Difference between revisions of "Aufgaben:Exercise 1.2Z: Sets of Digits"
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[[File:EN_Sto_Z_1_2_neu.png|right|frame|Sets of digits A, B, C]] | [[File:EN_Sto_Z_1_2_neu.png|right|frame|Sets of digits A, B, C]] |
Revision as of 17:11, 23 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,
- G = (\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 videoMengentheoretische Begriffe und Gesetzmäßigkeiten \Rightarrow.
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 solution 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.