Let \$A = \{1, 2, 3, 4, 6\}\$ and \$R\$ be the relation on \$A\$ defined by \$\{(a, b): a, \displaystyle b\in A\ , b\$ is exactly divisible by \$a\}\$(i) Write \$R\$ in roster form(ii) Find the domain of \$R\$(iii) Find the range of \$R\$

\$A = \{1, 2, 3, 4, 6\}\$\$ R = \{(a, b): a, \displaystyle b\in A, b \mbox{ is exactly divisible by } a\}\$(i) \$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\}\$(ii) Domain of \$R = \{1, 2, 3, 4, 6\}\$(iii) Range of \$R = \{1, 2, 3, 4, 6\}\$

Show that each of the relation \$R\$ in the set \$A=\left\{x\in Z: 0\leq x\leq 12\right\}\$, given by(i) \$R=\left\{(a, b):|a-b| \text{is a multiple of 4} \right\}\$(ii) \$R=\left\{(a, b):a=b\right\}\$is an equivalence relation. Find the set of all elements related to \$1\$ in each case.

(i)\$A=\left\{x\in Z:0\leq x\leq 12\right\}=\left\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\right\}\$\$R=\left\{(a, b):|a-b| \text{is a multiple of 4}\right\}\$For any element \$a\in A\$, we have \$(a, a)\in R\$ as \$|a-a|=0\$ is a multiple of \$4\$.\$\therefore R\$ is reflexive.Now, let \$(a, b)\in R\Rightarrow |a-b|\$ is a multiple of \$4\$.\$\Rightarrow |-(a-b)|=|b-a|\$ is a multiple of \$4\$.\$\Rightarrow (b, a)\in R\$\$\therefore \$ is symmetric.Now, let \$(a, b), (b, c)\in R\$\$\Rightarrow |a-b|\$ is a multiple of \$4\$ and \$|b-c|\$ is a multiple of \$4\$.\$\Rightarrow (a-b)\$ is a multiple of \$4\$ and \$(b-c)\$ is a multiple of \$4\$.\$\Rightarrow (a-c)=(a-b)+(b-c)\$ is a multiple of \$4\$.\$\Rightarrow |a-c|\$ is a multiple of \$4\$.\$\Rightarrow (a, c)\in R\$.\$\therefore R\$ is transitive.Hence, \$R\$ is an equivalence relation.The set of elements related to \$1\$ is \$\left\{1, 5, 9\right\}\$ since \$|1-1|=0\$ is a multiple of \$4\$,\$|5-1|=4\$ is a multiple of \$4\$, and \$|9-1|=8\$ is a multiple of \$4\$Hence, \$R\$ is an equivalence relation.(ii)\$A=\{x\in Z: 0\leq x\leq 12\}=\{0,1,2,3,4,......,11,12\}\$\$R=\left\{(a, b):a=b\right\}=\{(0,),(1,1),(2,2),........,(11,11),(12,12)\}\$For any element \$a\in A\$, we have \$(a, a)\in R\$, since \$a=a\$.\$\therefore R\$ is reflexive.Now, let \$(a, b)\in R\$.\$\Rightarrow a=b\$\$\Rightarrow b=a\$\$\Rightarrow (b, a)\in R\$\$\therefore R\$ is symmetric.Now, let \$(a, b)\in R\$ and \$(b, c)\in R\$.\$\Rightarrow a=b\$ and \$b=c\$\$\Rightarrow a=c\$\$\Rightarrow (a, c)\in R\$\$\therefore R\$ is transitive.Hence, \$R\$ is an equivalence relation.The elements in \$R\$ that are related to \$1\$ will be those elements from set \$A\$ which are equal to \$1\$.Hence, the set of elements related to \$1\$ is \$\left\{1\right\}\$.

Let \$A = \{1, 2, 3, .... , 14\}\$. Define a relation \$R\$ from \$A\$ to \$A\$ by \$R = \{(x, y): 3x - y = 0\$ where \$x, y \displaystyle \in A\}\$. Write down its domain, co-domain and range.

The relation \$R\$ from \$A\$ to \$A\$ is given as\$R = \{(x, y): 3x - y = 0 ; x, y\in A\}\$i.e., \$R = \{(x, y): 3x = y ; x, y\in A\}\$\$\displaystyle \therefore R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}\$The domain of \$R\$ is the set of all first elements of the ordered pairs in the relation\$\displaystyle \therefore \$ Domain of \$R = \{1, 2, 3, 4\}\$The whole set \$A\$ is the co-domain of the relation \$R\$\$\displaystyle \therefore \$ Codomain of \$R = A = \{1, 2, 3,....., 14\}\$The range of \$R\$ is the set of all second elements of the ordered pairs in the relation.\$\displaystyle \therefore \$ Range of \$R = \{3, 6, 9, 12\}\$

Relation \$R\$ in the set \$Z\$ of all integers defined as \$R=\left\{(x, y): (x-y) \, is \, an \, integer\right\}\$enter 1-reflexive and transitive but not symmetric 2-reflexive only 3-Transitive only 4-Equivalence 5-None

\$R=\left\{(x, y):x-y \, \text{is an integer} \right\}\$Now, for every \$x\in Z, (x, x)\in R\$ as \$x-x=0\$ is an integer.\$\therefore R\$ is reflexive.Now, for every \$x, y\in Z\$ if \$(x, y)\in R\$, then \$x-y\$ is an integer.\$\Rightarrow -(x-y)\$ is also an integer.\$\Rightarrow (y-x)\$ is an integer.\$\therefore (y, x)\in R\$\$\Rightarrow R\$ is symmetric.Now,Let \$(x, y)\$ and \$(y, z)\in R\$, where \$x, y, z\in Z\$.\$\Rightarrow (x-y)\$ and \$(y-z)\$ are integers.\$\Rightarrow x-z =(x-y)+(y-z)\$ is an integer.\$\therefore (x, z)\in R\$\$\therefore R\$ is transitive.Hence, \$R\$ is reflexive, symmetric, and transitive.

Show that the relation \$R\$ in the set \$\textbf{R}\$ of real numbers, defined as \$R=\left\{(a, b):a \leq b^2\right\}\$ is neither reflexive nor symmetric nor transitive.

\$R=\left\{(a, b): a\leq b^2\right\}\$It can be observed that \$\left(\displaystyle\frac{1}{2}, \frac{1}{2}\right)\notin R\$, since \$\displaystyle \frac{1}{2} > \left(\frac{1}{2}\right)^2=\frac{1}{4}\$.\$\therefore R\$ is not reflexive.Now, \$(1, 4)\in R\$ as \$1 (1.5)^2=2.25\$\$\therefore (3, 1.5)\notin R\$\$\therefore R\$ is not transitive.Hence, \$R\$ is neither reflexive, nor symmetric, nor transitive.

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Q: Let A = {1,2} and B = {3,4}. Find the number of relations from A to B.

Given, A = {1,2} and B = {3,4}.Number of elements in set A = \$n(A)=2\$Number of elements in set B = \$n(B)=2\$Number of relations from A to B = \$2^{n(A) \times n(B)}\$\$=2^{2\times 2}=2^4=16\$

Q: The relation \$R=\{(1, 1), (2, 2), (3, 3)\}\$ on the set \$\{1, 2, 3\}\$ is

Given Relation\$R = \{(1,1),(2,2), (3,3)\}\$Reflexive: If a relation has \$\{(a,b)\}\$ as its element, then it should also have \$\{(a,a), (b,b)\}\$ as its elements too.Symmetric: If a relation has \$(a,b)\$ as its element, then it should also have \$\{(b,a)\}\$ as its element too.Transitive: If a relation has \$\{(a,b), (b,c)\}\$ as its elements, then it should also have \$\{(a,c)\}\$ as its element too.Now, the given relation satisfies all these three properties.Therefore, its an equivalence relation.

Q: Let \$R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}\$ be a relation on the set \$A = \{1, 2, 3, 4\}\$. The relation \$R\$ is

Let \$R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}\$ be a relation on the set \$A = \{1, 2, 3, 4\}\$, then(a) Since \$(2, 4)\$ \$\in\$ \$R\$ and \$(2, 3)\$ \$\in\$ \$R\$, so \$R\$ is not a function.(b) Since \$(1, 3)\$ \$\in\$ \$R\$ and \$(3, 1) \$ \$\notin\$ \$R\$ but \$(1, 1)\$ \$\notin\$ \$R\$, so \$R\$ is not transitive:(c) Since \$(2, 3)\$ \$\in\$ \$R\$ but \$(3, 2)\$ \$\notin\$ \$'R\$, so \$R\$ is not symmetric.(d) Since \$(1, 1)\$ \$\notin\$ \$R\$, so \$R\$ is not reflexive.Hence, option C is correct.

Q: Let \$n(A) = n\$. Then the number of all relations on \$A\$ is

For any set \$A\$ such that \$n(A)=n\$then number of all relations on \$A\$ is \${ 2 }^{ n^{2} }\$As the total number of Relations that can be defined from a set \$A\$ to \$B\$ is the number of possible subsets of \$ A \times B\$. If \$n(A) = p\$ and \$n(B) = q\$ then \$n(A \times B) = pq\$ and the number of subsets of \$A \times B\$ = \$2^{pq}\$.

Q: Let \$A=\left \{ 1,2,3 \right \}\$. The total number of distinct relations that can be defined over \$A\$ is:

Total number of distinct binary relations over the set \$ A\$ will be\$2^{n^2}\$\$=2^{3^{2}}\$\$=2^{9}\$\$=512\$

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Let $A = {1, 2, 3}$ . The total number of distinct relations that can be defined over $A$ is:

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Introduction to Relations

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Equivalence Relations

Let A={1,2,3,4,6} and R be the relation on A defined by {(a,b):a,b∈A ,b is exactly divisible by a} (i) Write R in roster form (ii) Find the domain of R (iii) Find the range of R

Show that each of the relation R in the set A={x∈Z:0≤x≤12}, given by (i) R={(a,b):∣a−b∣is a multiple of 4} (ii) R={(a,b):a=b}is an equivalence relation. Find the set of all elements related to 1 in each case.

Let A={1,2,3,....,14}. Define a relation R from A to A by R={(x,y):3x−y=0 where x,y∈A}. Write down its domain, co-domain and range.

Let A = {1,2} and B = {3,4}. Find the number of relations from A to B.

The relation R={(1,1),(2,2),(3,3)} on the set {1,2,3} is

Relation R in the set Z of all integers defined as R={(x,y):(x−y)isaninteger} enter 1-reflexive and transitive but not symmetric 2-reflexive only 3-Transitive only 4-Equivalence 5-None

Let R={(1,3),(4,2),(2,4),(2,3),(3,1)} be a relation on the set A={1,2,3,4}. The relation R is

Let n(A)=n. Then the number of all relations on A is

Show that the relation R in the set R of real numbers, defined as R={(a,b):a≤b2} is neither reflexive nor symmetric nor transitive.

Let A={1,2,3}. The total number of distinct relations that can be defined over A is:

Let $A = {1, 2, 3, 4, 6}$ and $R$ be the relation on $A$ defined by ${(a, b) : a, b \in A, b$ is exactly divisible by $a}$
(i) Write $R$ in roster form
(ii) Find the domain of $R$
(iii) Find the range of $R$

Show that each of the relation $R$ in the set $A = {x \in Z : 0 \leq x \leq 12}$ , given by

(i) $R = {(a, b) : ∣ a - b ∣ is a multiple of 4}$

(ii) $R = {(a, b) : a = b}$
is an equivalence relation. Find the set of all elements related to $1$ in each case.

Let $A = {1, 2, 3, . . . ., 14}$ . Define a relation $R$ from $A$ to $A$ by $R = {(x, y) : 3 x - y = 0$ where $x, y \in A}$ . Write down its domain, co-domain and range.

The relation $R = {(1, 1), (2, 2), (3, 3)}$ on the set ${1, 2, 3}$ is

Relation $R$ in the set $Z$ of all integers defined as $R = {(x, y) : (x - y) i s a n i n t e g e r}$

enter 1-reflexive and transitive but not symmetric
2-reflexive only
3-Transitive only
4-Equivalence
5-None

Let $R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)}$ be a relation on the set $A = {1, 2, 3, 4}$ . The relation $R$ is

Let $n (A) = n$ . Then the number of all relations on $A$ is

Show that the relation $R$ in the set $R$ of real numbers, defined as $R = {(a, b) : a \leq b^{2}}$ is neither reflexive nor symmetric nor transitive.

Let $A = {1, 2, 3}$ . The total number of distinct relations that can be defined over $A$ is: