Understanding the Relation on Natural Numbers

The question asks us to examine a specific relation \( R \) defined on the set of natural numbers, denoted by \( \mathbb{N} \). The set of natural numbers \( \mathbb{N} \) usually starts with 1, i.e., \( \{1, 2, 3, \dots\} \). The relation \( R \) is defined as \( R = \{(a, b) : a, b \in \mathbb{N} \) and \( b \) is divisible by \( a\} \). This means that an ordered pair \( (a, b) \) belongs to the relation \( R \) if and only if \( a \) divides \( b \). We need to determine if this relation \( R \) is reflexive, symmetric, or transitive.

Checking for Reflexivity of the Relation

A relation \( R \) on a set \( A \) is said to be reflexive if for every element \( a \in A \), the ordered pair \( (a, a) \) is in the relation \( R \). In this problem, our set is \( \mathbb{N} \). So, we need to check if for every \( a \in \mathbb{N} \), \( (a, a) \in R \).

According to the definition of \( R \), \( (a, a) \in R \) means that the second element (\( a \)) is divisible by the first element (\( a \)). Is any natural number \( a \) divisible by itself?

Yes, for any natural number \( a \), \( a \) is divisible by \( a \) because \( a = 1 \times a \), and \( 1 \) is an integer (in fact, a natural number). For example:

• For \( a = 1 \), 1 is divisible by 1 (\( 1 = 1 \times 1 \)), so \( (1, 1) \in R \).
• For \( a = 5 \), 5 is divisible by 5 (\( 5 = 1 \times 5 \)), so \( (5, 5) \in R \).
• For any \( a \in \mathbb{N} \), \( a \) is divisible by \( a \).

Since \( (a, a) \in R \) for all \( a \in \mathbb{N} \), the relation \( R \) is reflexive.

Checking for Symmetry of the Relation

A relation \( R \) on a set \( A \) is said to be symmetric if for all elements \( a, b \in A \), whenever \( (a, b) \in R \), it must also be true that \( (b, a) \in R \). In this case, we need to check if for all \( a, b \in \mathbb{N} \), if \( b \) is divisible by \( a \), then \( a \) is also divisible by \( b \).

Let's consider an example. Let \( a = 2 \) and \( b = 4 \). Both are natural numbers.

• Is \( (2, 4) \in R \)? Yes, because 4 is divisible by 2 (\( 4 = 2 \times 2 \)).
• Now, is \( (4, 2) \in R \)? This would mean that 2 is divisible by 4. Is 2 divisible by 4? No, 2 cannot be written as \( 4 \times k \) for any natural number \( k \). The only way \( 2 = 4k \) would hold is if \( k = 2/4 = 1/2 \), which is not a natural number.

Since we found a pair \( (2, 4) \in R \) such that \( (4, 2) \notin R \), the condition for symmetry is not met. Therefore, the relation \( R \) is not symmetric.

Checking for Transitivity of the Relation

A relation \( R \) on a set \( A \) is said to be transitive if for all elements \( a, b, c \in A \), whenever \( (a, b) \in R \) and \( (b, c) \in R \), it must also be true that \( (a, c) \in R \). In this case, we need to check if for all \( a, b, c \in \mathbb{N} \), if \( b \) is divisible by \( a \) and \( c \) is divisible by \( b \), then \( c \) is also divisible by \( a \).

Let's assume \( (a, b) \in R \) and \( (b, c) \in R \), where \( a, b, c \in \mathbb{N} \).

• \( (a, b) \in R \) means \( b \) is divisible by \( a \). This can be written as \( b = ka \) for some natural number \( k \).
• \( (b, c) \in R \) means \( c \) is divisible by \( b \). This can be written as \( c = lb \) for some natural number \( l \).

Now we need to check if \( (a, c) \in R \), which means \( c \) is divisible by \( a \).

We have \( c = lb \) and we know \( b = ka \). Let's substitute the expression for \( b \) into the equation for \( c \):

\( c = l(ka) \)

Using the associative property of multiplication, we can write this as:

\( c = (lk)a \)

Since \( k \) and \( l \) are natural numbers, their product \( lk \) is also a natural number. Let \( m = lk \). Then \( c = ma \), where \( m \in \mathbb{N} \). This means that \( c \) is divisible by \( a \).

For example, let \( a = 2, b = 6, c = 18 \). All are natural numbers.

• Is \( (2, 6) \in R \)? Yes, 6 is divisible by 2 (\( 6 = 3 \times 2 \)).
• Is \( (6, 18) \in R \)? Yes, 18 is divisible by 6 (\( 18 = 3 \times 6 \)).
• Now, is \( (2, 18) \in R \)? Yes, 18 is divisible by 2 (\( 18 = 9 \times 2 \)).

This holds true for all \( a, b, c \in \mathbb{N} \) where \( a|b \) and \( b|c \). The result is that \( a|c \). Therefore, the relation \( R \) is transitive.

Summary of Relation Properties

Based on our analysis:

• The relation \( R \) is Reflexive.
• The relation \( R \) is Not Symmetric.
• The relation \( R \) is Transitive.

An equivalence relation is a relation that is reflexive, symmetric, and transitive. Since this relation is not symmetric, it is not an equivalence relation.

Thus, the relation \( R \) is reflexive, transitive but not symmetric.

Conclusion on Relation Properties

The relation \( R \) defined by \( (a, b) \in R \) if and only if \( b \) is divisible by \( a \) on the set of natural numbers \( \mathbb{N} \) is found to be reflexive and transitive, but not symmetric.

Revision Table: Key Concepts of Relations

Additional Information on Relations

Relations are fundamental concepts in mathematics, particularly in discrete mathematics and set theory. They describe connections between elements of sets. Understanding the properties like reflexivity, symmetry, and transitivity helps classify different types of relations.

• Partial Order Relation: A relation that is reflexive, antisymmetric, and transitive is called a partial order. The divisibility relation on \( \mathbb{N} \) is reflexive and transitive, and it is also antisymmetric (if \( a|b \) and \( b|a \) for \( a, b \in \mathbb{N} \), then \( a=b \)). Therefore, the divisibility relation on \( \mathbb{N} \) is a partial order relation.
• Equivalence Relation: As mentioned, a relation is an equivalence relation if it is reflexive, symmetric, and transitive. Equivalence relations partition a set into disjoint equivalence classes. Examples include the relation "is equal to" on any set, or "has the same age as" on a set of people.

Analyzing relations based on these properties is a key skill in understanding mathematical structures.