Question

Prove that the relation "less than" in the set of natural number is transitive but not reflexive and
symmetric.​

Answer 1

Answer withStep-by-step explanation:

We have a set of natural numbers

Let m , n, o are three natural numbers

1) Reflexive

(m,m) ≈ m<m which is not true as a number cant be less then by itself

Therefore less than is not reflexive

2) Symmetric

(m,n)≈ m<n ------(i)

(n,m)≈ n<m ------(ii)

from (i) and (ii) we can say that it is not possible

Therefore for (m,n) there does not exist (n,m)

Therefore it is not symmetric

3) Transitive

(m,n)≈ m<n ------(i)

(n,o)≈ n<0 ------(ii)

From (i) and (ii) m<n<o⇒ m<o i.e. ( m,o) exist

It is transitive

Hence proved that the less than relation on natural numbers is transitive but neither reflexive nor symmetric