Functions
1.0 Definitions
2.0 Relation
3.0 Types of Relation
4.0 Functions
5.0 Standard Real Functions and their Graphical Representation
5.10 Reciprocal Function
5.1 Constant Function
5.2 Identity Function
5.3 Modulus Function
5.4 Greatest Integer Function or Floor Function
5.5 Smallest Integer Function or Ceiling Function
5.6 Fractional Part Function
5.7 Signum Function
5.8 Exponential Function
5.9 Logarithmic Function
5.11 Square Root or Radical Function
5.12 Square Function
5.13 Cube Function
5.14 Cube Root Function
6.0 Operations on Real Functions
7.0 Types of Functions
8.0 Composition of a Function
9.0 Inverse of a Function
1.1 Cartesian Product of Sets
5.1 Constant Function
5.2 Identity Function
5.3 Modulus Function
5.4 Greatest Integer Function or Floor Function
5.5 Smallest Integer Function or Ceiling Function
5.6 Fractional Part Function
5.7 Signum Function
5.8 Exponential Function
5.9 Logarithmic Function
5.11 Square Root or Radical Function
5.12 Square Function
5.13 Cube Function
5.14 Cube Root Function
Let $A$ and $B$ be two non-empty sets. The sets of all ordered pairs $(a,b)$ such that $a \in A$ and $b \in B$ is called the cartesian product of the sets $A$ and $B$ and is denoted by $A \times B$.
$$A \times B = \{ (a,b):\;\;a \in A\;\;and\;\;b \in B\} $$
and
$$B \times A = \{ (b,a):\;\;a \in A\;\;and\;\;b \in B\} $$
$$If\;\;A = \emptyset \;\;or\;\;B = \emptyset ,\;\;then\;\;A \times B = \emptyset $$
Example 3. Write the Cartesian products as defined below for the sets
$A = \{ x:x\;is\;a\;odd\;natural\;number\;less\;than\;10\} $ and $B = \{ x:x\;is\;divisible\;by\;three\;and\;less\;than\;10\} $
i. $A \times B$
ii. $B \times A$
iii.$B \times B$
iv. $A \times A$
Solution:
$\begin{equation} \begin{aligned} From\;the\;definition, \\ A = \{ 1,3,5,7,9\} \\ B = \{ 3,6,9\} \\\end{aligned} \end{equation} $
$i.\;A \times B = \{ (1,3),(1,6),(1,9),(3,3),(3,6),(3,9),(5,3),(5,6),(5,9),(7,3),(7,6),(7,9),(9,3),(9,6),(9,9)\}$
$ii.\;A \times B =\{ (3,1),(3,3),(3,5),(3,7),(3,9),(6,1),(6,3),(6,5),(6,7),(6,9),(9,1),(9,3),(9,5),(9,7),(9,9)\}$
$iii.\;B \times B = \{ (3,3),(3,6),(3,9),(6,3),(6,6),(6,9),(9,3),(9,6),(9,9)\}$
$iv.\;A \times A = \{ (1,1),(1,3),(1,5),(1,7),(1,9),(3,1),(3,3),(3,5),(3,7),(3,9),(5,1),(5,3),(5,5),(5,7),(5,9),$
$(7,1),(7,3),(7,5),(7,7),(7,9),(9,1),(9,3),(9,5),(9,7),(9,9)\}$
Example 4. Find $A$ and $B$ if
i. $A \times B = \{ (1,a),(2,a),(1,e),(2,e),(3,a),(3,e)\} $
ii.$B \times A = \{ (a,e),(a,i),(o,e),(o,i),(u,e),(u,i)\} $
iii. $A \times B = \emptyset $
iv. $A \times B = B \times A = \{ (1,1)\} $
Solution: $In\;A \times B,\;the\;element,\;(a,b),\;is\;such\;that\;a \in A\;and\;b \in B$
$\begin{equation} \begin{aligned} i.\;A \times B = \{ (1,a),(2,a),(1,e),(2,e),(3,a),(3,e)\} \\ By\;definition,\; \\ A = \{ 1,2,3\} \;\;and\;\;B = \{ a,e\} \\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} ii.\;B \times A = \{ (a,e),(a,i),(o,e),(o,i),(u,e),(u,i)\} \\ By\;definition, \\ B = \;\{ a,o,u\} \;and\;A = \{ e,i\} \\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} iii.\;A \times B = \emptyset \\ Since,\;the\;cartesian\;product\;is\;a\;null\;set,\;either\;A\;is\;a\,null\;set\;or\;B\;is\;a\;null\;set\\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} iv.\;A \times B = B \times A = \{ (1,1)\} \\ A \times B = B \times A \Rightarrow A = B \\ A = B = \{ 1\} \\\end{aligned} \end{equation} $
Cartesian Product of Three Sets:
Let $A$, $B$ and $C$ be three sets. Then, $A \times B \times C$ is the set of all ordered triplets having first element from $A$, second element from $B$, and third element from $C$.
$$A \times B \times C = \{ (a,b,c):\;a \in A,\;\;b \in B\;\;and\;\;c \in C\} $$
Example 5. Write cartesian products of the following, $A = \{ a,e,i\} $, $B = \{ 2,4\} $ and $C = \{ x,y\} $
$i.A \times B \times C$ $ii.B \times A \times C$ $iii.B \times B \times C$
Solution: Given
$\begin{equation} \begin{aligned} A = \{ a,e,i\} \\ B = \{ 2,4\} \\ C = \{ x,y\} \\\end{aligned} \end{equation} $
$i.A \times B \times C$
$$\begin{equation} \begin{aligned} A \times B \times C = \{ (a,2,x),(a,2,y),(a,4,x),(a,4,y), \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(e,2,x),(e,2,y),(e,4,x),(e,4,y), \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(i,2,x),(i,2,y),(i,4,x),(i,4,y)\} \\\end{aligned} \end{equation} $$
$ii.\;B \times A \times C$
$$\begin{equation} \begin{aligned} B \times A \times C = \{ (2,a,x),(2,a,y),(4,a,x),(4,a,y), \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2,e,x),(2,e,y),(4,e,x),(4,e,y), \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2,i,x),(2,i,y),(4,i,x),(4,i,y)\} \\\end{aligned} \end{equation} $$
$iii.\;B \times B \times C$
$$\begin{equation} \begin{aligned} B \times B \times C = \{ (2,2,x),(2,2,y),(2,4,x),(2,4,y), \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(4,2,x),(4,2,y),(4,4,x),(4,4,y)\} \\\end{aligned} \end{equation} $$
Example 6. Find $A$, $B$ and $C$ from the following cartesian products given,
$i.$ $A \times B \times C = \{ (e,2,h),(e,2,s),(i,2,h),(i,2,s)\} $
$ii.$ $C \times B \times C = \{ (1,2,1),(1,3,1),(2,2,2),(2,3,2)\} $
$iii.$ $A \times B \times C = \emptyset $
Solution:
$i.$ $A \times B \times C = \{ (e,2,h),(e,2,s),(i,2,h),(i,2,s)\} $ $$\begin{equation} \begin{aligned} A = \{ e,i\} \\ B = \{ 2\} \\ C = \{ h,s\} \\\end{aligned} \end{equation} $$
$ii.$ $C \times B \times C = \{ (1,2,1),(1,3,1),(2,2,2),(2,3,2)\} $ $$\begin{equation} \begin{aligned} B = \{ 2,3\} \\ C = \{ 1,2\} \\\end{aligned} \end{equation} $$
$iii.$ $A \times B \times C = \emptyset $
Since the product is an empty set, either of the sets is an empty set, i.e. either $A = \emptyset $, or $B = \emptyset $ or $C = \emptyset $.
Number of Elements in the Cartesian Product of Two Sets:
Theorem: If $A$ and $B$ are two finite sets, then $n(A \times B) = n(A) \times n(B)$.
Proof: Let $A = \{ {a_1},{a_2},{a_3},....,{a_m}\} $ and $B = \{ {b_1},{b_2},{b_3},....,{b_n}\} $ be two sets having $m$ and $n$ elements respectively. Then,
$$\begin{equation} \begin{aligned} A \times B = [({a_1},{b_1}),({a_1},{b_2}),({a_1},{b_3}),......,({a_1},{b_n}) \\ \;\;\;\;\;\;\;\;\;\;\;\;({a_2},{b_1}),({a_2},{b_2}),({a_2},{b_3}),......,({a_2},{b_n}) \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots \;\;\;\;\;\;\;\;\;\; \vdots \;\;\;\;\;\;\;\;\;\; \vdots \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \vdots \\ \;\;\;\;\;\;\;\;\;\;\;\;({a_m},{b_1}),({a_m},{b_2}),({a_m},{b_3}),......,({a_m},{b_n})] \\\end{aligned} \end{equation} $$
Clearly, there are $m$ rows of order pairs and each row has $n$ distinct ordered pairs.
So, $A \times B$ has $mn$ elements.
Thus, $A = \{ {a_1},{a_2},{a_3},....,{a_m}\} $ and $B = \{ {b_1},{b_2},{b_3},....,{b_n}\} $
Some special cases:
- If either $A$ or $B$ is an infinite set, then $A \times B$ is an infinite set.
- If $A$, $B$, $C$ are finite sets, then $n(A \times B \times C) = n(A) \times n(B) \times n(C)$
Example 7. Find the number of elements in each of the following
| $\begin{equation} \begin{aligned} i.\;A = \{ 2,3,1,9\} \\ B = \{ a,v\} \\ n(A \times B) = ? \\\end{aligned} \end{equation} $ | $\begin{equation} \begin{aligned} ii.\;A = \{ x,y\} \\ n(A \times B) = 6 \\ n(B) = ? \\\end{aligned} \end{equation} $ | $\begin{equation} \begin{aligned} iii.\;n(A \times B) = 18 \\ n(B)\;and\;n(B) \\ (All\;possible\;values) \\\end{aligned} \end{equation} $ |
| $\begin{equation} \begin{aligned} iv.\;n(A \times B \times C) = 39 \\ n(A),\,\;n(B)\;and\;n(C) \\\end{aligned} \end{equation} $ | $v.\;n(A \times B \times C) = 0$ | $vi.\;n(A \times B \times C) = 1$ |
Solution:
$\begin{equation} \begin{aligned} i.\;A = \{ 2,3,1,9\} \\ B = \{ a,v\} \\ n(A \times B) = ? \\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} n(A) = 4 \\ n(B) = 2 \\ Thus, \\ n(A \times B) = n(A) \times n(B) \\ n(A \times B) = 2 \times 4 = 8 \\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} ii.\;A = \{ x,y\} \\ n(A \times B) = 6 \\ n(B) = ? \\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} n(B) = 2 \\ n(A \times B) = 6 \\ Since, \\ n(A \times B) = n(A) \times n(B) \\ n(A) = {{n(A \times B)} \over {n(B)}} = {6 \over 2} \\ Thus, \\ n(A) = 3 \\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} iii.\;n(A \times B) = 18 \\ n(B)\;and\;n(B) \\ (All\;possible\;values) \\\end{aligned} \end{equation} $
$\begin{equation} \begin{aligned} Since, \\ n(A \times B) = n(A) \times n(B) \\ n(A) \times n(B) = 18 \\\end{aligned} \end{equation} $
The possible sizes of the sets are,
| $n(A)$ | $n(B)$ |
| 1 | 18 |
| 2 | 9 |
| 3 | 6 |
| 6 | 3 |
| 9 | 2 |
| 18 | 1 |
$v.\;n(A \times B \times C) = 0$
$\begin{equation} \begin{aligned} iv.\;n(A \times B \times C) = 39 \\ n(A),\,\;n(B)\;and\;n(C) \\\end{aligned} \end{equation} $
$$\begin{equation} \begin{aligned} Since, \\ n(A \times B \times C) = n(A) \times n(B) \times n(C) \\ n(A) \times n(B) \times n(C) = 39 \\\end{aligned} \end{equation} $$
The possible sizes of the sets are,
| $n(A)$ | $n(B)$ | $n(C)$ |
| 1 | 3 | 13 |
| 1 | 13 | 3 |
| 3 | 13 | 1 |
| 13 | 3 | 1 |
| 13 | 1 | 3 |
| 3 | 1 | 13 |
| 1 | 1 | 39 |
| 1 | 39 | 1 |
| 39 | 1 | 1 |
$$\begin{equation} \begin{aligned} n(A \times B \times C) = 0 \Rightarrow A \times B \times C = \emptyset \\ Since,\,\;A \times B \times C = \emptyset \\ Either\;A\;or\;B\;or\;C\;is\;a\;null\;set \\ Thus,\;size\;of\;the\;sets\;cannot\;be\;defined. \\\end{aligned} \end{equation} $$
$vi.\;n(A \times B \times C) = 1$
$$\begin{equation} \begin{aligned} n(A \times B \times C) = 1 \\ \Rightarrow n(A) \times n(B) \times n(C) = 1 \\ \Rightarrow n(A) = n(B) = n(C) = 1 \\\end{aligned} \end{equation} $$
Graphical representation of Cartesian product of sets
If $A$ and $B$ are two non-empty sets, then the Cartesian $A \times B$ is represented as shown in figure.
Let $A = \{ p,q\} $ and $B = \{ r\} $
Therefore, number of elements in $A$ i.e., $n(A)=2$ and number of elements in $B$ i.e., $n(B)=1$
$$\therefore n(A \times B) = 2 \times 1 = 2$$
From the figure, we can conclude that the set of ordered pair is $$A \times B = \left\{ {p,q} \right\} \times \left\{ k \right\} = \left\{ {\left( {p,k} \right),\left( {q,k} \right)} \right\}$$
Diagrammatic representation of cartesian product of sets
If $A$ and $B$ are two non-empty sets, then the diagrammatic representation of $A \times B$ is as follows.
Step 1: Draw two disjoint Venn diagram for each of the two sets, $A$ and $B$.
Step 2: List down the elements of each vertically.
Step 3: Join every element in $A$ to every element in $B$.
