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Sagot :
To show that the relation [tex]\( a \sim b \)[/tex] defined by [tex]\( n \mid (a - b) \)[/tex] is an equivalence relation on the set of integers [tex]\( \mathbb{Z} \)[/tex], we need to verify that this relation satisfies the three properties of equivalence relations: reflexivity, symmetry, and transitivity.
### Reflexivity
An equivalence relation must be reflexive, meaning every element must be related to itself. For our specific relation [tex]\( a \sim b \)[/tex], we need to show [tex]\( a \sim a \)[/tex] for any integer [tex]\( a \)[/tex].
- [tex]\( a \sim a \)[/tex] if and only if [tex]\( n \mid (a - a) \)[/tex].
- Since [tex]\( a - a = 0 \)[/tex] and [tex]\( n \)[/tex] divides 0 (for any [tex]\( n \)[/tex]) by definition, we have [tex]\( n \mid 0 \)[/tex].
Thus, [tex]\( a \sim a \)[/tex] for all integers [tex]\( a \)[/tex], proving reflexivity.
### Symmetry
An equivalence relation must be symmetric, meaning if [tex]\( a \sim b \)[/tex], then [tex]\( b \sim a \)[/tex]. For our specific relation [tex]\( a \sim b \)[/tex], we need to show that if [tex]\( n \mid (a - b) \)[/tex], then [tex]\( n \mid (b - a) \)[/tex].
- Suppose [tex]\( a \sim b \)[/tex], which means [tex]\( n \mid (a - b) \)[/tex].
- By definition of divisibility, this implies there exists some integer [tex]\( k \)[/tex] such that [tex]\( a - b = kn \)[/tex].
- Consider the relation [tex]\( b - a \)[/tex]. We have [tex]\( b - a = - (a - b) \)[/tex].
Since [tex]\( a - b = kn \)[/tex], it follows that [tex]\( b - a = -kn \)[/tex]. Therefore, [tex]\( n \)[/tex] divides [tex]\( -kn \)[/tex], and since [tex]\( n \mid (kn) \)[/tex], it must also hold that [tex]\( n \mid (-kn) \)[/tex].
This establishes that [tex]\( b \sim a \)[/tex] when [tex]\( a \sim b \)[/tex], proving symmetry.
### Transitivity
An equivalence relation must be transitive, meaning if [tex]\( a \sim b \)[/tex] and [tex]\( b \sim c \)[/tex], then [tex]\( a \sim c \)[/tex]. For our specific relation [tex]\( a \sim b \)[/tex], we need to show that if [tex]\( n \mid (a - b) \)[/tex] and [tex]\( n \mid (b - c) \)[/tex], then [tex]\( n \mid (a - c) \)[/tex].
- Suppose [tex]\( a \sim b \)[/tex], meaning [tex]\( n \mid (a - b) \)[/tex]. Therefore, there exists some integer [tex]\( k_1 \)[/tex] such that [tex]\( a - b = k_1 n \)[/tex].
- Suppose [tex]\( b \sim c \)[/tex], meaning [tex]\( n \mid (b - c) \)[/tex]. Therefore, there exists some integer [tex]\( k_2 \)[/tex] such that [tex]\( b - c = k_2 n \)[/tex].
We need to show that [tex]\( n \mid (a - c) \)[/tex]:
- Consider [tex]\( a - c \)[/tex].
- By adding the two relations given, [tex]\( (a - b) + (b - c) = k_1 n + k_2 n \)[/tex].
This simplifies to:
[tex]\[ a - c = k_1 n + k_2 n = (k_1 + k_2) n \][/tex]
Since [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex] are integers, [tex]\( (k_1 + k_2) \)[/tex] is also an integer. Therefore, [tex]\( a - c \)[/tex] is some integer multiple of [tex]\( n \)[/tex], which means [tex]\( n \mid (a - c) \)[/tex].
Hence, if [tex]\( a \sim b \)[/tex] and [tex]\( b \sim c \)[/tex], then [tex]\( a \sim c \)[/tex], proving transitivity.
### Conclusion
Since the relation [tex]\( a \sim b \)[/tex] defined by [tex]\( n \mid (a - b) \)[/tex] satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on the set of integers [tex]\( \mathbb{Z} \)[/tex].
### Reflexivity
An equivalence relation must be reflexive, meaning every element must be related to itself. For our specific relation [tex]\( a \sim b \)[/tex], we need to show [tex]\( a \sim a \)[/tex] for any integer [tex]\( a \)[/tex].
- [tex]\( a \sim a \)[/tex] if and only if [tex]\( n \mid (a - a) \)[/tex].
- Since [tex]\( a - a = 0 \)[/tex] and [tex]\( n \)[/tex] divides 0 (for any [tex]\( n \)[/tex]) by definition, we have [tex]\( n \mid 0 \)[/tex].
Thus, [tex]\( a \sim a \)[/tex] for all integers [tex]\( a \)[/tex], proving reflexivity.
### Symmetry
An equivalence relation must be symmetric, meaning if [tex]\( a \sim b \)[/tex], then [tex]\( b \sim a \)[/tex]. For our specific relation [tex]\( a \sim b \)[/tex], we need to show that if [tex]\( n \mid (a - b) \)[/tex], then [tex]\( n \mid (b - a) \)[/tex].
- Suppose [tex]\( a \sim b \)[/tex], which means [tex]\( n \mid (a - b) \)[/tex].
- By definition of divisibility, this implies there exists some integer [tex]\( k \)[/tex] such that [tex]\( a - b = kn \)[/tex].
- Consider the relation [tex]\( b - a \)[/tex]. We have [tex]\( b - a = - (a - b) \)[/tex].
Since [tex]\( a - b = kn \)[/tex], it follows that [tex]\( b - a = -kn \)[/tex]. Therefore, [tex]\( n \)[/tex] divides [tex]\( -kn \)[/tex], and since [tex]\( n \mid (kn) \)[/tex], it must also hold that [tex]\( n \mid (-kn) \)[/tex].
This establishes that [tex]\( b \sim a \)[/tex] when [tex]\( a \sim b \)[/tex], proving symmetry.
### Transitivity
An equivalence relation must be transitive, meaning if [tex]\( a \sim b \)[/tex] and [tex]\( b \sim c \)[/tex], then [tex]\( a \sim c \)[/tex]. For our specific relation [tex]\( a \sim b \)[/tex], we need to show that if [tex]\( n \mid (a - b) \)[/tex] and [tex]\( n \mid (b - c) \)[/tex], then [tex]\( n \mid (a - c) \)[/tex].
- Suppose [tex]\( a \sim b \)[/tex], meaning [tex]\( n \mid (a - b) \)[/tex]. Therefore, there exists some integer [tex]\( k_1 \)[/tex] such that [tex]\( a - b = k_1 n \)[/tex].
- Suppose [tex]\( b \sim c \)[/tex], meaning [tex]\( n \mid (b - c) \)[/tex]. Therefore, there exists some integer [tex]\( k_2 \)[/tex] such that [tex]\( b - c = k_2 n \)[/tex].
We need to show that [tex]\( n \mid (a - c) \)[/tex]:
- Consider [tex]\( a - c \)[/tex].
- By adding the two relations given, [tex]\( (a - b) + (b - c) = k_1 n + k_2 n \)[/tex].
This simplifies to:
[tex]\[ a - c = k_1 n + k_2 n = (k_1 + k_2) n \][/tex]
Since [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex] are integers, [tex]\( (k_1 + k_2) \)[/tex] is also an integer. Therefore, [tex]\( a - c \)[/tex] is some integer multiple of [tex]\( n \)[/tex], which means [tex]\( n \mid (a - c) \)[/tex].
Hence, if [tex]\( a \sim b \)[/tex] and [tex]\( b \sim c \)[/tex], then [tex]\( a \sim c \)[/tex], proving transitivity.
### Conclusion
Since the relation [tex]\( a \sim b \)[/tex] defined by [tex]\( n \mid (a - b) \)[/tex] satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on the set of integers [tex]\( \mathbb{Z} \)[/tex].
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