Let A and B be two non-empty sets. The cartesian product
of A and B is denoted by A $\times$ B´ and is defined as the set of
all ordered pairs (a, b), where a $\in$ A and b $\in$ B.
Symbolically, A $\times$ B = {(a, b) : a $\in$ A and b $\in$ B}.
If there are three sets A, B and C, a $\in$ A, b $\in$ B and c $\in$ C,
then we form an ordered triplet (a, b, c). The set of all
ordered triplets (a, b, b) is called the cartesian product of
three sets A, B and C, i.e.,
A $\times$ B $\times$ C = {(a, b, c) : a $\in$ A, b $\in$ B and c $\in$ C}
An ordered pair and ordered triplet are also called
2-tuple and 3-tuple, respectively.
e.g., Let A = {1, 2, 3} and B = {x, y}
So, A $\times$ B = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)}
and B $\times$ A = {(x, 1), (y, 1), (x, 2), (y, 2), (x, 3), (y, 3)}
Note
- If A $\neq$ B, then A $\times$ B $\neq$ B $\times$ A.
- If A has p elements and B has q elements, then A $\times$ B has
pq elements.
- If A = $\phi$ and B = $\phi$, then A $\times$ B = $\phi$ .
- Cartesian product of n sets $A_1$, $A_2$, $A_3$, . . . , $A_n$ is the set of all n-tuples $a_1$, $a_2$, $a_3$, . . . , $a_n$, $a_i \in A_i$, i = 1, 2, 3, . . . , n and it is denoted by $A_1 \times A_2 \times A_3 \times . . . \times A_n$ or $\prod_{i=1}^{n}A_i$
Properties of Cartesian Product
If A, B and C are three sets, then
1. (a) A $\times$ (B $\cup$ C) = (A $\times$ B) $\cup$ (A $\times$ C) and
(b) A $\times$ (B $\cup$ C) = (A $\times$ B) $\cup$ (A $\times$ C)
2. A $\times$ (B - C) = (A $\times$ B) - (A $\times$ C)
3. A $\times$ B = B $\times$ A ⇔ A = B
4. If A $\subseteq$ B ⇒ A $\times$ A $\subseteq$ (A $\times$ B) $\cap$ (B $\times$ A)
5. If A $\subseteq$ B ⇒ A $\times$ C $\subseteq$ (B $\times$ C)
6. If A $\subseteq$ B and C $\subseteq$ D ⇒ A $\times$ C $\subseteq$ B $\times$ D
7. (A $\times$ B) $\cap$ (C $\times$ D) = (A $\cap$ C) $\times$ (B $\cap$ D)
8. A $\times$ (B' $\cup$ C')' = (A $\times$ B) $\cap$ (A $\times$ C)
9. A $\times$ (B' $\cap$ C')' = (A $\times$ B) $\cup$ (A $\times$ C)
10. If A and B have n common elements, then A $\times$ B and
B $\times$ A will have $n^2$ common elements.
Sample Problem 1
If the set A has 3 elements and the
set B = {3, 4, 5}, then find the number of elements in (A $\times$ B) is . . . . [NCERT]
(a) 8
(b) 9
(c) 7
(d) 10
Answer: (b)
Here, n(A) = 3, B = {3, 4, 5}, n(B) = 3
So, n(A $\times$ B) = n(A) $\times$ n(B) = 3 $\times$ 3 = 9
Sample Problem 2
If A and B are sets such that n(A $\times$ B) = 6 and A $\times$ B
contains (1, 2), (2, 1) and (3, 2), then find the sets A, B
and A $\times$ B.
Answer:
Since n(A) . n(B) = n(A $\times$ B) = 6, n(A) and
n(B) are divisors of 6.
Hence n(A) = 1 or 2 or 3 or 6.
Since (1, 2), (2, 1) and (3, 2) $\in$ A $\times$ B, 1, 2, 3 $\in$ A and
hence n(A) $\geqslant$ 3.
Also, 2, 1 $\in$ B and hence n(B) $\geqslant$ 2. Thus
n(A) = 3 and n(B) = 2.
Therefore, A = {1, 2, 3} and B =
{1, 2}, so that
A $\times$ B = = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
Sample Problem 3
If two sets A B and are having 99 elements in
common, then the number of elements common to
each of the sets A $\times$ B and B $\times$ A are
(a) $2^{99}$
(b) $99^2$
(c) 100
(d) 18
Answer: (b)
n[(A $\times$) B) $\cap$ (B $\times$ A)] = n[(A $\cap$) B) $\times$ (B $\cap$ A)]
= 99 $\times$ 99
= $99^2$
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