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 a_{1 }= a_{2} and b_{1} = b_{2}.

**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 (a_{1}, b_{1}) is different from (b_{1}, a_{1}).

**A × B = {(a _{1}, b_{1}), (a_{1 }, b_{2}), (a_{1} ,b_{3}), (a_{2}, b_{1}), (a_{2}, b_{2}), (a_{2}, b_{3}), (a_{3}, b_{1}), (a_{3}, b_{2}), (a_{3}, b_{3})}.**

## 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

## Cartesian Product in Set Relations Functions Examples

**Example 1: **If A = [(x, y) : x^{2} + y^{2} = 25] and B = [(x,y) : x^{2}+ 9y^{2} = 144], then how many points does A ∩ B contain?

**Solution:**

A = Set of all values (x, y) : x^{2} + y^{2} = 25 = 5^{2}

B = x^{2} / 144 + y^{2} / 16 = 1

i.e., x^{2} / (12)^{2} + y^{2 }/ (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 = {8^{n} − 7n −1 : n ∈ N} and Y = {49 (n−1) : n ∈ N}, then how is X and Y related?

**Solution:**

Since 8^{n} − 7n −1 = (7+1)^{n} − 7n − 1 = 7^{n} + ^{n}C_{1 }7^{n−1}+ ^{n}C_{2}7^{n−2 }+ ….. +^{n}C_{n−1 }7 + ^{n}C_{n}− 7n −1

= ^{n}C_{2}7^{2 }+ ^{n}C_{3}7^{3 }+….+^{n}C_{n}7^{n}, (^{n}C_{0 }= ^{n}C_{n}, ^{n}C_{1} = ^{n}C_{n−1} etc.)

= 49 [^{n}C_{2}+ ^{n}C_{3} (7) + …… + ^{n}C_{n}7^{n − 2}]

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

For n = 1, 8^{n }− 7n − 1 = 8 − 7 − 1 = 0;

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

∴ 8^{n }− 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 2^{2}, i.e., 4 elements in common. Hence, n [(A × B) ∩ (B × A)] = 4.

## Frequently Asked Questions

### What do you mean by a set in mathematics?

In mathematics, a set is a collection of well-defined objects.

### How do we represent a set?

A set can be represented by statement form, roaster form and set builder form.

### What do you mean by the Cartesian product of 2 sets?

The Cartesian product of two sets A and B, denoted A × B, is the set of all ordered pairs where a is in A and b is in B. A×B = {(a, b): a ∈ A and b ∈ B}.

### What is A×(B⋃C)?

A×(B⋃C) = (A×B) ⋃ (A×C).

### Is Cartesian product A×B commutative?

No. Cartesian product A×B is not commutative.

## Comments