An ordered pair is a pair of elements with a specific order, usually written as $(a,b)$. The order in which the elements are listed is crucial: $(a,b)≠(b, a)$ unless $a=b$. In set theory, the cartesian product of two sets is the product of two non-empty sets in an ordered way. The Cartesian product comprises two words – Cartesian and product. The word Cartesian is named after the French mathematician and philosopher René Descartes (1596-1650).
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In this article, we will cover the concept of the ordered pairs and cartesian products of two sets. This concept falls under the broader category of sets, relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.
A pair of elements grouped together in a particular order is known as an ordered pair.
e.g. : $(a, b),(3,5),(-1,0) \ldots$
The ordered pairs $(a, b)$ and $(b, a)$ are different. Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal. (i.e.) $(x, y)=(u, v)$ if and only if $x=u, y=v$.
The cartesian product of two non-empty sets $A$ and $B$ is the set of all ordered pairs $( x, y )$, where $x \in A$ and $y \in B$.
Symbolically, we write it as $\mathrm{A} \times \mathrm{B}$ and it is read as ' A cross B '.
$
A \times B=\{(a, b): a \in A, b \in B\}
$
For example, If $A=\{1,2\}$ and $B=\{a, b\}$
Then $A \times B=\{(1, a),(1, b),(2, a),(2, b)\}$
As we know, an empty set does not have any elements. The cardinality of an empty set or the size is also zero. The cartesian product of a set, say $A$ and the empty set $\phi$, is an empty set only.
$\mathrm{A} \times \phi=\phi$
1. Non-commutative: The cartesian product of two sets is non-commutative because the order of elements is not the same.
$\begin{aligned}
& A \times B \neq B \times A \text { in general. For instance: } A \times B=\{(a, b) \mid a \in A \text { and } b \in B\} \\
& B \times A=\{(b, a) \mid b \in B \text { and } a \in A\}
\end{aligned}$
$A \times B$ and $B \times A$ contain different pairs unless $\mathrm{A}=\mathrm{B}$ and even then, the order of elements in the pairs is different.
2. Cardinality: If $A$ has $m$ elements and $B$ has $n$ elements, then $A \times B$ has $\mathrm{m} \times \mathrm{n}$ elements.
$|A \times B|=|A| \cdot|B|$
3. Associativity with Triples: For sets $A, B,$ and $C$, the associative property is not applied due to unequal ordered.
$(A \times B) \times C \neq A \times(B \times C)$
However, both are sets of ordered pairs. Specifically:
$\begin{aligned}
& (A \times B) \times C=\{((a, b), c) \mid a \in A, b \in B, c \in C\} \\
& A \times(B \times C)=\{(a,(b, c)) \mid a \in A, b \in B, c \in C\}
\end{aligned}$
4. Distributive over Union: For sets $A, B,$ and $C$, the distribution property of sets follow such rule-
$A \times(B \cup C)=(A \times B) \cup(A \times C)(A \cup B) \times C=(A \times C) \cup(B \times C)$
If there are $p$ elements in $A$ and $q$ elements in $B$, then there will be $p q$ elements in $A \times B$, (i.e.), if $n(A)=p$ and $n(B)=q$, then $n(A \times B)=p q$.
If $A, B, C$, and $D$ are any four sets, then
1. $A \times (B \cup C)=(A \times B) \cup(A \times C)$
2. $A \times (B \cap C)=(A \times B) \cap(A \times C)$
3. $A \times (B-C)=(A \times B)-(A \times C)$
4. $(A \times B) \cap(C \times D)=(A \cap C) \times(B \times D)$
5. If $A \subseteq B$, then $(A \times C) \subseteq(B \times C)$
6. If $A \subseteq B$, then $A \times A \subseteq(A \times B) \cap(B \times A)$
7. If $A \subseteq B$ and $C \subseteq D$, then $A \times C \subseteq B \times D$
Example 1: If a set has $20$ elements. How many ordered pairs with distinct elements can be formed?
Solution:
As we learned
Let $A=\{1,2,3,20\}$
Let the ordered pair be $(a, b)$.
Selecting $2$ elements from $20$ elements is $_{20}C_{2}$
The number of ordered pairs $_{20}C_{2} =190$
Hence, the answer is $190$.
Example 2: Given the set $A=\{1,2,3\}$. How many ordered pairs can be formed with both the elements belonging to set $A$ ?
Solution:
Ordered pairs can be $(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2)$ and $(3,3)$
So, a total of $9$ such ordered pairs are possible.
Hence, the answer is $9$.
Example 3: What is the condition for $\mathrm{A} \times \mathrm{B}=\mathrm{B} \times \mathrm{A}$ being true?
1) Always true
2) Never true
3) When $A=B$
4) In most of the cases
Solution:
If $A=B$, then
$A \times B=A \times A$ and
$B \times A=A \times A$
So, both are equal.
Hence, $A \times B=B \times A$ is true if both sets are identical.
Hence, the answer is the option 3.
Example 4: If the ordered pair $(x-3, y+2)$ is the same as $(5,9)$, then what are the values of $x$ and $y$ ?
Solution:
Equality of Ordered Pair
$\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ are equal iff $a_1=a_2$ and $b_1=b_2$
So
$x - 3 = 5 y = 7$
$x = 8$
And
$y + 2 = 9$
Hence, the answer is $x=8$ and $y=7$.
Example 5: If $n(A)=5$ and $n(B)=7$ then $n(B \times A)$ equals
Solution:
$
\begin{aligned}
n(B \times A) & =n(B) \times n(A) \\
& =7 \times 5=35
\end{aligned}
$
Hence, the answer is $35$.
A pair of elements grouped together in a particular order is known as an ordered pair.
Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal.
The cartesian product of two non-empty sets $A$ and $B$ is the set of all ordered pairs ( $x, y$ ), where $\mathrm{x} \in \mathrm{A}$ and $\mathrm{y} \in \mathrm{B}$.
We know, that if $A$ and $B$ are finite sets, then $n(A \times B)=n(A) \cdot n(B)$
Here,
$40=\mathrm{n}(\mathrm{A}) \times \mathrm{n}(\mathrm{B})$
$\therefore n(B)=\frac{40}{4}=10$
As we learned
The cartesian product is a set of ordered pairs of $(a, b)$ where $a \in A$ and $b \in B$.
$A=\{1,2,3\} \text { and } B=\{a, b\}$
Then, $A \times B=\{(1, a) ;(1, b) ;(2, a) ;(2, b) ;(3, a) ;(3, b)\}$
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