Part 2: Proof that (A n C) – BC (A – B)n (C – B) Consider the sentences in the following scrambled list. By definition of intersection, x E (A – B) n (C – B). By definition of set difference x E An C and x ¢ B. By definition of intersection, x E A and x E C. By definition of set difference, x E A and x C. So by definition of set difference, x E A – B and x E C – B. By definition of intersection xE An C and x ¢ B. Hence both xE A and x € B and also x E C, and x & B. We prove part 2 by selecting appropriate sentences from the list and putting them in the correct order. 1. Suppose x E (A n C) – B. 2. By definition of intersection, x e (A – B) n (C – B). - 3. So by definition of set difference, x e A - B and x e C - B. 4. ---Select--- 5. ---Select--- 6. Hence both x e A and x ¢ B and also x e C, and x ¢ B. 7. Hence, (A n C) – BC (A – B) n (C – B) C by definition of subset.
Part 2: Proof that (A n C) – BC (A – B)n (C – B) Consider the sentences in the following scrambled list. By definition of intersection, x E (A – B) n (C – B). By definition of set difference x E An C and x ¢ B. By definition of intersection, x E A and x E C. By definition of set difference, x E A and x C. So by definition of set difference, x E A – B and x E C – B. By definition of intersection xE An C and x ¢ B. Hence both xE A and x € B and also x E C, and x & B. We prove part 2 by selecting appropriate sentences from the list and putting them in the correct order. 1. Suppose x E (A n C) – B. 2. By definition of intersection, x e (A – B) n (C – B). - 3. So by definition of set difference, x e A - B and x e C - B. 4. ---Select--- 5. ---Select--- 6. Hence both x e A and x ¢ B and also x e C, and x ¢ B. 7. Hence, (A n C) – BC (A – B) n (C – B) C by definition of subset.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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![### Part 2: Proof that \((A \cap C) − B \subseteq (A − B) \cap (C − B)\)
#### Consider the sentences in the following scrambled list:
1. By definition of intersection, \( x \in (A − B) \cap (C − B) \).
2. By definition of set difference, \( x \in A \cap C \) and \( x \notin B \).
3. By definition of intersection, \( x \in A \) and \( x \in C \).
4. By definition of set difference, \( x \in A \) and \( x \in C \).
5. So by definition of set difference, \( x \in A − B \) and \( x \in C − B \).
6. By definition of intersection \( x \in A \cap C \) and \( x \notin B \).
7. Hence, both \( x \in A \) and \( x \notin B \) and also \( x \in C \), and \( x \notin B \).
#### We prove part 2 by selecting appropriate sentences from the list and putting them in the correct order:
1. Suppose \( x \in (A \cap C) − B \).
2. By definition of intersection, \( x \in (A − B) \cap (C − B) \). [Incorrect]
3. So by definition of set difference, \( x \in A − B \) and \( x \in C − B \). [Incorrect]
4. [--- Select ---]
5. [--- Select ---]
6. Hence, both \( x \in A \) and \( x \notin B \) and also \( x \in C \), and \( x \notin B \).
7. Hence, \((A \cap C) − B \subseteq (A − B) \cap (C − B)\) by definition of subset.
#### Conclusion:
Since both subset relations have been proved, it follows by the definition of set equality that \( (A − B) \cap (C − B) = (A \cap C) − B \).
---
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### Explanation of Incorrect Steps:
1. Step 2: By definition of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F507ffaff-36d6-4c89-a9b8-d054ae79319d%2Fb3275099-500e-439c-be3b-f4798a50d93c%2Ffu8y6b5.png&w=3840&q=75)
Transcribed Image Text:### Part 2: Proof that \((A \cap C) − B \subseteq (A − B) \cap (C − B)\)
#### Consider the sentences in the following scrambled list:
1. By definition of intersection, \( x \in (A − B) \cap (C − B) \).
2. By definition of set difference, \( x \in A \cap C \) and \( x \notin B \).
3. By definition of intersection, \( x \in A \) and \( x \in C \).
4. By definition of set difference, \( x \in A \) and \( x \in C \).
5. So by definition of set difference, \( x \in A − B \) and \( x \in C − B \).
6. By definition of intersection \( x \in A \cap C \) and \( x \notin B \).
7. Hence, both \( x \in A \) and \( x \notin B \) and also \( x \in C \), and \( x \notin B \).
#### We prove part 2 by selecting appropriate sentences from the list and putting them in the correct order:
1. Suppose \( x \in (A \cap C) − B \).
2. By definition of intersection, \( x \in (A − B) \cap (C − B) \). [Incorrect]
3. So by definition of set difference, \( x \in A − B \) and \( x \in C − B \). [Incorrect]
4. [--- Select ---]
5. [--- Select ---]
6. Hence, both \( x \in A \) and \( x \notin B \) and also \( x \in C \), and \( x \notin B \).
7. Hence, \((A \cap C) − B \subseteq (A − B) \cap (C − B)\) by definition of subset.
#### Conclusion:
Since both subset relations have been proved, it follows by the definition of set equality that \( (A − B) \cap (C − B) = (A \cap C) − B \).
---
### Additional Help
- **Need Help?**
- [Read It]
- [Talk to a Tutor]
### Explanation of Incorrect Steps:
1. Step 2: By definition of
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