Cartesian Product in Set Relations Functions

The well-defined objects’ collection is termed a set. A set with finite elements is a finite set, whereas with infinite elements is an infinite set. A set can be represented in a roster and set builder form. The collection of ordered pairs, which consists of one object from each set, is a relation. It can be represented as a Cartesian product of two sets where all the elements have a common property. While plotting a graph, the x-coordinate is followed by the y-coordinate in an ordered way. In two non-empty sets, the first element is from set A, and the second element is from set B. The collection of such ordered pairs constitutes a Cartesian product. The ordered pairs are said to be equal if a₁ = a₂ and b₁ = b₂.

Example: Let A = {a, b, c} and B = {p,q}.

Then A × B = {(a, p), (a, q), (b, p),(b, q), (c, p), (c, q)}

Also, B × A = {(p, a), (p, b), (p, c), (q, a), (q, b), (q, c)}

Cartesian Product in Set Relations Functions Formula

The Cartesian product is denoted by A × B. The pair of (a₁, b₁) is different from (b₁, a₁).

A × B = {(a₁, b₁), (a₁ , b₂), (a₁ ,b₃), (a₂, b₁), (a₂, b₂), (a₂, b₃), (a₃, b₁), (a₃, b₂), (a₃, b₃)}.

Important Theorems on Cartesian Product of Sets

Theorem 1: For any three sets A, B, C.

Theorem 2: For any three sets A, B, C

Theorem 3: If A and B are any two non-empty sets, then

Theorem 4:

Theorem 5:

Theorem 6:

Theorem 7: For any sets A, B, C, D

Theorem 8: For any three sets A, B, C

Properties of Cartesian Product of Sets

Non-commutativity and Non-associativity

Let A, B, C, and D be sets.

The Cartesian product A × B is not commutative, 

because the ordered pairs are reversed unless at least one of the following conditions is satisfied:

• A is equal to B, or
• A or B is the empty set.

For example:

A = {1,2}; B = {3,4}

A × B = {1,2} × {3,4} = {(1,3), (1,4), (2,3), (2,4)}

B × A = {3,4} × {1,2} = {(3,1), (3,2), (4,1), (4,2)}

A = B = {1,2}

A × B = B × A = {1,2} × {1,2} = {(1,1), (1,2), (2,1), (2,2)}

A = {1,2}; B = ∅

A × B = {1,2} × ∅ = ∅

B × A = ∅ × {1,2} = ∅

The Cartesian product is not associative (unless one of the involved sets is empty). 

If for example A = {1}, then (A × A) × A = { ((1,1),1) } ≠ { (1,(1,1)) } = A × (A × A).

Also Read

Sets Relations and Functions

Functions and Its Types

Cartesian Product in Set Relations Functions Examples

Example 1: If A = [(x, y) : x² + y² = 25] and B = [(x,y) : x²+ 9y² = 144], then how many points does A ∩ B contain?

Solution:

A = Set of all values (x, y) : x² + y² = 25 = 5²

B = x² / 144 + y² / 16 = 1

i.e., x² / (12)² + y² / (4)² = 1.

Clearly, A ∩ B consists of four points.

Example 2: Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then what is R⁻¹ o R?

Solution:

We first find R⁻¹ 

We have R⁻¹ = {(5, 4) ; (4, 1) ; (6, 4) ; (6, 7) ; (7, 3)}.

We now obtain the elements of R⁻¹ o R. We first pick the element of R and then R⁻¹.

Since (4, 5) ∈ R and (5, 4) ∈ R⁻¹ , we have (4, 4) ∈ R⁻¹ oR

Similarly, (1, 4) ∈ R,(4, 1) ∈ R⁻¹ ⇒ (1,1) ∈ R⁻¹ o R

(4, 6) ∈ R, (6, 4) ∈ R⁻¹ ⇒ (4, 4) ∈ R⁻¹ o R,

(4, 6) ∈ R, (6, 7) ∈ R⁻¹ ⇒ (4, 7) ∈R⁻¹ o R

(7, 6) ∈ R, (6, 4) ∈ R⁻¹ ⇒ (7, 4) ∈ R⁻¹ o R,

(7, 6) ∈ R, (6, 7) ∈ R⁻¹ ⇒ (7, 7) ∈ R⁻¹ o R

(3, 7) ∈ R, (7, 3) ∈ R⁻¹ ⇒ (3, 3) ∈ R⁻¹ oR,

Hence, R⁻¹ o R = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.

Example 3: Let R be a relation on the set N be defined by {(x, y)| x, y i ^ N, 2x + y = 41}. Then R is _________ .

Solution:

On the set N of natural numbers, R = {(x, y) : x, y ∈ N , 2x + y = 41}.

Since (1, 1) ∉ R as 2.1 + 1 = 3 ≠ 41.

So, R is not reflexive.

(1, 39) ∈ R but (39, 1) ∉ R

So R is not symmetric.

(20, 1) (1, 39) ∈ R

But (20, 39) ∉ R.

So R is not transitive.

Example 4: If X = {8ⁿ − 7n −1 : n ∈ N} and Y = {49 (n−1) : n ∈ N}, then how is X and Y related?

Solution:

Since 8ⁿ − 7n −1 = (7+1)ⁿ − 7n − 1 = 7ⁿ + ⁿC₁ 7ⁿ⁻¹+ ⁿC₂7ⁿ⁻² + ….. +ⁿC_n−1 7 + ⁿC_n− 7n −1

= ⁿC₂7² + ⁿC₃7³ +….+ⁿC_n7ⁿ, (ⁿC₀ = ⁿC_n, ⁿC₁ = ⁿC_n−1 etc.)

= 49 [ⁿC₂+ ⁿC₃ (7) + …… + ⁿC_n7^{n − 2}]

∴ 8ⁿ − 7n − 1 is a multiple of 49 for n ≥ 2

For n = 1, 8ⁿ − 7n − 1 = 8 − 7 − 1 = 0;

For n = 2, 8ⁿ − 7n − 1 = 64 − 14 − 1 = 49

∴ 8ⁿ − 7n − 1 is a multiple of 49 for all n ∈ N.

∴ X contains elements which are multiples of 49, and clearly, Y contains all multiples of 49.

∴ X ⊆ Y.

Example 5: Let A = {1, 2, 3, 4, 5}; B = {2, 3, 6, 7}. Then what is the number of elements in (A × B) ∩ (B × A)?

Solution:

Here, A and B sets have 2 elements in common, so A × B and B × A have 2², i.e., 4 elements in common. Hence, n [(A × B) ∩ (B × A)] = 4.