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Let $R=\{(a, b): a, b, \in N$ and $a

Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.



N is the set of all the natural numbers.

N = {1, 2, 3, 4, 5, 6, 7…..}

$\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b}, \in \mathrm{N}$ and $\mathrm{a}<\mathrm{b}\}$

R = {(1, 2), (1, 3), (1, 4) …. (2, 3), (2, 4), (2, 5) ……}

For reflexivity,

A relation $R$ on $N$ is said to be reflexive if $(a, a) \in R$ for all a $€ N$.

But, here we see that $\mathrm{a}<\mathrm{b}$, so the two co-ordinates are never equal. Thus, the relation is not reflexive.

For symmetry,

A relation $R$ on $N$ is said to be symmetrical if $(a, b) \in R \dot{e}(b, a) \in R$

Here, $(a, b) \in R$ does not imply $(b, a) \in R$. Thus, it is not symmetric.

For transitivity,

A relation $R$ on $A$ is said to be transitive if $(a, b) \in R$ and $(b, c) \in R$ è $(a, c) \in R$ for all $(a$, b, c) $\in N$.

Let's take three values $a, b$ and $c$ such that $a


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