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Sets and Functions - Relations, Domain, Co-domain, and Range

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Ordered Pairs and Cartesian Product: An ordered pair (a,b)(a, b) is a pair of elements grouped together in a specific order, where (a,b)≠(b,a)(a, b) \neq (b, a) unless a=ba = b. The Cartesian Product A×BA \times B is the set of all possible ordered pairs where the first element belongs to set AA and the second to set BB. Visually, this can be represented as a rectangular grid of points on a coordinate plane, where the elements of AA are plotted on the horizontal axis and elements of BB on the vertical axis.

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Definition of a Relation: A relation RR from a non-empty set AA to a non-empty set BB is a subset of the Cartesian product A×BA \times B. The relation is derived by describing a relationship between the first element and the second element of the ordered pairs. Visually, a relation is often depicted using an 'Arrow Diagram' where elements of set AA and set BB are placed in two separate ovals, and arrows are drawn to connect related elements.

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Domain of a Relation: The set of all first elements of the ordered pairs in a relation RR from set AA to set BB is called the domain of the relation RR. Mathematically, Domain={x:(x,y)∈R}\text{Domain} = \{x : (x, y) \in R\}. In a visual mapping diagram, the domain consists of all elements in the first set (the source) that have at least one arrow originating from them.

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Range of a Relation: The set of all second elements in a relation RR from set AA to set BB is called the range of the relation. Formally, Range={y:(x,y)∈R}\text{Range} = \{y : (x, y) \in R\}. Visually, in an arrow diagram, the range is the specific collection of elements in the second set (the target) that are pointed to by arrows.

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Co-domain of a Relation: When a relation is defined from set AA to set BB, the entire set BB is called the co-domain of the relation RR. It is important to note that the Range⊆Co-domain\text{Range} \subseteq \text{Co-domain}. Visually, the co-domain is represented by the entire second oval or 'target' container, regardless of whether every element inside it is connected to an element in the domain.

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Total Number of Relations: If set AA has mm elements and set BB has nn elements, then the number of elements in A×BA \times B is mnmn. The total number of possible relations that can be defined from AA to BB is 2mn2^{mn}, because every relation is a subset of the Cartesian product set. This exponential growth illustrates how many different ways elements from two sets can be linked.

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Inverse Relation: For any relation RR from AA to BB, the inverse relation R−1R^{-1} is a relation from BB to AA defined by R−1={(b,a):(a,b)∈R}R^{-1} = \{(b, a) : (a, b) \in R\}. Visually, this is equivalent to taking an arrow diagram and reversing the direction of every single arrow. The domain of RR becomes the range of R−1R^{-1}, and the range of RR becomes the domain of R−1R^{-1}.

📐Formulae

n(A×B)=n(A)×n(B)n(A \times B) = n(A) \times n(B)

Total number of relations from A to B=2n(A)⋅n(B)\text{Total number of relations from } A \text{ to } B = 2^{n(A) \cdot n(B)}

R⊆(A×B)R \subseteq (A \times B)

Domain(R)={a∈A:(a,b)∈R for some b∈B}\text{Domain}(R) = \{a \in A : (a, b) \in R \text{ for some } b \in B\}

Range(R)={b∈B:(a,b)∈R for some a∈A}\text{Range}(R) = \{b \in B : (a, b) \in R \text{ for some } a \in A\}

R−1={(y,x):(x,y)∈R}R^{-1} = \{(y, x) : (x, y) \in R\}

💡Examples

Problem 1:

Let A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}. Define a relation RR from AA to AA by R={(x,y):y=x+1}R = \{(x, y) : y = x + 1\}. Write down the relation in roster form and find its domain, co-domain, and range.

Solution:

Step 1: Identify pairs satisfying y=x+1y = x + 1 where x,y∈Ax, y \in A. When x=1,y=2x=1, y=2; When x=2,y=3x=2, y=3; When x=3,y=4x=3, y=4; When x=4,y=5x=4, y=5; When x=5,y=6x=5, y=6; When x=6,y=7x=6, y=7 (But 7∉A7 \notin A, so this pair is excluded). Step 2: Write in roster form: R={(1,2),(2,3),(3,4),(4,5),(5,6)}R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}. Step 3: Extract the Domain: {1,2,3,4,5}\{1, 2, 3, 4, 5\}. Step 4: Extract the Range: {2,3,4,5,6}\{2, 3, 4, 5, 6\}. Step 5: The Co-domain is the entire set AA: {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}.

Explanation:

To solve this, we test each element of the first set in the given equation y=x+1y = x+1. If the resulting yy is also in the set, the pair is part of the relation. The domain is the set of starting values, and the range is the set of resulting values.

Problem 2:

Given A={x,y,z}A = \{x, y, z\} and B={1,2}B = \{1, 2\}. Find the total number of relations from AA to BB.

Solution:

Step 1: Find the number of elements in set AA: n(A)=3n(A) = 3. Step 2: Find the number of elements in set BB: n(B)=2n(B) = 2. Step 3: Calculate the number of elements in the Cartesian product A×BA \times B: n(A×B)=n(A)×n(B)=3×2=6n(A \times B) = n(A) \times n(B) = 3 \times 2 = 6. Step 4: Apply the formula for the number of relations: 2n(A×B)=26=642^{n(A \times B)} = 2^6 = 64.

Explanation:

Since every relation is a subset of the Cartesian product, the total number of relations is equal to the number of subsets of A×BA \times B, which is 2n(A×B)2^{n(A \times B)}.