Relations and Functions

We have,

R= $\{\left(a,b\right)\therefore \left|a-b\right|$ is even $\}$ is a relation in set A= $\{1,2,3,4,5\}$

For all $a\in A$ , $\left|a-a\right|=0$ is even.

So, $\left(a,a\right)\in R$ . Hence R is reflexive

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

$\left|a-b\right|$ is even

$\left|-b+a\right|$ is even $|$

$\left|-\left(b-a\right)\right|$ is even

$\left|b-a\right|$ is even

i.e., $\left(b,a\right)\in R$

Hence, R is symmetric.

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

We have $\left|a-b\right|$ is even

and $\left|b-c\right|$ is even

then, $\left|a-b\right|+\left|b-c\right|$ is even as even + even=even

$\left|a-b+b-c\right|$ is even

$\left|a-c\right|$ is even

$\therefore$ $\left(a,c\right)\in R$

So, R is transitive.

$\therefore$ R is an equivalence relation

All elements of [1,3,5] are odd positive numbers and its subset are odd and their difference given an even number. Hence, they are related to each other.

Similarly, all elements of [2,4] are even positive numbers and its subset are even and their difference gives an even number. Hence, they are related to each other.

However, elements of [1,3,5] are related to elements of [2,4] since the difference of the two subsets is never even.

(i) We have, $R=\{\left(x,y\right):3x-y=0\}$ a relation in set A= $\{1,2,3..........14\}$

For $x\in A,y=3x$ or $y\ne x$ i.e.,

$\left(x,x\right)$ does not exist in R

$\therefore$ R is not reflexive.

For $\left(x,y\right)\in R,y=3x$

Then $\left(y,x\right)x\ne 3y$

So $\left(y,x\right)\notin R$

$\therefore$ R is not symmetric

For $\left(x,y\right)\in R$ and $\left(y,z\right)\in R$ . We have

$y=3x$ and $z=3y$

Then $z=3\left(3x\right)=9x$

i.e., $\left(x,z\right)\notin R$

$\therefore$ R is not Transitive

(ii) We have,

R= $\{\left(x,y\right):y=x+5\&x<4\}$ is a relation in N

= $\{\left(1,1+5\right),\left(2,2+5\right),\left(3,3+5\right)\}$

= $\{\left(1,6\right),\left(2,7\right),\left(3,8\right)\}$

Clearly, R is not reflexive as $\left(x,x\right)\notin R$ and $x<4\&x\in N$

Also, R is not symmetric as $\left(1,6\right)\in R$ but $\left(6,1\right)\notin R$

And for $\left(x,y\right)\in R\left(y,z\right)\notin R$ . Hence, R is not Transitive.

(iii) R= $\{\left(x,y\right);y$ is divisible by x $\}$ is a relation in set

A= $\{1,2,3,4,5,6\}$

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

Hence, R is reflexive because $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\in R$ i.e., $\left(x,x\right)\in R$

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

And for $x,y,z\in A$ and $\left(x,y\right)\in R\&\left(y,z\right)\in R$

$\frac{y}{z}=n$ and $\frac{z}{y}=m$ where $n\&m\in N$

Then, $z=my=m(nx)=mn.x$

$\frac{z}{x}=m.n,m,mn\in N$

Hence, $\left(x,z\right)\in R$

$\therefore$ R is transitive

(iv) R= $\{\left(x,y\right):x-y$ is an integer $\}$ is a relation in set Z

For $x\in Z$

$x-x=0$ is an integer

So, $\left(x,x\right)\in R$ i.e., R is reflexive

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

$x-y$ is an integer

$-\left(y-x\right)$ is an integer

$\left(y-x\right)$ is an integer

So, $\left(y,x\right)\in R$ i.e., R is symmetric

For $x,y,z\in R$ and $\left(x,y\right)\in R\&\left(y,z\right)\in R$ We have,

$x-y$ is an integer

$y-z$ is an integer

So, $\left(x-y\right)+\left(y-z\right)$ is also an integer

$x-z$ is an integer

So, $\left(x,z\right)\in R$ i.e., R is transitive.

(v) (a) R= $\{\left(x,y\right):x$ and $y$ work at same place $\}$ in set A of human being.

For $x\in A$ we get

$x\&x$ work at same place

$\left(x,x\right)\in R$ So, R is reflexive.

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

$x\&y$ work at same place

$y\&x$ work at same place

$\left(y,x\right)\in R$ . So, R is symmetric.

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

$x\&y$ work at same place

$y\&z$ work at same place

$x\&z$ work at same place

i.e., $\left(x,z\right)\in R$ . So, R is Transitive.

(b) R= $\{\left(x,y\right):x\&y$ live in same locality $\}$

For $x\in A$ , we get,

$x\&x$ live in same locality

$\left(x,x\right)\in R$ So, R is reflexive.

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

$x\&y$ live in same locality

$y\&x$ live in same locality

i.e., $\left(y,x\right)\in R$ . So, R is symmetric.

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

$x,y$ live in same locality

$y,z$ live in same locality

$x,z$ live in same locality

So, $\left(x,z\right)\in R$ i.e., R is transitive.

(c) R= $\{\left(x,y\right):x$ is exactly 7 cm taller than y $\}$

For $x\in A$ ,

Height $\left(x\right)\ne 7+$ height of $\left(x\right)$

So, $\left(x,x\right)\notin R$ . i.e., R is not reflexive.

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

Height $\left(x\right)$ $=7+height\left(y\right)$

But $height\left(y\right)\ne 7+height\left(x\right)$

So, $\left(y,x\right)\in R$ i.e., R is not symmetric.

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