RELATIONS AND FUNCTIONS-Notes

Let \(A\) and \(B\) be two non-empty sets. The Cartesian product of \(A\) and \(B\), denoted by \(A\times B\), is defined as the set of all ordered pairs \((ab)\) such that \(a\in A\) and \(b\in B\). Symbolically, \[A\times B=\{(a,b): a\in A,\mathrm{\;} b\in B\}\]

Here, the order of elements in each pair is important; in general, \(\left(a,b)\neq(b,a\right)\).

If set \(A\) has \(m\) elements and set \(B\) has \(n\) elements, then the Cartesian product \(A\times B\) contains exactly \(m\times n\) ordered pairs. This shows that the size of the Cartesian product depends on the sizes of both sets. For example, if \(A=\{1,2\}\) and \(B=\{x,y,z\}\), then \(A\times B\) consists of six ordered pairs, each formed by combining one element of \(A\) with one element of \(B\).

It is important to note that, in general, \(A\times B\neq B\times A\), since the order of elements in ordered pairs differs. However, if \(A=B\), then the Cartesian product \(A\times A\) is called the Cartesian square of \(A\). In this case, each ordered pair has both elements from the same set.

If either of the sets \(A\) or \(B\) is empty, then their Cartesian product is also an empty set. Thus, \[A\times\emptyset=\emptyset \text{ and } \emptyset\times B=\emptyset\]

Geometrically, Cartesian products play a crucial role in describing points in the plane. For example, the set \(\mathbb{R}\times\mathbb{R}\) represents the collection of all ordered pairs of real numbers and corresponds to all points in the Cartesian plane.

The concept of the Cartesian product provides the necessary framework for defining relations as subsets of \(A\times B \) and functions as special kinds of relations. Therefore, understanding Cartesian products is essential for a clear and logical development of relations and functions.

Remarks

• Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal.
• If there are \(p\) elements in \(A\) and \(q\) elements in \(B\), then there will be \(pq\) elements in \(A \times B\), i.e., if \(n(A) = p\) and \(n(B) = q\), then \(n(A \times B) = pq\).
• If \(A\) and \(B\) are non-empty sets and either \(A\) or \(B\) is an infinite set, then so is \(A \times B\).
• \(A \times A \times A = \{(a, b, c) : a, b, c \in A\}\). Here (a, b, c) is called an ordered triplet.

In mathematics, the idea of a relation arises when we wish to describe a meaningful connection between elements of two sets. Instead of considering elements in isolation, relations allow us to formally express how elements of one set are associated with elements of another set. This concept forms a natural extension of the Cartesian product discussed earlier and serves as a foundational step toward the study of functions.

Let \(A\) and \(B\) be two non-empty sets. A relation from set \(A\) to set \(B\) is defined as any subset of the Cartesian product \(A \times B\). If \(R\) is a relation from \(A\) to \(B\), then \[ R \subseteq A \times B. \] If an ordered pair \((a,b)\) belongs to \(R\), we say that “\(a\) is related to \(b\)” and write it symbolically as \(a \, R \, b\), where \(a \in A\) and \(b \in B\).

The definition itself follows directly from the construction of the Cartesian product. Since \(A \times B\) contains all possible ordered pairs \((a,b)\) with \(a \in A\) and \(b \in B\), selecting some of these ordered pairs according to a specific rule naturally gives rise to a relation. Thus, every relation is fundamentally based on ordered pairing.

To understand the derivation of the total number of possible relations from a finite set \(A\) to a finite set \(B\), let us assume that \(A\) has \(m\) elements and \(B\) has \(n\) elements. Then the Cartesian product \(A \times B\) has \(m \times n\) ordered pairs. Any relation is a subset of this Cartesian product. The number of subsets of a set containing \(m n\) elements is given by \[ \begin{aligned} \boxed{\;\text{Number of relations} = 2^{mn}\;} \end{aligned} \] This result shows that even for small sets, the number of possible relations can be very large.

Relations can also be defined within a single set. If \(A\) is a non-empty set, then any subset of \(A \times A\) is called a relation on \(A\). Such relations are particularly important in mathematics, as they help describe properties like equality, order, and equivalence among elements of the same set.

An important theoretical aspect of relations is their description through a rule or condition. Often, a relation is expressed as \[ R = \{(a,b) \in A \times B \mid \text{a given condition holds}\}. \] This form highlights that a relation does not merely list pairs but captures a logical connection between elements of two sets.

From a conceptual point of view, relations provide a bridge between set theory and algebra. They allow us to systematically study associations, dependencies, and mappings. In fact, a function is nothing but a special type of relation with additional restrictions. Hence, mastering the idea of relations is essential for a clear understanding of functions, mappings, and many advanced structures encountered later in mathematics.

Image

A relation \(R\) from a non-empty set \(A\) to a non-empty set \(B\) is a subset of the cartesian product \(A \times B\). The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in \(A \times B\). The second element is called the image of the first element.

Domain

The set of all first elements of the ordered pairs in a relation \(R\) from a set \(A\) to a set \(B\) is called the domain of the relation \(R\).

Codomain

The set of all second elements in a relation \(R\) from a set \(A\) to a set \(B\) is called the range of the relation \(R\). The whole set \(B\) is called the codomain of the relation \(R\). Note that range \(\subset\) codomain.

In mathematics, the concept of a function emerges from the idea of a relation that assigns outputs to inputs in a precise and unambiguous manner. Functions are fundamental tools used to describe dependence of one quantity on another and form the core of mathematical modelling in algebra, geometry, and calculus.

Let \(A\) and \(B\) be two non-empty sets. A function from \(A\) to \(B\) is a relation \(f\) such that every element of \(A\) is associated with exactly one element of \(B\). Symbolically, a function is written as \[ f : A \rightarrow B, \] and for each \(a \in A\), there exists a unique \(b \in B\) such that \((a,b) \in f\). This element \(b\) is called the image of \(a\) under the function \(f\) and is denoted by \(f(a)\).

The necessity of uniqueness in the definition of a function can be understood logically. Suppose an element \(a \in A\) were related to two different elements \(b_1, b_2 \in B\). Then the output corresponding to \(a\) would be ambiguous, which contradicts the very purpose of a function as a rule of assignment. Hence, uniqueness of the image is an essential requirement.

The set \(A\) is called the domain of the function, while the set \(B\) is called the codomain. The set of all actual images of elements of \(A\) under \(f\) is called the range of the function and is a subset of the codomain. This can be expressed as \[ \text{Range}(f) = \{ f(a) \mid a \in A \} \subseteq B. \]

We now derive the number of possible functions from a finite set \(A\) to a finite set \(B\). Let \(A\) contain \(m\) elements and \(B\) contain \(n\) elements. For each element of \(A\), there are \(n\) independent choices of images in \(B\). Hence, the total number of functions is obtained as \[ \begin{aligned} \text{Number of functions} &= \underbrace{n \times n \times \cdots \times n}_{m \text{ times}} \\ &= n^m. \end{aligned} \] This result highlights the structured nature of functions in comparison to general relations.

An important theoretical result concerns the equality of functions. Two functions \(f : A \rightarrow B\) and \(g : A \rightarrow B\) are said to be equal if and only if they assign the same image to every element of the domain. That is, \[ \begin{aligned} f = g &\iff f(a) = g(a) \text{ for all } a \in A. \end{aligned} \] This condition follows directly from the definition of a function as a set of ordered pairs.

Functions may also be represented in multiple ways, such as by listing ordered pairs, by a verbal rule, or by an algebraic expression. Regardless of the form of representation, the defining property remains the same: each element of the domain must have one and only one image in the codomain.

After understanding the idea of a function as a rule of correspondence, it becomes necessary to study how functions can be combined and manipulated algebraically. The algebra of real functions deals with operations performed on functions whose domain and range consist of real numbers. These operations closely resemble the algebraic operations on real numbers and allow us to construct new functions from given ones in a systematic manner.

Let \(f\) and \(g\) be two real-valued functions defined on a common domain \(D \subseteq \mathbb{R}\). The sum, difference, and product of \(f\) and \(g\) are defined pointwise. That is, for every \(x \in D\), \[ \begin{aligned} (f+g)(x) &= f(x) + g(x), \\ (f-g)(x) &= f(x) - g(x), \\ (fg)(x) &= f(x)\, g(x) \end{aligned} \] Each of these definitions ensures that the resulting expressions assign a unique real number to every element of the domain.

The quotient of two real functions is defined with an additional restriction. If \(g(x) \neq 0\) for all \(x \in D\), then the quotient function is given by \[ \begin{aligned} \left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)}, \qquad x \in D. \end{aligned} \] If \(g(x) = 0\) for some values of \(x\), those values must be excluded from the domain of the quotient function. This condition ensures that the function remains well-defined.

We now examine the domain of algebraic combinations of functions. Although \(f\) and \(g\) may be defined on a common set \(D\), the domain of functions such as \(\frac{f}{g}\) may be smaller. Hence, the domain of any algebraic combination is taken as the largest possible subset of \(\mathbb{R}\) for which the defining expression is meaningful. This aspect is crucial while working with real functions.

The algebra of real functions satisfies several properties that closely parallel the properties of real numbers. For any real functions \(f\), \(g\), and \(h\) defined on a common domain, we have \[ \begin{aligned} f + g &= g + f \\\scriptsize \text{(comm} & \scriptsize\text{utativity)}, \\ f + (g + h) &= (f + g) + h \\\scriptsize \text{(associ} & \scriptsize\text{ativity)}, \\ f(g + h) &= fg + fh \\\scriptsize \text{(distr} & \scriptsize\text{ibutivity)} \end{aligned} \] These properties can be proved by evaluating both sides at an arbitrary element \(x\) in the domain and using the corresponding properties of real numbers.

For example, to prove commutativity of addition, let \(x\) be any element of the domain. Then \[ \begin{aligned} (f+g)(x) &= f(x) + g(x) \\ &= g(x) + f(x) \\ &= (g+f)(x) \end{aligned} \] Since this holds for all \(x\) in the domain, we conclude that \(f + g = g + f\).

An important aspect of the algebra of real functions is the existence of identity functions. The zero function, defined by \(z(x) = 0\) for all \(x\), acts as the additive identity, since \[ \begin{aligned} (f+z)(x) = f(x) \end{aligned} \] Similarly, the constant function \(u(x) = 1\) acts as the multiplicative identity, satisfying \[ \begin{aligned} (fu)(x) = f(x) \end{aligned} \]

Thus, the algebra of real functions provides a structured framework in which functions behave much like real numbers under basic operations. This framework is essential in simplifying expressions, analyzing functional relationships, and preparing the ground for more advanced concepts such as composite functions and inverse functions. In the NCERT syllabus, it plays a key role in strengthening the student’s understanding of functions as mathematical objects that can be combined, transformed, and studied systematically.

If \((x + 1, y – 2) = (3,1)\), find the values of \(x\) and \(y\).

Solution

Given that the ordered pairs \((x+1,\, y-2)\) and \((3,\,1)\) are equal, we use the fundamental property of ordered pairs, which states that two ordered pairs are equal if and only if their corresponding components are equal.

Accordingly, we equate the first components and the second components separately. This gives \[ \begin{aligned} x+1 &= 3,\\ y-2 &= 1 \end{aligned} \]

Solving the first equation, we obtain \[ \begin{aligned} x+1 &= 3 \\ x &= 2 \end{aligned} \]

Similarly, solving the second equation, we get \[ \begin{aligned} y-2 &= 1 \\ y &= 3 \end{aligned} \]

Hence, the required values of the variables are \[ (x,\, y) = (2,\, 3) \]

If \(P = \{a, b, c\}\) and \(Q = \{r\}\), form the sets \(P \times Q\) and \(Q \times P\). Are these two products equal?

Solution

Given \(P = \{a, b, c\}\) and \(Q = \{r\}\), we first form the Cartesian product \(P \times Q\). By definition, the Cartesian product consists of all ordered pairs whose first element belongs to \(P\) and whose second element belongs to \(Q\).

Thus, \[ \begin{aligned} P \times Q &= \{a, b, c\} \times \{r\} \\ &= \{(a,r), (b,r), (c,r)\} \end{aligned} \]

Next, we form the Cartesian product \(Q \times P\), which contains all ordered pairs whose first element belongs to \(Q\) and whose second element belongs to \(P\).

Hence, \[ \begin{aligned} Q \times P &= \{r\} \times \{a, b, c\} \\ &= \{(r,a), (r,b), (r,c)\} \end{aligned} \]

Since the ordered pairs in \(P \times Q\) differ from those in \(Q \times P\) in the order of their elements, we conclude that \[ P \times Q \neq Q \times P \]

If \(P = \{1, 2\}\), form the set \(P \times P \times P\).

Solution

Given \(P = \{1, 2\}\), we are required to form the Cartesian product \(P \times P \times P\). We proceed step by step by first evaluating the Cartesian product of \(P\) with itself.

The Cartesian product \(P \times P\) is given by \[ \begin{aligned} P \times P &= \{1,2\} \times \{1,2\} \\ &= \{(1,1), (1,2), (2,1), (2,2)\} \end{aligned} \]

Next, we form the Cartesian product of the set \(P \times P\) with \(P\). Each ordered pair in \(P \times P\) is combined with each element of \(P\) to form ordered triples.

Thus, \[ \begin{aligned} P \times P \times P &= \{(1,1), (1,2), (2,1), (2,2)\} \times \{1,2\} \\ &= \{(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)\} \end{aligned} \]

Hence, the set \(P \times P \times P\) consists of all ordered triples whose components are elements of the set \(P = \{1,2\}\).

Let \(A = \{1, 2\}\) and \(B = \{3, 4\}\). Find the number of relations from \(A\) to \(B\).

Solution

Given \(A = \{1, 2\}\) and \(B = \{3, 4\}\), we first note the number of elements in each set. Clearly, \[ \begin{aligned} n(A) &= 2, \\ n(B) &= 2 \end{aligned} \]

The total number of ordered pairs in the Cartesian product \(A \times B\) is obtained by multiplying the number of elements of \(A\) and \(B\). Hence, \[ \begin{aligned} n(A \times B) &= n(A)\, n(B) \\ &= 2 \times 2 \\ &= 4. \end{aligned} \]

A relation from \(A\) to \(B\) is any subset of \(A \times B\). Since a set with \(4\) elements has \(2^4\) subsets, the total number of relations from \(A\) to \(B\) is \[ \begin{aligned} \text{Number of relations} &= 2^{n(A \times B)} \\ &= 2^{4} \end{aligned} \]

Thus, the number of relations from the set \(A\) to the set \(B\) is \(2^{4}\).

Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) \(R = \{(2,1),(3,1), (4,2)\}\),
(ii) \(R = \{(2,2),(2,4),(3,3), (4,4)\}\)
(iii) \(R = \{(1,2),(2,3),(3,4), (4,5), (5,6), (6,7)\}\)

Solution

A relation is said to be a function if every element of the first component appears with exactly one element of the second component. That is, no element of the domain should have more than one image.

In relation (i), \[ \begin{aligned} R &= \{(2,1),(3,1),(4,2)\} \end{aligned} \] Here, the elements \(2\), \(3\), and \(4\) each appear once as first components and each is associated with a unique second component. Although the elements \(2\) and \(3\) have the same image \(1\), this does not violate the definition of a function. Hence, this relation is a function.

In relation (ii), \[ \begin{aligned} R &= \{(2,2),(2,4),(3,3),(4,4)\} \end{aligned} \] In this case, the element \(2\) appears twice as a first component and is related to two different elements, namely \(2\) and \(4\). Since one element of the domain has more than one image, the relation does not satisfy the defining condition of a function. Therefore, this relation is not a function.

In relation (iii), \[ \begin{aligned} R &= \{(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)\} \end{aligned} \] Here, each element of the first component appears exactly once and is associated with a unique second component. Hence, every element of the domain has one and only one image. Therefore, this relation is a function.

Let \(f(x) = \sqrt{x}\) and \(g(x) = x\) be two functions defined over the set of non negative real numbers. Find \((f + g) (x),\ (f – g) (x),\ (fg) (x)\) and \(\left(\frac{f}{g}\right)(x)\)

Solution

Given the functions \(f(x)=\sqrt{x}\) and \(g(x)=x\), both defined on the set of non-negative real numbers, we evaluate the required algebraic combinations of these functions using their definitions.

The sum of the two functions is obtained as \[ \begin{aligned} (f+g)(x) &= f(x)+g(x) \\ &= \sqrt{x}+x \end{aligned} \]

The difference of the two functions is given by \[ \begin{aligned} (f-g)(x) &= f(x)-g(x) \\ &= \sqrt{x}-x \end{aligned} \]

The product of the functions is \[ \begin{aligned} (fg)(x) &= f(x)\cdot g(x) \\ &= \sqrt{x}\cdot x \\ &= x^{3/2} \end{aligned} \]

The quotient of the functions is defined for those values of \(x\) for which \(g(x)\neq 0\). Hence, \[ \begin{aligned} \left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)} \\ &= \frac{\sqrt{x}}{x} \\ &= x^{-1/2}, \qquad x\neq 0 \end{aligned} \]

Thus, the expressions for \((f+g)(x)\), \((f-g)(x)\), \((fg)(x)\), and \(\left(\frac{f}{g}\right)(x)\) are obtained as above, with the quotient defined only for positive values of \(x\).