A relation from a set A to a set B is a subset of the Cartesian product A x B. When we say that an element a in A is related to an element b in B, we write aRb or (a, b) in R. Functions are a special kind of relation where every element of the domain has exactly one image in the codomain.
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Key Concepts
- Types of Relations
- Empty relation: R = Ø (no element of A is related to any element of A).
- Universal relation: R = A x A (every element is related to every element).
- Reflexive relation: (a, a) in R for every a in A.
- Symmetric relation: If (a, b) in R then (b, a) in R.
- Transitive relation: If (a, b) in R and (b, c) in R, then (a, c) in R.
- Equivalence relation: Reflexive + Symmetric + Transitive.
- Types of Functions
- One-one (Injective): f(a) = f(b) implies a = b. Different inputs give different outputs.
- Onto (Surjective): Every element of the codomain is the image of at least one element of the domain.
- Bijective: Both one-one and onto. A bijection has an inverse function.
Composition of Functions: If f: A → B and g: B → C, then gof: A → C is defined as gof(x) = g(f(x)).
Invertible Functions: A function f is invertible if and only if it is bijective. The inverse f-1 satisfies f-1(f(x)) = x.
Binary Operations: A binary operation · on set A is a function · : A x A → A. It is commutative if a · b = b · a, and associative if (a · b) · c = a · (b · c). An element e is the identity if a · e = e · a = a. Element b is the inverse of a if a · b = b · a = e.
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Worked Examples
Check whether R = {(a, b) : a - b is divisible by 3} on Z is an equivalence relation.
- Reflexive: a - a = 0 = 3 x 0, so (a, a) in R. Yes.
- Symmetric: If 3 | (a - b), then 3 | (b - a). Yes.
- Transitive: If 3 | (a - b) and 3 | (b - c), then 3 | (a - c). Yes.
So R is an equivalence relation.
Let f: R → R, f(x) = 2x + 3. Is f bijective?
- One-one: If 2x1 + 3 = 2x2 + 3, then x1 = x2. Yes.
- Onto: For any y in R, x = (y - 3)/2 gives f(x) = y. Yes.
So f is bijective.
If f(x) = x2 and g(x) = x + 1, find gof(x) and fog(x).
gof(x) = g(f(x)) = g(x2) = x2 + 1
fog(x) = f(g(x)) = f(x + 1) = (x + 1)2
Note: gof is not equal to fog in general.
Show that f: N → N defined by f(x) = x2 is one-one but not onto.
One-one: f(a) = f(b) => a2 = b2 => a = b (for natural numbers). Yes.
Not onto: 3 is a natural number but no natural number x satisfies x2 = 3. Not onto.
Find the identity element for the operation a · b = a + b - 2 on Q.
Let e be the identity. Then a · e = a => a + e - 2 = a => e = 2.
Check: a · 2 = a + 2 - 2 = a. So e = 2 is the identity element.
Let f: R → R be defined by f(x) = 3x - 5. Find f-1.
Let y = 3x - 5. Then x = (y + 5)/3. So f-1(y) = (y + 5)/3, i.e., f-1(x) = (x + 5)/3.
Check if the relation R on {1, 2, 3} defined as R = {(1,1),(2,2),(3,3),(1,2),(2,1)} is an equivalence relation.
Reflexive: (1,1),(2,2),(3,3) in R. Yes.
Symmetric: (1,2) in R and (2,1) in R. Yes.
Transitive: (1,2) and (2,1) are in R, so we need (1,1) — it is there. All checks pass. Yes.
R is an equivalence relation.
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Common mistakes
- Confusing codomain with range. Range is the actual set of outputs; codomain is the declared target set. A function is onto only if range = codomain.
- Forgetting that composition gof means g applied after f, i.e., g(f(x)).
- Checking only one condition (say, one-one) to call a function bijective — both conditions must hold.
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Summary
Relations generalise the idea of pairwise connections between elements. Functions are a special type of relation. Bijective functions have inverses. Composition and binary operations are central tools in this chapter.