<h3 id="successrelations--functions-br-lecture-1success"><Success>Relations & functions <br> lecture-1</Success></h3> <div style='float: left; width:50%;'> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Example</Success> <div style='float: left;text-align:left; width:99%;font-size:30px;'> <ul> <li> <p>$A = \{X,Y,Z\}$</p> </li> <li> <p>$B =$ {blue, red}</p> </li> <li> <p>$X$ - blue or $X$ - red</p> </li> <li> <p>$Y$ - blue or $Y$ - red</p> </li> <li> <p>$Z$ - blue or $Z$ - red</p> </li> </ul> <p>($X$,blue), ($X$,red), ($Y$,blue), ($Y$,red), ($Z$,blue), ($Z$,red)</p> </div> <div style='float: right; width:1%;font-size:30px;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b68cfa9c-489a-4328-8846-5adda5ecf800/public"></p> </div>

### <Success>Example</Success> <div style='float: left; width:99%;font-size:30px;'> <p>${\bold{Vehicles~ and~ their~ name~ plate}}$</p> <p>a∈A, b∈B, we obtained a pair (a,b).</p> <p>A = {D L, M P, U P, A P, T N}</p> <p>B = {01,02,03,….,09}.</p> <p>The possibility of a name plate for a person coming from (say) MP is MP03.</p> <p>In fact, given $a\in$ A & $b\in$ B, $(a, b)$ - says the following:</p> <p>$a$ - denotes the state</p> <p>$b$ - denotes the number of the vehicle.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Ordered pair</Success> <div style='float: left;text-align:left; width:99%;font-size:30px;'> <ul> <li>Given sets $A$ and $B$, and elements $a, b$ with $a \in A$ and $b \in B$, the pair $(a,b)$ is called an <strong>ordered pair</strong>.</li> </ul> <p>$\bold{Example:}$</p> <p>$\{(x, y) \mid x, y $ are real numbers and $ x^2+y^2=1 \}$</p> <p>This set represents a circle.</p> <p>Elements of this set or elements belonging to a circle are examples for ordered pairs.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Cartesian product</Success> <div style='float: left; width:99%;font-size:30px;'> <ul> <li>Let $A$ and $B$ be any two sets. The <strong>cartesian product</strong> of $A$ and $B$, denoted $A \times B$, is defined as</li> </ul> <p>$A \times B =\{(a, b) \mid a \in A \text { and } b \in B\} .$</p> <p>$\bold{Example:}$</p> <p>$A=\{x, y, z\}, \ B= \{\text{blue, red}\}$</p> <p>$A \times B$ = $\{(X,blue), (X,red), (Y,blue), (Y,red), (Z,blue), (Z,red)\}$.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Example</Success> <div style='float: left;text-align:left; width:99%;font-size:30px;'> <p>The set $S=\{(x, y) \mid x \ \& \ y \ \text{are real numbers and} \ x^2+y^2=1\}$ - circle.</p> <p>This set is not a cartesian product of two sets.</p> <p>$(0,1)$ is on the circle.</p> <p>Similarly, $(1,0)$ is also on the circle.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Example (continue)</Success> <div style='float: left;text-align:left; width:99%;font-size:30px;'> <ul> <li> <p>Suppose $S=A \times B$.</p> </li> <li> <p>Since $(0,1) \in A \times B \Rightarrow 1 \in B$</p> </li> <li> <p>Similarly $(1,0) \in A \times B \Rightarrow 1 \in A$</p> </li> <li> <p>Whereas the pair $(1,1) \notin S$, because</p> </li> </ul> <p>$ 2=1^2+1^2 \neq 1 \therefore(1,1) \notin S$</p> <p>$\therefore S$ is not a cartesian product of two sets.</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1868b22d-321a-40c3-dabf-055c0d2f8900/public"></p> </div>

### <Success>Remarks</Success> <div style='float: left;text-align:left; width:99%;font-size:30px;' <ul> <li>$A \times B$ and $B \times A$.</li> </ul> <p>$A \times B=\{(a, b) \mid a \in A$ and $b \in B\}$</p> <p>$B \times A=\{(b, a) \mid b \in B$ and $a \in A\}$</p> <p>$(a, b) \neq(b, a)$</p> <ul> <li>Given two elements $(a, b) \ \& \ (c, d)$ from $A \times B$,</li> </ul> <p>We say that $(a, b)=(c, d)$ if $a=c$ and $b=d$.</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/49ea610d-73fe-4411-7cfa-6a56cba26100/public"></p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;font-size:30px;'> <p>Suppose $\left(x-\frac{1}{2}+y, x+ \frac{1}{2} -y\right)=(2,3)$</p> <p>Find $x$ and $y$.</p> </div> ====== <h3 id="successsolutionsuccess"><Success>Solution</Success></h3> <div style='float: left; width:99%;font-size:30px;'> <p><strong>Given</strong></p> <p>$\left(x+y-\frac{1}{2}, x-y+\frac{1}{2}\right)=(2,3)$</p> <p>$\Rightarrow \ x+y-\frac{1}{2}=2$</p> <p>$\quad \underline { x-y+\frac{1}{2} =3} $</p> <p>$\quad 2 x =5 \Rightarrow x=\frac{5}{2}$</p> <p>$\Rightarrow x+y-\frac{1}{2}=2$<br> $\underline{-x+y-\frac{1}{2}=3}$<br> $ 2y-1 = -1 \Rightarrow y=0$</p> <p>$\therefore x=\frac{5}{2}$ and $y=0$</p> </div> <div style='float: right; width:1%;font-size:30px;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/5c0feb4d-0ba9-43b7-20be-c6da19f86d00/public"></p> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;font-size:30px;'> <p>Suppose $A$ is a set and we are given that $(0,1),(-1,0)$ is in $A \times A$.</p> <p>Find $A$.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;font-size:30px;'> <p>$(0,1) \in A \times A \Rightarrow 0 \in A$ and $1 \in A$</p> <p>Similarly, $(-1,0) \in A \times A$</p> <p>$\Rightarrow-1 \in A$ and $0 \in A$.</p> <p>$\therefore \{-1,0,1\}=A.$</p> <p>$\therefore A \times A= \biggl \{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),\\ (0,1),(1,-1),(1,0),(1,1)\biggl \}$</p> </div> ====== ### <Success>Remark</Success> <div style='float: left; width:99%;font-size:30px;'> <p>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</p> <p>in $A \times B$ is $p q$.</p> <p>$\therefore \#(A \times B)= \#(A) \cdot \#(B) .$</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d43cd6c8-f262-4b48-6463-b08eee772d00/public"></p> </div>

### <Success>Example</Success> <div style='float: left; width:99%;;font-size:30px;'> <p>$A=\{-1,0,1\}$.<br> $\#(A)=3$<br> $\\ \Rightarrow \#(A \times A)=\#(A) \cdot \#(A) \\ =3 \cdot 3=9.$<br> $A \times A=\{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),\\ (0,1), (1,-1),(1,0),(1,1)\}$</p> <p>$A \times A$ has got 9 elements.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;;font-size:30px;'> <p>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 $A \times A$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left; width:99%;;font-size:30px;'> <p>$\#(A \times A)=16$</p> <p>$\Rightarrow \#(A)=4$</p> <p>$(1,2) \in A \times A \Rightarrow 1,2 \in A$</p> <p>Similarly, $(3,4) \in A \times A \Rightarrow 3,4 \in A$</p> <p>$\therefore$ A = $\{1,2,3,4\}$</p> <p>$A \times A=\{(1,2),(1,3),(1,4),(1,1),(2,1) $</p> <p>$\{(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4) \}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Example</Success> <div style='float: left; width:99%;font-size:30px;'> <p>$\R^2$ - the two dimensional plane.</p> <p>$\R^2$ $=\{(x,y) \mid \ x \ \text{and} \ y \ \text{are real numbers}\}$</p> <p>$\R^2 = \R \times \R$</p> <p>$\R^2$\ $\{(0,0)\}$ $=\{(r,\theta) \mid \ r > 0 \ \& \ 0 \le \theta < 360^{\circ}\}$</p> <p>This set is a cartesian product of two sets.</p> <p>$=\{r \in \mathbb{R} \mid r>0\} \times \left\{\theta \mid 0 \leq \theta<360^{\circ}\right\}$</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/4e46aaac-6b80-46a4-06b8-b7ec911ccf00/public"></p> </div>

### <Success>Remark</Success> <div style='float: left; width:99%;font-size:30px;'> <p>If $A$ and $B$ are any two sets with atleast one of them being infinite, then $A \times B$ is also an infinite set.</p> </div> ====== ### <Success>Example</Success> <div style='float: left; width:99%;font-size:30px;'> <p>$A = \{a,b,c\}$</p> <p>$B = \{1,2,3 \cdots\}$</p> <p>$A \times B = \{(a,1), (a,2),\cdots, (b,1), (b,2), \cdots, (c,1), (c,2), \cdots \}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Example</Success> <div style='float: left; width:99%;font-size:30px;'> <p>$\mathbb{R}^3$ - three dimensional plane</p> <p>$\mathbb{R}^3 =\{(x, y, z) \mid x,y, \& \ z $ are real numbers $\}$</p> <p>$\mathbb{R}^3=\{(x,(y, z)) \mid x,y,\& \ z$ are real numbers $\}$</p> <p>$\mathbb{R}^3 = \mathbb{R} \times \mathbb{R}^2$</p> <p>$(x, y, z) =((x, y), z) $</p> <p>$\mathbb{R}^3 =\mathbb{R}^2 \times \mathbb{R}$</p> <p>$\mathbb{R}^3= \mathbb{R} \times \mathbb{R} \times \mathbb{R}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Cartesian product of 3 sets</Success> <div style='float: left; width:99%;font-size:30px;'> <p>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<br> $A_1 \times A_2 \times A_3 = \{(a _1, a _2,a _3) \mid a _1 \in A _1, a _2 \in A _2 \ \& \ a _3 \in A _3 \}$</p> <p>${\bold{Example:}}$</p> <p>$\mathbb{R}^3$ - the three dimensional plane</p> <p>$\mathbb{R}^3=\mathbb{R} \times \mathbb{R} \times \mathbb{R}$</p> <p>is an example for a cartesian product of 3 sets.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Problem</Success> <div style='float: left; width:99%;font-size:30px;'> <p>Suppose $(x+y+z, x-y-z, x+y-z) = (1,2,3)$</p> <p>Find $x,y$ & $z$.</p> <p>${\bold{Remark:}}$</p> <p>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</p> <p>$(a_1, a_2, a_3) = (a_1^{\prime}, a_2^{\prime}, a_3^{\prime})$</p> <p>$\Rightarrow a_1 = a_1^{\prime}, a_2^{\prime} = a_2^{\prime}$ $\&$ $a_3 = a_3^{\prime}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;font-size:30px;'> <p>Given $(x+y+z, x-y-z, x+y-z) = (1,2,3)$</p> <p>$x+y+z=1 \cdots (1)$</p> <p>$x-y-z=2 \cdots (2)$</p> <p>$x+y-z=3 \cdots (3)$</p> <p>Adding (1) and (2), we get $2x = 3 \Rightarrow x=\frac{3}{2}$</p> <p>Adding (1) and (3), we get $2x+2y=4$</p> <p>$\Rightarrow 2(\frac{3}{2})+2y=4$</p> <p>$\Rightarrow 2y=4-3=1$ $\Rightarrow y=\frac{1}{2}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Solution</Success> <div style='float: left; width:99%;font-size:30px;'> <p>$x+y+z=1$</p> <p>$\frac{3}{2} + \frac{1}{2} +z =1 \\ \Rightarrow 2+z = 1 \\ \Rightarrow z =-1$<br> $\therefore$ The values for $x,y$ & $z$ are $\frac{3}{2}, \frac{1}{2}$ & $-1$ respectively.</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/3cde2594-faa9-4efe-9887-7ef67f44fa00/public"></p> </div>

<div> <h2><Success>Thank you</Success></h2> </div>