Important Key points :

Example 1.1
If A = {1,3,5} and B = {2,3} then
(i) find A$\times$B and B$\times$A.
(ii) Is A×B = B×A? If not why?
(iii) Show that n(A×B) = n(B×A) = n(A)× n(B)

Solution :
Given that A = {1,3,5} and B = {2,3}.
(i) A×B = {1,3,5} × {2,3} = {(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)} ...(1)
B×A= {2,3} × {1,3,5} = {(2,1), (2,3), (2,5), (3,1), (3,3), (3,5)} ...(2)
(ii)From (1) and (2) we can conclude that A×B = B×A as (1,2) $\neq$(2,1) and (1, 3)$\neq$(3,1), etc.
(iii) n(A)=3; n (B) = 2.(no.of.elements in Set A and B)
From (1) and (2) we observe that, n (A×B) = n (B×A) = 6;
we see that, n (A) ×n (B) = 3 × 2 = 6 and n (B) × n (A) = 2×3 = 6
Hence, n (A×B) =n (B×A) = n(A) × n (B) = 6.
Thus, n (A×B) =n (B×A) = n(A) × n (B).

Example 1.2
If A×B = {(3,2), (3,4), (5,2), (5,4)} then find A and B.

Solution :
A×B ={(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A$\times$B }.
∴ A = {3,5}
B = {set of all second coordinates of elements of A$\times$B }.
∴ B = {2,4}
Thus A = {3,5} and B = {2,4}.

Example 1.3
Let A = {x $\in$N | 1B = {x $\in$W| 0 $\leq$ x < 2} and
C = {x $\in$N |x <3}. Then verify that
(i) A$\times$(B $\cup$ C) = (A $\times$ B) $\cup$ (A $\times$ C)
(ii) A $\times$ (B $\cap$ C) = (A $\times$ B) $\cap$ (A $\times$ C)

Solution :
A = {x $\in$ N| 1B = {x $\in$W| 0 $\leq$ x < 2} = {0,1}
C = {x $\in$N |x <3} ={1,2}
B $\cup$ C = {0,1} $\cup${1,2} = {0,1,2}
A×(B $\cup$C) =={2,3} $\times$ {0,1,2} = {(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)} ...(1)
A$\times$B = {2,3}$\times${0,1} = {(2,0),(2,1),(3,0),(3,1)}
A$\times$C = {2,3} $\times$ {1,2} = {(2,1),(2,2),(3,1),(3,2)}
(A×B)$\cup$(A×C) = {(2,0),(2,1),(3,0),(3,1)} , {(2,1),(2,2),(3,1),(3,2)}
$\hspace{0.5in}$ = {(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)} ...(2)
From (1) and (2), A×(B $\cup$C) = (A×B)$\cup$(A×C) is verified.
B $\cap$C = {0,1} $\cap${1,2} = {1}
A×(B $\cap$C) ={2, 3}×{1}={(2,1),(3,1)} ...(3)
A$\times$B = {2, 3}×{0,1} = {(2, 0),(2,1),(3, 0),(3,1)}
A$\times$C = {2, 3}×{1,2} = {(2,1),(2,2),(3,1),(3,2)}
(A×B)$\cap$(A×C) = {(2, 0),(2,1),(3, 0),(3,1)} $\cap${(2,1),(2,2),(3,1),(3,2)}
\hspace{0.5in}= {(2,1),(3,1)} ....(4)
From (3) and (4), A×(B $\cap$C) = (A×B) $\cap$(A×C) is verified.

Exercise 1.1

1. Find A × B, A × A and B × A
(i) A = {2, -2, 3} and B = {1, -4}
(ii) A = B = {p, q}
(iii) A – {m, n} ; B = Φ

Solution :
Given
(i) A = {2, -2, 3} and B = {1, -4}
A × B = {2,-2, 3} × {1,-4}
$\hspace{1cm}$= {(2, 1), (2, -4)(-2, 1) (-2, -4) (3, 1) (3,-4)}
A × A = {2,-2, 3} × {2,-2, 3}
$\hspace{1cm}$= {(2, 2)(2, -2)(2, 3)(-2, 2)(-2, -2)(-2, 3X3,2) (3,-2) (3,3)}
B × A = {1,-4} × {2,-2, 3}
$\hspace{1cm}$= {(1, 2)(1, -2)( 1, 3)(-4, 2) (-4,-2)(-4, 3)}
(ii) A = B = {p, q}
A × B = {p, q) × {p, q}
$\hspace{1cm}$= {(p,p),(p,q)(q,p)(q,q)}
A × A = {p,q) × (p,q)
$\hspace{1cm}$= {(p,p)(p,q)(q,p)(q,q)
B × A = {p,q} × {p,q}
$\hspace{1cm}$= {(p,p)(p,q)(q,p)(q,q)
(iii) A = {m, n} × B = Φ
Note: B = Φ or {}
A × B = {m, n) × { }
$\hspace{1cm}$= { )
A × A = {m, n) × (m, n)}
$\hspace{1cm}$= {(m, m)(m, n)(n, m)(n, n)}
B × A = { } × {m, n}
$\hspace{1cm}$= { }

2. Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.

Solution :
A = {1, 2, 3}, B = {2, 3, 5, 7}
A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3) , (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) , (5, 1), (5, 2), (5, 3), (7, 1), (7, 2) , (7, 3)}

3. If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4), (3,3) ,(3, 4)} find A and B.

Solution :
B × A = {(-2, 3)(-2, 4) (0, 3) (0, 4) (3, 3) (3,4)}
A = {set of all first coordinates of elements of A$\times$B }
$\Rightarrow$A = {3,4}
B = {set of all second coordinates of elements of A$\times$B }.
$\Rightarrow$B = {-2,0,3}

4. If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) $\cap$ (C $\times$ C).

Solution :
Given A = {5,6}, B = {4, 5, 6},C = {5, 6, 7}
A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} ……….. (1)
B × B = {(4, 4), (4, 5), (4, 6), (5, 4),(5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} …(2)
C × C = {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),(6, 7), (7, 5), (7, 6), (7, 7)} …(3)
(B × B) ∩ (C × C) = {(5, 5), (5,6), (6, 5), (6,6)} …(4)
From (1) and (4), A × A = (B × B) $\cap$ (C × C) It is proved.

5. Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check if (A $\cap$ C) $\times$ (B $\cap$ D) = (A $\times$ B) $\cap$ (C $\times$ D) is true?

Solution :
A = {1,2,3}, B = {2, 3, 5}, C = {3,4} D = {1,3,5}
A $\cap$ C = {1,2,3} $\cap$ {3,4}
$\hspace{1 cm}$= (3}
B $\cap$ D = {2,3, 5} $\cap$ {1,3,5}
$\hspace{1 cm}$= {3,5}
(A $\cap$ C) × (B $\cap$ D) = {3} × {3,5}
$\hspace{1 cm}$= {(3, 3)(3, 5)} ….(1)
A × B = {1,2,3} × {2,3,5}
$\hspace{1 cm}$= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5)}
C × D = {3,4} × {1,3,5}
$\hspace{1 cm}$= {(3,1) (3, 3) (3, 5) (4,1) (4, 3) (4, 5)}
(A × B) $\cap$ (C × D) = {(3, 3) (3, 5)} ….(2)
From (1) and (2) we get (A $\cap$ C) × (B $\cap$ D) = (A × B) $\cap$ (C × D) This is true.

6 .Let A = {x ∈ W | x < 2}, B = {x ∈ N |1 < x < 4} and C = {3, 5}. Verify that
(i) A × (B $\cup$ C) = (A × B) $\cup$ (A × C)
(ii) A × (B $\cap$ C) = (A × B) $\cap$ (A × C)
(iii) (A $\cup$ B) × C = (A × C) $\cup$ (B × C)

Solution :
A = {x ∈ W|x < 2} = {0,1}
B = {x ∈ N |1 < x < 4} = {2,3,4}
C = {3,5}
(i) A × (B $\cup$ C) = (A × B) $\cup$ (A × C)
B $\cup$ C = {2, 3, 4} $\cup$ {3, 5}
$\hspace{1cm}$= {2, 3, 4, 5}
A × (B $\cup$ C) = {(0, 2), (0, 3), (0,4), (0, 5), (1, 2) , (1, 3), (1, 4),(1, 5)}... (1)
(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
(A × B) $\cup$ (A × C)= {(0, 2), (0, 3), (0,4), (1, 2), (1, 3), (1, 4), (0, 5), (1, 5)} ….(2)
From (1) and (2), A × (B $\cup$ C) = (A × B) $\cup$ (A × C) Hence it is proved.

7. Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that
(i) (A $\cap$ B) × C = (A × c) $\cap$ (B × C)
(ii) A × (B – C) = (A × B) – (A × C)

Solution :
Answer:
Given A = {1,2, 3, 4, 5,6, 7}, B = {2, 3, 5,7}, C = {2}
(i) (A ∩ B) × C = (A × C) ∩ (B × C)
A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2, 3, 5, 7}
$\hspace{1cm}$= {2, 3, 5, 7}
(A ∩ B) × C = {2, 3, 5, 7} × {2}
$\hspace{1cm}$= {(2, 2) (3, 2) (5, 2) (7, 2)} ….(1)
A × C = {1,2, 3, 4, 5, 6, 7} × {2}
$\hspace{1cm}$= {(1,2) (2, 2) (3, 2) (4, 2) (5.2) (6, 2) (7, 2)}
B × C = {2, 3, 5, 7} × {2}
$\hspace{1cm}$= {(2, 2) (3, 2) (5, 2) (7, 2)}
(A × C) ∩ (B × C) = {(2, 2) (3, 2) (5, 2) (7, 2)} ….(2)
From (1) and (2) we get, (A ∩ B) × C = (A × C) ∩ (B × C)