Problem 6

Question

Prove that $$ A \cap B=A \Longleftrightarrow A \subseteq B . $$

Step-by-Step Solution

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Answer

$ A \cap B = A $ if and only if $ A \subseteq B $.

1Step 1: Understand the Given Statement

The statement to prove is a biconditional statement: $ A \cap B = A \Longleftrightarrow A \subseteq B $. This means we need to show both directions: $ A \cap B = A \implies A \subseteq B $ and $ A \subseteq B \implies A \cap B = A $.

2Step 2: Prove $ A \cap B = A \implies A \subseteq B $

Assume $ A \cap B = A $. By definition of intersection, $ A \cap B $ consists of all elements $ x $ that are in both $ A $ and $ B $. Therefore, every element of $ A $ is also in $ B $, which means $ A \subseteq B $.

3Step 3: Prove $ A \subseteq B \implies A \cap B = A $

Assume $ A \subseteq B $. This means every element of $ A $ is in $ B $ too. The intersection $ A \cap B $ consists of all elements that are in both $ A $ and $ B $, but since every element of $ A $ is in $ B $, it follows that $ A \cap B = A $.

4Step 4: Conclusion

Both directions of the biconditional statement are proved: if $ A \cap B = A $, then $ A \subseteq B $, and if $ A \subseteq B $, then $ A \cap B = A $. Thus, $ A \cap B = A \Longleftrightarrow A \subseteq B $ is true.

Key Concepts

Set IntersectionSubsetBiconditional Statement

Set Intersection

In set theory, the concept of intersection is fundamental. The intersection of two sets, denoted as $ A \cap B $, is a set containing all elements that are common to both $ A $ and $ B $. If an element is in both sets $ A $ and $ B $, it will also be in $ A \cap B $. This is what intersection means in mathematical terms. Basic features include:

Only elements present in both sets are included in the intersection.
If $ A \cap B $ results in an empty set, it means there are no shared elements.

For example, if $ A = \{1, 2, 3\} $ and $ B = \{2, 3, 4\} $, then $ A \cap B = \{2, 3\} $ because 2 and 3 are present in both sets. Understanding the intersection helps in analyzing how sets relate to each other, especially when proving statements involving subsets and other set operations.
It is a crucial part of the exercise where we need to explore how the intersection $ A \cap B = A $ interacts with the idea of subsets.

Subset

A subset is another central concept in set theory. We say $ A $ is a subset of $ B $, written as $ A \subseteq B $, if every element of $ A $ is also an element of $ B $. Subsets are about containment; if $ A \subseteq B $, then $ A $ is contained completely within $ B $.

If $ A \subseteq B $, then it's possible that $ A = B $, making them equal sets.
The empty set is a subset of every set.

For instance, if $ A = \{1, 2\} $ and $ B = \{1, 2, 3, 4\} $, then $ A \subseteq B $ because all elements of $ A $ (1 and 2) are found in $ B $. Subsets are crucial as they provide a framework for understanding and proving more complex relationships, like the one in this exercise, where $ A \cap B = A $ implies $ A \subseteq B $. In this case, the intersection being equal to $ A $ suggests $ A $ must lie entirely within $ B $.
This links the idea of intersection to subset definitions, forming the basis of the proof.

Biconditional Statement

A biconditional statement is a logical sentence that expresses "if and only if" conditions. Denoted as $ p \Longleftrightarrow q $, it means $ p $ if and only if $ q $. To prove a biconditional statement true, you must demonstrate both $ p \Rightarrow q $ (if $ p $, then $ q $) and $ q \Rightarrow p $ (if $ q $, then $ p $).

It creates a strong equivalence between two statements, implying that they always occur together.
If one part is false, the other must necessarily be false for the biconditional to hold.

In our exercise, we prove $ A \cap B = A \Longleftrightarrow A \subseteq B $. This requires showing:1. $ A \cap B = A $ leads to $ A \subseteq B $. If their intersection is all of $ A $, then $ A $ is necessarily within $ B $.2. $ A \subseteq B $ implies $ A \cap B = A $. Here, since all elements of $ A $ are in $ B $, their intersection with $ B $ is all of $ A $.Successfully arguing both directions provides a comprehensive understanding of the bidirectional relationship inherent in biconditional statements.

Problem 6

Problem 7

Other exercises in this chapter

Problem 6

Let $\boldsymbol{G}$ be a group satisfying $(\forall a \in G)$ $$ a^{2}=e $$ Prove that $\boldsymbol{G}$ is an abelian group.

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Problem 6

Let $f: \mathrm{R} \rightarrow \mathbb{R}$, be a function such that $\forall x \in] 0 ;+\infty[,$, $$ \left[f\left(x^{3}+1\right)\right]^{\sqrt{x}}=5 $$ fin

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Problem 7

(Putnam 1948) If $n$ is a positive integer, demonstrate that $$ \lfloor\sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4 n+2} \rrbracket $$

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Problem 7

Prove that for integer $\boldsymbol{n}>2$, $$ n^{n / 2}

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