Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Let's start by analyzing each option step-by-step to determine which one is not true.
### Option A: [tex]\( s_n \cup s_{2n} = s_{2n} \)[/tex]
Set [tex]\( s_n = \{100, 101, \ldots, 100 + n\} \)[/tex] and [tex]\( s_{2n} = \{100, 101, \ldots, 100 + 2n\} \)[/tex].
The union [tex]\( s_n \cup s_{2n} \)[/tex] combines all elements from both sets:
- Elements of [tex]\( s_n \)[/tex] are from 100 to [tex]\( 100 + n \)[/tex].
- Elements of [tex]\( s_{2n} \)[/tex] are from 100 to [tex]\( 100 + 2n \)[/tex].
Since [tex]\( 100 + 2n \)[/tex] is always greater than [tex]\( 100 + n \)[/tex], the union of these sets will be exactly [tex]\( s_{2n} \)[/tex], i.e., [tex]\( s_n \cup s_{2n} = s_{2n} \)[/tex].
Thus, Option A is true.
### Option B: If [tex]\( n > m \)[/tex], then [tex]\( s_n \cap s_m = s_n \)[/tex]
Suppose [tex]\( n > m \)[/tex]. Then:
- [tex]\( s_n = \{100, 101, \ldots, 100 + n \} \)[/tex]
- [tex]\( s_m = \{100, 101, \ldots, 100 + m \} \)[/tex]
Since [tex]\( n > m \)[/tex], all elements of [tex]\( s_m \)[/tex] are also in [tex]\( s_n \)[/tex]. Therefore, the intersection [tex]\( s_n \cap s_m \)[/tex] contains exactly the elements of [tex]\( s_m \)[/tex] only, not [tex]\( s_n \)[/tex].
Thus, Option B is false.
### Option C: [tex]\( s_{1000} = \{100, 101, \ldots, 1100\} \)[/tex]
- [tex]\( s_{1000} = \{100, 101, \ldots, 100 + 1000\} = \{100, 101, \ldots, 1100\} \)[/tex]
This statement is directly equal to the definition of [tex]\( s_{1000} \)[/tex].
Thus, Option C is true.
### Option D: [tex]\( s_{60} - s_{30} = \{131, 132, \ldots, 160\} \)[/tex]
- [tex]\( s_{60} = \{100, 101, \ldots, 160\} \)[/tex]
- [tex]\( s_{30} = \{100, 101, \ldots, 130\} \)[/tex]
The difference [tex]\( s_{60} - s_{30} \)[/tex] removes elements that are in [tex]\( s_{30} \)[/tex] from [tex]\( s_{60} \)[/tex], leaving:
- Elements in [tex]\( s_{60} \)[/tex] but not in [tex]\( s_{30} \)[/tex] are [tex]\( 131, 132, \ldots, 160 \)[/tex].
Thus, Option D is true.
### Conclusion
By analyzing the provided options, the correct answer (the statement that is not true) is:
[tex]\[ \boxed{\text{B}} \][/tex]
### Option A: [tex]\( s_n \cup s_{2n} = s_{2n} \)[/tex]
Set [tex]\( s_n = \{100, 101, \ldots, 100 + n\} \)[/tex] and [tex]\( s_{2n} = \{100, 101, \ldots, 100 + 2n\} \)[/tex].
The union [tex]\( s_n \cup s_{2n} \)[/tex] combines all elements from both sets:
- Elements of [tex]\( s_n \)[/tex] are from 100 to [tex]\( 100 + n \)[/tex].
- Elements of [tex]\( s_{2n} \)[/tex] are from 100 to [tex]\( 100 + 2n \)[/tex].
Since [tex]\( 100 + 2n \)[/tex] is always greater than [tex]\( 100 + n \)[/tex], the union of these sets will be exactly [tex]\( s_{2n} \)[/tex], i.e., [tex]\( s_n \cup s_{2n} = s_{2n} \)[/tex].
Thus, Option A is true.
### Option B: If [tex]\( n > m \)[/tex], then [tex]\( s_n \cap s_m = s_n \)[/tex]
Suppose [tex]\( n > m \)[/tex]. Then:
- [tex]\( s_n = \{100, 101, \ldots, 100 + n \} \)[/tex]
- [tex]\( s_m = \{100, 101, \ldots, 100 + m \} \)[/tex]
Since [tex]\( n > m \)[/tex], all elements of [tex]\( s_m \)[/tex] are also in [tex]\( s_n \)[/tex]. Therefore, the intersection [tex]\( s_n \cap s_m \)[/tex] contains exactly the elements of [tex]\( s_m \)[/tex] only, not [tex]\( s_n \)[/tex].
Thus, Option B is false.
### Option C: [tex]\( s_{1000} = \{100, 101, \ldots, 1100\} \)[/tex]
- [tex]\( s_{1000} = \{100, 101, \ldots, 100 + 1000\} = \{100, 101, \ldots, 1100\} \)[/tex]
This statement is directly equal to the definition of [tex]\( s_{1000} \)[/tex].
Thus, Option C is true.
### Option D: [tex]\( s_{60} - s_{30} = \{131, 132, \ldots, 160\} \)[/tex]
- [tex]\( s_{60} = \{100, 101, \ldots, 160\} \)[/tex]
- [tex]\( s_{30} = \{100, 101, \ldots, 130\} \)[/tex]
The difference [tex]\( s_{60} - s_{30} \)[/tex] removes elements that are in [tex]\( s_{30} \)[/tex] from [tex]\( s_{60} \)[/tex], leaving:
- Elements in [tex]\( s_{60} \)[/tex] but not in [tex]\( s_{30} \)[/tex] are [tex]\( 131, 132, \ldots, 160 \)[/tex].
Thus, Option D is true.
### Conclusion
By analyzing the provided options, the correct answer (the statement that is not true) is:
[tex]\[ \boxed{\text{B}} \][/tex]
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.