Pair of entries 'a' and 'b' which are separated by a comma and are enclosed within the bracket is called an ordered pair, where 'a' is the first element and 'b' is the second element.
The order, in which the entries are written in an ordered pair, is specific.
Ordered pair: A pair of entries grouped in a particular order, which are separated by a comma and enclosed within brackets.
If two ordered pairs are equal, then their corresponding first elements and second elements are equal.
If (a, b) = (1, 4) ⇒ a = 1 and b = 4.
Cartesian product of sets
Consider X = {a, b, m}, Y = {f, c, d, h}
Product of sets X and Y: The product of sets means to pair the elements of sets together in a systematic manner.
Cartesian product of sets X and Y is symbolically denoted by X cross Y.
X x Y = {(a,f),(a,c),(a,d),(a,h),(b,f),(b,c),(b,d),(b,h),(m,f),(m,c),(m,d),(m,h)}
Example
Let X = {♣,♦,♠,♥} and Y = {2,3,4,5,6,7,8,9,10,J,Q,K,A}
X × Y |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Jack(J) |
Queen(Q) |
King(K) |
Ace(A) |
Club(♣) |
(♣,2) |
(♣,3) |
(♣,4) |
(♣,5) |
(♣,6) |
(♣,7) |
(♣,8) |
(♣,9) |
(♣,10) |
(♣,J) |
(♣,Q) |
(♣,K) |
(♣,A) |
Diamond (♦) |
(♦,2) |
(♦,3) |
(♦,4) |
(♦,5) |
(♦,6) |
(♦,7) |
(♦,8) |
(♦,9) |
(♦,10) |
(♦,J) |
(♦,Q) |
(♦,K) |
(♦,A) |
Spade(♠) |
(♠,2) |
(♠,3) |
(♠,4) |
(♠,5) |
(♠,6) |
(♠,7) |
(♠,8) |
(♠,9) |
(♠,10) |
(♠,J) |
(♠,Q) |
(♠,K) |
(♠,A) |
Heart(♥) |
(♥,2) |
(♥,3) |
(♥,4) |
(♥,5) |
(♥,6) |
(♥,7) |
(♥,8) |
(♥,9) |
(♥,10) |
(♥,J) |
(♥,Q) |
(♥,K) |
(♥,A) |
The Cartesian product of two non-empty sets X and Y can be represented in the form of tabular cells or the point of intersection of perpendicular lines.
The Cartesian product of sets X and Y will result in the set of all 52 playing cards as shown in the tabular form.
The Cartesian product of two non-empty finite sets P and Q is the set of all ordered pairs of elements from P and Q.
In set-builder form, P x Q = {(p,q):p ∈ P, q ∈ Q}·
If either P = ∅ or Q = ∅ then P x Q = ∅·
If either P or Q is infinite, then P x Q = infinite·
If n (P) = m and n (Q) = n, then n ( P x Q ) = mn
Multiple Products:
If A_{1}, A_{2}, A_{3} ... A_{n} is a finite family of sets, then
A_{1} × A_{2} ×A_{3} ... ×A_{n}={(a_{1}, a_{2}, a_{3},…a_{n}):a_{1} ϵ A_{1}, a_{2} ϵA_{2}…a_{n} ϵA_{n}}
A × A× A = {(a, b, c): a, b. c ϵ A}
R × R = {(a, b): a, b ϵ R}
R × R × R = {(a, b, c): a, b. c ϵ R}
Commutative Property
Consider A = {4, 2} and B = {3, 1}
A x B = {(4,3),(4,1),(2,3),(2,1)}
B x A = {(3,4),(3,2),(1,4),(1,2)}
⇒ A x B ≠ B x A
This implies the Cartesian product is not commutative.
But if A and B are equal sets, then the Cartesian product of A and B is the same as the Cartesian product of B and A.
Distributive Property
A x (B ∪ C) = (A x B) ∪ (A x C)
Let A = {1, 2}, B = {3, 4, 5} and C = {5, 6}
B ∪ C = {3,4,5,6}
LHS = A x (B ∪ C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6)}
A × B = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
RHS = (A x B) ∪ (A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6)}
Hence, both the right hand side and the left hand side are sets consisting of the same ordered pairs.
Hence, this property is verified.
The other two results of the distributive property,
I. A x (B ∩ C) = (A x B) ∩ (A x C)
II. A x (B - C) = (A x B) - (A x C)
Pair of entries 'a' and 'b' which are separated by a comma and are enclosed within the bracket is called an ordered pair, where 'a' is the first element and 'b' is the second element.
The order, in which the entries are written in an ordered pair, is specific.
Ordered pair: A pair of entries grouped in a particular order, which are separated by a comma and enclosed within brackets.
If two ordered pairs are equal, then their corresponding first elements and second elements are equal.
If (a, b) = (1, 4) ⇒ a = 1 and b = 4.
Cartesian product of sets
Consider X = {a, b, m}, Y = {f, c, d, h}
Product of sets X and Y: The product of sets means to pair the elements of sets together in a systematic manner.
Cartesian product of sets X and Y is symbolically denoted by X cross Y.
X x Y = {(a,f),(a,c),(a,d),(a,h),(b,f),(b,c),(b,d),(b,h),(m,f),(m,c),(m,d),(m,h)}
Example
Let X = {♣,♦,♠,♥} and Y = {2,3,4,5,6,7,8,9,10,J,Q,K,A}
X × Y |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Jack(J) |
Queen(Q) |
King(K) |
Ace(A) |
Club(♣) |
(♣,2) |
(♣,3) |
(♣,4) |
(♣,5) |
(♣,6) |
(♣,7) |
(♣,8) |
(♣,9) |
(♣,10) |
(♣,J) |
(♣,Q) |
(♣,K) |
(♣,A) |
Diamond (♦) |
(♦,2) |
(♦,3) |
(♦,4) |
(♦,5) |
(♦,6) |
(♦,7) |
(♦,8) |
(♦,9) |
(♦,10) |
(♦,J) |
(♦,Q) |
(♦,K) |
(♦,A) |
Spade(♠) |
(♠,2) |
(♠,3) |
(♠,4) |
(♠,5) |
(♠,6) |
(♠,7) |
(♠,8) |
(♠,9) |
(♠,10) |
(♠,J) |
(♠,Q) |
(♠,K) |
(♠,A) |
Heart(♥) |
(♥,2) |
(♥,3) |
(♥,4) |
(♥,5) |
(♥,6) |
(♥,7) |
(♥,8) |
(♥,9) |
(♥,10) |
(♥,J) |
(♥,Q) |
(♥,K) |
(♥,A) |
The Cartesian product of two non-empty sets X and Y can be represented in the form of tabular cells or the point of intersection of perpendicular lines.
The Cartesian product of sets X and Y will result in the set of all 52 playing cards as shown in the tabular form.
The Cartesian product of two non-empty finite sets P and Q is the set of all ordered pairs of elements from P and Q.
In set-builder form, P x Q = {(p,q):p ∈ P, q ∈ Q}·
If either P = ∅ or Q = ∅ then P x Q = ∅·
If either P or Q is infinite, then P x Q = infinite·
If n (P) = m and n (Q) = n, then n ( P x Q ) = mn
Multiple Products:
If A_{1}, A_{2}, A_{3} ... A_{n} is a finite family of sets, then
A_{1} × A_{2} ×A_{3} ... ×A_{n}={(a_{1}, a_{2}, a_{3},…a_{n}):a_{1} ϵ A_{1}, a_{2} ϵA_{2}…a_{n} ϵA_{n}}
A × A× A = {(a, b, c): a, b. c ϵ A}
R × R = {(a, b): a, b ϵ R}
R × R × R = {(a, b, c): a, b. c ϵ R}
Commutative Property
Consider A = {4, 2} and B = {3, 1}
A x B = {(4,3),(4,1),(2,3),(2,1)}
B x A = {(3,4),(3,2),(1,4),(1,2)}
⇒ A x B ≠ B x A
This implies the Cartesian product is not commutative.
But if A and B are equal sets, then the Cartesian product of A and B is the same as the Cartesian product of B and A.
Distributive Property
A x (B ∪ C) = (A x B) ∪ (A x C)
Let A = {1, 2}, B = {3, 4, 5} and C = {5, 6}
B ∪ C = {3,4,5,6}
LHS = A x (B ∪ C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6)}
A × B = {(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
RHS = (A x B) ∪ (A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6)}
Hence, both the right hand side and the left hand side are sets consisting of the same ordered pairs.
Hence, this property is verified.
The other two results of the distributive property,
I. A x (B ∩ C) = (A x B) ∩ (A x C)
II. A x (B - C) = (A x B) - (A x C)