/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Beweisen Sie das Distributivgese... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 20

Beweisen Sie das Distributivgesetz: $$ M_{1} \cup\left(M_{2} \cap M_{3}\right)=\left(M_{1} \cup M_{2}\right) \cap\left(M_{1} \cup M_{3}\right) $$

Short Answer

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Question: Prove the Distributive Law of sets using set theory. Answer: The Distributive Law of sets states that for any sets M1, M2, and M3: \(M_{1} \cup\left(M_{2} \cap M_{3}\right)=\left(M_{1} \cup M_{2}\right) \cap\left(M_{1} \cup M_{3}\right)\). To prove this, we show that both sides of the equation are subsets of each other, i.e., \(M_{1} \cup (M_{2} \cap M_{3}) \subseteq (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\) and \((M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3}) \subseteq M_{1} \cup (M_{2} \cap M_{3})\). By proving these subset relations, we confirm the Distributive Law for sets.

Step by step solution

01

Prove the first subset relation

We want to prove that: $$ M_{1} \cup (M_{2} \cap M_{3}) \subseteq (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3}) $$ Let's take an arbitrary element \(x\) from \(M_{1} \cup (M_{2} \cap M_{3})\). It means either \(x \in M_{1}\) or \(x \in M_{2} \cap M_{3}\). Case 1: If \(x \in M_{1}\), then \(x \in (M_{1} \cup M_{2})\) and \(x \in (M_{1} \cup M_{3})\). Hence, \(x \in (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\). Case 2: If \(x \in M_{2} \cap M_{3}\), then \(x \in M_{2}\) and \(x \in M_{3}\). So, \(x \in (M_{1} \cup M_{2})\) and \(x \in (M_{1} \cup M_{3})\). Hence, \(x \in (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\). In both cases, we have shown that for an arbitrary element \(x \in M_{1} \cup (M_{2} \cap M_{3})\), \(x \in (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\). Therefore, we can conclude that \(M_{1} \cup (M_{2} \cap M_{3}) \subseteq (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\).
02

Prove the second subset relation

We want to prove that: $$ (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3}) \subseteq M_{1} \cup (M_{2} \cap M_{3}) $$ Let's take an arbitrary element \(x\) from \((M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\). This means that \(x \in (M_{1} \cup M_{2})\) and \(x \in (M_{1} \cup M_{3})\). Now, we can proceed with two sub-cases: Sub-case 1: If \(x \in M_{1}\), then \(x \in M_{1} \cup (M_{2} \cap M_{3})\). Sub-case 2: If \(x \notin M_{1}\), then \(x \in M_{2}\) (due to \(x \in (M_{1} \cup M_{2})\)) and \(x \in M_{3}\) (due to \(x \in (M_{1} \cup M_{3})\)). Thus, \(x \in M_{2} \cap M_{3}\), which implies \(x \in M_{1} \cup (M_{2} \cap M_{3})\). In both sub-cases, we have shown that for an arbitrary element \(x \in (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\), \(x \in M_{1} \cup (M_{2} \cap M_{3})\). Therefore, we can conclude that \((M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3}) \subseteq M_{1} \cup (M_{2} \cap M_{3})\).
03

Conclusion

We have proven that: 1. \(M_{1} \cup (M_{2} \cap M_{3}) \subseteq (M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3})\) 2. \((M_{1} \cup M_{2}) \cap (M_{1} \cup M_{3}) \subseteq M_{1} \cup (M_{2} \cap M_{3})\) By combining these two results, we can conclude that: $$ M_{1} \cup\left(M_{2} \cap M_{3}\right)=\left(M_{1} \cup M_{2}\right) \cap\left(M_{1} \cup M_{3}\right) $$ This confirms the Distributive Law for sets.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Subset Relation
Understanding the subset relation is crucial when dealing with sets in mathematics. A subset is a set where every element is also a part of another set. Think of it like a toy box (subset) within a playroom (bigger set); every toy in the box is also in the room, but not necessarily the other way around. For example, if we have sets A and B, we would say A is a subset of B if every element in A can also be found in B, symbolically expressed as A ⊆ B.

In the context of proving the Distributive Law for sets, the subset relationship helps us to understand how elements are shared or not shared between sets. We start by considering any element we might encounter in the union of sets and examining which other sets this element must belong to. This logical exploration, driven by the subset relations, allows us to establish the truth of the Distributive Law step by step. By showing that both sides of the equation hold the same elements, we affirm their equivalence, much like confirming that two lists of ingredients will result in the same recipe.
Arbitrary Element
An 'arbitrary element' in set theory is like a mystery box鈥攊t could be anything. When we select an arbitrary element from a set, we're saying, 'Let's take any one thing from this set without any specific preference.' It's our wild card, used to prove statements that must hold true for all potential cases. For example, picking an arbitrary element x from set M implies that 'x' can represent any one of the elements in M, and whatever we prove about 'x' will apply to all elements in M.

This approach is like a universal fitting room in a clothing store鈥攊f a shirt fits well on any random customer (our arbitrary element), it's guaranteed to fit all customers. Similarly, by taking an arbitrary element from our set in the Distributive Law and showing it fits in the corresponding unions and intersections, we demonstrate a rule that applies across the entire set, therefore bolstering our argument for the Distributive Law's validity.
Set Theory
Set theory is the bedrock of modern mathematics, much like how atoms are the building blocks of matter. It's about understanding collections of objects鈥攌nown as 'sets'鈥攁nd how they interact with one another through operations like unions, intersections, and differences. Picture set theory as the grammar of mathematics; it provides the rules and structure for how mathematical objects work together.

When we talk about operations like union (combining elements of sets) and intersection (finding common elements), set theory gives us the language and rules to describe these interactions. In the exercise where we prove the Distributive Law of sets, we're using set theory's principles to show how different groups of elements (sets) come together in specific ways, much like the pieces of a puzzle fitting together to create a complete picture. It's a foundational concept that not only applies to mathematics but to understanding any system that deals with grouped elements or categories.

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Most popular questions from this chapter

An einer Weggabelung in der W眉ste leben zwei Br眉der, die vollkommen gleich aussehen, zwischen denen es aber einen gewaltigen Unterschied gibt: Der eine sagt immer die Wahrheit, der andere l眉gt immer. Schon halb verdurstet kommt man zu dieser Weggabelung und wei脽 genau: Einer der beiden Wege f眉hrt zu einer Oase, der andere hingegen immer tiefer in die W眉ste hinein. Man darf aber nur einem der Br眉der (man weiB nicht, welcher es ist) genau eine Frage stellen. Was muss man fragen, um sicher den Weg zur Oase zu finden?

Wir sind im Text nicht explizit auf den Unterschied zwischen Aussagen und Aussageformen eingegangen. W盲hrend wir Aussagen als feststellende S盲tze definiert haben, die einen eindeutigen Wahrheitswert \(w\) oder \(f\) haben, sind Aussageformen S盲tze, deren Wahrheitswert sich vorerst nicht bestimmen l盲sst, weil sie noch eine oder mehrere freie Variable beinhalten. Beispiele f眉r Aussageformen w盲ren ,,Die Zahl \(x\) ist ungerade" oder "Monarch \(x\) regierte l盲nger als 20 Jahre", wobei \(x\) jeweils die freie Variable bezeichnet. Ersetzt man in einer Aussageform die freien Variablen durch passende Objekte oder bindet die Variablen durch Quantoren, erh盲lt man Aussagen. 脺berpr眉fen Sie, ob es sich bei den folgenden S盲tzen um Aussagen, Aussageformen oder keines der beiden handelt: (a),\(x\) ist ungerade" mit \(x=2\) (b),\(x\) ist ungerade" mit \(x=3\) (c) \(\forall x \in \mathbb{R}: 1 /\left(1+x^{2} y^{2}\right) \leq 1\) (d) \(\forall(x, y) \in \mathbb{R}^{2}: 1 /\left(1+x^{\overline{2}} y^{2}\right) \leq 1\)

Welche der folgenden Aussagen sind richtig? F眉r alle \(x \in \mathbb{R}\) gilt: 1.,\(x>1\) ist hinreichend f眉r \(x^{2}>1\)." 2\. \(x>1\) ist notwendig f眉r \(x^{2}>1 .^{4}\) 3.,\(x \geq 1\) ist hinreichend f眉r \(x^{2}>1 .\) 4.,\(x \geq 1\) ist notwendig f眉r \(x^{2}>1\)."

tafel ge werden: In der koniunktiven Normalform als Koniunktion von \(\mathrm{~ W i d e n t}\) Disjunktionen der beteiligten Variablen bzw. ihrer Negationen, und in der disjunktiven Normalform als Disjunktion von entsprechenden Konjunktionen. Dies ist in der Digitalelektronik sehr praktisch, weil es eine automatisierbare M枚glichkeit darstellt, zu jeder Wahrheitstafel einen 盲quivalenten logischen Ausdruck und damit eine Schaltung zu konstruieren. Wir betrachten nun die beiden Wahrheitstafeln \begin{tabular}{|ll|l|} \hline\(A\) & \(B\) & \(G\) \\ \hline\(w\) & \(w\) & \(w\) \\\ \(w\) & \(f\) & \(f\) \\ \(f\) & \(w\) & \(f\) \\ \(f\) & \(f\) & \(w\) \\ \hline \end{tabular} \begin{tabular}{|lll|l|} \hline\(A\) & \(B\) & \(C\) & \(H\) \\ \hline\(w\) & \(w\) & \(w\) & \(w\) \\ \(w\) & \(w\) & \(f\) & \(f\) \\ \(w\) & \(f\) & \(w\) & \(f\) \\ \(w\) & \(f\) & \(f\) & \(f\) \\ \(f\) & \(w\) & \(w\) & \(w\) \\ \(f\) & \(w\) & \(f\) & \(f\) \\ \(f\) & \(f\) & \(w\) & \(w\) \\ \(f\) & \(f\) & \(f\) & \(w\) \\ \hline \end{tabular} F眉r die Aussage \(G\) lautet die disjunktive Normalform $$ G \Leftrightarrow((A \wedge B) \vee((\neg A) \wedge(\neg B))) $$ die konjunktive $$ G \Leftrightarrow(((\neg A) \vee B) \wedge(A \vee(\neg B))) $$ Bestimmen Sie nun diese beiden Normalformen f眉r die Aussage \(H\). Gibt es ein Kriterium, f眉r welche Art von Wahrheitstafel welche Normalform vorzuziehen ist, wenn man einen m枚glichst einfachen Ausdruck erhalten will? Lassen sich die so erhaltenen Ausdr眉cke noch weiter vereinfachen?

Wir betrachten die Teilmengen \(X, Y\) und \(Z\) von \(\mathbb{R}\). Verneinen Sie die Aussage $$ \forall x \in X \exists y \in Y \forall z \in Z: x \cdot y
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