Ch 01 - Relations and Functions12th Mathematics

• Definition: A relation $R$ from set $A$ to set $B$ is an arbitrary subset of the Cartesian product $A \times B$.
• Notation: $R \subseteq A \times B$. If an element $a$ is related to $b$, we write $aRb$ or $(a, b) \in R$.

• Definition: A relation $R$ in a set $A$ is an empty relation if no element of $A$ is related to any element of $A$.
• Notation: $R = \emptyset \subset A \times A$.

• Definition: A relation $R$ in a set $A$ is a universal relation if each element of $A$ is related to every element of $A$.
• Notation: $R = A \times A$.

• Identity Relation ($I_A$): Contains ONLY the diagonal pairs $(a,a)$. No other pairs are allowed. $I_A = \{(a,a): a \in A\}$.
• Reflexive Relation: MUST contain all diagonal pairs $(a,a)$ to satisfy the core rule, but it MAY also contain other distinct non-diagonal pairs $(a,b)$.
• Distinction Conclusion: Every Identity Relation is a reflexive relation, but a reflexive relation is NOT necessarily an Identity Relation.

Component Anatomy:

• $a$: Any arbitrary element belonging to the set $A$.
• $R$: The relation defined on the set.

Boundary Conditions / Domain:

• The condition must hold true for every single element $a \in A$. If even one element maps to something else but not itself, it is not reflexive.

Component Anatomy:

• $(a, b)$: An ordered pair that already exists in the relation $R$.
• $(b, a)$: The reversed ordered pair that must also exist in $R$.

Boundary Conditions / Domain:

• $a, b \in A$.
• The check is conditional: it is only required if $(a, b)$ is already present in $R$.

Component Anatomy:

• $(a, b)$ and $(b, c)$: Two linked ordered pairs existing in the relation $R$.
• $(a, c)$: The direct link pair that must exist to satisfy transitivity.

Boundary Conditions / Domain:

• $a, b, c \in A$.
• This is a conditional implication: if the premise is true, the conclusion must follow.

• Concept: A relation is only non-transitive if you can find the premise $(a,b)$ and $(b,c)$ in $R$, but the conclusion $(a,c)$ is missing.
• Conclusion: If $R$ contains $(a,b)$ but absolutely no pair starting with $b$, the relation is transitive by default because the condition to disprove it cannot be met.

• The Flawed Logic: A common student trap is thinking: 'If $(a,b) \in R$, symmetry gives $(b,a) \in R$. Transitivity on $(a,b)$ and $(b,a)$ gives $(a,a) \in R$, so it must be reflexive!'
• The Trap (Boundary Condition): This logic only proves $(a,a) \in R$ for elements $a$ that are already related to some $b$. If an element $a \in A$ is completely isolated and does not appear in ANY ordered pair in $R$, the chain never starts, and $(a,a)$ will be missing.
• Conclusion: Reflexivity cannot be assumed; it must ALWAYS be checked independently for every single element in the parent set.

• Definition: A relation $R$ on a set $A$ is an equivalence relation if it satisfies three properties simultaneously:
  1. Reflexive: $(a, a) \in R$
  2. Symmetric: $(a, b) \in R \implies (b, a) \in R$
  3. Transitive: $(a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R$

• Smallest Equivalence Relation: The Identity Relation ($I_A = \{(a,a): a \in A\}$). It contains the absolute minimum number of elements required to satisfy reflexivity, and trivially satisfies symmetry and transitivity since there are no non-diagonal elements to trigger violations.
• Largest Equivalence Relation: The Universal Relation ($R = A \times A$). It links every element to every other element, easily satisfying all three properties by sheer brute force of inclusion.

• Definition: For an equivalence relation $R$ on set $A$, the equivalence class of an element $a$, denoted by $[a]$, is the set of all elements in $A$ related to $a$.
• Property: $[a] = \{x \in A: (x, a) \in R\}$.
• Partition: Equivalence classes are mutually disjoint or identical, and their union forms the entire set $A$.

An equivalence relation on a set $X$ induces a partition. This partition must satisfy three strict axioms:

• Axiom 1 (Non-Empty): All equivalence classes are non-empty subsets of $X$.
• Axiom 2 (Mutually Exclusive / Disjoint): Any two equivalence classes are either exactly identical or completely disjoint. Mathematically, $[a] \cap [b] = \emptyset$ if $a \not\sim b$.
• Axiom 3 (Exhaustive): The union of all distinct equivalence classes perfectly reconstructs the entire original set $X$. Mathematically, $\bigcup [a] = X$.

• Concept: The number of distinct equivalence relations on a set of $n$ elements is given by the Bell number $B_n$.
• Values to Memorize:
  • $B_1 = 1$
  • $B_2 = 2$
  • $B_3 = 5$
  • $B_4 = 15$
  • $B_5 = 52$

Component Anatomy:

• $(a,b)$ and $(c,d)$: Elements belonging to $\mathbb{N} \times \mathbb{N}$.
• $ad = bc$: The equivalence rule, which essentially models the equality of fractions $\frac{a}{b} = \frac{c}{d}$.

Boundary Conditions / Domain:

• $a, b, c, d \in \mathbb{N}$. Since the domain is natural numbers, denominators can never be zero.

Component Anatomy:

• The goal is to prove $(a,b) R (c,d)$ and $(c,d) R (e,f) \implies (a,b) R (e,f)$.

Boundary Conditions / Domain:

• Variables belong to $\mathbb{N}$. Division is safely permitted conceptually during the proof because $b, d, f \neq 0$.

Component Anatomy:

• $m$: The cardinality (number of elements) of the domain set $A$.
• $n$: The cardinality of the codomain set $B$.

Boundary Conditions / Domain:

• Sets $A$ and $B$ must be finite non-empty sets.

Component Anatomy:

• $n$: The total number of elements in the set $A$.

Boundary Conditions / Domain:

• Set $A$ is a finite set containing $n$ elements.

Component Anatomy:

• $n$: The total number of elements in the set $A$.

Boundary Conditions / Domain:

• Set $A$ is a finite set containing $n$ elements.

Component Anatomy:

• $f(x_1)$ and $f(x_2)$: The respective images of elements $x_1$ and $x_2$ under function $f$.
• $x_1, x_2$: Elements in the domain $X$.

Boundary Conditions / Domain:

• $x_1, x_2 \in X$.
• Alternatively, the contrapositive must hold: $x_1 \neq x_2 \implies f(x_1) \neq f(x_2)$.

• Concept: A visual/graphical method to determine if a continuous function is one-one.
• Execution: Draw horizontal lines across the entire coordinate plane.
• Condition: If every horizontal line intersects the graph of the function at most one point, the function is one-one (injective).

Component Anatomy:

• $y$: An arbitrary element in the codomain $Y$.
• $x$: An element in the domain $X$.

Boundary Conditions / Domain:

• $y \in Y, x \in X$.
• The Range of $f$ must exactly equal the entire Codomain $Y$.

• Concept: A graphical method to determine if a function $f: X \to Y$ is onto, heavily reliant on the explicitly defined Codomain.
• Execution: Draw horizontal lines $y = c$ for every value $c$ lying strictly within the defined Codomain $Y$.
• Condition: If every such horizontal line intersects the graph at at least one point, the function is onto (surjective).
• Trap: The test is completely meaningless without knowing the codomain. A graph might 'look' onto $\mathbb{R}$, but if the codomain is restricted artificially, the boundary conditions change.

• Definition: A function $f: X \to Y$ is bijective if it explicitly satisfies two conditions:
  1. One-One (Injective): Distinct inputs map to distinct outputs.
  2. Onto (Surjective): Every element in the codomain has a pre-image in the domain.
• Significance: Bijective functions represent a perfect 1:1 mapping and are the only functions that possess an inverse ($f^{-1}$).

Component Anatomy:

• $f(x) = \frac{ax+b}{cx+d}$: A standard rational linear function.
• $y$: An arbitrary element in the defined codomain.

Boundary Conditions / Domain:

• Domain: $\mathbb{R} - \{-\frac{d}{c}\}$. The denominator cannot be zero.
• Codomain constraint for surjectivity: $\mathbb{R} - \{\frac{a}{c}\}$.

Component Anatomy:

• $x = \frac{dy - b}{a - cy}$: The isolated domain element mapped from codomain element $y$.
• $a, c$: The leading coefficients of the original rational function $f(x) = \frac{ax+b}{cx+d}$.

Boundary Conditions / Domain:

• Domain Constraint: $x \neq -\frac{d}{c}$.
• Codomain Constraint: $y \neq \frac{a}{c}$. The function must explicitly exclude the horizontal asymptote from its codomain to be perfectly surjective.

Component Anatomy:

• $m$: The number of elements in the domain set $A$.
• $n$: The number of elements in the codomain set $B$.

Boundary Conditions / Domain:

• $A$ and $B$ must be finite, non-empty sets.

Component Anatomy:

• $m$: Elements in domain $A$.
• $n$: Elements in codomain $B$.
• $^nP_m$: Mathematical permutation of $n$ items taken $m$ at a time.

Boundary Conditions / Domain:

• $m \le n$. If the domain is larger than the codomain ($m > n$), it is impossible to map each element uniquely (Pigeonhole Principle), so the answer is strictly 0.

Component Anatomy:

• $n$: The cardinality of the codomain set $B$.
• $m$: The cardinality of the domain set $A$.
• $k$: The index of summation representing the number of excluded elements in the codomain.

Boundary Conditions / Domain:

• $m \ge n$. If the domain has fewer elements than the codomain ($m < n$), it is mathematically impossible to map onto all elements, and the number of onto functions is exactly $0$.

Component Anatomy:

• $n^m$: Total possible functions without constraints.
• $\binom{n}{1}$: Ways to choose 1 codomain element to completely exclude.

Boundary Conditions / Domain:

• This algorithm only yields a non-zero, meaningful result if the domain is greater than or equal to the codomain ($m \ge n$).

Component Anatomy:

• $m$: The total number of elements in the domain set $A$.
• $2$: The fixed number of elements in the codomain set $B$.

Boundary Conditions / Domain:

• $m \ge 2$. If $m < 2$, no onto function can exist.

Component Anatomy:

• $n$: The cardinality of the set $A$.
• $n!$: n factorial.

Boundary Conditions / Domain:

• A bijection from $A$ to $B$ is only mathematically possible if both sets have the exact same cardinality ($|A| = |B| = n$).

• Definition: For two functions $f: A \to B$ and $g: B \to C$, their composition is defined as $g \circ f: A \to C$.
• Mapping: $(g \circ f)(x) = g(f(x))$ for all $x \in A$.
• Mechanism: The output of the inner function $f$ serves as the direct input for the outer function $g$.

• Rule: Function composition is generally not commutative.
• Distinction: $g \circ f \neq f \circ g$.
• Explanation: $(g \circ f)(x) = g(f(x))$ applies $f$ first, then $g$. Conversely, $(f \circ g)(x) = f(g(x))$ applies $g$ first. The domains, codomains, and rules of mapping are fundamentally different depending on the order.

Component Anatomy:

• $\text{Domain}(f)$: The set of valid initial inputs for the inner function.
• $\text{Domain}(g)$: The set of valid inputs for the outer function.

Boundary Conditions / Domain:

• The output of $f$ must not violate any domain restrictions of $g$ (e.g., negative numbers under a square root or zero in a denominator).

• If $g \circ f$ is One-One (Injective): The inner function $f$ MUST be one-one. (Proof logic: If $f$ wasn't one-one, two inputs would collapse into one, and $g$ could never un-collapse them. However, $g$ need not be one-one).
• If $g \circ f$ is Onto (Surjective): The outer function $g$ MUST be onto. (Proof logic: The final output comes entirely from $g$. If $g$ can't reach a value, $g \circ f$ can't either. However, $f$ need not be onto).
• Corollary: If $g \circ f$ is bijective, $f$ is guaranteed to be one-one and $g$ is guaranteed to be onto.

Component Anatomy:

• $f$: The original function mapping from $X$ to $Y$.
• $g$: The inverse function mapping from $Y$ to $X$.
• $I_X, I_Y$: The identity functions on sets $X$ and $Y$.

Boundary Conditions / Domain:

• If such a $g$ exists, it is strictly unique and specifically denoted as $f^{-1}$.

• Theorem: A function $f: X \to Y$ is invertible if and only if $f$ is a bijective function.
• Why?
  • If it is not one-one, multiple inputs map to a single output, making reverse mapping ambiguous.
  • If it is not onto, some elements in $Y$ will have no destination in $X$ when reversing.

Component Anatomy:

• $f$ and $g$: Two invertible functions with compatible domains and codomains.

Boundary Conditions / Domain:

• Both $f$ and $g$ must individually be bijections.
• Notice the order strictly flips! You unwrap the outermost operation first.