# Let R = {(a, b) : a, b, ϵ N and a < b}.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…..}

R = {(a, b) : a, b, ϵ N and a < 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) є R for all a є N.

But, here we see that a < 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) є R è(b, a) є R

Here, (a, b) є R does not imply (b, a) є R . Thus, it is not symmetric.

For transitivity,

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

Let’s take three values a, b and c such that a < b < c. So, (a, b) є R and (b, c) є R è (a, c) є R. Thus, it is transitive.

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