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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Given Relation $R$ in the set $A=\{1,2,3,...,13,14\}$ defined as $R=\{(x,y):3x-y=0\}$. Which of the following is true?

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  • 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$\; a_1,a_2,a_3 \in A$
Given $A=\{1,2,3....13,14\}$ and $R=\{(x.y):3x-y=0\}$,
$R=\{(x.y):3x-y=0\}$ is reflexive if $(a,a) \in R\;$ for every $\; a\in\;A$
$\Rightarrow$ If $y=x$ $\rightarrow3x-x= 2x \neq 0$
$\Rightarrow$ $(a,a) \not \in R\;$, therefore $R$ is not reflexive.
We can verify this w/ a simple substitution:
Let $x = 3 \Rightarrow 3x - y = 0 \Rightarrow 9 = y \Rightarrow (3,9) \not \in R$
$R=\{(x.y):3x-y=0\}$ is symmetric, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
$\Rightarrow$ If $(x,y) \in R \Rightarrow 3x - y = 0 \Rightarrow 3x= y$
$\Rightarrow$ If $(y,x) \in R \Rightarrow 3y -x = 0 \Rightarrow 3y = x$
Both of these cannot be true, therefore, $R$ is not symmetric, because if $(x,y) \in R$, then $(y,x) \not \in R $
We can verify this w/ a simple substitution:
Let $x = 1, y =3,$ $(x,y) \in R, \Rightarrow 3x - y = 0 \Rightarrow 3 \times 1 - \ 3 = 0$, which is true..
However, $(y,x) \in R, \Rightarrow 3y - x = 0 \Rightarrow 3 \ times 3 - 1 = 8$ which is $\neq 0$
$R=\{(x.y):3x-y=0\}$ is transitive, if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
$\Rightarrow$ If $(x,y) \in R \Rightarrow 3x - y = 0 \Rightarrow 3x= y$
$\Rightarrow$ If $(y,z) \in R \Rightarrow 3y -z = 0 \Rightarrow 3y = z$
$\Rightarrow 3 (3x) = z \rightarrow 9x = z$
However, for $R$ to be transtitive, $(x,z)\in R \rightarrow 3x - z = 0$, i.e, $3x = z$
Therefore $R$ cannot be transitive since z can't be $=3y$ and $=9y$.
We can verify this w/ a simple substitution:
Let $x=1, y=3, z=9$.
$\Rightarrow$ If $(x,y) \in R \Rightarrow 3x - y = 0 \Rightarrow 3 \times 1 = 3$
$\Rightarrow$ If $(y,z) \in R \Rightarrow 3y -z = 0 \Rightarrow 3 \times 3 = 9$
However, $(x,z)\in R \rightarrow 3x - z = 0 \Rightarrow 3 \times 1 \neq 9$.
answered Mar 7, 2013 by balaji.thirumalai
 

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