IDNLearn.com offers a user-friendly platform for finding and sharing answers. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.
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].
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
