RELATIONS AND FUNCTIONS-Exercise 2.1
Maths - Exercise
Q1. If \(\left(\dfrac{x}{1} + 1,y-\dfrac{2}{3}\right)=\left(dfrac{5}{3},\dfrac{1}{3}\right)\), find the values of x and y.
Solution
We are given the equality of two ordered pairs. Hence, their corresponding coordinates must be equal. Thus, \[ \begin{aligned} \left(\dfrac{x}{3}+1,\, y-\dfrac{2}{3}\right) &= \left(\dfrac{5}{3},\, \dfrac{1}{3}\right). \end{aligned} \]
Equating the first coordinates, we get \[ \begin{aligned} \dfrac{x}{3}+1 &= \dfrac{5}{3} \\ \dfrac{x}{3} &= \dfrac{5}{3}-1 \\ \dfrac{x}{3} &= \dfrac{2}{3} \\ x &= 2 \end{aligned} \]
Equating the second coordinates, we obtain \[ \begin{aligned} y-\dfrac{2}{3} &= \dfrac{1}{3} \\ y &= \dfrac{1}{3}+\dfrac{2}{3} \\ y &= \dfrac{3}{3} = 1 \end{aligned} \]
Therefore, the required values of \(x\) and \(y\) are \[ \Rightarrow (x,y) = (2,1) \]
Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).
Solution
We are given that the set \(A\) contains 3 elements and the set \(B=\{3,4,5\}\). Hence, the number of elements in each set can be written as \[ \begin{aligned} n(A) &= 3 \\ n(B) &= 3 \end{aligned} \]
The number of elements in the Cartesian product of two finite sets is equal to the product of the number of elements in each set. Therefore, \[ \begin{aligned} n(A \times B) &= n(A)\times n(B) \\ &= 3 \times 3 \\ &= 9 \end{aligned} \]
Thus, the Cartesian product \(A \times B\) contains \(9\) elements.
Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Solution
We are given the two sets \[ \begin{aligned} G &= \{7,8\},\\ H &= \{5,4,2\} \end{aligned} \]
The Cartesian product \(G \times H\) consists of all ordered pairs \((g,h)\) such that \(g \in G\) and \(h \in H\). Therefore, \[ \begin{aligned} G \times H &= \{7,8\} \times \{5,4,2\} \\ &= \{(7,5),(7,4),(7,2),\\ &(8,5),(8,4),(8,2)\} \end{aligned} \]
Similarly, the Cartesian product \(H \times G\) consists of all ordered pairs \((h,g)\) such that \(h \in H\) and \(g \in G\). Hence, \[ \begin{aligned} H \times G &= \{5,4,2\} \times \{7,8\} \\ &= \{(5,7),(5,8),(4,7),\\&(4,8),(2,7),(2,8)\} \end{aligned} \]
Q4. State whether each of the following statements are true or false. If the statement
is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered
pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.
Solution
(i) We are given \(P=\{m,n\}\) and \(Q=\{n,m\}\). The Cartesian product \(P\times Q\) consists of all ordered pairs \((x,y)\) such that \(x\in P\) and \(y\in Q\). Thus, \[ \begin{aligned} P\times Q &= \{m,n\}\times\{n,m\} \\ &= \{(m,n),(m,m),(n,n),(n,m)\} \end{aligned} \] Hence, the given statement is false. The correct statement is \(P\times Q=\{(m,n),(m,m),(n,n),(n,m)\}\).
(ii) If \(A\) and \(B\) are non-empty sets, then there exists at least one element \(x\in A\) and at least one element \(y\in B\). Therefore, the ordered pair \((x,y)\) belongs to \(A\times B\). Hence, \[ \begin{aligned} A\times B \neq \varnothing \end{aligned} \] and it is a set of ordered pairs \((x,y)\) such that \(x\in A\) and \(y\in B\). Thus, the given statement is true.
(iii) We are given \(A=\{1,2\}\) and \(B=\{3,4\}\). Since the intersection of any set with the empty set is the empty set, \[ \begin{aligned} B\cap \varnothing &= \varnothing \end{aligned} \] Now, the Cartesian product of any set with the empty set is also the empty set. Hence, \[ \begin{aligned} A\times (B\cap \varnothing) &= A\times \varnothing \\ &= \varnothing \end{aligned} \] Therefore, the given statement is true.
Q5. If A = {–1, 1}, find A × A × A.
Solution
We are given the set \[ \begin{aligned} A=\{-1,1\} \end{aligned} \]
First, we find the Cartesian product of \(A\) with itself: \[ \begin{aligned} A\times A &= \{-1,1\}\times\{-1,1\} \\ &= \{(-1,-1),(-1,1),(1,-1),(1,1)\} \end{aligned} \]
Now, the Cartesian product \(A\times A\times A\) is obtained by taking the Cartesian product of \(A\times A\) with \(A\): \[ \begin{aligned} A\times A\times A &= \{(-1,-1),(-1,1),(1,-1),(1,1)\}\times\{-1,1\} \\ &= \{(-1,-1,-1),(-1,-1,1),(-1,1,-1),\\&(-1,1,1),(1,-1,-1),(1,-1,1),\\&(1,1,-1),(1,1,1)\} \end{aligned} \]
Q6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.
Solution
We are given the Cartesian product \[ \begin{aligned} A\times B=\{(a,x),(a,y),(b,x),(b,y)\} \end{aligned} \]
From the ordered pairs in \(A\times B\), the first components are \(a\) and \(b\). Hence, \[ \begin{aligned} A=\{a,b\} \end{aligned} \]
Similarly, the second components of the ordered pairs are \(x\) and \(y\). Therefore, \[ \begin{aligned} B=\{x,y\} \end{aligned} \]
Q7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × C is a subset of B × D.
Solution
We are given the sets \[ \begin{aligned} A&=\{1,2\} \\ B&=\{1,2,3,4\} \\ C&=\{5,6\} \\ D&=\{5,6,7,8\} \end{aligned} \]
(i) First, we find the intersection of \(B\) and \(C\): \[ \begin{aligned} B\cap C=\varnothing \end{aligned} \] Hence, \[ \begin{aligned} A\times (B\cap C) &= \{1,2\}\times \varnothing \\ &= \varnothing \end{aligned} \]
Now, we evaluate \(A\times B\) and \(A\times C\): \[ \begin{aligned} A\times B &= \{1,2\}\times\{1,2,3,4\} \\ &= \{(1,1),(1,2),(1,3),(1,4),\\&(2,1),(2,2),(2,3),(2,4)\} \end{aligned} \] \[ \begin{aligned} A\times C &= \{1,2\}\times\{5,6\} \\ &= \{(1,5),(1,6),\\&(2,5),(2,6)\} \end{aligned} \] Since there is no common ordered pair in the above two sets, \[ \begin{aligned} (A\times B)\cap(A\times C)=\varnothing \end{aligned} \] Thus, \[ \begin{aligned} A\times(B\cap C)=(A\times B)\cap(A\times C) \end{aligned} \] Hence, part (i) is verified.
(ii) Now, we find \(A\times C\): \[ \begin{aligned} A\times C&=\{(1,5),(1,6),\\&(2,5),(2,6)\} \end{aligned} \] Also, \[ \begin{aligned} B\times D=\{1,2,3,4\}\times\{5,6,7,8\} \end{aligned} \] Each element of \(A\times C\) has its first component in \(B\) and its second component in \(D\). Therefore, \[ \begin{aligned} A\times C \subseteq B\times D \end{aligned} \] Hence, part (ii) is verified.
Q8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
Solution
We are given the sets \[ \begin{aligned} A&=\{1,2\} \\ B&=\{3,4\} \end{aligned} \]
The Cartesian product \(A\times B\) is the set of all ordered pairs \((a,b)\) such that \(a\in A\) and \(b\in B\). Hence, \[ \begin{aligned} A\times B &= \{1,2\}\times\{3,4\} \\ &= \{(1,3),(1,4),(2,3),(2,4)\} \end{aligned} \]
Since set \(A\) has 2 elements and set \(B\) has 2 elements, the number of elements in their Cartesian product is \[ \begin{aligned} n(A) &= 2 \\ n(B) &= 2 \\ n(A\times B) &= 2\times 2 \\ &= 4 \end{aligned} \]
The number of subsets of a set containing 4 elements is given by \[ \begin{aligned} 2^{4} &= 16 \end{aligned} \]
Thus, \(A\times B\) has 16 subsets. These subsets are \[ \begin{aligned} \varnothing,\{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\},\{(1,3),(1,4)\},\{(1,3),(2,3)\},\\ \{(1,3),(2,4)\},\{(1,4),(2,3)\},\{(1,4),(2,4)\},\{(2,3),(2,4)\},\\ \{(1,3),(1,4),(2,3)\},\{(1,3),(1,4),(2,4)\}, \{(1,3),(2,3),(2,4)\},\\ \{(1,4),(2,3),(2,4)\},\{(1,3),(1,4),(2,3),(2,4)\} \end{aligned} \]
Q9. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.
Solution
We are given that the number of elements in the sets are \[ \begin{aligned} n(A)=3 \\ n(B)=2 \end{aligned} \] and the ordered pairs \((x,1),(y,2),(z,1)\) belong to \(A\times B\), where \(x,y\) and \(z\) are distinct elements
Since the first components of the given ordered pairs come from set \(A\), we observe that the elements of \(A\) are \(x,y\) and \(z\). Thus, \[ \begin{aligned} A=\{x,y,z\} \end{aligned} \]
Similarly, the second components of the ordered pairs come from set \(B\). These are \(1\) and \(2\). Hence, \[ \begin{aligned} B=\{1,2\} \end{aligned} \]
Therefore, the required sets are \(A=\{x,y,z\}\) and \(B=\{1,2\}\)
Q10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.
Solution
We are given that the Cartesian product \(A\times A\) has 9 elements. Hence, \[ \begin{aligned} n(A\times A)=9 \end{aligned} \] This implies that \[ \begin{aligned} n(A)^2=9 \end{aligned} \] and therefore \(n(A)=3\)
It is also given that the ordered pairs \((-1,0)\) and \((0,1)\) belong to \(A\times A\). Thus, the elements \(-1,0\) and \(1\) must all belong to the set \(A\). Since \(A\) has exactly three elements, we obtain \[ \begin{aligned} A=\{-1,0,1\} \end{aligned} \]
Now, we write the complete Cartesian product \(A\times A\): \[ \begin{aligned} A\times A &= \{-1,0,1\}\times\{-1,0,1\} \\ &= \{(-1,-1),(-1,0),(-1,1),\\(0,-1),(0,0),(0,1),\\(1,-1),(1,0),(1,1)\} \end{aligned} \]
Thus, the set \(A\) is \(\{-1,0,1\}\) and the remaining elements of \(A\times A\) are all the ordered pairs listed above
