Overview
This unit introduces relations and functions, fundamental concepts that connect algebra to real-world modelling. You will study how pairs of elements from two sets relate, how relations are represented, and how certain relations become functions when each input has a unique output. The unit covers types of relations such as reflexive, symmetric, transitive and equivalence relations, and shows how equivalence relations partition a set into classes. You will learn various ways to represent functions: set notation, mapping diagrams, graphs, formulas, and inverse relations. Important function types—one-to-one (injective), onto (surjective) and bijective—are defined and tested. Operations on functions (addition, multiplication, composition) and the concept of inverse functions are developed, with conditions for existence and methods to find inverses. Special classes of functions—polynomial, rational, exponential, logarithmic, trigonometric and piecewise—are introduced with domain and range considerations. Finally, this unit connects functions to calculus ideas by preparing you to study limits and continuity, and includes practice in identifying domain, range, and inverse relations. Mastery of this unit is essential for solving equations, modelling situations, and understanding higher-level mathematics used in physics, economics, and engineering.
Learning Objectives
- Define relations and functions precisely using set notation and ordered pairs.
- Classify relations as reflexive, symmetric, transitive and determine when they form equivalence relations.
- Represent relations and functions using roster form, set-builder form, mapping diagrams and graphs.
- Determine domain, codomain and range of given relations and functions correctly.
- Test and prove whether a relation is a function and whether a function is one-to-one, onto or bijective.
- Perform algebraic operations on functions and compose functions, and find domains of results.
- Find inverse functions where they exist and verify inverse relationships.
- Apply knowledge of functions to model and solve problems involving equations and mappings.
Topics in this chapter
18 topics · tap a topic title to jump straight to it.
Introduction to Relations
What is a relation? A relation between two sets A and B is any subset of the Cartesian product A × B. That means a relation is a collection of ordered pairs (a, b) where a is from A and b is from B. Relations describe how elements of one set are associated with elements of another.
Notation and examples. We write R ⊂ A × B to indicate that R is a relation from A to B. If (x, y) belongs to R, we may write xRy. For example, if A = {1,2,3} and B = {a,b}, a relation R could be {(1,a), (2,b)}. Relations may be finite or infinite depending on the sets involved.
Ways to present a relation. Relations are shown in several formats: as a list of ordered pairs (roster), as set-builder notation (for instance R = {(x,y) : y = x+1}), as a mapping diagram with arrows, or as a graph in the plane when A and B are numbers. Each representation gives different insight: the roster form is concrete, set-builder is concise, diagrams show structure, and graphs show geometric behaviour.
Domain, codomain and range. For a relation R ⊂ A × B, the domain is the subset of A consisting of those elements that appear in some pair of R. The codomain is B when stated as part of the relation, and the range (or image) is the subset of B of elements that actually appear paired. Distinguish codomain from range: codomain is given, range is produced by the relation.
Properties to observe. Relations may be arbitrary, so they need not have any special properties. Early practice asks you to identify which pairs belong, list domain and range, and move between different representations. This groundwork prepares you for function-specific rules and for studying special relations later.
- A = {1,2,3}, B = {x,y}. R = {(1,x),(3,y)} illustrates a relation with domain {1,3} and range {x,y}.
- R = {(x,y) : x,y real and y = x^2} is a relation represented by the parabola in the plane.
- A mapping diagram: A = {1,2}, B = {a,b,c}; arrows 1→a, 1→b, 2→c produce R = {(1,a),(1,b),(2,c)}.
- Relation 'divides' on A = {1,2,3,4}: R = {(a,b): a divides b} is a relation on A.
- Relation R is a subset: R ⊂ A × B
- Domain(R) = {a ∈ A : ∃b ∈ B with (a,b) ∈ R}
- Range(R) = {b ∈ B : ∃a ∈ A with (a,b) ∈ R}
Functions: Definition and Basic Examples
What is a function? A function f from a set A to a set B, written f: A → B, is a special type of relation in which every element of A has exactly one associated element in B. This single-output requirement is what distinguishes functions from general relations. When we say f(a) = b we mean the ordered pair (a,b) belongs to the relation f.
Formal meaning and everyday view. Formally, f ⊂ A × B with the property that for every a ∈ A there exists a unique b ∈ B with (a,b) ∈ f. Intuitively, a function is like a rule or machine: you give an input from A and it returns one output in B. If any input could give more than one output, the relation is not a function.
Representations of functions. Functions can be presented in several ways. The roster form lists ordered pairs explicitly for small finite sets. Set-builder notation gives a rule, like f = {(x,y) : y = x^2, x ∈ R}. Mapping diagrams show elements of domain and codomain with arrows; every domain element must have exactly one outgoing arrow. For real-valued functions of a real variable, graphs in the plane are common; the vertical line test helps check the single-output property on graphs.
Domain, codomain and range in functions. The domain is the set of permitted inputs; the codomain is the set the outputs are declared to lie in; the range (or image) is the set of actual outputs produced. For example, f: R → R given by f(x)=x^2 has domain R, codomain R, but range [0,∞). Always be precise: domain and codomain are part of the function’s definition and affect properties like surjectivity.
Common examples and special functions. Linear functions f(x)=mx+c provide the simplest nontrivial examples; polynomials, rationals, exponentials and trigonometric functions are other common families. Constant functions map every input to the same output and are functions though not injective (unless domain has one element). The identity function id_A maps each element to itself and always satisfies the function property.
Tests and pitfalls. Given a list of pairs check uniqueness of outputs for each input. In mapping diagrams ensure each domain node has one arrow. On the xy-plane use the vertical line test: if any vertical line intersects the graph at more than one point, it does not represent a function. Always state domain explicitly when given a formula—expressions such as 1/x or √(x) impose domain restrictions.
- f: {1,2,3} → {a,b} with pairs (1,a),(2,a),(3,b) is a function; domain {1,2,3}, range {a,b}.
- f(x) = x^2 as f: R → R is a function whose range is [0,∞).
- f(x) = 1/x with domain R \ {0} is a function; check domain to avoid division by zero.
- Function f: A → B means f ⊂ A × B and ∀a ∈ A, ∃! b ∈ B such that (a,b) ∈ f
- Range(f) = f(A) = {f(a) : a ∈ A}
One-to-one (Injective) Functions
Definition and meaning. A function f: A → B is injective (one-to-one) if distinct inputs always map to distinct outputs. Formally, f is injective if f(x1) = f(x2) implies x1 = x2. This condition ensures no two different elements of A collide to the same element of B. In practical terms, injectivity means you can recover the input uniquely from the output on the image of f.
Why injectivity matters. Injective functions are important because they allow a left-inverse: there exists a function g: f(A) → A such that g(f(x)) = x for every x in A. Invertibility on the whole codomain requires bijectivity, but injectivity is the half that guarantees uniqueness when reversing mapping on the image. Many constructions in algebra and analysis require injective maps to embed one set into another without loss of information.
Methods to test injectivity. For finite sets, a function is injective iff the number of distinct outputs equals the number of inputs; equivalently |A| ≤ |B| is necessary and when images count equals |A| it is sufficient. For functions given by formulas, use algebraic manipulation: assume f(x1)=f(x2) and deduce x1=x2. Graphically, a function of real variable is injective if every horizontal line meets its graph at most once; this is the horizontal line test.
Common examples and non-examples. Linear functions f(x)=mx+c with m ≠ 0 are injective on R since solving m x1 + c = m x2 + c gives x1 = x2. The function x^2 on R is not injective because x and -x share same square; however restricting the domain to [0,∞) makes it injective. Exponential functions a^x (a>0, a≠1) are injective on R because a^x is strictly monotone.
Construction and restriction. If a function fails to be injective on a large domain, often restricting its domain to a region of monotonicity yields injectivity. For example cosine is not injective on R but is injective on [0,π] if endpoints are included carefully. Recognising monotone intervals is a practical method to produce injective restrictions.
Proof style required in exams. When asked to prove injectivity give a clear argument: either algebraic (assume equality and deduce equality of inputs), or use monotonicity/horizontal line test with justification. Provide counterexamples to show non-injectivity. State domains precisely when presenting results about injectivity.
- f(x) = 3x - 5 on R is injective since 3x1-5 = 3x2-5 ⇒ x1 = x2.
- h(x) = x^2 on R is not injective because h(2)=h(-2)=4.
- Restriction: h: [0,∞) → [0,∞), h(x)=x^2 is injective.
- Injective: f(x1)=f(x2) ⇒ x1 = x2
- Horizontal line test: each horizontal line intersects graph at most once
Onto (Surjective) Functions
Definition and intuition. A function f: A → B is surjective (onto) if every element of B is the image of at least one element of A. In formal terms f(A) = B: the range equals the codomain. Surjectivity ensures that for every b in B there exists an a in A with f(a) = b, so no value in the codomain is left unused.
Why surjectivity matters. Surjective functions guarantee existence of pre-images for equations f(x) = b for any b in the codomain. This property is essential when solving equations, constructing right-inverses, and when mapping between algebraic structures where covering the entire target set is required. In many contexts, surjectivity ensures that a transformation is onto its intended target space.
Testing for surjectivity. For finite sets, check cardinalities: surjectivity requires |range| = |B|. For algebraic functions of a real variable, attempt to solve y = f(x) for x in terms of y and check if for any y in B a solution x exists within the domain. Use calculus or monotonicity where appropriate: a continuous strictly monotone function from R to R is surjective onto an interval that may be all R depending on limits at plus and minus infinity.
Examples and counterexamples. The exponential function e^x: R → (0,∞) is surjective onto (0,∞) because every positive real number has a real logarithm; but it is not surjective onto R since negatives and zero are not images. The cubic f(x) = x^3 from R to R is surjective since every real y has a real cube root. Polynomial functions of even degree with positive leading coefficient are not surjective onto R because they are bounded below.
Constructing surjections and restrictions. A function may be made surjective by changing its codomain to equal its range. For example, consider f(x)=x^2 from R to [0,∞): with codomain chosen as [0,∞) it becomes surjective. Inverse functions on entire codomain require the function to be bijective, so surjectivity is half of that requirement.
Exam approach. When asked to show surjectivity, present either a constructive method (given any y, produce x with f(x)=y) or show algebraically that all codomain elements are hit. If not surjective, provide a specific element of the codomain with no pre-image as a counterexample. Always state domains and codomains clearly in answers.
- f(x) = x^3 from R to R is surjective: for any y take x = y^(1/3).
- f(x) = e^x from R to R is not surjective, but e^x: R → (0,∞) is surjective.
- f: {1,2,3} → {a,b,c} with outputs {a,b,c} is surjective.
- Surjective: f(A) = B
- To test: ∀y ∈ B, ∃x ∈ A s.t. f(x) = y
Bijective Functions and Invertibility
Definition and consequences. A function f: A → B is bijective if it is both injective and surjective. Bijectivity establishes a perfect one-to-one correspondence between A and B: each a ∈ A maps to a unique b ∈ B and every b ∈ B has exactly one pre-image. For finite sets this implies |A| = |B| and a pairing without leftovers.
Inverse functions. A bijection has an inverse function f^{-1}: B → A defined by f^{-1}(b) = a if and only if f(a) = b. The inverse reverses the mapping: f^{-1}(f(a)) = a for every a ∈ A and f(f^{-1}(b)) = b for every b ∈ B. Existence of a two-sided inverse is equivalent to bijectivity: injective gives a left-inverse, surjective gives a right-inverse, both together give a full inverse.
Finding inverses algebraically. For a function given by a formula y = f(x), find the inverse by solving the equation for x in terms of y. After obtaining x = g(y), swap variable names to write f^{-1}(x) = g(x), but always verify domain and range: the domain of f^{-1} is the range of f, and the range of f^{-1} is the domain of f. Always check your candidate inverse by composing: f(f^{-1}(x)) = x and f^{-1}(f(x)) = x on the appropriate domains.
Graphical property. Graphs of inverse functions are reflections across the line y = x. If (a,b) lies on the graph of f then (b,a) lies on the graph of f^{-1}. Use this fact to visualise inverses and to check symmetry when sketching graphs.
Common examples and restrictions. Linear functions with nonzero slope are bijective from R to R. The cubic function x^3 is bijective on R with inverse cube root. The square function x^2 is not bijective on R but becomes bijective if we restrict domain to [0,∞) with codomain [0,∞); its inverse is √x. When a given function is not bijective on its full domain, find natural restrictions to make it bijective if an inverse is required.
Exam style proofs and checks. To prove a function is bijective either show injective and surjective separately or construct the inverse explicitly and verify it satisfies the composition identities. State domains and ranges clearly and justify any restrictions used. Provide composition checks to demonstrate correctness of the inverse.
- f(x) = 2x+3 on R is bijective; inverse f^{-1}(x) = (x-3)/2.
- f(x) = x^3 on R is bijective; inverse f^{-1}(x) = x^(1/3).
- g(x) = x^2 is not bijective on R but is bijective on [0,∞) with inverse √x.
- \[If f is bijective then f^{-1} exists and f^{-1}(f(x)) = x\]\[f(f^{-1}(y)) = y\]
- To find inverse: solve y = f(x) for x
Composition of Functions
Definition and meaning. If f: A → B and g: B → C are functions then the composition g ∘ f: A → C is defined by (g ∘ f)(x) = g(f(x)). Composition means first apply f to the input x to obtain f(x) in B, then apply g to that result to obtain g(f(x)) in C. Composition chains two processes into one mapping from A directly to C.
Order and non-commutativity. Composition is not generally commutative: g ∘ f will usually differ from f ∘ g because the order of applying functions matters. For composition to be defined, the codomain of the first function must align with the domain of the second; when working with specific formulas check that the intermediate values lie in the correct domain.
Domains of composed functions. The domain of g ∘ f is the set of x ∈ Domain(f) for which f(x) ∈ Domain(g). Practically, when composing algebraic expressions, first restrict x so f(x) is defined, then further restrict so that g(f(x)) is defined. This careful handling of domains is essential in board answers to avoid including illegal points.
Associativity and identity. Composition is associative: for functions f, g, h with appropriate domains and codomains, h ∘ (g ∘ f) = (h ∘ g) ∘ f. Identity functions id_A act neutrally: f ∘ id_A = f and id_B ∘ f = f. Associativity allows us to compose several functions without worrying about bracketing.
Algebraic manipulation and examples. To compute (g ∘ f)(x) algebraically, substitute f(x) into g. For example with f(x)=2x+1 and g(x)=x^2, (g∘f)(x) = (2x+1)^2. Be cautious when division or roots are involved: if f(x) produces zero which becomes denominator in g, exclude those x. When functions are given piecewise perform substitution separately on each relevant piece.
Inverse relation with composition. If f is bijective with inverse f^{-1}, then f^{-1} ∘ f = id and f ∘ f^{-1} = id on appropriate sets. Also, (g ∘ f)^{-1} = f^{-1} ∘ g^{-1} when inverses exist. These identities are useful in finding inverses of composed functions and in transforming equations.
Graphical interpretation. Graphically composition corresponds to horizontal and vertical transforms: applying f alters x to u = f(x), then g maps u to y. Visualising three horizontal levels (x, u, y) and arrows between them clarifies the two-step mapping process. Practice with linear, quadratic and rational examples to build fluency and attention to domain issues.
- f(x)=2x+1, g(x)=x^2. Then (g∘f)(x) = g(2x+1) = (2x+1)^2 and (f∘g)(x)=2x^2+1 — different functions.
- If f(x)=1/x (domain R\{0}) and g(x)=x+2, then (g∘f)(x)=1/x + 2; domain excludes x=0.
- Identity: id(x)=x; for any f, f∘id = f.
- (g∘f)(x) = g(f(x))
- Domain(g∘f) = {x ∈ Domain(f) : f(x) ∈ Domain(g)}
- Associative: h∘(g∘f) = (h∘g)∘f
Operations on Functions: Sum, Product, Quotient
Definitions and construction. Given two functions f and g defined on a common domain D, we can form new functions by combining values pointwise: (f+g)(x)=f(x)+g(x), (f-g)(x)=f(x)-g(x), (fg)(x)=f(x)g(x) and (f/g)(x)=f(x)/g(x) provided g(x)≠0. These operations let us build more complex functions from simpler building blocks and are used widely in algebra and modelling.
Domains and care with definitions. The domain of f+g, f-g and fg is the intersection of domains of f and g because both must be defined at x. For the quotient f/g additionally exclude points where g(x)=0. When combining functions with differing domains always state the intersection explicitly and then list further exclusions for division or other restrictions (e.g., square roots appearing after algebraic manipulation).
Algebraic simplification with domain notes. Simplify combined expressions algebraically when possible, but never ignore domain exclusions. For example if f(x) = (x^2-1)/(x-1) and g(x)=x-1 then fg simplifies pointwise to x+1 provided x ≠ 1 (because original f undefined at x=1). State the simplified formula and also mention excluded points separately — boards require precision about domain even if cancellation occurs.
Applications and interpretation. Pointwise operations model many scenarios: totals (sum of cost and tax), net effect (difference), scaling (product with a factor), and rates (quotient). For instance if distance d(x) and time t(x) are functions of a variable x, the speed function is v(x)=d(x)/t(x) where t(x)≠0. Seek units and feasibility when modelling.
Continuity and limits (brief). Sum and product of continuous functions are continuous. Quotient of continuous functions is continuous wherever denominator is nonzero. This fact is useful when dealing with limits and continuity in calculus; however for algebraic problems emphasise domain and algebraic simplification first.
Practice techniques. When performing operations, write domain intersections, simplify algebraic expressions, and check for additional restrictions. For piecewise-defined functions perform operations piecewise. For rational combinations find common denominators when adding or subtracting. In exams show all steps: domain, algebra, final simplified expression and domain of final function.
- f(x)=x+1, g(x)=x-1 ⇒ (f+g)(x)=2x, domain R.
- f(x)=1/x, g(x)=x ⇒ (f/g)(x)=(1/x)/x=1/x^2 with domain R\{0}.
- f(x)=x/(x-1), g(x)=(x-1) ⇒ (f·g)(x)=x with domain excluding x=1 because original f undefined.
- (f+g)(x)=f(x)+g(x)
- (f-g)(x)=f(x)-g(x)
- (fg)(x)=f(x)g(x)
- (f/g)(x)=f(x)/g(x) where g(x)≠0
Inverse Relations and Relation Inversion
Inverse relation — definition. For any relation R ⊂ A × B the inverse relation R^{-1} is formed by swapping the ordered pair components: R^{-1} = {(b,a) : (a,b) ∈ R}. The inverse relation runs from B to A and reverses all associations present in R. This construction applies to all relations, whether functions or not.
When inverse is a function. If R is a function f: A → B then R^{-1} is a function exactly when f is injective: only then does each b in the image correspond to a unique a. If f is bijective then R^{-1} is not just a relation but the well-defined inverse function f^{-1}: B → A. If f is not injective the inverse relation may assign one element of B to many elements of A, so it fails to be a function.
Basic properties. The inverse relation reverses domain and range: Domain(R^{-1}) = Range(R) and Range(R^{-1}) = Domain(R). Taking inverse twice returns the original relation: (R^{-1})^{-1} = R. Composition reverses order under inversion: (S ∘ R)^{-1} = R^{-1} ∘ S^{-1} when these compositions are defined. These properties are straightforward and follow from the pair-swapping definition.
How to find inverse relations. For a relation given as a list of pairs, form the inverse by swapping coordinates in each pair. For a function given by an equation y = f(x), attempt to solve for x in terms of y; if you can express x = g(y) and the solution gives a unique x for each y in the range, then f^{-1}(y) = g(y) describes the inverse function. After solving, exchange symbols y and x to put the inverse in the common function notation f^{-1}(x).
Domain and range adjustments. The domain of the inverse function equals the range of the original; ensure you explicitly state these in answers. When solving algebraically some values might be extraneous or lost due to squaring or cancellation, so check and restrict domain appropriately. For example f(x)=x^2 on R gives inverse relation mapping positive numbers to two pre-images; restricting f to [0,∞) gives a true inverse f^{-1}(x)=√x with domain [0,∞).
Graphical viewpoint. Geometrically, inverse relations reflect the set of points about the line y = x. For a bijective real function its graph and the graph of its inverse are mirror images across y=x. For non-function inverse relations the reflected set will violate the vertical line test and show multiple points with same x-coordinate.
Applications and cautions. Inverse relations undo processes and are widely used in solving equations and reversing maps. Always check uniqueness and domain carefully: swapping coordinates mechanically is not enough to claim the inverse is a function — injectivity must be shown or an appropriate restriction must be applied.
- R = {(1,a),(2,b),(3,c)} ⇒ R^{-1} = {(a,1),(b,2),(c,3)} and R^{-1} is a function.
- f(x)=x^2 on R ⇒ inverse relation includes (4,2) and (4,-2) so inverse is not a function on R unless restricted.
- If f is bijective, f^{-1} exists and f^{-1}(f(x)) = x for all x in domain.
- \[R^{-1} = {(b,a) : (a,b) ∈ R}\]
- \[Domain(R^{-1}) = Range(R)\]\[Range(R^{-1}) = Domain(R)\]
Equivalence Relations and Partitions
Definition and motivation. An equivalence relation ~ on a set A is a relation that is reflexive, symmetric and transitive. These three properties capture when elements are to be considered ‘equivalent’ with respect to some criterion. Equivalence relations formalise grouping by a chosen property and allow us to treat equivalent items as indistinguishable in certain mathematical contexts.
Equivalence classes. For any a ∈ A the equivalence class [a] is defined as [a] = {x ∈ A : x ~ a}. Each element belongs to exactly one equivalence class: two classes are either identical or disjoint. The collection of all equivalence classes forms a partition of A, meaning the classes are nonempty, pairwise disjoint and their union is the entire set A.
Constructing partitions from relations and vice versa. Every equivalence relation produces a partition of A into equivalence classes. Conversely, any partition of A into disjoint nonempty subsets defines an equivalence relation: declare x ~ y if and only if x and y lie in the same block of the partition. This correspondence is useful in counting, modular arithmetic, and forming quotient sets like Z_n from integers modulo n.
Examples to understand structure. Congruence modulo n is the standard example: a ≡ b (mod n) if n divides (a−b). The equivalence classes are residue classes 0,1,...,n−1 containing integers with the same remainder on division by n. Another example: on the set of triangles, declare two triangles equivalent if they are congruent; classes collect all triangles congruent to each other.
Properties and quick tests. To prove a relation is an equivalence relation check three properties: reflexive (a~a for all a), symmetric (a~b ⇒ b~a) and transitive (a~b, b~c ⇒ a~c). To describe the partition produced, pick representative elements and list their classes; show classes cover the whole set and do not overlap.
Use of quotient sets. The set of equivalence classes, denoted A/~, is important: its elements are the blocks of the partition. Working with A/~ simplifies problems by identifying equivalent elements. For instance arithmetic modulo n works in the quotient Z/nZ where each element is an equivalence class under congruence modulo n.
Exam style answers. When asked to prove a relation is equivalence give clear checks for reflexive, symmetric, transitive. When asked for classes list them explicitly and show they partition the set. State representatives and argue disjointness using the definition of the relation.
- On Z, define a~b iff a - b is divisible by 5; classes are {...,-10,-5,0,5,10,...}, {...,-9,-4,1,6,11,...}, etc.
- On set {1,2,3,4} partition {{1,2},{3,4}} defines equivalence: 1~2, 3~4 and reflexive pairs.
- Equality is an equivalence relation; each equivalence class is a single element.
- Reflexive: ∀x, x~x
- Symmetric: x~y ⇒ y~x
- Transitive: x~y and y~z ⇒ x~z
Relations on a Set: Reflexive, Symmetric, Transitive
Understanding the three basic properties. For a relation R on a set A the three fundamental properties to check are reflexive, symmetric and transitive. Reflexive means every element is related to itself: (a,a) ∈ R for all a in A. Symmetric means whenever (a,b) is in R the reversed pair (b,a) is also in R. Transitive means whenever (a,b) and (b,c) are in R then (a,c) must also be in R. Each property constrains the pattern of ordered pairs in a different way.
How to test each property concretely. To test reflexivity list all elements of A and verify the diagonal pairs are in R. To test symmetry pick a generic pair (a,b) ∈ R and check whether (b,a) ∈ R; a single counterexample suffices to disprove symmetry. For transitivity examine all chains (a,b),(b,c) in R and test whether (a,c) is included; again a single violating chain disproves transitivity. For infinite sets use algebraic descriptions of R and prove properties logically rather than by listing.
Examples and counterexamples. Equality relation = is reflexive, symmetric and transitive. The relation ≤ on R is reflexive and transitive but not symmetric (1 ≤ 2 holds but 2 ≤ 1 does not). The “is sibling of” relation is symmetric (if A is sibling of B then B is sibling of A) but not reflexive (a person is not their own sibling) and not transitive in general. The empty relation on a nonempty set is symmetric and transitive but not reflexive; the full relation A × A is reflexive, symmetric and transitive.
Directed graph (digraph) viewpoint. Visualise relations on small finite sets with digraphs: vertices are elements and directed edges represent ordered pairs. Reflexivity appears as loops at every vertex. Symmetry is visible as pairs of arrows in opposite directions between nodes. Transitivity implies that if there is a path of length two from a to c through b, then there must be a direct arrow a→c; in digraphs transitivity often leads to many edges.
Exercises to develop intuition. Given relations on small sets, practice checking properties and producing minimal changes to achieve a property (e.g., add diagonal pairs to make reflexive). When asked to prove or disprove a property, give either a general proof using definitions or a single counterexample showing the property fails. Clear statements and precise reasoning are required in exam answers.
- R = {(x,y) ∈ Z×Z : x=y} is reflexive, symmetric and transitive.
- R = { (x,y): x ≤ y } on R is reflexive and transitive but not symmetric.
- For A={1,2}, R={(1,2)} fails reflexivity, symmetry and transitivity.
- Reflexive: ∀x ∈ A, (x,x) ∈ R
- Symmetric: (x,y) ∈ R ⇒ (y,x) ∈ R
- Transitive: (x,y) ∈ R and (y,z) ∈ R ⇒ (x,z) ∈ R
Partial Orders and Posets (Brief Introduction)
Definition and core idea. A partial order on a set P is a relation ≤ that is reflexive, antisymmetric and transitive. Antisymmetry means that if a ≤ b and b ≤ a then a = b. A set P with such a relation is called a partially ordered set or poset. Partial orders capture situations where some elements can be compared and others cannot; they formalise 'less than' in contexts where not every pair is comparable.
Examples and interpretation. The usual ≤ on numbers is a total order because every pair is comparable. By contrast, the divisibility relation on positive integers is a partial order: for example 2 divides 6 so 2 ≤ 6 in this order, but 2 and 3 are incomparable because neither divides the other. The subset relation on the power set of a set is another typical poset: two distinct subsets may or may not be comparable by inclusion.
Comparability, chains and antichains. In a poset some elements form chains, where each pair is comparable; these behave like totally ordered subsets. An antichain is a set of elements no two of which are comparable. Recognising chains and antichains helps in scheduling problems and combinatorics: for instance, tasks with precedence constraints form a partial order and a schedule must respect these relations.
Hasse diagrams for visualisation. For finite posets Hasse diagrams are an effective visual tool. Place elements as points and draw an upward line from a to b when a < b and there is no element c with a < c < b (a cover relation). Omit arrows and transitive edges; the vertical placement suggests order. Hasse diagrams make minimal and maximal elements, and immediate successors, easy to identify.
Special elements and features. A minimal element has no smaller element besides itself; a maximal element has no larger element. A greatest element is one greater than all others, and a least element is less than all others; these need not exist. In divisibility, 1 is least but there may be several maximal elements if the set is restricted. Lattices are posets where every two elements have a least upper bound and greatest lower bound—this is a further structure built from posets.
Exercises and exam tasks. Common tasks include verifying antisymmetry from list of pairs; drawing the Hasse diagram; finding minimal, maximal, least and greatest elements; and deciding comparability of elements. For construction tasks you may be asked to extend a poset to a total order or to find chains of maximum length. Present answers with justification: show which pairs force antisymmetry, explain missing comparabilities, and draw a clear Hasse diagram with labels.
- Divisibility on {1,2,3,4,6} is a poset; 2 divides 4 so 2 ≤ 4; 2 and 3 are incomparable.
- Power set P({a,b}) ordered by ⊂ yields poset with Hasse diagram: ∅ at bottom, {a},{b} in middle, {a,b} at top.
- ≤ on R is a total order.
- Antisymmetric: (a ≤ b and b ≤ a) ⇒ a = b
- Poset = (P, ≤) where ≤ is reflexive, antisymmetric, transitive
Functions from R to R: Domain and Range Problems
Domain identification. For functions defined by formulas, determine domain by excluding values that make denominators zero, arguments of even roots negative, or other illegal operations. For piecewise definitions, list domain pieces explicitly. For composed functions check inner functions first to determine allowable x.
Range determination. Finding the set of output values often requires algebraic manipulation or analysis. For simple polynomials, range may be entire R or bounded below/above depending on degree and leading coefficient. For rational functions find horizontal/vertical asymptotes and consider limits to decide values excluded. For functions like y = x^2 range is [0,∞); for y = e^x range is (0,∞).
Techniques for range. Solving y = f(x) for x in terms of y and looking for values of y that allow real solutions is a common method. For continuous functions the intermediate value theorem helps: if f is continuous on an interval and takes values a and b, it takes all values between a and b. For algebraic expressions complete the square, use monotonicity, or use calculus extrema (if allowed) to find minimum/maximum values.
Examples of restrictions. f(x) = √(x-1) has domain [1,∞) and range [0,∞) because the square root gives nonnegative outputs. g(x)=1/(x^2-4) has domain R\{2,-2}; range excludes 0 since 1/(x^2-4)=0 has no solution; check behaviour near asymptotes to find all possible y-values.
Composition and operations. When functions are combined, find domain as intersection and exclude further points where intermediate values cause problems. For range of combined functions substitute and manipulate carefully; clear algebraic presentation and justification are essential for board answers.
Exam style presentation. State domain clearly, show algebraic steps or limiting arguments to support range, and indicate any excluded points explicitly. Use diagrams where helpful and present final domain and range in standard interval notation with endpoints described as included or excluded.
- Find domain of f(x)=√(2x-4)/(x-3): domain x ≥ 2 and x ≠ 3 ⇒ [2,∞)\{3}.
- Range of f(x)=x^2 on R is [0,∞).
- Range of g(x)=1/(x^2+1) is (0,1].
- Domain(f±g) = Domain(f) ∩ Domain(g)
- Domain(f∘g) = {x ∈ Domain(g) : g(x) ∈ Domain(f)}
Piecewise-defined Functions and Absolute Value
What are piecewise functions? Piecewise functions are defined by different formulas on different parts of the domain. Each piece is valid on a specified interval or set, and the overall function is the union of these pieces. Because behaviour can change at boundaries, examine continuity and values at those points carefully.
Absolute value as a basic piecewise case. The absolute value function |x| is defined piecewise by |x| = x for x ≥ 0 and |x| = −x for x < 0. This gives the familiar V-shaped graph with vertex at the origin. Many problems reduce to solving or analysing expressions involving | · | by splitting into the appropriate sign cases and handling each case algebraically.
Techniques for solving and analysing. When solving equations or inequalities with piecewise definitions, split into the cases determined by the domain pieces (for absolute value, by sign). Solve in each case and then collect solutions that satisfy the case condition. When studying continuity at boundary points evaluate the left-hand and right-hand limits and compare with the defined value; equal limits mean continuity at that point.
Invertibility and monotonicity. Piecewise functions may be invertible if each piece is monotone and the pieces together make the whole function one-to-one. For example f(x)=|x| is not invertible on R, but restricting to [0,∞) it becomes invertible with inverse f^{-1}(x)=x on [0,∞). When asked to find inverse of a piecewise function, find inverse on each monotone piece and assemble with the corresponding restricted domains and ranges for the inverse.
Graphing and endpoints. For each piece sketch the graph on its interval and mark endpoints as filled or open depending on inclusion. Check for corners or cusps at boundaries: absolute value has a corner at x=0 where derivative changes sign and so is not differentiable there. Use piecewise sketches to visualise continuity and extremal behaviour.
Common exam tasks. Expect to be asked to solve |f(x)| = a, sketch piecewise functions, test continuity at junctions, or find inverses by restricting domain. Provide casewise solutions, show checking of case conditions, and state final answer in interval form as required.
- f(x)= { x+1, x≤0; 2x-1, x>0 } — test continuity at x=0: left limit =1, right limit = -1 so discontinuous.
- Solve |x-3| = 5 ⇒ x-3 = 5 or x-3 = -5 ⇒ x = 8 or x = -2.
- f(x)=|x| on [0,∞) is invertible with inverse f^{-1}(x)=x.
- |x| = { x, x ≥ 0; -x, x < 0 }
- To invert piecewise: find inverse on each monotone piece and restrict codomain accordingly
Binary Relations Represented by Matrices and Digraphs
Matrix representation for finite sets. For a finite set A = {a1,a2,...,an}, any relation R on A can be encoded by an n×n matrix M with entries 0 or 1. Set M_{ij} = 1 if (a_i,a_j) ∈ R, otherwise M_{ij} = 0. This adjacency matrix translates relation problems into matrix calculations and is particularly useful for computing compositions and reachability for finite relations.
Directed graphs (digraphs). A digraph visualises the relation: vertices correspond to elements of A and there is a directed edge from vertex i to vertex j when M_{ij} = 1. Self-related elements appear as loops. Digraphs make reflexivity, symmetry and transitivity visible: symmetry corresponds to mutual two-way edges, reflexivity to loops at each vertex and transitivity to the existence of direct edges whenever two-step paths exist.
Boolean matrix multiplication and composition. Composition of relations corresponds to Boolean matrix multiplication. If M_R and M_S represent relations R and S, then the (i,k) entry of the matrix for S ∘ R equals 1 if there exists j such that M_R(i,j) = 1 and M_S(j,k) = 1. In Boolean terms this uses OR and AND instead of addition and multiplication. This gives a mechanical method to compute compositions and powers of relations for finite sets, helpful in problems about reachability or transitive closure.
Transitive closure and Warshall’s idea. The transitive closure R^+ of R is the smallest transitive relation containing R, capturing reachability by any number of steps. For finite sets one can compute powers of the adjacency matrix and combine them (using Boolean OR) or apply an algorithmic procedure similar to Warshall’s algorithm to find reachability between all pairs. For small matrices compute M, M^2, M^3 etc. until no new ones appear.
Applications and exam practice. Convert relations between roster form, matrix and digraph representations for small sets. Use matrix operations to find compositions and transitive closures and interpret results in terms of paths in the digraph. In answers show correspondence between matrix entries and pairs (a_i,a_j) explicitly to avoid ambiguity.
- A={1,2,3}, R={(1,2),(2,3)} ⇒ adjacency matrix [[0,1,0],[0,0,1],[0,0,0]].
- Digraph: vertices 1,2,3 with arrows 1→2 and 2→3 visualises R.
- Composition R∘R yields (1,3) since 1→2 and 2→3, reflected by matrix product.
- Matrix M_R: M_R(i,j)=1 if (a_i,a_j)∈R else 0
- Boolean product: (M∘N)(i,k)=∨_j (M(i,j) ∧ N(j,k))
Graphs of Functions: Sketching and Transformations
Purpose of sketching. Sketching graphs helps visualise functions, understand domain and range, and observe key features such as intercepts, symmetry, extrema and asymptotic behaviour. For exams a clear labelled sketch with key points often earns marks even when full algebraic detail is long.
Standard graphs to master. Know the shape and features of basic functions: y = x (straight line), y = x^2 (parabola with vertex at origin), y = x^3 (odd cubic shape), y = 1/x (hyperbola with asymptotes), y = e^x (increasing exponential approaching 0 as x→−∞), y = ln x (inverse of e^x), and basic trigonometric curves. For each know intercepts, symmetries (even/odd), and end behaviour.
Transformations and their effect. Transformations change graphs in predictable ways. Horizontal translation x→x−h shifts right by h, vertical translation adds k. Vertical scaling y→a f(x) stretches (|a|>1) or compresses (0<|a|<1). Horizontal scaling by b in f(bx) compresses if |b|>1. Reflections y→−f(x) and y→f(−x) reflect across x-axis and y-axis respectively. Combining transformations requires careful order: horizontal changes act on input before vertical ones on output.
Asymptotes and rational functions. Rational functions commonly have vertical asymptotes at zeros of the denominator and horizontal or oblique asymptotes determined by degrees of numerator and denominator. Sketch near asymptotes by checking limits as x approaches vertical asymptote from both sides and as x→±∞. Mark holes (removable discontinuities) where factors cancel but original function undefined.
Piecewise graphs and absolute value. For piecewise-defined functions sketch each part separately on its interval and indicate filled or open endpoints. For absolute value functions draw V-shaped graphs and apply transformations. Check continuity at boundaries and show any cusps or corners clearly.
Composition and combined graphs. When functions are added or multiplied, sketching can rely on component graphs to estimate combined shape. For composition, visualise two-step mapping: input to intermediate then to output. For inverse functions, reflect the original graph across the line y = x to obtain the inverse graph; label corresponding points to demonstrate the reflection property.
Exam technique. Always label axes, mark scale, label key points (intercepts, vertex, asymptotes) and state domain and range. A neat sketch with justification (e.g., symmetry, intercepts, limits) is preferred over an unlabeled curve. Use transformations language to explain how the sketch was obtained from a known template.
- Graph y = (x-1)^2 + 2: parabola shifted right by 1 and up by 2; vertex at (1,2).
- y = 1/x: hyperbola with vertical asymptote x=0 and horizontal asymptote y=0 in quadrants I and III.
- y = e^x: increasing curve crossing y-axis at 1 and approaching y=0 as x→-∞.
- Translation: y = f(x-h)+k shifts right by h and up by k
- Reflection: y = -f(x) reflects across x-axis; y = f(-x) reflects across y-axis
Special Function Types: Polynomial, Rational, Exponential, Logarithmic
Polynomial functions. A polynomial function is an expression of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_0 where coefficients are real numbers and n is a nonnegative integer. Its domain is all real numbers. The degree n determines end behaviour: even degree with positive leading coefficient rises to +∞ at both ends, odd degree has opposite directions. Real roots are zeros of the polynomial; multiplicity matters: an odd multiplicity root crosses the x-axis, an even multiplicity root touches and turns back. Factor theorem and synthetic division help find factors and sketch behavior.
Rational functions. Rational functions are ratios of two polynomials p(x)/q(x) with q not identically zero. Domain excludes zeros of q. Vertical asymptotes occur at non-cancelled zeros of q, while cancelled common factors produce removable holes. Horizontal asymptotes are decided by degrees: if degree numerator < degree denominator, horizontal asymptote y=0; if degrees equal, asymptote is ratio of leading coefficients; if numerator degree is one more, expect an oblique asymptote found by division. Study limits as x→±∞ and near poles to understand range and sketching.
Exponential and logarithmic functions. Exponential functions of the form a^x, with a>0 and a≠1, have domain R and range (0,∞). They are monotonic: increasing for a>1, decreasing for 0 Behaviour, transforms and applications. All these classes respond predictably to shifts and scales: f(x-h)+k shifts and scales affect graphs in standard ways. Polynomials model smooth approximations, rationals model quantities with asymptotic limits, exponentials model growth/decay in populations and finances, and logarithms linearise multiplicative processes. In exam problems identify domain and range, intercepts, asymptotes, symmetries and critical points where needed. Practical tips for exams. For polynomials use factorisation to find zeros and sign changes; for rationals perform division to reveal asymptotes and holes; for exponentials and logs check monotonicity and use inverses to solve equations. Present domain restrictions clearly, sketch graphs with key points and asymptotes, and state range explicitly. These practices earn full marks in board-style questions.
- Polynomial p(x)=x^3-3x has roots at x=0,±√3; sketch sign changes and cubic shape.
- Rational r(x)=(x^2-1)/(x-1)=x+1 with hole at x=1 because original denominator zero; domain excludes 1.
- Exponential f(x)=2^x is increasing with range (0,∞); inverse log2 x defined for x>0.
- Polynomial: p(x)=a_n x^n + ... + a_0
- Logarithm inverse: y = a^x ⇔ x = log_a y
- Change of base: log_b x = log_c x / log_c b
Binary Operations Viewed as Functions
Binary operations as functions. A binary operation * on a set S is a function from S × S to S: * : S × S → S. Representing operations this way emphasises that each ordered pair (a,b) in S × S gets a unique output a*b in S. Examples include addition and multiplication on numbers and are central objects in algebraic structures.
Properties and their function interpretation. Key properties like associativity ((a*b)*c = a*(b*c)), commutativity (a*b = b*a), identity element (e such that e*a = a*e = a) and inverses (for each a there exists a^{-1} with a*a^{-1} = e) are naturally expressed and checked using the functional viewpoint. For example, associativity concerns how the ternary function values behave when arguments are grouped differently; think of repeated application of the binary function.
Cayley tables and finite examples. For a finite set S a binary operation is conveniently shown by a Cayley table: rows and columns index S and each cell gives the product. The table verifies closure (outputs remain in S) and allows checking of commutativity (symmetry of the table), identity (a row/column matching input order), and existence of inverses. Use tables in exam problems to demonstrate properties or to construct operations with desired characteristics.
Pointwise operations on function spaces. If f and g are functions from X to S and S has a binary operation *, then we can form a new function f * g defined by (f * g)(x) = f(x) * g(x). This pointwise binary operation builds algebraic structure on sets of functions, important in functional analysis and algebraic contexts.
Applications and exam tasks. Typical questions ask to verify whether a given operation is associative, commutative, or has an identity. Provide clear checking for all possible ordered triples or produce counterexamples to show failure. For finite sets produce full Cayley tables; for infinite sets give proofs using algebraic properties. Present answers with definitions stated and properties verified step by step.
- Addition + on integers is a binary operation Z × Z → Z that is associative and commutative with identity 0.
- Define * on {a,b} by table: a*a=a, a*b=b, b*a=b, b*b=a. Check associativity by testing all triples.
- Pointwise product of functions f,g: (fg)(x)=f(x)g(x) defines binary operation on set of real-valued functions.
- * : S × S → S
- (f * g)(x) = f(x) * g(x) for functions f,g : X → S
Applications: Modelling with Functions and Relations
Functions as models. Functions convert real situations into mathematical language. For example population models P(t)=P_0 e^{kt} describe continuous growth or decay, cost functions C(x)=fixed + variable·x model production cost, and demand functions relate price to quantity. When modelling always specify domain (e.g., time ≥ 0, quantity whole units) and interpret the range in context (e.g., costs nonnegative).
Relations in data and databases. Relations appear in classification and database contexts: a relation student→grade can be many-to-one if many students share a grade. Understanding injective and surjective mappings helps design keys and constraints: a primary key requires a one-to-one mapping from records to identifier values to ensure uniqueness.
Equivalence relations for grouping. Equivalence relations are used to group objects with a shared property. For example clustering exam scores into grade categories uses equivalence classes; modular arithmetic groups integers by remainder and simplifies computations in number theory. In counting problems partitions induced by equivalence relations often simplify combinatorial reasoning.
Composition models sequential processes. Composition of functions naturally models processes with stages: if f maps raw input to an intermediate result and g maps intermediate to final product, then g∘f maps raw to final directly. This chaining is common in physics (transformations of units), chemistry (reaction steps), and computing (pipeline of functions).
Problem solving strategy. Translate the situation into functions or relations, state domain and codomain, and write algebraic expressions clearly. Solve equations, check for extraneous solutions, and interpret results back in context. Use graphs to visualise behaviour and asymptotes to check long-run trends. Always comment on feasibility and units when answering modelling questions.
Example applied tasks. Expect exam-style problems like: find domain and range for a model, restrict a function to make it invertible, or determine whether a mapping in a database is injective. Show all steps and interpret final answers in the context provided.
- Population model P(t)=1000 e^{0.02t} has domain t≥0 and models exponential growth.
- Tax function defined piecewise: 0≤x≤250000 ⇒ tax=0, 250000<x≤500000 ⇒ tax=0.05(x-250000), etc.
- Database relation: student→roll number is many-to-one if multiple students share a roll (which is usually disallowed); ideal is one-to-one mapping.
Key Concepts
- Relation
- A subset of the Cartesian product A × B consisting of ordered pairs that associate elements of A to elements of B.
- Function
- A relation in which each element of the domain is paired with exactly one element of the codomain.
- Domain
- The set of all inputs for which a relation or function is defined.
- Range (Image)
- The set of outputs actually produced by a relation or function from its domain.
- Codomain
- The set that contains all possible outputs as declared by the function definition.
- Injective (One-to-one)
- A function where distinct inputs always produce distinct outputs.
- Surjective (Onto)
- A function whose range equals its codomain so every codomain element is hit.
- Bijective
- A function that is both injective and surjective, hence invertible.
- Inverse Function
- A function that reverses the mapping of a bijection, swapping inputs and outputs.
- Composition
- Applying one function after another, defined by (g∘f)(x)=g(f(x)).
- Equivalence Relation
- A relation that is reflexive, symmetric and transitive, partitioning the set into classes.
- Partition
- A division of a set into disjoint nonempty subsets whose union is the whole set.
- Reflexive
- A relation property where every element relates to itself.
- Symmetric
- A relation property where if a relates to b then b relates to a.
- Transitive
- A relation property where a~b and b~c imply a~c.
- Partial Order (Poset)
- A set with a reflexive, antisymmetric and transitive relation.
- Adjacency Matrix
- A zero-one matrix representing a finite relation with entries showing presence of ordered pairs.
- Hasse Diagram
- A graphical representation of a finite poset showing cover relations without transitive edges.
- Vertical Line Test
- A visual test for functions in the plane: no vertical line should meet the graph more than once.
Practice Questions
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List the domain and range of the relation R = {(1,2),(2,3),(3,2)} / सम्बन्ध R = {(1,2),(2,3),(3,2)} का क्षेत्र और मान क्या हैं?
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Domain is {1,2,3} and range is {2,3}. In English: each first component 1,2,3 appears, so domain = {1,2,3}; the outputs seen are 2 and 3, so range = {2,3}. / क्षेत्र {1,2,3} है और मान {2,3} है। हिन्दी में: पहले अवयवों 1,2,3 प्रकट होते हैं अतः क्षेत्र = {1,2,3} है; प्राप्त मान 2 और 3 हैं अतः मान = {2,3}।
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Decide whether f: R → R defined by f(x)=x^2 is injective, surjective or bijective. / f: R → R जिसे f(x)=x^2 द्वारा परिभाषित किया गया है, क्या यह एक-एक, ऊपर-ऊपर (onto) या दोनों है?
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f is not injective because f(2)=f(-2)=4, so two different inputs give the same output. f is not surjective onto R because negative real numbers are not squares and hence are not in the range; the range is [0,∞). Therefore f is not bijective on R. / f एक-एक नहीं है क्योंकि f(2)=f(-2)=4; यह onto (सर्ववर्ती) भी नहीं है क्योंकि नकारात्मक वास्तविक संख्याएँ f के मान में नहीं आतीं (मान [0,∞) है); अतः f R पर bijective नहीं है।
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If f(x)=2x+3, find f^{-1}(x). State domain and range. / यदि f(x)=2x+3 है, तो f^{-1}(x) ज्ञात कीजिए। क्षेत्र और मान बताइए।
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Solve y = 2x + 3 for x: x = (y - 3)/2. So f^{-1}(x) = (x - 3)/2. Since f: R → R is linear with slope 2 ≠ 0, both domain and range of f are all real numbers R; likewise domain and range of f^{-1} are R. / y = 2x+3 को x के लिए हल करने पर x = (y-3)/2 मिलता है। अतः f^{-1}(x) = (x-3)/2। f का क्षेत्र और मान दोनों R हैं; इसी तरह f^{-1} का क्षेत्र और मान भी R हैं।
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For functions f(x)=x^2 (domain R) and g(x)=x+1, compute (g∘f)(x) and (f∘g)(x). / f(x)=x^2 (क्षेत्र R) और g(x)=x+1 के लिए (g∘f)(x) और (f∘g)(x) निकालिए।
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(g∘f)(x) = g(f(x)) = f(x) + 1 = x^2 + 1. (f∘g)(x) = f(g(x)) = (x+1)^2 = x^2 + 2x + 1. Thus g∘f and f∘g are distinct functions: one is x^2+1, the other x^2+2x+1. / (g∘f)(x) = x^2 + 1 और (f∘g)(x) = (x+1)^2 = x^2 + 2x + 1। ये अलग-अलग फलन हैं।
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Give an example of an equivalence relation on the set Z of integers and describe its equivalence classes. / पूर्णांकों के समूह Z पर एक समतुल्य सम्बन्ध का उदाहरण दीजिए और उसकी समतुल्य वर्गों का वर्णन कीजिए।
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One common example is congruence modulo n. Define a ≡ b (mod n) if n divides (a−b). This relation is reflexive (a−a = 0 is divisible by n), symmetric (if a−b divisible by n then b−a is also), and transitive (if a−b and b−c are divisible by n then a−c is divisible by n), so it is an equivalence relation. Its equivalence classes are the residue classes modulo n: each class collects all integers that leave the same remainder on division by n. For example with n = 3 the classes are [0] = {...,−6,−3,0,3,6,...}, [1] = {...,−5,−2,1,4,7,...} and [2] = {...,−4,−1,2,5,8,...}. These classes partition Z into n disjoint sets whose union is Z. / एक सामान्य उदाहरण modulo n के अनुसार समरूपता है: परिभाषा a ≡ b (mod n) जब n, (a−b) को विभाजित करे। यह प्रतिबिम्बात्मक, सममित और पारगम्य है, अतः समतुल्य सम्बन्ध है। इसके समतुल्य वर्ग भागशेष वर्ग होते हैं, जो समान शेष देने वाली पूर्णांकों का समूह हैं। उदाहरणार्थ n=3 पर ऊपर दिये गए तीन वर्ग Z को विभाजित करते हैं।
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Let A={1,2,3}. Define R on A by R={(1,1),(2,2),(3,3),(1,2)}. Is R reflexive, symmetric, transitive? / A={1,2,3} पर R={(1,1),(2,2),(3,3),(1,2)} परिभाषित है। क्या R प्रतिबिम्बात्मक (reflexive), सममित (symmetric), और पारगम्य (transitive) है?
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R is reflexive because (1,1),(2,2),(3,3) are included so every element relates to itself. R is not symmetric: (1,2) ∈ R but (2,1) ∉ R, so symmetry fails. For transitivity check all chains: we have (1,2) and (2,2) in R which imply (1,2) and that is present; (1,1) and (1,2) imply (1,2) present; other possible chains do not lead to a missing pair. Thus no counterexample to transitivity exists and R is transitive. Summarising: reflexive yes, symmetric no, transitive yes. / R प्रतिबिम्बात्मक है क्योंकि सभी (1,1),(2,2),(3,3) मौजूद हैं। यह सममित नहीं है क्योंकि (1,2) है पर (2,1) नहीं है। पारगम्य (transitive) है क्योंकि जो भी आवश्यक चेनें बनती हैं उनसे उत्पन्न जोड़े पहले से R में हैं; इसलिए transitive है।
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Find the domain of f(x)=√(x-1)/(x^2-4). / f(x)=√(x-1)/(x^2-4) का क्षेत्र निर्धारित कीजिए।
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We require the square root argument to be nonnegative: x−1 ≥ 0 ⇒ x ≥ 1. Also denominator must be nonzero: x^2−4 ≠ 0 ⇒ x ≠ 2 and x ≠ −2. Combining with x ≥ 1 excludes x = −2 automatically; we must exclude x = 2 as well. Therefore domain is [1,2) ∪ (2,∞). / √ के भीतर x−1 ≥0 चाहिए अतः x ≥1। हराने हेतु x^2−4 ≠0 ⇒ x ≠ ±2। x ≥1 eliminates −2, पर x=2 को हटाना होगा। अतः क्षेत्र [1,2) ∪ (2,∞) है।
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Show that if f and g are bijections then g∘f is a bijection. / सिद्ध कीजिए कि यदि f और g दोनों bijection हैं तो g∘f भी bijection होगा।
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Proof: Let f: A → B and g: B → C be bijections. First show injective: suppose (g∘f)(x1) = (g∘f)(x2). Then g(f(x1)) = g(f(x2)). Since g is injective, f(x1) = f(x2). Since f is injective, x1 = x2. Thus g∘f is injective. Next show surjective: take any z ∈ C. Since g is surjective, there exists y ∈ B with g(y) = z. Since f is surjective, there exists x ∈ A with f(x) = y. Then (g∘f)(x) = g(f(x)) = g(y) = z, so every z in C has a preimage in A and g∘f is surjective. Being both injective and surjective, g∘f is bijective. / प्रमाण: f: A→B तथा g: B→C दोनों bijection हैं। injective के लिये यदि g(f(x1))=g(f(x2)) तो g की injectivity से f(x1)=f(x2) और f की injectivity से x1=x2। अतः g∘f injective है। surjective के लिये किसी z∈C के लिये g के कारण y∈B मिलता है और f के कारण x∈A मिलता है जिससे g(f(x))=z। इसलिए g∘f surjective है। दोनों गुणों से g∘f bijection है।
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Compute inverse of f(x) = (3x-4)/(2x+1) where domain excludes x = -1/2, and state domain of inverse. / f(x) = (3x-4)/(2x+1) का inverse निकालिए (x = -1/2 को छोड़कर क्षेत्र) और inverse का क्षेत्र बताइए।
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Set y = (3x−4)/(2x+1). Solve for x: y(2x+1) = 3x − 4 ⇒ 2xy + y = 3x − 4 ⇒ bring x terms together: 2xy − 3x = −4 − y ⇒ x(2y − 3) = −(4 + y) ⇒ x = −(4 + y)/(2y − 3). Replacing y by x for function notation gives f^{-1}(x) = −(4 + x)/(2x − 3). The domain of f^{-1} excludes x values that make denominator zero, so 2x − 3 ≠ 0 ⇒ x ≠ 3/2. Also note that f was undefined at x = −1/2 so f(−1/2) does not appear in the range; that value is not in domain of inverse. Therefore domain of f^{-1} is R \ {3/2} (more precisely the range of f, which equals R except possibly the value corresponding to the excluded original point). / y=(3x−4)/(2x+1) हल करने पर x = −(4+y)/(2y−3) मिलता है, अतः f^{-1}(x)= −(4+x)/(2x−3)。 inverse का क्षेत्र वह x जहाँ 2x−3 ≠ 0 यानी x ≠ 3/2। मूल फलन x=−1/2 पर परिभाषित नहीं था, इसलिए वह मान inverse के क्षेत्र में नहीं आता।
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Given relation R on {1,2,3} with adjacency matrix [[1,0,1],[0,1,0],[0,1,1]], list the pairs in R and draw the digraph. / मैट्रिक्स [[1,0,1],[0,1,0],[0,1,1]] वाले {1,2,3} पर सम्बन्ध R के जोड़े लिखिए और इसका निर्देशित ग्राफ बनाइए।
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Interpret rows as first component and columns as second. Entries 1 at (1,1) and (1,3) give pairs (1,1) and (1,3). Entry (2,2) gives (2,2). Entries (3,2) and (3,3) give (3,2) and (3,3). Thus R = {(1,1),(1,3),(2,2),(3,2),(3,3)}. The digraph has vertices 1,2,3; loops at 1, 2 and 3 (for (1,1),(2,2),(3,3)); an arrow from 1 to 3; and an arrow from 3 to 2. / पंक्तियों को पहला अवयव और स्तम्भों को दूसरा अवयव मानकर 1 के स्थान पर (1,1) तथा (1,3), 2 पर (2,2), और 3 पर (3,2),(3,3) मिलते हैं। अतः R = {(1,1),(1,3),(2,2),(3,2),(3,3)}। निर्देशित ग्राफ में वर्टिस 1,2,3 होंगे, प्रत्येक पर लूप और तीर 1→3 तथा 3→2 होंगे।
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