Question

The relation 'less than' in the set of natural numbers is $A. Only symmetric$
B. Only transitive$C. Only reflexive$
D. Equivalence relation$$$$

Answer 1

We recall the definitions of reflexive, transitive and symmetric relations. We define the relation of ‘less than’ in natural number set R=\left\\{ \left( a,b \right):a < b,a\in \mathsf{\mathbb{N}},b\in \mathsf{\mathbb{N}} \right\\} and check whether$a < a$, $a < b\Rightarrow b < c$,$a < b,a < c\Rightarrow a < c$ is true for all $a,b,c\in \mathsf{\mathbb{N}}$to check reflexive, symmetric and transitive respectively which simultaneously give equivalence. $$$$

Complete step by step answer:
We know that a relation $R$ from set $A$ to set $B$ is a set of ordered pairs written as $R:A\to B$ which takes its elements from the set A\times B=\left\\{ \left( a,b \right):a\in A,b\in B \right\\}. If an element $a$ is related to $B$ then we can write as $aRb$ or $a\tilde{\ }b$. If the relation is defined on one set $A$ then $R:A\to A$ takes elements from the set A\times A=\left\\{ \left( a,b \right):a\in A,b\in A \right\\}
A binary relation $R$ defined on the set A which means $R:A\to A$ is called reflexive if $a\tilde{\ }a$for all$a\in A$. We can write in symbols,
$a\tilde{\ }a\Rightarrow \left( a,a \right)\in R$
The relation $R$ is symmetric if $a\tilde{\ }B\Rightarrow b\tilde{\ }a$ for all $b\in A$. We can write in symbols,
$\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$
The relation $R$ is transitive if $a\tilde{\ }B,b\tilde{\ }c\Rightarrow a\tilde{\ }c$ for all $c\in A$. We can write it in symbols,
$\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$
The relation is equivalent when $R$ is reflexive, symmetric and transitive. We are given the relation in the question 'less than' in the set of natural numbers. We define the relation of less than in the natural number set $\mathsf{\mathbb{N}}$ as $$R=\left\\{ \left( a,b \right):a < b,a\in \mathsf{\mathbb{N}},b\in \mathsf{\mathbb{N}} \right\\}$$ We see that if $R$ has to be reflexive then for all $a\in \mathsf{\mathbb{N}}$ we have$a\tilde{\ }a$ which means $a$ has to be less than itself that is $a < a$ which is not possible. So $R$ is not reflexive.
We see that if $R$ has to be symmetric then for all $a,b\in \mathsf{\mathbb{N}}$ we have $a\tilde{\ }b\Rightarrow b\tilde{\ }a$ or $a < b\Rightarrow b < a$ but we see that if $a>b$ then we cannot have $b>a$. So $R$ is not symmetric. We see that if $R$ has to be transitive then for all $a,b\in \mathsf{\mathbb{N}}$ we have $a\tilde{\ }b,b\tilde{\ }c\Rightarrow a\tilde{\ }c$which means $ a < b,b < c\Rightarrow a < c$ which is true because the natural number set $\mathsf{\mathbb{N}}$ is an ordered set. So $R$ is transitive.

So, the correct answer is “Option B”.

Note: We note that we call set an ordered set when elements of the set are arranged in ascending or descending order. If the elements are related as $a\tilde{\ }b\Rightarrow b\tilde{\ }a$ implies $a=b$ we call the relation antisymmetric. The relations “greater than” is also only transitive but the relation “less than equal to” is equivalence.