Show that the relation


Show that the relation $R=\{(a, b): a>b\}$ on $N$ is transitive but neither reflexive nor symmetric.


We have, $R=\{(a, b): a>b\}$ relation defined on $N$.


We observe that, any element a $\in N$ cannot be greater than itself.

$\Rightarrow(a, a) \notin R \forall a \in N$

$\Rightarrow R$ is not reflexive.

Let $(a, b) \in R \forall a, b \in N$

$\Rightarrow \mathrm{a}$ is greater than $\mathrm{b}$

But $b$ cannot be greater than $a$ if $a$ is greater than $b$.

$\Rightarrow(b, a) \notin R$

For e.g., we observe that $(5,2) \in R$ i.e $5>2$ but $2 \ngtr 5 \Rightarrow(2,5) \notin R$

$\Rightarrow R$ is not symmetric

Let $(a, b) \in R$ and $(b, c) \in R \forall a, b, c \in N$

$\Rightarrow \mathrm{a}>\mathrm{b}$ and $\mathrm{b}>\mathrm{c}$

$\Rightarrow a>c$

$\Rightarrow(a, c) \in R$

For e.g., we observe that

$(5,4) \in R \Rightarrow 5>4$ and $(4,3) \in R \Rightarrow 4>3$

And we know that $5>3 \therefore(5,3) \in \mathrm{R}$

$\Rightarrow \mathrm{R}$ is transitive.

Thus, $R$ is transitive but not reflexive not symmetric.


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