A relation in mathematics defines the relationship between two different sets of information. The result from \(g\) is a number in \((0,\infty)\). Solve for \(x\). Form the two composite functions \(f\circ g\) and \(g\circ f\), and check whether they both equal to the identity function: \[\displaylines{ \textstyle (f\circ g)(x) = f(g(x)) = 2 g(x)+1 = 2\left[\frac{1}{2}(x-1)\right]+1 = x, \cr \textstyle (g\circ f)(x) = g(f(x)) = \frac{1}{2} \big[f(x)-1\big] = \frac{1}{2} \left[(2x+1)-1\right] = x. \cr}\] Be sure you describe \(g^{-1}\) properly. In general, \(f^{-1}(D)\) means the preimage of the subset \(D\) under the function \(f\). A relation is any association or link between elements of one set, called the domain or (less formally) the set of inputs, and another set, called the range or set of outputs. CS340-Discrete Structures Section 4.1 Page 5 Properties of Binary Relations: R is reflexive x R x for all x∈A Every element is related to itself. share | cite | improve this question | follow | edited Jun 12 at 10:38. Example 2: Give an example of an Equivalence relation. Find the inverse of each of the following bijections. 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 $n < 0$. A relation R on set A is called Irreflexive if no $a \in A$ is related to a (aRa does not hold). 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. 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. In mathematics, relations and functions are the most important concepts. Therefore, we can say, ‘A set of ordered pairs is defined as a rel… 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. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. 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 $x\geq0$. 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. 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. 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. Nonetheless, \(g^{-1}(\{3\})\) is well-defined, because it means the preimage of \(\{3\}\). This makes the notation \(g^{-1}(3)\) meaningless. Define Discrete Mathematics Function. We write f(a) = b to denote the assignment of b to an element a of A by the function f. Composite Functions. \cr}\] Next, we determine the formulas in the two ranges. In class 11 and class 12, we have studied the important ideas which are covered in the relations and function. we need to find until . For it to be well-defined, every element \(b\in B\) must have a unique image. The notation \(f^{-1}(\{3\})\) means the preimage of the set \(\{3\}\). Yes, if \(f :A \to B\) and \(g : B \to C\) are functions and \(g \circ f\) is onto, then \(g\) must be onto. \cr}\] The details are left to you as an exercise. Then, because \(f^{-1}\) is the inverse function of \(f\), we know that \(f^{-1}(b)=a\). \(f(a_1) \in B\) and \(f(a_2) \in B.\) Let \(b_1=f(a_1)\) and \(b_2=f(a_2).\) Substituting into equation 5.5.3, \[g(b_1)=g(b_2).\] Be sure to specify their domains and codomains. discrete-mathematics elementary-set-theory relations function-and-relation-composition. Define Discrete Mathematics Function. In this case, we find \(f^{-1}(\{3\})=\{5\}\). The objects in a set are called theelements, ormembersof the set. Describe three relations from the real world that can be expressed as mathematical relations. & if $x\leq 3$, \cr \mbox{???} Exercise \(\PageIndex{10}\label{ex:invfcn-10}\). So the reflexive closure of is . 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 Let us refine this idea into a more concrete definition. 2 converse inverse? 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. In this case, it is often easier to start from the “outside” function. 947 6 6 gold badges 16 16 silver badges 30 30 bronze badges $\endgroup$ add a comment | 1 Answer Active Oldest Votes. \(w:{\mathbb{Z}}\to{\mathbb{Z}}\), \(w(n)=n+3\). Hence, addition and subtraction are opposite operations. So, subtraction is the opposite of addition. If \(n=2m\), then \(n\) is even, and \(m=\frac{n}{2}\). 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. If \(f :{A}\to{B}\) is bijective, then \(f^{-1}\circ f=I_A\) and \(f\circ f^{-1}=I_B\). This idea will be very important for our section on Infinite Sets and Cardinality. Example: Let A={a,b,c} and B={1,2,3}. 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. 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. 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 \(f(x)=y\). An example can be found in the numbers 2 and 3 in Example 7.4.4. 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. & if $x > 3$. Jan Tristan Milan Jan Tristan Milan. A relation R on set A is called Symmetric if $xRy$ implies $yRx$, $\forall x \in A$ and $\forall y \in A$. Chapter 9 Relations in Discrete Mathematics 1. Example \(\PageIndex{3}\label{eg:invfcn-03}\). Find the reflexive, symmetric, and transitive closure of R. Solution – For the given set, . 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. Example 8. Featured on Meta “Question closed” notifications experiment results and graduation Therefore, \(f^{-1}\) is a well-defined function. 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. Now, since \(f\) is one-to-one, we know \(a_1=a_2\) by definition of one-to-one. To find the inverse function of \(f :{\mathbb{R}}\to{\mathbb{R}}\) defined by \(f(x)=2x+1\), we start with the equation \(y=2x+1\). Many … A binary relation R from set x to y (written as $xRy$ or $R(x,y)$) is a subset of the Cartesian product $x \times y$. We conclude that \(f\) and \(g\) are inverse functions of each other. \cr}\] We need to consider two cases. Prove or give a counter-example. Simplify your answer as much as possible. 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. R is transitive x R y and y R z implies x R z, for all x,y,z∈A Example: i<7 … Another Composition Example I Prove that f 1 f = I where I is the identity function. Be sure to write the final answer in the form \(f^{-1}(y) = \ldots\,\). Previously, we have already discussed Relations and their basic types. A relation can be represented using a directed graph. For example, the converse of the relation 'child of' is the relation 'parent of'. This relation between A and C denotes the indirect or the composite relation. For the symmetric closure we need the inverse of , which is. Put your math smarts to the challenge with the assistance of this interactive quiz and printable worksheet on relation in math. Evaluate \(f(g(f(0)))\). Example 1: The addition means to find the sum, and subtraction means taking away. In formal terms, if X and Y are sets and L ⊆ X × Y is a relation from X to Y, then L T is the relation defined so that y L T x if and only if x L y. Welcome to this course on Discrete Mathematics. The functions \(f :{\mathbb{R}}\to{\mathbb{R}}\) and \(g :{\mathbb{R}}\to{\mathbb{R}}\) are defined by \[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… Richard Mayr (University of Edinburgh, UK) Discrete Mathematics. 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. The relations we will deal with are very important in discrete mathematics, and are known as equivalence relations ... but a simple examination or understanding of this idea will be interesting in its application to equivalence relations) For example, 2 ≡ 0 (mod 2), since the remainder on dividing 2 by 2 is in fact 0, as is the remainder on dividing 0 by 2. R is a partial order relation if R is reflexive, antisymmetric and transitive. \(f :{\mathbb{R}}\to{(0,1)}\), \(f(x)=1/(x^2+1)\); \(g :{(0,1)}\to{(0,1)}\), \(g(x)=1-x\). However, since \(g \circ f\) is onto, we know \(\exists a \in A\) such that \((g \circ f)(a) = c.\) This means \(g(f(a))=c\). For a bijective function \(f :{A}\to{B}\), \[f^{-1}\circ f=I_A, \qquad\mbox{and}\qquad f\circ f^{-1}=I_B,\]. Example – Let be a relation on set with . Chapter 1.1-1.3 10 / 21 \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 $x\geq0$. It encodes the information of relation: an element x is related to an element y, if and only if the pair (x, y) belongs to the set. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. Matrices in Discrete Mathematics and its Applications 1. If a function \(f :A \to B\) is a bijection, we can define another function \(g\) that essentially reverses the assignment rule associated with \(f\). So let us see a few examples to understand what is going on. The functions \(g,f :{\mathbb{R}}\to{\mathbb{R}}\) are defined by \(f(x)=1-3x\) and \(g(x)=x^2+1\). Relations may exist between objects of the same set or between objects of two or more sets. ” (iv) What is difference between Tautology, Contradiction and Contingency? Show that the functions \(f,g :{\mathbb{R}}\to{\mathbb{R}}\) defined by \(f(x)=2x+1\) and \(g(x)=\frac{1}{2}(x-1)\) 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 $x\geq0$. In the morning assembly at schools, students are supposed to stand in a queue in ascending order of the heights of all the students. Discrete Mathematics WEN-CHING LIEN Department of Mathematics National Cheng Kung University 2008 WEN-CHING LIEN Discrete Mathematics. The notation \(f^{-1}(3)\) means the image of 3 under the inverse function \(f^{-1}\). It works like connecting two machines to form a bigger one, see first figure below. which is what we want to show. Example problem on Composition of Relations. Example – What is the composite of the relations and where is a relation from to with and is a relation from to with ? The concepts are used to solve the problems in different chapters like probability, differentiation, integration, and so on. inverse: If it is not raining, then I will go to town. 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. Naturally, if a function is a bijection, we say that it is bijective. \cr}\], \[f(x) = 3x+2, \qquad\mbox{and}\qquad g(x) = \cases{ x^2 & if $x\leq5$, \cr 2x-1 & if $x > 5$. This article examines the concepts of a function and a relation. Let us look at some examples to understand the meaning of inverse. Example − The relation $R = \lbrace (a, a), (b, b) \rbrace$ on set $X = \lbrace a, b \rbrace$ is reflexive. 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 Let \(I_A\) and \(I_B\) denote the identity function on \(A\) and \(B\), respectively. Determine \(f\circ g\) and \(g\circ f\). https://www.tutorialspoint.com/.../discrete_mathematics_relations.htm 12- Composition OR Product Of Functions In Discrete Mathematics In HINDI ... Discrete Math 2.3.3 Inverse Functions and Composition of Functions - … We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Writing \(n=f(m)\), we find \[n = \cases{ 2m & if $m\geq0$, \cr -2m-1 & if $m < 0$. 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 \((g\circ f)(x)=g(f(x))\). 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. Lifetime Access! 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 \(f(x)=y\). Functions find their application in various fields like representation of the Relations. \cr}\], \[n = \cases{ 2m & if $m\geq0$, \cr -2m-1 & if $m < 0$. Composition of functions is a special case of composition of relations. Verify that \(f :{\mathbb{R}}\to{\mathbb{R}^+}\) defined by \(f(x)=e^x\), and \(g :{\mathbb{R}^+}\to{\mathbb{R}}\) defined by \(g(x)=\ln x\), are inverse functions of each other. Exercise \(\PageIndex{6}\label{ex:invfcn-06}\), The functions \(f,g :{\mathbb{Z}}\to{\mathbb{Z}}\) are defined by \[f(n) = \cases{ 2n-1 & if $n\geq0$ \cr 2n & if $n < 0$ \cr} \qquad\mbox{and}\qquad g(n) = \cases{ n+1 & if $n$ is even \cr 3n & if $n$ is odd \cr}\] Determine \(g\circ f\), (a) \({g\circ f}:{\mathbb{Z}}\to{\mathbb{Q}}\), \((g\circ f)(n)=1/(n^2+1)\), (b) \({g\circ f}:{\mathbb{R}}\to{(0,1)}\), \((g\circ f)(x)=x^2/(x^2+1)\), Exercise \(\PageIndex{8}\label{ex:invfcn-08}\). Here, the function \(f\) can be any function. 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). That is, express \(x\) in terms of \(y\). A binary relation from A to B is a subset of a Cartesian product A x B. The problem does not ask you to find the inverse function of \(f\) or the inverse function of \(g\). Exercise \(\PageIndex{5}\label{ex:invfcn-05}\). Let R is a relation on a set A, that is, R is a relation from a set A to itself. The inverse of a bijection \(f :{A} \to {B}\) is the function \(f^{-1}: B \to A\) with the property that. 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) = $\lbrace 2, 9 \rbrace , Ran(R) = \lbrace 1, 3, 7 \rbrace$. 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. Since \(f\) is a piecewise-defined function, we expect its inverse function to be piecewise-defined as well. R is a partial order relation if R is reflexive, antisymmetric and transitive. Both have to do with some sort of ordering of the elements in a set. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Discrete Mathematics - Functions - A Function assigns to each element of a set, exactly one element of a related set. 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. Prove or give a counter-example. 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. 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. \cr}\], \[g \circ f: \mathbb{R} \to \mathbb{R}, \qquad (g \circ f)(x)=3x^2+1\], \[f \circ g: \mathbb{R} \to \mathbb{R}, \qquad (f \circ g)(x)=(3x+1)^2\]. \[\begin{array}{|c||*{8}{c|}} \hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \alpha(x)& g & a & d & h & b & e & f & c \\ \hline \end{array}\] Find its inverse function. Types of Relation. In this section, we will get ourselves familiar with composite functions. We do not need to find the formula of the composite function, as we can evaluate the result directly: \(f(g(f(0))) = f(g(1)) = f(2) = -5\). First, we need to find the two ranges of input values in \(f^{-1}\). Over 6.5 hours of Learning! If \(g^{-1}(\{3\})=\{1,2,5\}\), we know \(g(1)=g(2)=g(5)=3\). Define Composition of Relations. Solving for \(x\), we find \(x=\frac{1}{2}\,(y-1)\). 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. \((f\circ g)(y)=f(g(y))=y\) for all \(y\in B\). Since every element in set \(C\) does have a pre-image in set \(B\), by the definition of onto, \(g\) must be onto. Example − The relation $R = \lbrace (1, 2), (2, 1), (3, 2), (2, 3) \rbrace$ on set $A = \lbrace 1, 2, 3 \rbrace$ is symmetric. Make a truth table for all. & if $x > 3$. Definition Of Matrix • A matrix is a rectangular array of numbers. Nevertheless, it is always a good practice to include them when we describe a function. Submitted by Prerana Jain, on August 17, 2018 . 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 \(g\circ f\). 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. On A Graph . If \(f^{-1}(3)=5\), we know that \(f(5)=3\). If \(f :A \to B\) and \(g : B \to C\) are functions and \(g \circ f\) is onto, must \(f\) be onto? The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Missed the LibreFest? This lesson explains the concept of composite functions. Generally an n-ary relation R between sets $A_1, \dots ,\ and\ A_n$ is a subset of the n-ary product $A_1 \times \dots \times A_n$. Also, R R is sometimes denoted by R 2. \cr}\] In this example, it is rather obvious what the domain and codomain are. A relation R on set A is called Anti-Symmetric if $xRy$ and $yRx$ implies $x = y \: \forall x \in A$ and $\forall y \in A$. Let R be a relation defined on the set A such that. (Beware: some authors do not use the term codomain(range), and use the term range inst… Suppose \(f :{A}\to{B}\) and \(g :{B}\to{C}\). Next, it is passed to \(g\) to obtain the final result. First, \(f(x)\) is obtained. You'll meet many others as you learn more! It starts with an element \(y\) in the codomain of \(f\), and recovers the element \(x\) in the domain of \(f\) such that \(f(x)=y\). 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. Let \(f :{A}\to{B}\) be a bijective function. In this article, we will learn about the relations and the different types of relation in the discrete mathematics. “Set Theory, Relations and Functions” form an integral part of Discrete Math. To find the algebraic description of \((g\circ f)(x)\), we need to compute and simplify the formula for \(g(f(x))\). A relation is an Equivalence Relation if it is reflexive, symmetric, and transitive. Find the inverse of the function \(r :{(0,\infty)}\to{\mathbb{R}}\) defined by \(r(x)=4+3\ln x\). We note that, in general, \(f\circ g \neq g\circ f\). How do the converse, contrapositive, and inverse relate to p !q ? This defines an ordered relation between the students and their heights. \cr}\], \[f(n) = \cases{ -2n & if $n < 0$, \cr 2n+1 & if $n\geq0$. Universal Relation Exercise \(\PageIndex{11}\label{ex:invfcn-11}\). Exercise \(\PageIndex{3}\label{ex:invfcn-03}\). The resulting expression is \(f^{-1}(y)\). (a) \({u^{-1}}:{\mathbb{Q}}\to{\mathbb{Q}}\), \(u^{-1}(x)=(x+2)/3\), Exercise \(\PageIndex{2}\label{ex:invfcn-02}\). Therefore, the inverse function is \[{f^{-1}}:{\mathbb{R}}\to{\mathbb{R}}, \qquad f^{-1}(y)=\frac{1}{2}\,(y-1).\] It is important to describe the domain and the codomain, because they may not be the same as the original function. Example − The relation $R = \lbrace (1, 2), (2, 3), (1, 3) \rbrace$ on set $A = \lbrace 1, 2, 3 \rbrace$ is transitive. \(f :{\mathbb{Q}-\{2\}}\to{\mathbb{Q}^*}\), \(f(x)=1/(x-2)\); \(g :{\mathbb{Q}^*}\to{\mathbb{Q}^*}\), \(g(x)=1/x\). 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