Q1. Let \(A = \{1, 2, 3,...,14\}\). Define a relation \(R\) from \(A\) to \(A\) by \(R = \{(x, y) : 3x – y = 0, \text{ where } x, y ∈ A\}\). Write down its domain, codomain and range.

Solution

We are given \[ \begin{aligned} A=\{1,2,3,\ldots,14\} \end{aligned} \] and the relation \[ \begin{aligned} R=\{(x,y):3x-y=0,\ x,y\in A\} \end{aligned} \]

From the given condition, \[ \begin{aligned} 3x-y=0 \\ 3x=y \end{aligned} \] Thus, for each \(x\in A\), the corresponding value of \(y\) must be \(3x\) and also belong to \(A\)

Taking values of \(x\) such that \(3x\le 14\), we obtain \[ \begin{aligned} R=\{(1,3),(2,6),(3,9),(4,12)\} \end{aligned} \]

Hence, the domain of the relation is the set of first components \[ \begin{aligned} D=\{1,2,3,4\} \end{aligned} \]

The range of the relation is the set of second components \[ \begin{aligned} \text{Range}=\{3,6,9,12\} \end{aligned} \]

Since the relation is defined from \(A\) to \(A\), the codomain is \[ \begin{aligned} \text{Codomain}=\{1,2,3,\ldots,14\} \end{aligned} \]

Q2. Define a relation \(R\) on the set \(\mathbb{N}\) of natural numbers by \(R = \{(x, y) : y = x + 5, x\text{ is a natural number less than } 4; x, y \in \mathbb{N}\}\). Depict this relationship using roster form. Write down the domain and the range.

Solution

The relation is defined on the set of natural numbers by \[ \begin{aligned} R=\{(x,y): y=x+5,\ x\in\mathbb{N},\ x<4,\ x,y\in\mathbb{N}\} \end{aligned} \]

Since \(x\) is a natural number less than \(4\), the possible values of \(x\) are \(1,2\) and \(3\). Substituting these values in \(y=x+5\), we obtain \[ \begin{aligned} R=\{(1,6),(2,7),(3,8)\} \end{aligned} \]

Hence, the domain of the relation is the set of all first components \[ \begin{aligned} D=\{1,2,3\} \end{aligned} \]

The range of the relation is the set of all second components \[ \begin{aligned} \text{Range}=\{6,7,8\} \end{aligned} \]

Q3. \(A = \{1, 2, 3, 5\}\) and \(B = \{4, 6, 9\}\). Define a relation \(R\) from \(A\) to \(B\) by \(R = {(x, y):\text{ the difference between} x \text{ and } y\text{ is odd }; x \in A, y \in B\}\). Write \(R\) in roster form.

Solution

We are given the sets \[ \begin{aligned} A=\{1,2,3,5\} \\ B=\{4,6,9\} \end{aligned} \] and the relation \[ \begin{aligned} R=\{(x,y): \text{the difference between } x \text{ and } y \text{ is odd},\ x\in A,\ y\in B\} \end{aligned} \]

The difference between two numbers is odd when one of them is even and the other is odd. Using this condition, we examine each possible ordered pair \((x,y)\) with \(x\in A\) and \(y\in B\)

The valid ordered pairs satisfying the given condition are \[ \begin{aligned} R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\} \end{aligned} \]

Q5. Let \(A = \{1, 2, 3, 4, 6\}\). Let \(R\) be the relation on A defined by \(\{(a, b): a , b \in A, b\text{ is exactly divisible by }a\}\).
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.

Solution

We are given the set \[ \begin{aligned} A=\{1,2,3,4,6\} \end{aligned} \] and the relation \[ \begin{aligned} R=\{(a,b): a,b\in A,\ b \text{ is exactly divisible by } a\} \end{aligned} \]

To write \(R\) in roster form, we list all ordered pairs \((a,b)\) such that \(b\) is a multiple of \(a\) and both belong to set \(A\). Thus, we obtain \[ \begin{aligned} R=\{&(1,1),(1,2),(1,3),(1,4),(1,6),\\ &(2,2),(2,4),(2,6),\\ &(3,3),(3,6),\\ &(4,4),(6,6)\} \end{aligned} \]

The domain of the relation is the set of all first components of the ordered pairs in \(R\). Hence, \[ \begin{aligned} D=\{1,2,3,4,6\} \end{aligned} \]

The range of the relation is the set of all second components of the ordered pairs in \(R\). Therefore, \[ \begin{aligned} \text{Range}=\{1,2,3,4,6\} \end{aligned} \]

Q6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

Solution

The relation is given by \[ \begin{aligned} R=\{(x,x+5): x\in\{0,1,2,3,4,5\}\} \end{aligned} \]

From the definition of the relation, the first components of the ordered pairs are the elements of the given set. Hence, the domain of the relation is \[ \begin{aligned} D=\{0,1,2,3,4,5\} \end{aligned} \]

Substituting each value of \(x\) into \(x+5\), we obtain the corresponding second components \(5,6,7,8,9\) and \(10\). Therefore, the range of the relation is \[ \begin{aligned} \text{Range}=\{5,6,7,8,9,10\} \end{aligned} \]

Q7. Write the relation \(R = \{(x, x^3) : x\text{ is a prime number less than }10\}\) in roster form.

Solution

The relation is defined as \[ \begin{aligned} R=\{(x,x^{3}): x \text{ is a prime number less than } 10\} \end{aligned} \]

The prime numbers less than \(10\) are \(2,3,5\) and \(7\). Substituting these values of \(x\) into \(x^{3}\), we obtain \[ \begin{aligned} R=\{(2,8),(3,27),(5,125),(7,343)\} \end{aligned} \]

Q8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Solution

We are given the sets \[ \begin{aligned} A=\{x,y,z\} \\ B=\{1,2\} \end{aligned} \] Hence, \[ \begin{aligned} n(A)=3 \\ n(B)=2 \end{aligned} \]

The Cartesian product of the sets \(A\) and \(B\) contains \[ \begin{aligned} n(A\times B)=n(A)\times n(B) \\ =3\times 2 \\ =6 \end{aligned} \] elements

A relation from \(A\) to \(B\) is any subset of the Cartesian product \(A\times B\). Therefore, the number of relations from \(A\) to \(B\) is equal to the number of subsets of a set having \(6\) elements \[ \begin{aligned} \text{Number of relations}=2^{6} \\ =64 \end{aligned} \]

Q9. Let \(R\) be the relation on \(\mathbb{Z}\) defined by \(R = \{(a,b): a, b \in \mathbb{Z}, a – b \text{ is an integer}\}\). Find the domain and range of R.

Solution

The relation is defined on the set of integers by \[ \begin{aligned} R=\{(a,b): a,b\in \mathbb{Z},\ a-b \text{ is an integer}\} \end{aligned} \]

Since the difference of any two integers is always an integer, the given condition is satisfied for every ordered pair \((a,b)\) where \(a,b\in \mathbb{Z}\). Hence, the relation includes all possible ordered pairs of integers, that is, \[ \begin{aligned} R=\mathbb{Z}\times \mathbb{Z} \end{aligned} \]

Therefore, the domain of the relation, being the set of all first components, is \[ \begin{aligned} D=\mathbb{Z} \end{aligned} \]

Similarly, the range of the relation, being the set of all second components, is \[ \begin{aligned} \text{Range}=\mathbb{Z} \end{aligned} \]