrace$ on set $S = \lbrace 1, 2, 3 \rbrace$, it can be represented by the following graph −, The Empty Relation between sets X and Y, or on E, is the empty set $\emptyset$, The Full Relation between sets X and Y is the set $X \times Y$, The Identity Relation on set X is the set $\lbrace (x, x) | x \in X \rbrace$, The Inverse Relation R' of a relation R is defined as − $R' = \lbrace (b, a) | (a, b) \in R \rbrace$, Example − If $R = \lbrace (1, 2), (2, 3) \rbrace$ then $R' $ will be $\lbrace (2, 1), (3, 2) \rbrace$, A relation R on set A is called Reflexive if $\forall a \in A$ is related to a (aRa holds). If a partial ordering has the additional property that for any two distinct elements \(a\) and \(b\), either \(a\prec b\) or \(b\prec a\) (hence, any pair of distinct elements are comparable), we call the relation a total ordering. Example − The relation $R = \lbrace (1, 1), (2, 2), (3, 3), (1, 2), (2,1), (2,3), (3,2), (1,3), (3,1) \rbrace$ on set $A = \lbrace 1, 2, 3 \rbrace$ is an equivalence relation since it is reflexive, symmetric, and transitive. Then, throwing two dice is an example of an equivalence relation. If a function \(f :A \to B\) is a bijection, we can define another function \(g\) that essentially reverses the assignment rule associated with \(f\). Similarly, R 3 = R 2 R = R R R, and so on. \cr}\] Find its inverse. The notation \(f^{-1}(\{3\})\) means the preimage of the set \(\{3\}\). Certificate of Completion for your Job Interviews! Relations may exist between objects of the same set or between objects of two or more sets. An example is given demonstrating how to work algebraically with composite functions and another example involves an application that uses the composition of functions. Jan Tristan Milan Jan Tristan Milan. This defines an ordered relation between the students and their heights. Hence, \(\mathbb{R}\) is the domain of \(f\circ g\). Example: Nonetheless, \(g^{-1}(\{3\})\) is well-defined, because it means the preimage of \(\{3\}\). There are many types of relation which is exist between the sets, 1. A binary relation from A to B is a subset of a Cartesian product A x B. Since  \(b_1=b_2\) we have \(f(a_1)=f(a_2).\) discrete-mathematics relations function-and-relation-composition. Submitted by Prerana Jain, on August 17, 2018 . Thus we have demonstrated if \((g\circ f)(a_1)=(g\circ f)(a_2)\) then \(a_1=a_2\) and therefore by the definition of one-to-one, \(g\circ f\) is one-to-one. Two special relations occur frequently in mathematics. find the composition of functions; define the inverse of a function; ... At most of the universities, a undergraduate-level course in discrete mathematics is a required part of pursuing a computer science degree. Discrete Mathematical Structures Q.1 Write short Answers (i) Explain Equivalence Relation (ii) Define Recursive Function with an example (iii) Find the Converse, Contrapositive and Inverse of the following implication “ If today is Thursday, then I have a test today. To check whether \(f :{A}\to{B}\) and \(g :{B}\to{A}\) are inverse of each other, we need to show that. In general, \(f^{-1}(D)\) means the preimage of the subset \(D\) under the function \(f\). Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. \(f :{\mathbb{Z}}\to{\mathbb{Z}}\), \(f(n)=n+1\); \(g :{\mathbb{Z}}\to{\mathbb{Z}}\), \(g(n)=2-n\). Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. In mathematics (specifically set theory), a binary relation over sets X and Y is a subset of the Cartesian product X × Y; that is, it is a set of ordered pairs (x, y) consisting of elements x in X and y in Y. Find the inverse function of \(f :{\mathbb{Z}}\to{\mathbb{N}\cup\{0\}}\) defined by \[f(n) = \cases{ 2n & if $n\geq0$, \cr -2n-1 & if $n < 0$. R is symmetric x R y implies y R x, for all x,y∈A The relation is reversable. By definition of composition of functions, we have \[g(f(a_1))=g(f(a_2)).\]  Solution: Begin by replacing the function notation g (x) with y. g (x) = x 2 + 1 y = x 2 + 1 w h e r e x ≥ 0. If \(f :A \to B\) and \(g : B \to C\) are functions and \(g \circ f\) is onto, must \(f\) be onto? Find the inverse function of \(g :{\mathbb{R}}\to{\mathbb{R}}\) defined by \[g(x) = \cases{ 3x+5 & if $x\leq 6$, \cr 5x-7 & if $x > 6$. \cr}\] In this example, it is rather obvious what the domain and codomain are. Therefore, we can continue our computation with \(f\), and the final result is a number in \(\mathbb{R}\). Assume \(f,g :{\mathbb{R}}\to{\mathbb{R}}\) are defined as \(f(x)=x^2\), and \(g(x)=3x+1\). \cr}\], \[g \circ f: \mathbb{R} \to \mathbb{R}, \qquad (g \circ f)(x)=3x^2+1\], \[f \circ g: \mathbb{R} \to \mathbb{R}, \qquad (f \circ g)(x)=(3x+1)^2\]. No. So, subtraction is the opposite of addition. For each ordered pair (x, y) in the relation R, there will be a directed edge from the vertex ‘x’ to vertex ‘y’. Do not forget to include the domain and the codomain, and describe them properly. Determine \(f\circ g\) and \(g\circ f\). \cr}\] Determine \(f\circ g\), Let \(\mathbb{R}^*\) denote the set of nonzero real numbers. A matrix with m rows and n columns is called an m x n matrix. The images of the bijection \({\alpha}:{\{1,2,3,4,5,6,7,8\}}\to{\{a,b,c,d,e,f,g,h\}}\) are given below. A relation R on set A is called Anti-Symmetric if $xRy$ and $yRx$ implies $x = y \: \forall x \in A$ and $\forall y \in A$. For example, the converse of the relation 'child of' is the relation 'parent of'. We can also use an arrow diagram to provide another pictorial view, see second figure below. For two distinct sets, A and B, having cardinalities m and n respectively, the maximum cardinality of a relation R from A to B is mn. Some people mistakenly refer to the range as the codomain(range), but as we will see, that really means the set of all possible outputs—even values that the relation does not actually use. It starts with an element \(y\) in the codomain of \(f\), and recovers the element \(x\) in the domain of \(f\) such that \(f(x)=y\). Suppose \((g\circ f)(a_1)=(g\circ f)(a_2)\) for some \(a_1,a_2 \in A.\)  WMST \(a_1=a_2.\) The proof of \(f\circ f^{-1} = I_B\) procceds in the exact same manner, and is omitted here. We have the following results. Example − The relation $R = \lbrace (1, 2), (2, 3), (1, 3) \rbrace$ on set $A = \lbrace 1, 2, 3 \rbrace$ is transitive. Legal. The number of vertices in the graph is equal to the number of elements in the set from which the relation has been defined. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Partial Orderings Let R be a binary relation on a set A. R is antisymmetric if for all x,y A, if xRy and yRx, then x=y. Another Composition Example I Prove that f 1 f = I where I is the identity function. Determine \(h\circ h\). Define Discrete Mathematics Function. \(f :{\mathbb{Z}}\to{\mathbb{N}}\), \(f(n)=n^2+1\); \(g :{\mathbb{N}}\to{\mathbb{Q}}\), \(g(n)=\frac{1}{n}\). The relationship from the elements of one set X to elements of another set Y is defined as function or mapping, which is represented as f:X→Y. If \(f :{A}\to{B}\) is bijective, then \(f^{-1}\circ f=I_A\) and \(f\circ f^{-1}=I_B\). \cr}\] Next, we determine the formulas in the two ranges. we need to find until . Discrete Mathematics WEN-CHING LIEN Department of Mathematics National Cheng Kung University 2008 WEN-CHING LIEN Discrete Mathematics. You job is to verify that the answers are indeed correct, that the functions are inverse functions of each other. 2 CS 441 Discrete mathematics for CS M. Hauskrecht Functions • Definition: Let A and B be two sets.A function from A to B, denoted f : A B, is an assignment of exactly one element of B to each element of A. Show that it is a bijection, and find its inverse function, hands-on Exercise \(\PageIndex{2}\label{he:invfcn-02}\). Then, applying the function \(g\) to any element \(y\) from the codomain \(B\), we are able to obtain an element \(x\) from the domain \(A\) such that \(f(x)=y\). \[\begin{array}{|c||*{8}{c|}} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \alpha(x)& g & a & d & h & b & e & f & c \\ \hline \end{array}\] Find its inverse function. How to find \(f^{-1}\) Composite Function; Identity Function relates to Inverse Functions ; Summary and Review; Exercises ; A bijection (or one-to-one correspondence) is a function that is both one-to-one and onto. IntroductionIntroduction Relationships between elements of setsRelationships between elements of … Example \(\PageIndex{2}\label{eg:invfcn-02}\), The function \(s :{\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]}\to{[-1,1]}\) defined by \(s(x)=\sin x\) is a bijection. \(f(a) \in B\) and \(g(f(a))=c\); let \(b=f(a)\) and now there is a \(b \in B\) such that \(g(b)=c.\) The interval \((0,\infty)\) contains positive numbers only, so it is a subset of \(\mathbb{R}^*\). Interchange x and y. x = y 2 + 1 w h e r e y ≥ 0. Then \(f \circ g : \{2,3\} \to \{5\}\) is defined by  \(\{(2,5),(3,5)\}.\)  Clearly \(f \circ g\) is onto, while \(f\) is not onto. Let \(A\) and \(B\) be finite sets. 12- Composition OR Product Of Functions In Discrete Mathematics ... Discrete Math 2.3.3 Inverse Functions and Composition of Functions - Duration: 9:48. A relation R on set A is called Irreflexive if no $a \in A$ is related to a (aRa does not hold). collection of declarative statements that has either a truth value \"true” or a truth value \"false It works like connecting two machines to form a bigger one, see first figure below. We write f(a) = b to denote the assignment of b to an element a of A by the function f. Welcome to this course on Discrete Mathematics. Given \(B' \subseteq B\), the composition of two functions \(f :{A}\to{B'}\) and \(g :{B}\to{C}\) is the function \(g\circ f :{A}\to{C}\) defined by \((g\circ f)(x)=g(f(x))\). \cr}\], \[g(x) = \cases{ 3x+5 & if $x\leq 6$, \cr 5x-7 & if $x > 6$. If a function \(f :A \to B\) is a bijection, we can define another function \(g\) that essentially reverses the assignment rule associated with \(f\). Example 3: All functions are relations, but not all relations are functions. 947 6 6 gold badges 16 16 silver badges 30 30 bronze badges $\endgroup$ add a comment | 1 Answer Active Oldest Votes. In this course you will learn the important fundamentals of Discrete Math – Set Theory, Relations, Functions and Mathematical Induction with the help of 6.5 Hours of content comprising of Video Lectures, Quizzes and Exercises.Discrete Math is the real world mathematics. where \(i_A\) and \(i_B\) denote the identity function on \(A\) and \(B\), respectively. Chapters 2 and 9 3 / 74. Example : Let R be a relation defined as given below. Definition of modular arithmetic via an equivalence relation; properties of addition, multiplication, and exponentation (mod n); Euclid's algorithm, binary MOD and DIV functions, multiplicative inverses (mod p). Set operations in programming languages: Issues about data structures used to represent sets and the computational cost of set operations. The images under \({\alpha^{-1}}:{\{a,b,c,d,e,f,g,h\}}\to {\{1,2,3,4,5,6,7,8\}}\) are given below. Therefore, the inverse function is defined by \(f^{-1}:\mathbb{N} \cup \{0\} \to \mathbb{Z}\) by: \[f^{-1}(n) = \cases{ \frac{2}{n} & if $n$ is even, \cr -\frac{n+1}{2} & if $n$  is odd. It is a set of ordered pairs where the first member of the pair belongs to the first set and the second member of the pair belongs second sets. Example − The relation $R = \lbrace (a, a), (b, b) \rbrace$ on set $X = \lbrace a, b \rbrace$ is reflexive. We conclude that \(f\) and \(g\) are inverse functions of each other. It is defined by \[(g\circ f)(x) = g(f(x)) = 5f(x)-7 = \cases{ 5(3x+1)-7 & if $x < 0$, \cr 5(2x+5)-7 & if $x\geq0$. Exercise \(\PageIndex{12}\label{ex:invfcn-12}\). Be sure to write the final answer in the form \(f^{-1}(y) = \ldots\,\). The images for \(x\leq1\) are \(y\leq3\), and the images for \(x>1\) are \(y>3\). In this case, it is often easier to start from the “outside” function. In this course you will learn the important fundamentals of Discrete Math – Set Theory, Relations, Functions and Mathematical Induction with the help of 6.5 Hours of content comprising of Video Lectures, Quizzes and Exercises. \cr}\], \[\begin{array}{|c||*{8}{c|}} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \alpha(x)& g & a & d & h & b & e & f & c \\ \hline \end{array}\], \[\begin{array}{|c||*{8}{c|}} \hline x & a & b & c & d & e & f & g & h \\ \hline \alpha^{-1}(x)& 2 & 5 & 8 & 3 & 6 & 7 & 1 & 4 \\ \hline \end{array}\], \[f(n) = \cases{ 2n-1 & if $n\geq0$ \cr 2n & if $n < 0$ \cr} \qquad\mbox{and}\qquad g(n) = \cases{ n+1 & if $n$ is even \cr 3n & if $n$ is odd \cr}\], 5.4: Onto Functions and Images/Preimages of Sets, Identity Function relates to Inverse Functions, \(f^{-1}(y)=x \iff y=f(x),\) so write \(y=f(x)\), using the function definition of \(f(x).\). \cr}\], \[f^{-1}(x) = \cases{ \mbox{???} Solve for y. x = y 2 + 1 x − 1 = y 2 ± x − 1 = y. How do the converse, contrapositive, and inverse relate to p !q ? A relation R on set A is called Transitive if $xRy$ and $yRz$ implies $xRz, \forall x,y,z \in A$. R is transitive x R y and y R z implies x R z, for all x,y,z∈A Example: i<7 … Relations. The Pigeonhole Principle, illustrated by some pure number theoretic results. For two relations P (from A to B) and Q (from B to C), we can define the composition R of P and Q; We write the composition R of P and Q as R = P∘Q Describe three relations from the real world that can be expressed as mathematical relations. Given functions \(f :{A}\to{B}'\) and \(g :{B}\to{C}\) where \(B' \subseteq B\) , the composite function, \(g\circ f\), which is pronounced as “\(g\) after \(f\)”, is defined as \[{g\circ f}:{A}\to{C}, \qquad (g\circ f)(x) = g(f(x)).\] The image is obtained in two steps. We are now ready to present our answer: \(f \circ g: \mathbb{R} \to \mathbb{R},\) by: In a similar manner, the composite function \(g\circ f :{\mathbb{R}^*} {(0,\infty)}\) is defined as \[(g\circ f)(x) = \frac{3}{x^2}+11.\] Be sure you understand how we determine the domain and codomain of \(g\circ f\). Find the inverse of the function defined by g (x) = x 2 + 1 where x ≥ 0. A relation R on set A is called Symmetric if $xRy$ implies $yRx$, $\forall x \in A$ and $\forall y \in A$. Example − The relation $R = \lbrace (a, b), (b, a) \rbrace$ on set $X = \lbrace a, b \rbrace$ is irreflexive. Writing \(n=f(m)\), we find \[n = \cases{ 2m & if $m\geq0$, \cr -2m-1 & if $m < 0$. Combining Relation: Suppose R is a relation from set A to B and S is a relation from set B to C, the combination of both the relations is the relation which consists of ordered pairs (a,c) where a Є A and c Є C and there exist an element b Є B for which (a,b) Є R and (b,c) Є S. Make a truth table for all. Exercise \(\PageIndex{1}\label{ex:invfcn-01}\). \cr}\]. Also, R R is sometimes denoted by R 2. If \(f :A \to B\) and \(g : B \to C\) are functions and \(g \circ f\) is onto, must \(g\) be onto? \(f :{\mathbb{R}}\to{[\,1,\infty)}\),\(f(x)=x^2+1\); \(g :{[\,1,\infty)}\to {[\,0,\infty)}\) \(g(x)=\sqrt{x-1}\). Therefore, we can say, ‘A set of ordered pairs is defined as a rel… Solve for \(x\). If the ordered pair of G is reversed, the relation also changes. \(v:{\mathbb{Q}-\{1\}}\to{\mathbb{Q}-\{2\}}\), \(v(x)=\frac{2x}{x-1}\). \(f :{\mathbb{Q}-\{10/3\}}\to{\mathbb{Q}-\{3\}}\),\(f(x)=3x-7\); \(g :{\mathbb{Q}-\{3\}}\to{\mathbb{Q}-\{2\}}\), \(g(x)=2x/(x-3)\). Function ‘f’ is a relation on X and Y such that for each x∈X, there exists a unique y∈Y such that (x,y)∈R. Cartesian product denoted by *is a binary operator which is usually applied between sets. Define Composition of Relations. If there are two sets A and B, and relation R have order pair (x, y), then −, The domain of R, Dom(R), is the set $\lbrace x \:| \: (x, y) \in R \:for\: some\: y\: in\: B \rbrace$, The range of R, Ran(R), is the set $\lbrace y\: |\: (x, y) \in R \:for\: some\: x\: in\: A\rbrace$, Let, $A = \lbrace 1, 2, 9 \rbrace $ and $ B = \lbrace 1, 3, 7 \rbrace$, Case 1 − If relation R is 'equal to' then $R = \lbrace (1, 1), (3, 3) \rbrace$, Dom(R) = $\lbrace 1, 3 \rbrace , Ran(R) = \lbrace 1, 3 \rbrace$, Case 2 − If relation R is 'less than' then $R = \lbrace (1, 3), (1, 7), (2, 3), (2, 7) \rbrace$, Dom(R) = $\lbrace 1, 2 \rbrace , Ran(R) = \lbrace 3, 7 \rbrace$, Case 3 − If relation R is 'greater than' then $R = \lbrace (2, 1), (9, 1), (9, 3), (9, 7) \rbrace$, Dom(R) = $\lbrace 2, 9 \rbrace , Ran(R) = \lbrace 1, 3, 7 \rbrace$. This means given any element \(b\in B\), we must be able to find one and only one element \(a\in A\) such that \(f(a)=b\). Example problem on Composition of Relations. Given \(f :{A}\to{B}\) and \(g :{B}\to{C}\), if both \(f\) and \(g\) are one-to-one, then \(g\circ f\) is also one-to-one. Welcome to this course on Discrete Mathematics. \cr}\] Find its inverse function. Form the two composite functions \(f\circ g\) and \(g\circ f\), and check whether they both equal to the identity function: \[\displaylines{ \textstyle (f\circ g)(x) = f(g(x)) = 2 g(x)+1 = 2\left[\frac{1}{2}(x-1)\right]+1 = x, \cr \textstyle (g\circ f)(x) = g(f(x)) = \frac{1}{2} \big[f(x)-1\big] = \frac{1}{2} \left[(2x+1)-1\right] = x. If \(f^{-1}(3)=5\), we know that \(f(5)=3\). If a function \(g :{\mathbb{Z}}\to{\mathbb{Z}}\) is many-to-one, then it does not have an inverse function. Example 8. For the function ‘f’, X is the domain or pre-image and Y is the codomain of image. In the book Advanced Calculus by Shlomo and Sternberg (Chapter 0, Section 6), the inverse of an relation is defined as follows: "The inverse $ R^{-1} $, of a relation R is the set of ordered pairs More precisely, start with \(g\), and write the intermediate answer in terms of \(f(x)\), then substitute in the definition of \(f(x)\) and simplify the result. In an inverse function, the domain and the codomain are switched, so we have to start with \(f^{-1}:\mathbb{N} \cup \{0\} \to \mathbb{Z}\) before we describe the formula that defines \(f^{-1}\). Then, applying the function \(g\) to any element \(y\) from the codomain \(B\), we are able to obtain an element \(x\) from the domain \(A\) such that \(f(x)=y\). This follows from direct computation: \[(f\circ I_A)(a) = f(I_A(a)) = f(a).\] The proofs of \(I_B\circ f=f\) and (b)–(d) are left as exercises. 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