Difference between revisions of "Aufgaben:Exercise 1.2Z: Sets of Digits"
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− | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Set_Theory_Basics}} |
− | [[File: | + | [[File:EN_Sto_Z_1_2_neu.png|right|frame|Sets of digits: A, B, C]] |
− | + | Let the universal set G be the set of all digits between $1 and 9$. Given are the following subsets: | |
− | + | :$$A = \big[\text{digits} \leqslant 3\big],$$ | |
− | $$ B = [ | + | :$$ B = \big[\text{digits divisible by 3}\big],$$ |
− | $$ C = [ | + | :$$ C = \big[\text{digits 5, 6, 7, 8}\big].$$ |
− | + | Besides these, let other sets be defined: | |
− | $$D = (A \cap \ | + | :$$D = (A \cap \overline B) \cup (\overline A \cap B),$$ |
− | $$E = (A \cup B) \cap (\ | + | :$$E = (A \cup B) \cap (\overline A \cup \overline B), $$ |
− | $$F = (A \cup C) \cap \ | + | :$$F = (A \cup C) \cap \overline B, $$ |
− | $$ | + | :$$H = (\overline A \cap \overline C) \cup (A \cap B \cap C).$$ |
− | + | ||
− | + | First consider which digits belong to the sets D, E, F and H and then answer the following questions. <br>Justify your answers in terms of set theory. | |
− | === | + | |
+ | |||
+ | |||
+ | |||
+ | Hints: | ||
+ | *The task belongs to the chapter [[Theory_of_Stochastic_Signals/Set_Theory_Basics|Set Theory Basics]]. | ||
+ | |||
+ | *The topic of this chapter is illustrated with examples in the (German language) learning video | ||
+ | |||
+ | :[[Mengentheoretische_Begriffe_und_Gesetzmäßigkeiten_(Lernvideo)|Mengentheoretische Begriffe und Gesetzmäßigkeiten]] ⇒ "Set Theoretical Concepts and Laws". | ||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which of the following statements are correct? |
|type="[]"} | |type="[]"} | ||
− | - A | + | - A and B are disjoint sets. |
− | + A | + | + A and C are disjoint sets. |
− | - B | + | - B and C are disjoint sets. |
− | { | + | {Which of the following statements are correct? |
|type="[]"} | |type="[]"} | ||
− | - | + | - The union A∪B∪C gives the universal set G. |
− | + | + | + The complementary set to A∩B∩C gives the universal set G. |
− | { | + | {Which of the following statements are correct? |
|type="[]"} | |type="[]"} | ||
− | + | + | + The complementary sets of D and E are identical. |
− | + F | + | + F is a subset of the complementary set of B. |
− | - | + | - The sets B, C and D form a complete system. |
− | + | + | + The sets A, C and H form a complete system. |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | : | + | For the other sets defined in the problem holds: |
− | : | + | |
− | : | + | :$$ D = (A \cap \overline B) \cup (\overline A \cap B) |
− | :''' | + | =\big[\{1, 2, 3\} \cap \{1, 2, 4, 5, 7, 8\}\big] \cup \big[\{4, 5, 6, 7, 8, 9\} \cap \{3, 6, 9\}\big] = \{1, 2, 6, 9\},$$ |
− | + | ||
− | + | :$$ E = (A \cup B) \cap (\overline A \cup \overline B) = (A \cap \overline A) \cup (A \cap \overline B) \cup (\overline A \cap B) \cup (\overline A \cap \overline B) = (A \cap \overline B) \cup (\overline A \cap B) = D = \{1, 2, 6, 9\},$$ | |
− | : | + | |
+ | :$$F = (A \cup C) \cap \overline B = \{1, 2, 3, 5, 6, 7, 8\} \cap \{1, 2, 4, 5, 7, 8\} = \{1, 2, 5, 7, 8\},$$ | ||
+ | |||
+ | :H=(ˉA∩¯C)∪(A∩B∩C)=(¯A∩¯C)∪ϕ={4,9}. | ||
+ | |||
+ | '''(1)''' Only the <u>proposed solution 2</u> is correct: | ||
+ | * A and C have no common element. | ||
+ | * A and B each contain a 3. | ||
+ | * B and C each contain a 6. | ||
+ | |||
+ | |||
+ | |||
+ | '''(2)''' Correct is the <u>proposed solution 2</u>: | ||
+ | *No digit is contained in A, B and C at the same time ⇒ A∩B∩C=ϕ ⇒ ¯A∩B∩C=¯ϕ=G. | ||
+ | *The first proposition, on the other hand, is wrong. It is missing a 4. | ||
+ | |||
+ | |||
+ | |||
+ | '''(3)''' Correct are the <u>proposed solutions 1, 2 and 4</u>: | ||
+ | *The first proposal is correct: The sets D and E contain exactly the same elements and thus also their complementary sets. | ||
+ | *The second proposal is also correct: In general, i.e. for any X and B the following holds: $(X \cap \overline B) \subset \overline B \ \Rightarrow WithX = A \cup CitfollowsthatF \subset \overline B$. | ||
+ | *The last proposal is also correct: A={1,2,3}, C={5,6,7,8} and H={4,9} form a "complete system". | ||
+ | *The third suggestion, on the other hand, is wrong because B and C are not disjoint. | ||
+ | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
− | [[Category: | + | [[Category:Theory of Stochastic Signals: Exercises|^1.2 Set Theory Basics |
^]] | ^]] |
Latest revision as of 15:13, 25 November 2021
Let the universal set G be the set of all digits between 1 and 9. Given are the following subsets:
- A=[digits⩽
- B = \big[\text{digits divisible by 3}\big],
- C = \big[\text{digits 5, 6, 7, 8}\big].
Besides these, let other sets be defined:
- D = (A \cap \overline B) \cup (\overline A \cap B),
- E = (A \cup B) \cap (\overline A \cup \overline B),
- F = (A \cup C) \cap \overline B,
- H = (\overline A \cap \overline C) \cup (A \cap B \cap C).
First consider which digits belong to the sets D, E, F and H and then answer the following questions.
Justify your answers in terms of set theory.
Hints:
- The task belongs to the chapter Set Theory Basics.
- The topic of this chapter is illustrated with examples in the (German language) learning video
- Mengentheoretische Begriffe und Gesetzmäßigkeiten \Rightarrow "Set Theoretical Concepts and Laws".
Questions
Solution
For the other sets defined in the problem holds:
- D = (A \cap \overline B) \cup (\overline A \cap B) =\big[\{1, 2, 3\} \cap \{1, 2, 4, 5, 7, 8\}\big] \cup \big[\{4, 5, 6, 7, 8, 9\} \cap \{3, 6, 9\}\big] = \{1, 2, 6, 9\},
- E = (A \cup B) \cap (\overline A \cup \overline B) = (A \cap \overline A) \cup (A \cap \overline B) \cup (\overline A \cap B) \cup (\overline A \cap \overline B) = (A \cap \overline B) \cup (\overline A \cap B) = D = \{1, 2, 6, 9\},
- F = (A \cup C) \cap \overline B = \{1, 2, 3, 5, 6, 7, 8\} \cap \{1, 2, 4, 5, 7, 8\} = \{1, 2, 5, 7, 8\},
- H = (\bar A \cap \overline C) \cup (A \cap B \cap C) = (\overline A \cap \overline C) \cup \phi = \{4, 9\}.
(1) Only the proposed solution 2 is correct:
- A and C have no common element.
- A and B each contain a 3.
- B and C each contain a 6.
(2) Correct is the proposed solution 2:
- No digit is contained in A, B and C at the same time ⇒ A \cap B \cap C = \phi ⇒ \overline{A \cap B \cap C} = \overline{\phi} = G.
- The first proposition, on the other hand, is wrong. It is missing a 4.
(3) Correct are the proposed solutions 1, 2 and 4:
- The first proposal is correct: The sets D and E contain exactly the same elements and thus also their complementary sets.
- The second proposal is also correct: In general, i.e. for any X and B the following holds: (X \cap \overline B) \subset \overline B \ \Rightarrow With X = A \cup C it follows that F \subset \overline B.
- The last proposal is also correct: A = \{1, 2, 3\}, C = \{5, 6, 7, 8\} and H = \{4, 9\} form a "complete system".
- The third suggestion, on the other hand, is wrong because B and C are not disjoint.