• $A = \{X,Y,Z\}$

• $B =$ {blue, red}

• $X$ - blue or $X$ - red

• $Y$ - blue or $Y$ - red

• $Z$ - blue or $Z$ - red

($X$,blue), ($X$,red), ($Y$,blue), ($Y$,red), ($Z$,blue), ($Z$,red)

${\bold{Vehicles~ and~ their~ name~ plate}}$

a∈A, b∈B, we obtained a pair (a,b).

A = {D L, M P, U P, A P, T N}

B = {01,02,03,….,09}.

The possibility of a name plate for a person coming from (say) MP is MP03.

In fact, given $a\in$ A & $b\in$ B, $(a, b)$ - says the following:

$a$ - denotes the state

$b$ - denotes the number of the vehicle.

• Given sets $A$ and $B$, and elements $a, b$ with $a \in A$ and $b \in B$, the pair $(a,b)$ is called an ordered pair.

$\bold{Example:}$

$\{(x, y) \mid x, y$ are real numbers and $x^2+y^2=1 \}$

This set represents a circle.

Elements of this set or elements belonging to a circle are examples for ordered pairs.

• Let $A$ and $B$ be any two sets. The cartesian product of $A$ and $B$, denoted $A \times B$, is defined as

$A \times B =\{(a, b) \mid a \in A \text { and } b \in B\} .$

$\bold{Example:}$

$A=\{x, y, z\}, \ B= \{\text{blue, red}\}$

$A \times B$ = $\{(X,blue), (X,red), (Y,blue), (Y,red), (Z,blue), (Z,red)\}$.

The set $S=\{(x, y) \mid x \ \& \ y \ \text{are real numbers and} \ x^2+y^2=1\}$ - circle.

This set is not a cartesian product of two sets.

$(0,1)$ is on the circle.

Similarly, $(1,0)$ is also on the circle.

• Suppose $S=A \times B$.

• Since $(0,1) \in A \times B \Rightarrow 1 \in B$

• Similarly $(1,0) \in A \times B \Rightarrow 1 \in A$

• Whereas the pair $(1,1) \notin S$, because

$2=1^2+1^2 \neq 1 \therefore(1,1) \notin S$

$\therefore S$ is not a cartesian product of two sets.

$A \times B=\{(a, b) \mid a \in A$ and $b \in B\}$

$B \times A=\{(b, a) \mid b \in B$ and $a \in A\}$

$(a, b) \neq(b, a)$

• Given two elements $(a, b) \ \& \ (c, d)$ from $A \times B$,

We say that $(a, b)=(c, d)$ if $a=c$ and $b=d$.

Suppose $\left(x-\frac{1}{2}+y, x+ \frac{1}{2} -y\right)=(2,3)$

Find $x$ and $y$.

Given

$\left(x+y-\frac{1}{2}, x-y+\frac{1}{2}\right)=(2,3)$

$\Rightarrow \ x+y-\frac{1}{2}=2$

$\quad \underline { x-y+\frac{1}{2} =3}$

$\quad 2 x =5 \Rightarrow x=\frac{5}{2}$

$\Rightarrow x+y-\frac{1}{2}=2$
$\underline{-x+y-\frac{1}{2}=3}$
$2y-1 = -1 \Rightarrow y=0$

$\therefore x=\frac{5}{2}$ and $y=0$

Suppose $A$ is a set and we are given that $(0,1),(-1,0)$ is in $A \times A$.

Find $A$.

$(0,1) \in A \times A \Rightarrow 0 \in A$ and $1 \in A$

Similarly, $(-1,0) \in A \times A$

$\Rightarrow-1 \in A$ and $0 \in A$.

$\therefore \{-1,0,1\}=A.$

$\therefore A \times A= \biggl \{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),\\ (0,1),(1,-1),(1,0),(1,1)\biggl \}$

If $A$ & $B$ are any two sets with number of elements of $A$ as $p$ and with number of elements of $B$ as $q$, then the no. of elements

in $A \times B$ is $p q$.

$\therefore \#(A \times B)= \#(A) \cdot \#(B) .$

$A=\{-1,0,1\}$.
$\#(A)=3$
$\\ \Rightarrow \#(A \times A)=\#(A) \cdot \#(A) \\ =3 \cdot 3=9.$
$A \times A=\{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),\\ (0,1), (1,-1),(1,0),(1,1)\}$

$A \times A$ has got 9 elements.

Set $A$ be a set whose cartesian product with itself has got $16$ elements. Suppose we know that $(1,2)$ and $(3,4)$ are in $A \times A$, find $A$ and $A \times A$

$\#(A \times A)=16$

$\Rightarrow \#(A)=4$

$(1,2) \in A \times A \Rightarrow 1,2 \in A$

Similarly, $(3,4) \in A \times A \Rightarrow 3,4 \in A$

$\therefore$ A = $\{1,2,3,4\}$

$A \times A=\{(1,2),(1,3),(1,4),(1,1),(2,1)$

$\{(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4) \}$

$\R^2$ - the two dimensional plane.

$\R^2$ $=\{(x,y) \mid \ x \ \text{and} \ y \ \text{are real numbers}\}$

$\R^2 = \R \times \R$

$\R^2$\ $\{(0,0)\}$ $=\{(r,\theta) \mid \ r > 0 \ \& \ 0 \le \theta < 360^{\circ}\}$

This set is a cartesian product of two sets.

$=\{r \in \mathbb{R} \mid r>0\} \times \left\{\theta \mid 0 \leq \theta<360^{\circ}\right\}$

If $A$ and $B$ are any two sets with atleast one of them being infinite, then $A \times B$ is also an infinite set.

$A = \{a,b,c\}$

$B = \{1,2,3 \cdots\}$

$A \times B = \{(a,1), (a,2),\cdots, (b,1), (b,2), \cdots, (c,1), (c,2), \cdots \}$

$\mathbb{R}^3$ - three dimensional plane

$\mathbb{R}^3 =\{(x, y, z) \mid x,y, \& \ z$ are real numbers $\}$

$\mathbb{R}^3=\{(x,(y, z)) \mid x,y,\& \ z$ are real numbers $\}$

$\mathbb{R}^3 = \mathbb{R} \times \mathbb{R}^2$

$(x, y, z) =((x, y), z)$

$\mathbb{R}^3 =\mathbb{R}^2 \times \mathbb{R}$

$\mathbb{R}^3= \mathbb{R} \times \mathbb{R} \times \mathbb{R}$

Let $A_1, A_2$ and $A_3$ be three sets. Then the cartesian product of $A_1, A_2$ and $A_3$, denoted $A_1 \times A_2 \times A_3$ is defined as
$A_1 \times A_2 \times A_3 = \{(a _1, a _2,a _3) \mid a _1 \in A _1, a _2 \in A _2 \ \& \ a _3 \in A _3 \}$

${\bold{Example:}}$

$\mathbb{R}^3$ - the three dimensional plane

$\mathbb{R}^3=\mathbb{R} \times \mathbb{R} \times \mathbb{R}$

is an example for a cartesian product of 3 sets.

Suppose $(x+y+z, x-y-z, x+y-z) = (1,2,3)$

Find $x,y$ & $z$.

${\bold{Remark:}}$

If $A_1, A_2,$ and $A_3$ are any 3 sets, and if $(a_1, a_2, a_3)$ $\&$ $(a_1^{\prime}, a_2^{\prime}, a_3^{\prime}) \in A_1 \times A_2 \times A_3$ and

$(a_1, a_2, a_3) = (a_1^{\prime}, a_2^{\prime}, a_3^{\prime})$

$\Rightarrow a_1 = a_1^{\prime}, a_2^{\prime} = a_2^{\prime}$ $\&$ $a_3 = a_3^{\prime}$

Given $(x+y+z, x-y-z, x+y-z) = (1,2,3)$

$x+y+z=1 \cdots (1)$

$x-y-z=2 \cdots (2)$

$x+y-z=3 \cdots (3)$

Adding (1) and (2), we get $2x = 3 \Rightarrow x=\frac{3}{2}$

Adding (1) and (3), we get $2x+2y=4$

$\Rightarrow 2(\frac{3}{2})+2y=4$

$\Rightarrow 2y=4-3=1$ $\Rightarrow y=\frac{1}{2}$

$x+y+z=1$

$\frac{3}{2} + \frac{1}{2} +z =1 \\ \Rightarrow 2+z = 1 \\ \Rightarrow z =-1$
$\therefore$ The values for $x,y$ & $z$ are $\frac{3}{2}, \frac{1}{2}$ & $-1$ respectively.

Thank you