Ch. - 2 Relations and Function

Ch. - 2 Relations and Function

Chapter 2 Ex.2.1 Question 1

If \(\begin{align}\left( {\frac{x}{3} + 1,\;y - \frac{2}{3}} \right) = \left( {\frac{5}{3},\frac{1}{3}} \right), \end{align}\) find the values of \(x\) and \(y.\)

Solution

It is given that \(\begin{align}\left( {\frac{x}{3} + 1,\;y - \frac{2}{3}} \right) = \left( {\frac{5}{3},\frac{1}{3}} \right) \end{align}\).

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore, \(\begin{align}\frac{x}{3} + 1 = \frac{5}{3}\end{align}\) and \(\begin{align}y - \frac{2}{3} = \frac{1}{3} \end{align}\)

\(\begin{align}\frac{x}{3} + 1 &= \frac{5}{3} \\\Rightarrow \frac{x}{3} &= \frac{5}{3} - 1\\ \Rightarrow \frac{x}{3} &= \frac{{5 - 3}}{3}\\ \Rightarrow \frac{x}{3} &= \frac{2}{3}\\ \Rightarrow x &= 2 \end {align}\qquad \) and \(\begin{align}\qquad y - \frac{2}{3} &= \frac{1}{3}\\ \Rightarrow y &= \frac{1}{3} + \frac{2}{3}\\ \Rightarrow y &= 1\end{align}\)

Hence, \(x = 2\) and \(y = 1\)

Chapter 2 Ex.2.1 Question 2

If the set \(A\) has \(3\) elements and the set \(B = \{ 3,4,5\} \), then find the number of elements in \(\left( {A \times B} \right)\)?

Solution

It is given that set \(A\) has \(3\) elements and the set \(B = \{ 3,4,5\} \)

Number of elements in set \(A,\) \(n\left( A \right) = 3\)

Number of elements in set \(B,\) \(n\left( B \right) = 3\)

Number of elements in \(\left( {A \times B} \right)\;=\;\) (Number of elements in \(A\)) \( \times \) (Number of elements in \(B\))

\[\begin{align}n\left( {A \times B} \right) &= n\left( A \right) \times n\left( B \right)\\ &= 3 \times 3\\ &= 9\end{align}\]

Thus, the number of elements in \(\left( {A \times B} \right)\) is \(9.\)

Chapter 2 Ex.2.1 Question 3

If  \(G = \{ 7,8\} \) and \(H = \{ 5,4,2\} \), find \(G \times H\) and \(H \times G\).

Solution

It is given that \(G = \{ 7,8\} \) and \(H = \{ 5,4,2\}\)

We know that the Cartesian product \(P \times Q\) of two non-empty sets \(P\) and \(Q\) is defined as

\[P \times Q = \left\{ {\left( {p,q} \right):p \in P,q \in Q} \right\}\]

Therefore,

\[G \times H = \left\{ {\left( {7,5} \right),\left( {7,4} \right),\left( {7,2} \right),\left( {8,5} \right),\left( {8,4} \right),\left( {8,2} \right)} \right\}\]

\[H \times G = \left\{ {\left( {5,7} \right),\left( {5,8} \right),\left( {4,7} \right),\left( {4,8} \right),\left( {2,7} \right),\left( {2,8} \right)} \right\}\]

Chapter 2 Ex.2.1 Question 4

State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If \(P = \left\{ {m,n} \right\}\) and \(Q = \left\{ {n,m} \right\}\), then \(P \times Q = \left\{ {\left( {m,n} \right),\left( {n,m} \right)} \right\}\)

(ii) If \(A\) and \(B\) are non-empty sets, then \(\left( {A \times B} \right)\) is a non-empty set of ordered pairs \(\left( {x,y} \right)\) such that \(x \in A\) and \(y \in B\).

(iii) If \(A = \left\{ {1,2} \right\}\), \(B = \left\{ {3,4} \right\}\), then \(A \times \left( {B \cap \phi } \right) = \phi \)

Solution

(i) False, \(P \times Q = \left\{ {\left( {m,n} \right),\left( {m,m} \right),\left( {n,n} \right),\left( {n,m} \right)} \right\}\)

(ii) True

(iii) True

Chapter 2 Ex.2.1 Question 5

If \(A = \left\{ { - 1,1} \right\}\), find \(A \times A \times A\).

Solution

It is known that for any non-empty set \(A,\) \(A \times A \times A\) is defined as

\[A  \times A \times A = \left\{ {\left( {a,b,c} \right):a,b,c \in A} \right\}\]

It is given that \(A = \left\{ { - 1,1} \right\}\)

Therefore,

\(A \times A \times A = \left\{ {\left( { - 1, - 1, - 1} \right),\left( { - 1, - 1,1} \right),\left( { - 1,1, - 1} \right),\left( { - 1,1,1} \right),\left( {1, - 1, - 1} \right),\left( {1, - 1,1} \right),\left( {1,1, - 1} \right),\left( {1,1,1} \right)} \right\}\)

Chapter 2 Ex.2.1 Question 6

If \(A \times B = \left\{ {\left( {a,x} \right),\left( {a,y} \right),\left( {b,x} \right),\left( {b,y} \right)} \right\}\). Find \(A\) and \(B\).

Solution

It is given that \(A \times B = \left\{ {\left( {a,x} \right),\left( {a,y} \right),\left( {b,x} \right),\left( {b,y} \right)} \right\}\)

We know that the Cartesian product of two non-empty sets \(P\) and \(Q\) is defined as

\[P \times Q = \left\{ {\left( {p,q} \right):p \in P,q \in Q} \right\}\]

Therefore, \(A\) is the set of all first elements and \(B\) is the set of all second elements.

Thus, \(A = \left\{ {a,b} \right\}\) and \(B = \left\{ {x,y} \right\}\)

Chapter 2 Ex.2.1 Question 7

Let \(A = \left\{ {1,2} \right\}\), \(B = \left\{ {1,2,3,4} \right\}\), \(C = \left\{ {5,6} \right\}\) and \(D = \left\{ {5,6,7,8} \right\}\). Verify that

(i) \(A \times \left( {B \cap C} \right) = \left( {A \times B} \right) \cap \left( {A \times C} \right)\).

(ii) \(A \times C\) is a subset of \(B \times D\).

Solution

(i) To verify: \(A \times \left( {B \cap C} \right) = \left( {A \times B} \right) \cap \left( {A \times C} \right)\)

We have \(B \cap C = \left\{ {1,2,3,4} \right\} \cap \left\{ {5,6} \right\} = \phi \)

\[\begin{align}LHS &= A \times \left( {B \cap C} \right)\\ &= A \times \phi \\& = \phi\end{align}\]

Now,

\[\begin{align}A \times B &= \left\{ {\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}\\\\A \times C &= \left\{ {\left( {1,5} \right),\left( {1,6} \right),\left( {2,5} \right),\left( {2,6} \right)} \right\}\end{align}\]

Therefore,

\[\begin{align}RHS &= \left( {A \times B} \right) \cap \left( {A \times C} \right)\\ &= \phi\end{align}\]

Therefore, \(\text{L.H.S.} = \text{R.H.S}\)

Hence, \(A \times \left( {B \cap C} \right) = \left( {A \times B} \right) \cap \left( {A \times C} \right)\)

(ii) To verify: \(A \times C\) is a subset of \(B \times D\)

We have

\[\begin{align}A \times C &= \left\{ {\left( {1,5} \right),\left( {1,6} \right),\left( {2,5} \right),\left( {2,6} \right)} \right\}\\\\B \times D &= \left\{ \begin{array}{l}\left( {1,5} \right),\left( {1,6} \right),\left( {1,7} \right),\left( {1,8} \right),\left( {2,5} \right),\left( {2,6} \right),\left( {2,7} \right),\left( {2,8} \right),\\\left( {3,5} \right),\left( {3,6} \right),\left( {3,7} \right),\left( {3,8} \right),\left( {4,5} \right),\left( {4,6} \right),\left( {4,7} \right),\left( {4,8} \right)\end{array} \right\} \end{align}\]

We can observe that all the elements of set \(A \times C\) are the elements of set \(B \times D\). Therefore, \(A \times C\) is a subset of \(B \times D\).

Chapter 2 Ex.2.1 Question 8

Let \(A = \left\{ {1,2} \right\}\) and \(B = \left\{ {3,4} \right\}\). Write \(A \times B\). How many subsets will \(A \times B\) have? List them.

Solution

\(A = \left\{ {1,2} \right\}\) and \(B = \left\{ {3,4} \right\}\)

Therefore, \(A \times B = \left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}\)

Hence, \(n\left( {A \times B} \right) = 4\)

We know that if \(C\) is a set with \(n\left( C \right) = m\), then \(n\left[ {P\left( C \right)} \right] = {2^m}\).

Therefore, the set \(A \times B\) has \({2^4} = 16\) subsets. These are

\[\left\{ \begin{array}{l}\phi ,\left\{ {\left( {1,3} \right)} \right\},\left\{ {\left( {1,4} \right)} \right\},\left\{ {\left( {2,3} \right)} \right\},\left\{ {\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,3} \right),\left( {1,4} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {2,3} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,4} \right),\left( {2,3} \right)} \right\},\left\{ {\left( {1,4} \right),\left( {2,4} \right)} \right\},\left\{ {\left( {2,3} \right),\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,4} \right)} \right\},\\\left\{ {\left( {1,3} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\},\left\{ {\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\},\left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}\end{array} \right\}\]

Chapter 2 Ex.2.1 Question 9

Let \(A\) and \(B\) be two sets such that \(n\left( A \right) = 3\) and \(n\left( B \right) = 2\). If \(\left( {x,1} \right),\left( {y,2} \right),\left( {z,1} \right)\) are in \(A \times B\), find \(A\) and \(B,\) where \(x,\; y\) and \(z\) are distinct elements.

Solution

It is given that \(n\left( A \right) = 3\) and \(n\left( B \right) = 2\); and \(\left( {x,1} \right),\left( {y,2} \right),\left( {z,1} \right)\) are in \(A \times B\).

We know that

\(\rm{A} =\) Set of first elements of the ordered pair elements of \(A \times B.\)

\(\rm{B} =\) Set of second elements of the ordered pair elements of \(A \times B.\)

Therefore, \(x,\; y, \) and \(z\) are the elements of  \(A;\) \(1\) and \(2\) are the elements of \(B.\)

Since \(n\left( A \right) = 3\) and \(n\left( B \right) = 2\),

It is clear that \(A = \left\{ {x,y,z} \right\}\) and \(B = \left\{ {1,2} \right\}.\)

Chapter 2 Ex.2.1 Question 10

The Cartesian product \(A \times A\) has \(9\) elements among which are found \(\left( { - 1,0} \right)\) and \(\left( {0,1} \right)\). Find the set \(A\) and the remaining elements of \(A \times A\).

Solution

We know that if \(n\left( A \right) = p\) and \(n\left( B \right) = q\), then \(n\left( {A \times B} \right) = pq.\)

Therefore, \(n\left( {A \times A} \right) = n\left( A \right) \times n\left( A \right)\)

It is given that \(n\left( {A \times A} \right) = 9\)

\[n\left( A \right) \times n\left( A \right) = 9\]

Hence,

\[n\left( A \right) = 3\]

The ordered pairs \(\left( { - 1,0} \right)\) and \(\left( {0,1} \right)\) are two of the nine elements of \(A \times A\).

We know that \(A \times A = \left\{ {\left( {a,a} \right):a \in A} \right\}\).

Therefore, \(–1,\; 0,\) and \(1\) are elements of \(A.\)

Since \(n\left( A \right) = 3,\) it is clear that \(A = \left\{ { - 1,0,1} \right\}.\)

The remaining elements of set \(A \times A\) are \(\left( { - 1, - 1} \right),\left( { - 1,1} \right),\left( {0, - 1} \right),\left( {0,0} \right),\, \left( {1, - 1} \right),\left( {1,0} \right),\) and \(\left( {1,1} \right)\)

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