Relations and Functions

The given relation in set A of points in a plane is

R= $\{\left(P,Q\right):$ distance of point P from origin=distance of point Q from $origin\}$

If O is the point of origin

R= $\left\{\left(P,Q\right):PO=QO\right\}$

Then, for $P\in A$ we have PO=PO

So,  $\left(P,P\right)\in R$

i.e., P is reflexive

for,  $P,Q\in A$ and $\left(P,Q\right)\in R$ we have

PO=QO

QO=PO i.e.,  $\left(Q,P\right)\in R$

i.e., R is symmetric

for $P,Q,S\in A$ and $\left(P,Q\right)\&\left(Q,S\right)\in R$

PO=QO and QO=SO

PO=SO

i.e.,  $\left(P,S\right)\in R$

so, R is transitive

Hence, R is an equivalence relation

For a point $P\ne \left(o,o\right)$ the set of all points related to P i.e., distance from origin to the points are equal is a circle with center at origin (o, o) by the definition of circle

Let A= $\left\{a,b,c\right\}$

(i) R= $\left\{\left(a,b\right),\left(b,a\right)\right\}$ is a relation in set A

So, $\left(a,b\right)\in R$ and $\left(b,a\right)\in R\to$ Symmetric

$\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R,\left(b,a\right)\in R$ but $\left(a,a\right)\notin R$ $\to$ not transitive

(ii) R= $\left\{\left(a,b\right),\left(b,c\right),\left(a,c\right)\right\}$ is a relation in set A

So, $\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R$ but $\left(b,a\right)\notin R\to$ not symmetric

$\left(a,b\right)\in R\&\left(b,c\right)\in R$ and also $\left(a,c\right)\in R\to$ transitive

(iii) R= $\left\{\left(a,a\right),\left(b,b\right),\left(c,c\right),\left(a,b\right),\left(b,a\right),\left(a,c\right),\left(c,a\right)\right\}$

So, $\left(a,a\right),\left(b,b\right),\left(c,c\right)\in R\to$ Reflexive

$\left(a,b\right)\in R\to \left(b,a\right)\in R\to$ Symmetric

$\left(a,c\right)\in R\to \left(c,a\right)\in R$

$\left(b,a\right)\in R$ and $\left(a,c\right)\in R$

But $\left(b,c\right)\notin R\to$ not transitive

(iv) R= $\left\{\left(a,a\right),\left(b,b\right),\left(c,c\right),\left(a,b\right),\left(b,c\right),\left(a,c\right)\right\}$ is s relation in set A

So, $\left(a,a\right),\left(b,b\right),\left(c,c\right)\in R\to$ reflexive

$\left(a,b\right)\&\left(b,c\right)\in R$ so, $\left(a,c\right)\in R\to$ transitive

$\left(a,b\right)\in R$ but $\left(b,a\right)\notin R\to$ not symmetric

(v) R= $\left\{\left(a,a\right),\left(a,b\right),\left(b,a\right)\right\}$

So, $\left(b,b\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R$ and $\left(b,a\right)\in R\to$ symmetric

And $\left(a,b\right)\in R\&\left(b,a\right)\in R$

and also $\left(a,a\right)\in R\to$ transitive

3. Given, G = {7, 8} and H = {5, 4, 2}

By the definition of the Cartesian product,

G *H = { (x, y): x∈G and y = ∈ H}

= { (7, 5), (7, 4), (7, 2), (8, 5), (8,4), (8,2)}

H* G = { (x, y): x∈ H and y ∈G}

= { (5, 7), (5, 8), (4,7), (4, 8), (2, 7), (2,8)}

We have,

R= $\{\left(x,y\right):x\&y$ have same number of pages $\}$ is a relation in set of A of all books in

For $\left(x,y\right)\in R\&x,y\in A$

As x=y=same no. of pages

Then,  $\left(x,x\right)\in R$

Hence, R is reflexive.

For $\left(x,y\right)\in R$ and $x,y\in A$

Also,  $\left(y,x\right)\in R$ ,  $\therefore x=y$

Hence, R is symmetric.

For $x,y,z\in A$ and $\left(x,y\right)\in R$ and $\left(y,z\right)\in R$

x=y and y=z

x=z

i.e.,  $\left(x,z\right)\in R$

hence, R is also transitive

$\therefore$ R is an equivalence relation.

We have,

R= $\{\left(1,2\right),\left(2,1\right)\}$ is a relation in set $\{1,2,3\}$

Then, as $\left(1,1\right)\notin R$ and $\left(2,2\right)\notin R$

So, R is not reflective

As $\left(1,2\right)\in R$ and $\left(2,1\right)\in R$

So, R is symmetric

And as $\left(1,2\right)\in R,\left(2,1\right)\in R$ but $\left(1,1\right)\notin R$

So, R is not transitive.

We have,

R= $\{\left(a,b\right):a\le b^3\}$ is a relation in R.

For, $\left(a,b\right)\in R$ and $a=\frac{1}{2}$ we can write

$a\le a^3$ => $\frac{1}{2}\le {\left(\frac{1}{2}\right)}^3$ => $\frac{1}{2}\le \frac{1}{8}$ which is not true.

So, R is not reflexive.

For $\left(a,b\right)=\left(1,2\right)\in R$ we have,

$a\le b^3$ => $1\le 2^3$ => $1\le 8$ is true.

So, $\left(1,2\right)\in R$

But $2\le 1^3$ => $2\le 1$ is not true

So, $\left(2,1\right)\notin R$ and $\left(b,a\right)\notin R$

Hence, R is not symmetric.

For, $\left(a,b\right)=\left(10,4\right)$ and $\left(b,c\right)=\left(4,2\right)\in R$

$10\le 4^3$ => $10\le 64$ is true=> $\left(10,4\right)\in R$

$4\le 2^3$ => $4\le 8$ is true=> $\left(4,2\right)\in R$

But $10\le 2^3$ => $10\le 8$ is not true=> $\left(10,2\right)\notin R$

Hence, for $\left(a,b\right),\left(b,c\right)\in R,\left(a,c\right)\notin R$

So, R is not transitive.

We have, R= $\{\left(a,b\right):a\le b\}$ is a relation in R.

For,  $a\in R$ ,

$a\le b$ but $b\le a$ is not possible i.e.,  $\left(b,a\right)\notin R$

Hence, R is not symmetric.

For $\left(a,b\right)\in R\&\left(b,c\right)\in R$ and $a,b,c\in R$

$a\le b$ and $b\le c$

So,  $a\le c$

i.e.,  $\left(a,c\right)\in R$

$\therefore$ R is transitive.

We have,

R= $\{\left(a,b\right):b=a+1\}$ is a relation in set $\{1,2,3,4,5,6\}$

So, R= $\{\left(1,2\right),\left(2,3\right),\left(3,4\right),\left(4,5\right),\left(5,6\right),\left(6,7\right)\}$

As,  $\left(1,1\right)\notin R$ , R is not reflexive

As,  $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$ , R is not symmetric

And as $\left(1,2\right)$ & $\left(2,3\right)\in R$ but $\left(1,3\right)\notin R$

Hence, R is not transitive.