91Ó°ÊÓ

In mathematics, a relation is termed reflexive if every element in the set relates to itself. This means for any element, say \( a \), we must have \( a \sim a \).
In the given context, the relation \( R \) is defined on the set of integers \( \mathbb{Z} \) by \( m \sim_{R} m \) where \( m \) divides itself.
This observation is straightforward because any integer divided by itself results in 1, which is an integer. Simply put, for any integer \( m \), \( m \div m = 1 \). Therefore, \( m \) divides itself, satisfying the reflexivity criterion.
In summary, a reflexive relation ensures that every element is related to itself within the set. This property is easily identifiable when operating with integers, as self-division leaves a remainder of zero, solidifying the reflexivity of \( R \).

• Any integer \( m \) satisfies \( m \sim_{R} m \) because \( m \) divides itself.
• This division results in the integer 1, confirming the reflexive property.

Transitivity is a property of relations that deals with elements connecting through a chain.
If \( a \sim b \) and \( b \sim c \), then we must have \( a \sim c \). This chain-like structure ensures a consistent relationship carries through a sequence.
For our case, transitivity of \( R \) on integers is observed as follows: if \( m \sim_{R} n \) (meaning \( m \) divides \( n \)), and \( n \sim_{R} p \) (meaning \( n \) divides \( p \)), then indeed \( m \sim_{R} p \). Here's why:
- Assume \( n = km \) for some integer \( k \). This means \( m \) divides \( n \).
- Also, let \( p = ln \) for some integer \( l \). This means \( n \) divides \( p \).
- From these, \( p = l(km) = (lk)m \). Thus, \( m \) divides \( p \), proving transitivity.
In essence, if you can walk through the connection from one element to another, then back to the first through a second element, it's a clear sign of transitivity. The relation here sustains this by linking divisions through intermediate values.

• Start from \( m \sim_{R} n \) (\( m \) divides \( n \)).
• Follow to \( n \sim_{R} p \) (\( n \) divides \( p \)).
• Conclude \( m \sim_{R} p \) (\( m \) divides \( p \)), demonstrating transitivity.

The concept of symmetry in relations introduces a two-way relationship.
For symmetry, if \( a \sim b \), then it must also be true that \( b \sim a \). In simple terms, both directions of the connection must hold true for symmetry.
However, in the relation \( R \) defined as "m divides n," symmetry doesn't hold. While it may be true for some numbers, it's not universally applicable.
To understand better, let's consider an example:

• Suppose \( m = 2 \) and \( n = 4 \).
• Clearly, \( m \sim_{R} n \) as 2 divides 4.
• Conversely, \( n \ot\sim_{R} m \) because 4 does not divide 2 equally.

This proves that while 2 can divide 4, 4 fails to divide 2 without leaving a fraction. Hence, the relation isn't symmetric.
In essence, symmetry requires a kind of reciprocation that doesn't exist in the divisor relationship. This lack of balance in the dividing property shines a light on why symmetry isn't satisfied in this case.