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Question: Let N be the set of all natural numbers and let R be a relation in N’ defined by \(R=\left\\{ \left(...

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 factor of b} \right\\} . Then, show that R is reflexive and transitive but not symmetric.

Explanation

Solution

Hint:To show that R is reflexive, prove that for all a∈N(a,a)∈Ra\in N\left( a,a \right)\in R . To show that R is not symmetric, find an example such that (a,b)∈R\left( a,b \right)\in R and (b,a)∉R\left( b,a \right)\notin R . To show that R is transitive, prove that for every a,b,c∈A,a,b,c\in A, if (a,b)∈R\left( a,b \right)\in R and (b,a)∈R\left( b,a \right)\in R, then (a,c)∈R\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, (a,a)∈R\left( a,a \right)\in R .
Thus, R on a set ‘A’ is not reflexive if there exists an element a∈Aa\in A such that (a,a)∉R\left( a,a \right)\notin R.
Symmetric Relation: - A relation R on a set ‘A’ is said to be symmetric relation if (a,b)∈R\left( a,b \right)\in R then (b,a)\left( b,a \right)must be belong to R. i.e. (a,b)∈R⇒(b,a)∈R\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R For all a,b∈Aa,b\in A.
Transitive relation:- A relation R on ‘A’ is said to be a transitive relation If (a,b)∈R\left( a,b \right)\in R and (b,c)∈R\left( b,c \right)\in R then (a,c)∈R\left( a,c \right)\in R.
I.e. (a,b)∈R\left( a,b \right)\in R and (b,c)∈R⇒(a,c)∈R\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 factor 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∈N,(a,a)∈Ra\in N,\left( a,a \right)\in R .
We know that every natural number is a factor ‘a’ of itself. So, for any ‘a’ belonging to N, (a,a) will satisfy the condition that “’a’ is a factor of ‘a’”. So, (a,a)∈R\left( a,a \right)\in R.
Thus (a,a)∈R∀a∈N\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∈Na,b,c\in N and (a,b)∈R\left( a,b \right)\in R and (b,c)∈R\left( b,c \right)\in R. To show that R is transitive on N, we have to show that (a,c)∈R\left( a,c \right)\in R .
(a,b)∈R⇒\left( a,b \right)\in R\Rightarrow “’a’ is a factor of ‘b’”
And (b,c)∈R⇒\left( b,c \right)\in R\Rightarrow “’b’ is a factor of ‘c’”
As ‘a’ is a factor of ‘b’, ‘b’ can be written as aK1a{{K}_{1}} where ‘K1{{K}_{1}} ‘ is any integer
i.e. b=aK1b=a{{K}_{1}} ………………… (1)
Similarly as ‘b’ is a factor of ‘c’, ‘c’ can be written as bK2b{{K}_{2}}, where K2{{K}_{2}} is any integer.
i.e. c=bK2c=b{{K}_{2}} ………………. (2)
Let us put b=(aK1)b=\left( a{{K}_{1}} \right) in this equation using equation (1)

& \Rightarrow c=\left( a{{K}_{1}} \right)\left( {{K}_{2}} \right) \\\ & \Rightarrow c=a\left( {{K}_{1}}{{K}_{2}} \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. $\Rightarrow c=a\left( k \right)$ . As ‘c’ can be written as a product of ‘a’ and ‘K’, we can say that “’a’ is a factor 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 show that R is not symmetric. To show that R is not symmetric, we have to find an example such that $\left( a,b \right)\in R$ and $\left( b,a \right)\notin R$ $a,b\in N$ and let us take a=2 and b=4. $a,b\in N$ [Both ‘2’ and ‘4’ are natural numbers] $\left( a,b \right)\in R$ [As ‘2’ is a factor of ‘4’] But $\left( b,a \right)\notin R$ [As ‘4’ is not a factor of ‘2’] 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.