Recall that the concepts of relations, functions, domain, co-domain, and range are fundamental to algebra and calculus. In Class XII, we extend these concepts to study specific types of relations and functions, composition of functions, invertible functions, and binary operations.
A relation $R$ from a set $A$ to a set $B$ is mathematically defined as an arbitrary subset of the Cartesian product $A \times B Concordance$.
Relations can be classified based on their structural properties on a set $A$. Let $R$ be a relation in a set $A$ (i.e., a subset of $A \times A$).
| Relation Type | Mathematical Condition | Systemic Role & Limits |
|---|---|---|
| Empty Relation | $R = \phi \subset A \times A$ | No element of $A$ is related to any element of $A$. |
| Universal Relation | $R = A \times A$ | Each element of $A$ is related to every element of $A$. |
| Reflexive Relation | $(a, a) \in R, \forall a \in A$ | Every element is related to itself. |
| Symmetric Relation | $(a_1, a_2) \in R \implies (a_2, a_1) \in R, \forall a_1, a_2 \in A$ | Order of elements in the relation is reversible. |
| Transitive Relation | $(a_1, a_2) \in R \text{ and } (a_2, a_3) \in R \implies (a_1, a_3) \in R, \forall a_1, a_2, a_3 \in A$ | Relational connectivity is chain-transmissible. |
| Equivalence Relation | Reflexive, Symmetric, and Transitive | Partitions the set into disjoint equivalence classes. |
Given an arbitrary equivalence relation $R$ in a set $X$, $R$ divides $X$ into mutually disjoint subsets $A_i$ called partitions or subdivisions of $X$ satisfying:
The subsets $A_i$ are called equivalence classes. The equivalence class containing an element $a$ is denoted by $[a]$.
Mathematical Problem Statement: Show that the relation $R$ in the set $\mathbf{Z}$ of integers given by $R = \{(a, b) : 2 \text{ divides } a - b\}$ is an equivalence relation.
Step-by-Step Algorithmic Solution Trace:
- Reflexivity: For any $a \in \mathbf{Z}$, $a - a = 0$, which is divisible by $2$. Thus, $(a, a) \in R$ for all $a \in \mathbf{Z}$. Hence, $R$ is reflexive.
- Symmetry: Let $(a, b) \in R$. This $\implies 2$ divides $a - b$, so $a - b = 2k$ for some integer $k$. Then, $b - a = -2k = 2(-k)$, which means $2$ divides $b - a$. Thus, $(b, a) \in R$. Hence, $R$ is symmetric.
- Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. This means $a - b = 2k$ and $b - c = 2m$ for some integers $k, m$. Adding these equations gives $a - c = (a - b) + (b - c) = 2k + 2m = 2(k + m)$. Since $k + m$ is an integer, $2$ divides $a - c$, implying $(a, c) \in R$. Hence, $R$ is transitive.
Final Evaluation Output: Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation in $\mathbf{Z}$.
Functions are special types of relations. We classify functions based on how elements of the domain map to the co-domain.
| Function Type | Mathematical Condition | Systemic Role & Evaluation Limits |
|---|---|---|
| One-One (Injective) | $f(x_1) = f(x_2) \implies x_1 = x_2, \forall x_1, x_2 \in X$ | Distinct elements in the domain have distinct images in the co-domain. |
| Many-One | $\exists x_1, x_2 \in X \text{ such that } x_1 \neq x_2 \text{ but } f(x_1) = f(x_2)$ | Two or more distinct elements in the domain map to the same image. |
| Onto (Surjective) | $\forall y \in Y, \exists x \in X \text{ such that } f(x) = y$ | Every element in the co-domain is the image of at least one element in the domain (Range = Co-domain). |
| One-One and Onto (Bijective) | Both injective and surjective | Establishes a perfect one-to-one correspondence between domain and co-domain. |
Mathematical Problem Statement: Prove that the function $f : \mathbf{R} \rightarrow \mathbf{R}$, given by $f(x) = 2x$, is one-one and onto.
Step-by-Step Algorithmic Solution Trace:
- One-One (Injectivity): Let $x_1, x_2 \in \mathbf{R}$ such that $f(x_1) = f(x_2)$. This $\implies 2x_1 = 2x_2$. Dividing both sides by $2$, we get $x_1 = x_2$. Thus, $f$ is one-one.
- Onto (Surjectivity): Let $y \in \mathbf{R}$ (co-domain). We want to find $x \in \mathbf{R}$ (domain) such that $f(x) = y$. Setting $2x = y$ gives $x = \frac{y}{2}$. Since $y \in \mathbf{R}$, $x = \frac{y}{2} \in \mathbf{R}$. Evaluating $f(x)$, we get $f\left(\frac{y}{2}\right) = 2\left(\frac{y}{2}\right) = y$. Thus, $f$ is onto.
Final Evaluation Output: The function $f(x) = 2x$ is bijective (both one-one and onto).
We can combine functions using composition and define the inverse of a function under specific conditions.
Let $f : A \rightarrow B$ and $g : B \rightarrow C$ be two functions. The composition of $f$ and $g$, denoted by $g \circ f$, is defined as the function $g \circ f : A \rightarrow C$ given by:
$$g \circ f(x) = g(f(x)), \quad \forall x \in A$$A function $f : X \rightarrow Y$ is defined to be invertible if there exists a function $g : Y \rightarrow X$ such that:
$$g \circ f = I_X \quad \text{and} \quad f \circ g = I_Y$$where $I_X$ and $I_Y$ are identity functions on sets $X$ and $Y$ respectively. The function $g$ is called the inverse of $f$ and is denoted by $f^{-1}$.
Mathematical Problem Statement: Let $f : \mathbf{N} \rightarrow Y$ be a function defined as $f(x) = 4x + 3$, where $Y = \{y \in \mathbf{N} : y = 4x + 3 \text{ for some } x \in \mathbf{N}\}$. Show that $f$ is invertible. Find the inverse.
Step-by-Step Algorithmic Solution Trace:
- Let $y$ be an arbitrary element of $Y$. By definition of $Y$, $y = 4x + 3$ for some $x \in \mathbf{N}$.
- Solve for $x$ in terms of $y$: $y - 3 = 4x \implies x = \frac{y - 3}{4}$.
- Define a function $g : Y \rightarrow \mathbf{N}$ by $g(y) = \frac{y - 3}{4}$.
- Compute $g \circ f(x)$: $g(f(x)) = g(4x + 3) = \frac{(4x + 3) - 3}{4} = \frac{4x}{4} = x = I_{\mathbf{N}}(x)$.
- Compute $f \circ g(y)$: $f(g(y)) = f\left(\frac{y - 3}{4}\right) = 4\left(\frac{y - 3}{4}\right) + 3 = (y - 3) + 3 = y = I_Y(y)$.
Final Evaluation Output: Since $g \circ f = I_{\mathbf{N}}$ and $f \circ g = I_Y$, the function $f$ is invertible, and its inverse is $f^{-1}(y) = \frac{y - 3}{4}$.
Historical Chronicle: Evolution of the Concept of Function