# Test whether the following relations

Question:

Test whether the following relations R1, R2, and R3 are (i) reflexive (ii) symmetric and (iii) transitive:

(i) $R_{1}$ on $Q_{0}$ defined by $(a, b) \in R_{1} \Leftrightarrow a=1 / b$.

(ii) $R_{2}$ on $Z$ defined by $(a, b) \in R_{2} \Leftrightarrow|a-b| \leq 5$

(iii) $R_{3}$ on $R$ defined by $(a, b) \in R_{3} \Leftrightarrow a^{2}-4 a b+3 b^{2}=0$.

Solution:

(i) Reflexivity:
Let a be an arbitrary element of R1. Then,

$a \in R_{1}$

$\Rightarrow a \neq \frac{1}{a}$ for all $a \in \mathrm{Q}_{0}$

So, $R_{1}$ is not reflexive.

Symmetry:

Let $(a, b) \in R_{1}$. Then,

$(a, b) \in R_{1}$

$\Rightarrow a=\frac{1}{b}$

$\Rightarrow b=\frac{1}{a}$

$\Rightarrow(b, a) \in R_{1}$

So, $R_{1}$ is symmetric.

Transitivity:
Here,

$(a, b) \in R_{1}$ and $(b, c) \in R_{2}$

$\Rightarrow a=\frac{1}{b}$ and $b=\frac{1}{c}$

$\Rightarrow a=\frac{1}{\frac{1}{c}}=c$

$\Rightarrow a \neq \frac{1}{c}$

$\Rightarrow(a, c) \notin R_{1}$

So, $R_{1}$ is not transitive.

(ii)
Reflexivity:

Let $a$ be an arbitrary element of $R_{2}$. Then,

$a \in R_{2}$

$\Rightarrow|a-a|=0 \leq 5$

So, $R_{1}$ is reflexive.

Symmetry:

Let $(a, b) \in R_{2}$

$\Rightarrow|a-b| \leq 5$

$\Rightarrow|b-a| \leq 5$                      $[$ Since, $|a-b|=|b-a|]$

$\Rightarrow(b, a) \in R_{2}$

So, $R_{2}$ is symmetric.

Transitivity:

Let $(1,3) \in R_{2}$ and $(3,7) \in R_{2}$

$\Rightarrow|1-3| \leq 5$ and $|3-7| \leq 5$

But $|1-7| \not \leq 5$

$\Rightarrow(1,7) \notin R_{2}$

So, $R_{2}$ is not transitive.

(iii)
Reflexivity: Let a be an arbitrary element of R3. Then,

$a \in R_{3}$

$\Rightarrow a^{2}-4 a \times a+3 a^{2}=0$

So, $R_{3}$ is reflexive.

Symmetry:

Let $(a, b) \in R_{3}$

$\Rightarrow a^{2}-4 a b+3 b^{2}=0$

But $b^{2}-4 b a+3 a^{2} \neq 0$ for all $a, b \in R$

So, $R_{3}$ is not symmetric.

Transitivity:

$(1,2) \in R_{3}$ and $(2,3) \in R_{3}$

$\Rightarrow 1-8+6=0$ and $4-24+27=0$

But $1-12+9 \neq 0$

So, $R_{3}$ is not transitive.