371/MR00. példa / Matematika - Megoldott feladatok gyűjteménye

MR-371 / 31. példa

Adott az S halmaz: undefined. Határozza meg az A és B halmazokat,

undefined $B = \{y | y \in S \land (\frac{y^{2}}{2} - y) \in S\}$

majd a következő halmazokat: undefined

A = \{x | x \in S \land \frac{2 x}{12 - x} \in S\}

x = 0 \to \frac{2 \cdot 0}{12 - 0} = \frac{0}{12} = 0 \in S

x = 1 \to \frac{2 \cdot 1}{12 - 1} = \frac{2}{11} \notin S

x = 2 \to \frac{2 \cdot 2}{12 - 2} = \frac{4}{10} = \frac{2}{5} \notin S

x = 3 \to \frac{2 \cdot 3}{12 - 3} = \frac{6}{9} = \frac{2}{3} \notin S

x = 4 \to \frac{2 \cdot 4}{12 - 4} = \frac{8}{8} = 1 \in S

x = 5 \to \frac{2 \cdot 5}{12 - 5} = \frac{10}{7} \notin S

x = 6 \to \frac{2 \cdot 6}{12 - 6} = \frac{12}{6} = 2 \in S

x = 7 \to \frac{2 \cdot 7}{12 - 7} = \frac{14}{5} \notin S

x = 8 \to \frac{2 \cdot 8}{12 - 8} = \frac{16}{4} = 4 \in S

x = 9 \to \frac{2 \cdot 9}{12 - 9} = \frac{18}{3} = 6 \in S

A = \{0, 4, 6, 8, 9\}

B = \{y | y \in S \land (\frac{y^{2}}{2} - y) \in S\}

y = 0 \to \frac{0^{2}}{2} - 0 = 0 - 0 = 0 \in S

y = 1 \to \frac{1^{2}}{2} - 1 = \frac{1}{2} - 1 = \frac{1 - 2}{2} = - \frac{1}{2} \notin S

y = 2 \to \frac{2^{2}}{2} - 2 = \frac{4}{2} - 2 = 2 - 2 = 0 \in S

y = 3 \to \frac{3^{2}}{2} - 3 = \frac{9}{2} - 3 = \frac{9 - 6}{2} = \frac{3}{2} \notin S

y = 4 \to \frac{4^{2}}{2} - 4 = \frac{16}{2} - 4 = 8 - 2 = 6 \in S

y = 5 \to \frac{5^{2}}{2} - 5 = \frac{25}{2} - 5 = \frac{25 - 10}{2} = \frac{15}{2} \notin S

y = 6 \to \frac{6^{2}}{2} - 6 = \frac{36}{2} - 6 = 18 - 6 = 12 \notin S

y = 7 \to \frac{7^{2}}{2} - 7 = \frac{49}{2} - 7 = \frac{49 - 14}{2} = \frac{35}{2} \notin S

y = 8 \to \frac{8^{2}}{2} - 8 = \frac{64}{2} - 8 = 32 - 8 = 24 \notin S

y = 9 \to \frac{9^{2}}{2} - 9 = \frac{81}{2} - 9 = \frac{81 - 18}{2} = \frac{63}{2} \notin S

B = \{0, 2, 4\}

A \cup B = \{0, 4, 6, 8, 9\} \cup \{0, 2, 4\}

A \cup B = \{0, 2, 4, 6, 8, 9\}

A \cap B = \{0, 4, 6, 8, 9\} \cap \{0, 2, 4\}

A \cap B = \{0, 4\}

A ∖ B = \{0, 4, 6, 8, 9\} ∖ \{0, 2, 4\}

A ∖ B = \{6, 8, 9\}

B ∖ A = \{0, 2, 4\} ∖ \{0, 4, 6, 8, 9\}

B ∖ A = \{2\}

P (A ∖ B) = \{\emptyset, \{6\}, \{8\}, \{9\}, \{6, 8\}, \{6, 9\}, \{8, 9\}, \{6, 8, 9\}\}

A \cup B = \{0, 2, 4, 6, 8, 9\}

A \cup B = \{0, 2, 4, 6, 8, 9\}

A \cup B = \{0, 2, 4, 6, 8, 9\} A \cap B = \{0, 4\} A ∖ B = \{6, 8, 9\} B ∖ A = \{0, 2, 4\} ∖ \{0, 4, 6, 8, 9\} P (A ∖ B) = \{\emptyset, \{6\}, \{8\}, \{9\}, \{6, 8\}, \{6, 9\}, \{8, 9\}, \{6, 8, 9\}\}

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