01 · Explore
Introduction to Relations
A mathematical relation describes a connection between elements of two sets, abstracting the everyday concept of 'relationship' into a formal set-theoretic framework.
In mathematics, a relation R from a set A to a set B is defined as any subset of the Cartesian product A × B. If an ordered pair (a, b) belongs to R, we say that 'a' is related to 'b' under R, often written as a R b.
A relation R in a set A is called an empty relation if no element of A is related to any element of A, i.e., R = ∅ ⊂ A × A. Conversely, R is called a universal relation if each element of A is related to every element of A, i.e., R = A × A.
Both the empty relation and the universal relation are sometimes called trivial relations. For a relation defined on a single set A, we study R as a subset of A × A to understand how elements within the same set interact based on specific rules.
| Relation Type | Definition | Example in Set A = {1, 2, 3, 4} |
|---|---|---|
| Empty | R = ∅ | R = {(a, b) : a - b = 10} |
| Universal | R = A × A | R = {(a, b) : |a - b| ≥ 0} |
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On A = {2, 4, 6}, let a R b mean a + b = 15. Why is R empty?
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02 · Explore
Reflexive, Symmetric, and Transitive Relations
To classify complex relations, we look for three fundamental properties: reflexivity, symmetry, and transitivity.
A relation R on set A is Reflexive if every element is related to itself. Formally, (a, a) ∈ R for every a ∈ A. If even one element is not related to itself, the relation is not reflexive.
A relation is Symmetric if the direction of the relationship doesn't matter. If (a, b) ∈ R, then (b, a) must also be in R. For example, 'is a sibling of' is symmetric, but 'is taller than' is not.
A relation is Transitive if a chain of relationships implies a direct one. If (a, b) ∈ R and (b, c) ∈ R, then (a, c) must be in R. This is similar to the logic: if A = B and B = C, then A = C.
Perpendicular Lines Relation
R = {(L1, L2) : L1 ⊥ L2}
Take all straight lines in one plane. Perpendicularity is symmetric: L1 ⊥ L2 implies L2 ⊥ L1. It is not reflexive because no line is perpendicular to itself. It is not transitive: if L1 ⊥ L2 and L2 ⊥ L3 in that plane, L1 and L3 have parallel directions (and may coincide), so L1 is not perpendicular to L3.
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Question
Is the relation R = {(1, 1), (2, 2), (1, 2)} on set {1, 2} transitive?
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03 · Explore
Equivalence Relations and Classes
An equivalence relation is a powerful tool that groups 'similar' elements together, effectively partitioning a set into distinct categories.
A relation R on set A is an Equivalence Relation if and only if it is reflexive, symmetric, and transitive simultaneously. Such relations behave like the 'equality' operator, grouping elements that share a specific property.
Equivalence relations divide a set into mutually disjoint subsets called Equivalence Classes. Every element in a class is related to every other element in that same class, and no element is related to anyone outside its class.
The union of all equivalence classes equals the original set A. For example, if we define a relation on integers where 'a R b' if (a - b) is even, the set is partitioned into two classes: even integers [0] and odd integers [1].
Congruence of Triangles
R = {(T1, T2) : T1 ≅ T2}
1. Reflexive: Every triangle is congruent to itself. 2. Symmetric: If T1 ≅ T2, then T2 ≅ T1. 3. Transitive: If T1 ≅ T2 and T2 ≅ T3, then T1 ≅ T3. Since all three hold, it is an equivalence relation.
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Question
What are the three properties required for a relation to be an equivalence relation?
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04 · Explore
Types of Functions: Injective and Many-One
Functions are special relations where every input has exactly one output. We classify them based on how they map domain elements to the co-domain.
A function f: X → Y is One-One (or Injective) if distinct elements in the domain have distinct images in the co-domain. Mathematically, f(x1) = f(x2) must imply x1 = x2. If two different inputs produce the same output, the function is Many-One.
To prove a function is one-one, we assume f(x1) = f(x2) and use algebraic manipulation to show x1 = x2. To disprove it, we find a counter-example where two different inputs yield the same result.
For example, f(x) = x² defined on the set of Natural numbers (N) is one-one because if x1² = x2², then x1 = x2 (since natural numbers are positive). However, on the set of Real numbers (R), it is many-one because f(1) = 1 and f(-1) = 1.
| Type | Meaning | Example on R → R |
|---|---|---|
| One-one | Different inputs have different outputs | f(x) = 5x + 1 |
| Many-one | Some different inputs have the same output | g(x) = x²; g(2) = g(−2) = 4 |
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Is f: R → R, f(x) = 5x + 1, one-one? Show why.
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05 · Explore
Surjective and Bijective Functions
Beyond how inputs are mapped, we also look at whether the entire co-domain is 'covered' by the function's outputs.
A function f: X → Y is Onto (or Surjective) if every element in the co-domain Y is the image of at least one element in the domain X. In other words, the Range of the function must equal the Co-domain.
A function is One-One and Onto (or Bijective) if it satisfies both conditions. Bijective functions represent a perfect 'one-to-one correspondence' between two sets.
For finite sets, a function from a set to itself is one-one if and only if it is onto. This property is a characteristic difference between finite and infinite sets; for instance, f(x) = 2x on natural numbers is one-one but not onto.
Checking Surjectivity
f: R → R, f(x) = 2x
To check if it is onto, let y be any real number in the co-domain. We need to find x such that 2x = y. Solving for x gives x = y/2. Since y/2 is always a real number, every y has a pre-image x. Thus, f is onto.
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What is a bijective function?
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06 · Explore
Composition of Functions
Composition allows us to combine two functions by using the output of the first as the input for the second.
Given two functions f: A → B and g: B → C, the composition gof: A → C is defined by (gof)(x) = g(f(x)). Note that for gof to be defined, the range of f must be a subset of the domain of g.
Composition is generally not commutative, meaning fog is usually not equal to gof. The order in which functions are applied matters significantly.
Consider f(x) = cos x and g(x) = 3x². Then (gof)(x) = g(cos x) = 3(cos x)² = 3 cos² x. However, (fog)(x) = f(3x²) = cos(3x²). Since 3 cos² x ≠ cos(3x²) for most x, gof ≠ fog.
The Process of Composition
- 1
Input x
Start with an element x from set A.
- 2
Apply f
Calculate f(x), which lands in set B.
- 3
Apply g
Use f(x) as the input for g to get g(f(x)) in set C.
The composition gof maps elements directly from set A to set C by passing through the intermediate results of f.
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Question
For f: R → R, f(x) = x² + 2, and g: R → R, g(x) = 3x − 1, find (g ∘ f)(x).
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07 · Explore
Invertible Functions
An invertible function is one that can be 'undone,' allowing us to return to the original input from the output.
A function f: X → Y is Invertible if there exists a function g: Y → X such that gof = Ix (identity on X) and fog = Iy (identity on Y). The function g is called the inverse of f, denoted as f⁻¹.
A fundamental theorem states that a function is invertible if and only if it is bijective (both one-one and onto). This provides a clear path to prove invertibility without always finding the inverse formula first.
To find the inverse formula, we set y = f(x) and solve the equation for x in terms of y. The resulting expression x = g(y) defines the inverse function.
Finding an Inverse
f: R → R, f(x) = 4x + 3
1. Set y = 4x + 3. 2. Solve for x: y - 3 = 4x, so x = (y - 3)/4. 3. Define g(y) = (y - 3)/4. 4. Check: g(f(x)) = ((4x + 3) - 3)/4 = 4x/4 = x. Also f(g(y)) = 4((y - 3)/4) + 3 = y - 3 + 3 = y. Thus, f is invertible.
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Question
What condition must a function satisfy to have an inverse?
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08 · Explore
Special Functions and Their Properties
Several standard functions serve as important examples and counter-examples when studying relations and functions.
For f: R → R, the Greatest Integer Function f(x) = [x] returns the largest integer less than or equal to x. It is not one-one because [1.2] = [1.8] = 1; it is not onto R because no non-integer real output is possible.
For f: R → R, the Modulus Function f(x) = |x| returns the absolute value of x. It is not one-one because |1| = |−1| = 1, and it is not onto R because negative real numbers have no pre-image.
The Signum Function f(x) is defined as 1 if x > 0, 0 if x = 0, and -1 if x < 0. It is neither one-one nor onto when defined from R to R, as multiple positive inputs map to 1 and values like 2 are never reached.
| Function | Notation | One-One? | Onto (R to R)? |
|---|---|---|---|
| Identity | f(x) = x | Yes | Yes |
| Modulus | f(x) = |x| | No | No |
| Greatest Integer | f(x) = [x] | No | No |
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Why is the modulus function f(x) = |x| not one-one on the set of real numbers?
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