Cartesian Product in Set Relations Functions

The well-defined objects’ collection is termed as 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 constitute 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

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

(i) A×(BC)=(A×B)(A×C)(ii) A×(BC)=(A×B)(A×C)(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)\\

Theorem 2: For any three sets A, B, C A×(BC)=(A×B)(A×C)A\times (B-C)=(A\times B)-(A\times C)

Theorem 3: If A and B are any two non-empty sets, then A×B=B×AA=BA\times B=B\times A\Leftrightarrow A=B

Theorem 4: If AB,A\subseteq B, then A×A(A×B)(B×A)A\times A\subseteq (A\times B)\cap (B\times A)

Theorem 5: If AB,A\subseteq B, then A×CB×CA\times C\subseteq B\times C for any set C.

Theorem 6: IfAB,A\subseteq B, and CD,C\subseteq D, then A×CB×DA\times C\subseteq B\times D

Theorem 7: For any sets A, B, C, D (A×B)(CD)=(AC)×(BD)(A\times B)\cap (C\cup D)=(A\cap C)\times (B\cap D)

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

(i) A×(B×C)=(A×B)(A×C)(ii) A×(BC)=(A×B)(A×C)(i) \ A\times (B’\times C’)’=(A\times B)\cap (A\times C)\\ (ii) \ A\times (B’\cap C’)’=(A\times B)\cup (A\times C)

Properties Of Cartesian Product Of Sets

i.A×(BC)=(A×B)(A×C)ii.A×(BC)=(A×B)(A×C)iii.A×(BC)=(A×B)(A×C)iv.(A×B)(C×D)=(AC)×(BD)v. if AB then A×CB×Cvi.A×BB×Avii.(A×B)×CA×(B×C)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)\\ iii. A \times (B – C) = (A \times B) – (A \times C)\\ iv. (A \times B) \cap ( C \times D) =(A \cap C) \times (B \cap D)\\ v.\text \ if \ A \subset B \text \ then \ A \times C \subset B \times C\\ vi. A \times B \ne B \times A \\ vii.(A \times B) \times C \ne A \times (B \times C)

Non-commutativity and non-associativity

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

The Cartesian product A × B is not commutative, A×BB×A,{\displaystyle A\times B\neq B\times A,}

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). (A×B)×CA×(B×C){\displaystyle (A\times B)\times C\neq A\times (B\times C)}

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 of 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 + ….. +nCn−1 7 + nCn− 7n −1

= nC272 + nC373 +….+nCn7n, (nC0 = nCn, nC1 = nCn−1 etc.)

= 49 [nC2+ nC3 (7) + …… + nCn7n − 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.