Question

Let N be the set of all natural numbers and let R be a relation in N, defined by R=\left\\{ \left( a,b \right):\text{a is a multiple of b} \right\\} . Show that R is reflexive and transitive but not symmetric.

Answer 1

Hint:To show that R is reflexive, prove that for all $a\in N\left( a,a \right)\in R$ . To show that R is not symmetric, find an example such that $\left( a,b \right)\in R$ and $\left( b,a \right)\notin R$ . To show that R is transitive prove that for every $a,b,c\in A,$ if $\left( a,b \right)\in R$ and $\left( b,a \right)\in R$, then $\left( a,c \right)\in R$.

Complete step-by-step answer:
Let ‘A’ be a set then Reflexivity, Symmetry and transitivity of a relation on set ‘A’ is defined as follows:
Reflexive relation: - A relation R on a set ‘A’ is said to be reflexive if every element of A is related to itself i.e. for every element say (a) in set A, $\left( a,a \right)\in R$ .
Thus, R on a set ‘A’ is not reflexive if there exists an element $a\in A$ such that $\left( a,a \right)\notin R$.
Symmetric Relation: - A relation R on a set ‘A’ is said to be symmetric relation if $\left( a,b \right)\in R$ then $\left( b,a \right)$must be belong to R. i.e. $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$ For all $a,b\in A$.
Transitive relation:- A relation R on ‘A’ is said to be a transitive relation If $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$.
I.e. $\left( a,b \right)\in R$ and $\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
Given set is N: set of all natural numbers and relation R defined by R=\left\\{ \left( a,b \right):\text{a is a multiple of b} \right\\}. We have to show that R is reflexive and transitive but not symmetric.
Let us first show that R is reflexive. To show that R is reflexive, we need to show that for every $a\in N,\left( a,a \right)\in R$ .
We know that every natural number is a multiple of itself. So, for any ‘a’ belonging to N, (a,a) will satisfy the condition that “’a’ is a multiple of ‘a’”. So, $\left( a,a \right)\in R$.
Thus $\left( a,a \right)\in R\forall a\in N$.
So, R is reflexive on set N.
Next, let us show that r is transitive.
Let $a,b,c\in N$ and $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$. To show that R is transitive on N, we have to show that $\left( a,c \right)\in R$ .
$\left( a,b \right)\in R\Rightarrow$ “’a’ is a multiple of ‘b’” ⟹’b’ is a factor of ‘a’.
And $\left( b,c \right)\in R\Rightarrow$ “’b’ is a multiple of ‘c’” ⟹ ‘c’ is a factor of ‘b’.
As ‘b’ is a factor of ‘a’, ‘a’ can be written as $b{{K}_{1}}$ where ‘${{K}_{1}}$ ‘ is any integer
i.e. $a=b{{K}_{1}}$ ………………… (1)
$\Rightarrow b={a}/{{{K}_{1}}}\;$
Similarly as ‘c’ is a factor of ‘b’, ‘b’ can be written as ${{K}_{2}}c$, where ${{K}_{2}}$ is any integer.
i.e. $b={{K}_{2}}c$ ………………. (2)
Let us put $b=\left( {a}/{{{K}_{1}}}\; \right)$ in this equation using equation (1)

& \Rightarrow {a}/{{{K}_{1}}}\;=\left( {{K}_{2}} \right)\left( c \right) \\\ & \Rightarrow a=\left( {{K}_{1}}{{K}_{2}} \right)\left( c \right) \\\ \end{aligned}$$ ${{K}_{1}}$ is an integer and $${{K}_{2}}$$ is also an integer. So, their product will also be integer. Let ${{K}_{1}}{{K}_{2}}=K$ where K is an integer. $\Rightarrow a=\left( k \right)\left( c \right)$ . As ‘a’ can be written as a product of ‘c’ and ‘K’, we can say that “’c’ is a factor of ‘a’” and “’a’ is a multiple of ‘c’”. So, $\left( a,c \right)\in R$. Here, if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R\forall a,b,c\in N$ So, R is transitive on N. Let us take a=4 and b=2. $a,b\in N$ [Both ‘4’ and ‘2’ are natural numbers] $\left( a,b \right)\in R$ [As ‘4’ is a multiple of ‘2’] But $\left( b,a \right)\notin R$ [As ‘2’ is not a multiple of ‘4’] Thus R is not symmetric. Note: To prove that R is transitive, be careful that if $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$ then $\left( a,c \right)\in R$ . If there is no pair such that $\left( a,b \right)\in R$ and $\left( b,c \right)\in R$. Then we don’t need to check, R will always be transitive in this case.We can also check for reflexive if a=20,b=5 $a,b\in N$,$\left( a,b \right)\in R$ [As ‘20’ is a multiple of ‘5’] ,But $\left( b,a \right)\notin R$ [As ‘5’ is not a multiple of ‘20’] Thus R is not symmetric.