Let R be a relation on the set N of natural numbers defined by ‘nRm ⟺ n is a factor of m’. Then which one of the following is correct?
R is reflexive, transitive but not symmetric
Analyzing the Factor Relation on Natural Numbers
We are given a relation R defined on the set of natural numbers N. The relation is given by \( nRm \iff n \) is a factor of \( m \). We need to determine if this relation R is reflexive, symmetric, or transitive.
Let's examine each property for the relation \( nRm \iff n \) is a factor of \( m \), where \( n, m \in N \).
Reflexivity of the Factor Relation
A relation R on a set A is reflexive if for every element \( a \in A \), \( aRa \) holds. In our case, the set is N and the relation is R. We need to check if for every \( n \in N \), \( nRn \) holds.
\( nRn \) means \( n \) is a factor of \( n \).
For any natural number \( n \), \( n = n \times 1 \). Since 1 is a natural number, \( n \) is indeed a factor of itself.
For example, for \( n=5 \), 5 is a factor of 5 because \( 5 = 5 \times 1 \).
Thus, the relation R is reflexive.
Symmetry of the Factor Relation
A relation R on a set A is symmetric if for every \( a, b \in A \), whenever \( aRb \) holds, \( bRa \) also holds. In our case, we need to check if for every \( n, m \in N \), whenever \( nRm \) holds, \( mRn \) also holds.
\( nRm \) means \( n \) is a factor of \( m \).
\( mRn \) means \( m \) is a factor of \( n \).
If \( n \) is a factor of \( m \), does it necessarily mean \( m \) is a factor of \( n \)? Let's consider a counterexample.
Let \( n=2 \) and \( m=4 \). Here, \( n, m \in N \).
- Is \( nRm \)? Is 2 a factor of 4? Yes, because \( 4 = 2 \times 2 \). So, \( 2R4 \) holds.
- Is \( mRn \)? Is 4 a factor of 2? No, because there is no natural number \( k \) such that \( 2 = 4 \times k \). So, \( 4R2 \) does not hold.
Since we found a case where \( nRm \) holds but \( mRn \) does not hold, the relation R is not symmetric.
Transitivity of the Factor Relation
A relation R on a set A is transitive if for every \( a, b, c \in A \), whenever \( aRb \) holds and \( bRc \) holds, \( aRc \) also holds. In our case, we need to check if for every \( n, m, p \in N \), whenever \( nRm \) holds and \( mRp \) holds, \( nRp \) also holds.
- \( nRm \) means \( n \) is a factor of \( m \). This implies \( m = k \times n \) for some natural number \( k \).
- \( mRp \) means \( m \) is a factor of \( p \). This implies \( p = l \times m \) for some natural number \( l \).
Now let's substitute the expression for \( m \) from the first condition into the second condition:
\( p = l \times m \)
\( p = l \times (k \times n) \)
\( p = (l \times k) \times n \)
Since \( k \) and \( l \) are natural numbers, their product \( l \times k \) is also a natural number. Let \( j = l \times k \). Then \( p = j \times n \), where \( j \in N \).
This means \( n \) is a factor of \( p \).
So, if \( nRm \) and \( mRp \) hold, then \( nRp \) also holds.
For example, let \( n=2, m=6, p=18 \). All are natural numbers.
- Is \( nRm \)? Is 2 a factor of 6? Yes, \( 6 = 3 \times 2 \).
- Is \( mRp \)? Is 6 a factor of 18? Yes, \( 18 = 3 \times 6 \).
- Is \( nRp \)? Is 2 a factor of 18? Yes, \( 18 = 9 \times 2 \).
This example supports the transitivity. Our logical derivation confirms it for all natural numbers.
Thus, the relation R is transitive.
Summary of Properties
Based on our analysis:
- R is reflexive.
- R is not symmetric.
- R is transitive.
Therefore, the relation R is reflexive and transitive but not symmetric. It is not an equivalence relation because it is not symmetric (an equivalence relation must be reflexive, symmetric, and transitive).
| Property | Holds for R? | Reason / Counterexample |
|---|---|---|
| Reflexive (\( nRn \)) | Yes | \( n = n \times 1 \), 1 is a natural number. |
| Symmetric (If \( nRm \), then \( mRn \)) | No | Take \( n=2, m=4 \). 2 is a factor of 4, but 4 is not a factor of 2. |
| Transitive (If \( nRm \) and \( mRp \), then \( nRp \)) | Yes | If \( m=kn \) and \( p=lm \), then \( p=l(kn) = (lk)n \), so \( n \) is a factor of \( p \). |
Conclusion on the Factor Relation
The relation R defined as '\( nRm \iff n \) is a factor of \( m \)' on the set of natural numbers N is reflexive and transitive, but it is not symmetric.
Revision Table: Relation Properties
Understanding the different types of relations is key in set theory and discrete mathematics. Here's a quick review of the properties discussed:
| Property | Definition on Set A | Condition |
|---|---|---|
| Reflexive | For every element \( a \in A \), \( (a, a) \in R \) | Every element is related to itself. |
| Symmetric | For every \( a, b \in A \), if \( (a, b) \in R \), then \( (b, a) \in R \) | If \( a \) is related to \( b \), then \( b \) is related to \( a \). |
| Transitive | For every \( a, b, c \in A \), if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \) | If \( a \) is related to \( b \), and \( b \) to \( c \), then \( a \) is related to \( c \). |
| Equivalence Relation | A relation that is Reflexive, Symmetric, and Transitive. | Partitions the set into disjoint equivalence classes. |
| Partial Order Relation | A relation that is Reflexive, Antisymmetric, and Transitive. | Often represents an ordering or hierarchy (like the factor relation). Antisymmetric means if \( (a,b) \in R \) and \( (b,a) \in R \), then \( a=b \). The factor relation is antisymmetric on N: if \( n \) is a factor of \( m \) and \( m \) is a factor of \( n \), then \( n=m \). |
Additional Information: Types of Relations and Examples
Relations are fundamental concepts in mathematics. They describe connections between elements of sets. Here are some more details and examples:
- Binary Relation: A relation R on a set A is a subset of \( A \times A \). If \( (a, b) \in R \), we write \( aRb \).
- Natural Numbers (N): The set of positive integers \( \{1, 2, 3, \dots\} \). Note that some definitions include 0, but the context usually clarifies. Here, based on typical factor definitions, we consider \( \{1, 2, 3, \dots\} \).
- Factor: A natural number \( n \) is a factor of a natural number \( m \) if \( m \) can be divided by \( n \) with no remainder, or if there exists a natural number \( k \) such that \( m = n \times k \).
- Example of a Symmetric Relation: Relation 'is equal to' on N. If \( n=m \), then \( m=n \). This is also reflexive and transitive, so it's an equivalence relation.
- Example of a Relation that is not Transitive: Relation 'is one less than' on N. \( nRm \iff n = m-1 \). \( 3R4 \) is true (3 is one less than 4). \( 4R5 \) is true (4 is one less than 5). But \( 3R5 \) is false (3 is not one less than 5). So, it's not transitive.
The factor relation \( nRm \iff n \) is a factor of \( m \) is a classic example of a partial order relation on the set of natural numbers.
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Select the answer using the code given below.
