Cartesian Product: Definition, Properties & Examples

Cartesian Product – A Cartesian product is based on a set of ordered sets. It’s the set of all feasible ordered combinations that includes one member from each of those sets. Do you remember words such as axes (\(x\)-axis, \(y\)-axis), origin, and others while plotting a graph paper? For example, \(\left({2,3} \right)\) represents a value of \(2\) on the (\(x\)-axis and \(3\) on the \(y\)-axis, which is not the same as \(\left( {3,2} \right).\) The order of representation is fixed, with the value of the \(x\) coordinate coming first, followed by the value of the \(y\) coordinate. The Cartesian product is the ordered product of two elements, such as \(x\) and \(y.\) Let us know about the Cartesian product in detail.

What is Cartesian Product of Sets?

Suppose \(A\) is a set of two colours and \(B\) is a set of three objects, i.e., \(A = \left\{{{\text{red,blue}}} \right\}\) and \(B = \left\{{b,c,s} \right\}\)
Where \(b,c\) and \(s\) represent a belt, coat and shirt, respectively.

How many pairs of objects can be created from these two sets? If we proceed in a very orderly manner, we can see that there will be distinct pairs as given below: \(\left( {{\text{red}},b} \right),\left( {{\text{red}},c} \right),\left( {{\text{red}},s} \right)\left( {{\text{blue}},b} \right),\left( {{\text{blue}},c} \right),\left( {{\text{blue}},s} \right)\)
Thus, we get \(6\) distinct objects. The word Cartesian product is made of two words, i.e., Cartesian and product. René Descartes, a French mathematician and philosopher has coined the term Cartesian.

The cartesian product of \(2\) non-empty sets \(A\) and \(B\) is the set of all possible ordered pairs where the first component is from \(A\) and the second component is from \(B.\) The resultant collection of ordered pairs is denoted by \(A \times B.\)
If \(A\) and \(B\) are two non-empty sets, the Cartesian product of \(A\) and \(B\) is the set, denoted by
\(A \times B = \{ \left({a,b} \right):\left({a \in A} \right)\) and \(\left( {b \in B} \right)\)

The following points need special attention. The Cartesian product of \(2\) sets is a set, and the elements of that set are ordered pairs. In each ordered pair, the first component is an element of \(A,\) and the second component is an element of \(B.\)
If either \(A\) or \(B\) is the null set, then \(A \times B\) will also be empty set, i.e., \(A \times B = \phi .\)
From the above illustration, we note that \(A \times B = \left\{ {\left( {{\text{red}},b} \right),\left( {{\text{red}},c} \right),\left( {{\text{red}},s} \right)\left( {{\text{blue}},b} \right),\left( {{\text{blue}},c} \right)\left( {{\text{blue}},s} \right)} \right\}\)

Let us take another example
If \(A = \left\{{x,y,z} \right\}\) and \(B = \left\{{1,2,3} \right\}\) then \(A \times B = \left\{{\left({x,1} \right),\left({x,2} \right),\left({x,3} \right),\left({y,1} \right),\left({y,2} \right),\left({y,3} \right),\left({z,1} \right),\left({z,2} \right),\left({z,3} \right)} \right\}\)

The cartesian product can be easily obtained using a cartesian product calculator, which can be searched on google. There you just need to make an entry in set \(A\) and set \(B,\) then click on a button cartesian product. You will get \(A \times B\) in a moment.

Some points to note are:

1. If there are \(n\) elements and \(m\) elements in set \(A\) and set \(B\) respectively, then there will be \(nm\) elements in set \(A \times B.\)
2. If \(A\) and \(B\) are non-empty sets and either \(A\) or \(B\) is an infinite set, the number of elements in \(A \times B\) is infinite.
3. \(A \times B \times C = \left\{ {\left( {a,b,c} \right):a \in A,b \in B\,{\text{and}}\,c \in C} \right\}\). Here \(\left({a,b,c} \right)\) is called an ordered triplet.

Properties of Cartesian Product

The properties of the Cartesian product are as follows:

1. The Cartesian product is non-commutative:
\(A \times B \ne B \times A\)
It means the order of multiplication plays an important role in finding the cartesian product.
2. \(A \times B = B \times A,\) if only \(A = B\)
3. \(A \times B = \phi ,\) if either \(A = \phi \) or \(B = \phi \)
4. The Cartesian product is associative:
\(\left({A \times B} \right) \times C = A \times \left({B \times C}\right)\). It means the cartesian product of the three-set is the same, i.e., it doesn’t depend upon which bracket is multiplied first as the final result will be the same.
5. Distributive property over a set intersection:
\(A \times \left({B \cap C} \right) = \left({A \times B} \right) \cap \left({A \times C} \right)\)
6. Distributive property over set union:
\(A \times \left({B \cup C} \right) = \left({A \times B} \right) \cup \left({A \times C} \right)\)
7. If \(A \subseteq B,\) then \(A \times C \subseteq B \times C\) for any set \(C.\)

Cartesian Product Example

Consider the two sets:
\(A = \left\{{DL,MP,KA} \right\},\) where \(DL,MP,KA\) represent Delhi,
Madhya Pradesh and Karnataka, respectively and \(B = \left\{{01,02,03} \right\}\) represent codes for the licence plates of vehicles issued
by \(DL,MP\) and \(KA.\)

If the three states of Delhi, Madhya Pradesh, and Karnataka were to create codes for car licence plates, they would have to start with an element from set \(A\) .what are the pairs available from these sets and how many such pairs will be possible?

The available pairs are \(\left({DL,01} \right),\left({DL,02} \right),\left({DL,03} \right),\left({MP,01} \right),\left({MP,02} \right),\left({MP,03} \right),\left({KA,01} \right),\left({KA,02} \right),\left({KA,03} \right)\) and the product of set⁡ \(A\) and set \(B\) is given by \(A \times B\)
\( = \left\{{\left({DL,01} \right),\left({DL,02} \right),\left({DL,03} \right),\left({MP,01} \right),\left({MP,02} \right),\left({MP,03} \right),\left({KA,01} \right),\left({KA,02} \right),\left({KA,03} \right)} \right\}\)

It can be seen that there will be \(9\) such pairs in the Cartesian product since there are three elements in sets \(A\) and \(B\) each. This gives us a total of \(9\) possible codes. It’s also worth noting that the pairing of these aspects is critical. For example, the code \(\left({DL,01} \right)\) will not be the same as the code \(\left({01,DL} \right).\)

Now consider another two sets \(A = \left\{{{a_1},{a_2}} \right\}\) and \(B = \left\{{{b_1},{b_2},{b_3},{b_4}}\right\}\)\(A \times B = \left\{{\left({{a_1},{b_1}} \right),\left({{a_1},{b_2}} \right),\left({{a_1},{b_1}} \right),\left( {{a_1},{b_4}} \right),\left({{a_2},{b_1}} \right),\left({{a_2},{b_2}} \right),} \right.\left.{,\left({{a_2},{b_3}} \right),\left({{a_2},{b_4}} \right)} \right\}\)

The \(8\) ordered pairs thus formed can represent the position of points in the plane if \(A\) and \(B\) are subsets of the set of real numbers.

It is clear that the point in the position \(\left({{a_1},{b_2}} \right)\) will be distinct from the point in the position \(\left({{b_2},{a_1}} \right).\)

Solved Examples – Cartesian Product

Let us understand the concept through binomial distribution examples and solutions.

Q.1. If \(P = \left\{{a,b,c} \right\}\) and \(Q = \left\{ r \right\},\) form the sets \(P \times Q\) and \(Q \times P.\) Are these two products equal?
Ans: By the definition of the cartesian product,
\(P \times Q = \left\{{\left({a,r} \right),\left({b,r} \right),\left({c,r} \right)} \right\}\) and \(Q \times P = \left\{{\left({r,a} \right),\left({r,b} \right),\left({r,c} \right)} \right\}\)
Since, by the definition of equality of ordered pairs, the pair \(\left({a,r} \right)\) is not equal to the pair \(\left({r,a} \right),\)
We conclude that \(P \times Q \ne Q \times P\)
However, the number of elements in each set will be the same.

Q.2. Let \(A = \left\{{1,2,3} \right\},B = \left\{{3,4} \right\}\) and \(C = \left\{{4,5,6} \right\}.\) Find
(i) \(A \times \left({B \cap C} \right)\)
(ii) \(\left({A \times B} \right) \cap \left({A \times C} \right)\)
(iii) \(A \times \left({B \cup C} \right)\)
(iv) \(\left({A \times B} \right) \cup \left({A \times C} \right)\)
Ans: (i) By the definition of the intersection of two sets, \(\left({B \cap C} \right) = \left\{ 4 \right\}.\)
Therefore, \(A \times \left({B \cap C} \right) = \left\{{\left({1,4} \right),\left({2,4} \right),\left({3,4} \right)} \right\}\)
(ii) Now \(\left({A \times B} \right) = \left\{{\left({1,3} \right),\left({1,4} \right),\left({2,3} \right),\left({2,4} \right),\left({3,3} \right),\left({3,4} \right)} \right\}\)
and \(\left({A \times C} \right) = \left\{{\left({1,4} \right),\left({1,5} \right),\left({1,6} \right),\left({2,4} \right),\left({2,5}\right),\left({2,6}\right),\left({3,3} \right),\left({3,4} \right),\left({3,5} \right),\left({3,6} \right)} \right\}\)
Therefore, \(\left({A \times B} \right) \cap \left({A \times C} \right) = \left\{{\left({1,4} \right),\left({2,4} \right),\left({3,4} \right)} \right\}\)
(iii) Since, \(\left({B \cup C} \right) = \left\{ {3,4,5,6} \right\},\) we have \(A \times \left({B \cup C} \right) = \left\{{\left({1,3} \right),\left({1,4} \right),\left({1,4} \right),\left({1,6} \right),\left({2,3} \right),\left({2,4} \right),\left({2,5} \right),\left({2,6}\right),\left({3,4} \right),\left({3,5} \right),\left({3,6} \right)} \right\}\)
(iv) Using the sets \(A \times B\) and \(A \times C\) from part (ii) above, we obtain \(\left({A \times B} \right) \cup \left({A \times B} \right) \cup \left({A \times C} \right) = \left\{{\left({1,3} \right),\left({1,4} \right),\left({1,5} \right),\left({1,6} \right),\left({2,3} \right),\left({2,4} \right),\left({2,5} \right)} \right.\) \(\left.{\left({2,6} \right),\left({3,4} \right),\left({3,5} \right),\left({3,6} \right)} \right\}\)

Q.3. If \(P = \left\{{1,2} \right\},R = \left\{{3,4} \right\}\) and \(S = \left\{{5,6} \right\},\) then find set \(P \times R \times S\)
Ans: We have, \(P \times R = \left\{{\left({1,3} \right),\left({1,4} \right),\left({2,3} \right),\left({2,4} \right)} \right\}\)
So, \(P \times R \times S = \left\{{\left({1,3,5} \right),\left({1,3,6} \right),\left({1,4,5}\right),\left({1,4,6}\right),\left({2,3,5} \right),\left({2,3,6} \right),\left({2,4,5} \right),\left({2,4,6} \right)} \right\}\)

Q.4. If \(R\) is the set of all real numbers, what do the cartesian products \(R \times R\) and \(R \times R \times R\) represent?
Ans: The Cartesian product \(R \times R\) i.e. the set \(R \times R = \left\{{\left({x,y} \right):x,y\in R}\right\}\) represents the coordinates of all the points in two-dimensional space and the cartesian product \(R \times R \times R\) i.e. the set \(R \times R \times R = \left\{{\left({x,y,z} \right):x,y,z \in {\mathbf{R}}} \right\}\) represents the coordinates of all the points in three-dimensional space.

Q.5. If \(A \times B = \left\{{\left({p,q} \right),\left({p,r} \right),\left({m,q} \right),\left({m,r} \right)} \right\},\) find \(A\) and \(B.\)
Ans: Since, \(A \times B = \left\{{\left({p,q} \right),\left({p,r} \right),\left({m,q} \right),\left({m,r} \right)} \right\}\)
So,
\(A = \) set of first elements \( = \left\{{p,m} \right\}\)
\(B = \) set of second elements \( = \left\{{q,r} \right\}\)

Summary

In this article, we learnt about the Cartesian product, its properties and examples. We understood that the Cartesian product of two sets is a set. A set consists of a collection of elements. In this case, the elements of a Cartesian product are the ordered pairs. We should treat an ordered pair as a single object that consists of two other objects in a specified order. The use of a cartesian product is to find out the set of all possible ordered pairs from given sets.

FAQs

The following questions will help students understand the cartesian product in a better way.

Q.1. How do you find a Cartesian product?
Ans: If \(A\) and \(B\) are two non-empty sets, the Cartesian product of \(A\) and \(B\) is the set \(A \times B = \left\{{\left({a,b} \right):\left({a \in A} \right)and\,\left({b \in B} \right)} \right\}.\)
To find the cartesian product between two sets, we find out the set of all possible ordered pairs, with the first element coming from the first set and the second element coming from the second set.

Q.2. What is the use of a Cartesian product?
Ans: The basic use of a cartesian product is to find out the set of all possible ordered pairs from given sets. One real-life application is that most computers show images as a raster of dots known as pixels, each of which can be addressed using its coordinates. These coordinates are ordered pairs and so Cartesian product elements. We colour in the dots in the raster that depicts the Cartesian product that corresponds to elements of a subset of a Cartesian product to display a picture.

Q.3. What is the Cartesian product of \(3\) sets?
Ans: The cartesian product of three non-empty sets \(A,B\) and \(C\) is the set of all possible ordered pairs where the first component is from \(A\) and the second component is from \(B\) and the third component is from \(C.\)The resultant collection of the ordered triplet is denoted by \(A \times B \times C\)
If \(A,B\) and \(C\) are three non-empty sets, the Cartesian product of three sets is the set, denoted by
\(A \times B \times C = \left\{{\left({a,b,c} \right):a \in A,b \in B\,{\text{and}}\,c \in C} \right\}\)

Q.4. What is the Cartesian product of \(A \times B \times C\)?
Ans: If \(A,B\) and \(C\) are three non-empty sets, the Cartesian product
\(A \times B \times C = \left\{{\left({a,b,c} \right):a \in A,b \in B\,{\text{and}}\,c \in C} \right\}\)
For example:
If \(A = \left\{{1,2} \right\},B = \left\{{3,4} \right\}\) and \(C = \left\{{5,6} \right\}\) then \(A \times B \times C = \left\{{\left({1,3,5} \right),\left({1,3,6} \right),\left({1,4,5} \right),\left({1,4,6} \right),\left({2,3,5} \right),\left({2,3,6} \right),\left({2,4,5} \right),\left({2,4,6} \right)} \right\}\)

Q.5. What is the Cartesian product of functions?
Ans: If \(f\left( x \right) = x;0 < x < 3,x \in N\) and \(g\left( x \right) = {x^2};0 < x < 3,x \in N\) then Cartesian product \(f\left( x \right) \times g\left( x \right) = \left\{ {\left( {1,\,1} \right),\,\left( {1,\,4} \right),\,\left( {2,\,1} \right),\,\left( {2,\,4} \right)} \right\}\)

We hope this detailed article on the cartesian product will be helpful to you.