# Let $R$ be relation defined on the set of natural number $N$ as follows:$R=\{(x,y):x\;\;\;N,y\;\;\;N,2x+y=41\}.$Find the domain and range of the relation R.Also verify whether R is reflexive,symmetric and transitive.

$\begin{array}{1 1} \text{Domain of R={1,2,3.....20} Range of R ={1,3,.......,37,39} R is not transitive, not symmetric, not reflexive} \\ \text{Domain of R={1,2,3.....20} Range of R ={1,3,.......,37,39} R is transitive, not symmetric, not reflexive} \\ \text{Range of R={1,2,3.....20} Domain of R ={1,3,.......,37,39} R is not transitive, not symmetric, not reflexive} \\ \text{Range of R={1,2,3.....20} Domain of R ={1,3,.......,37,39} R is transitive, not symmetric, not reflexive} \end{array}$

Toolbox:
• A relation R in a set A is called $\mathbf{ reflexive},$ if $(a,a) \in R\;$ for every $\; a\in\;A$
• A relation R in a set A is called $\mathbf{symmetric}$, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
• A relation R in a set A is called $\mathbf{transitive},$ if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
• Domain of R is the possible values x can take to satisfy the condition $2x+y=41$ and $x \in N\;y \in N$
• Range of R is the possible values of y for all values of x in domain R. Satisfying $2x+y=41\; x \in N \;y \in N$
Given $R: \{(x,y):x \in N\; y \in N,2x+y=41\}$

$2x+y=41$

$y=41-2x$

Since $y \in N \qquad y > 0$

$41-2x >0$

$-2x >-41$

$2x <41$

$x <\frac {41}{2}$

Since $x \in N.$ x can take values $=\{1,2,3....20\}$

Domain of $R=\{1,2,3.....20\}$

Range of R is $y=41-2x \qquad for\; x=\{1,2,3....20\}$

Range of $R =\{1,3,.......,37,39\}$

When $x=1\qquad y=41-2=39$

When $x=2 \;y=41-2x \times 2=39$

When $x=20\; y=40-2 \times 20=1$

Domain of $f=\{1,2.....20\}$

Range of $f =\{1,3......37,39\}$

R is not reflexive

Since (1,1) does not satisfy

$2x+y=41$

$2 \times 1+1 \neq 41$

R is not symmetric since $(1,39) \in R$ satisfies $2x+y=41\; but (39,1)\; \notin R$

Since $2 \times 39+1 \neq 41$

but $2 \times 1+39=41$

Solution:R is not transitive since (1,39) $\in$ but no values for x=39. satisfies the given relation

edited Mar 27, 2013 by meena.p