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Let $R$ be a relation defined on $N$ as a $R$ b is $2 a+3 b$ is a multiple of $5, a, b \in N$. Then $R$ is
not reflexive
transitive but not symmetric
symmetric but not transitive
an equivalence relation
Solution
$a R a \Rightarrow 5 a$ is multiple it 5
So reflexive
$a R b \Rightarrow 2 a +3 b =5 \alpha$,
Now b R a
$2 b+3 a=2 b+\left(\frac{5 \alpha-3 b}{2}\right) \cdot 3$
$=\frac{15}{2} \alpha-\frac{5}{2} b=\frac{5}{2}(3 \alpha-b)$
$=\frac{5}{2}(2 a+2 b-2 \alpha)$
$=5(a+b-\alpha)$
Hence symmetric
$\text { a R b } \quad \Rightarrow 2 a+3 b=5 \alpha \text {. }$
$\text { b R c } \quad \Rightarrow 2 b+3 c=5 \beta$
$\text { Now } \quad 2 a+5 b+3 c=5(\alpha+\beta)$
$\Rightarrow 2 a +5 b +3 c =5(\alpha+\beta)$
$\Rightarrow 2 a +3 c =5(\alpha+\beta- b )$
$\Rightarrow a R c$
Hence relation is equivalence relation.