5. Let x and y be irrational numbers such that x-y is also irrational. Define sets A and B by A = {x +r:reQ} and B = {y+r:reQ). Prove that the sets A and B have no elements in common. be the set of all numbers of the form a t b/2 where a and b are arbitrary

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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**5. Let \( x \) and \( y \) be irrational numbers such that \( x - y \) is also irrational. Define sets \( A \) and \( B \) by \( A = \{ x + r : r \in \mathbb{Q} \} \) and \( B = \{ y + r : r \in \mathbb{Q} \} \). Prove that the sets \( A \) and \( B \) have no elements in common.**

**Explanation:**

The problem involves showing that two sets constructed by adding rational numbers to given irrational numbers are disjoint, under the condition that the difference between the irrational numbers is also irrational. Here’s the detailed reasoning:

1. **Definitions of Sets:**
   - Set \( A = \{ x + r : r \in \mathbb{Q} \} \): This represents all numbers formed by adding any rational number \( r \) to the irrational number \( x \).
   - Set \( B = \{ y + r : r \in \mathbb{Q} \} \): This represents all numbers formed by adding any rational number \( r \) to the irrational number \( y \).

2. **Condition:**
   - \( x - y \) is irrational.

3. **Proof Strategy:**
   - Assume for contradiction that there exists some element common to both sets \( A \) and \( B \).
   - Then there exist rational numbers \( r_1 \) and \( r_2 \) such that \( x + r_1 = y + r_2 \).
   - Rearranging gives \( x - y = r_2 - r_1 \), which implies that \( x - y \) is rational, a contradiction.

Therefore, the sets \( A \) and \( B \) have no elements in common.
Transcribed Image Text:**5. Let \( x \) and \( y \) be irrational numbers such that \( x - y \) is also irrational. Define sets \( A \) and \( B \) by \( A = \{ x + r : r \in \mathbb{Q} \} \) and \( B = \{ y + r : r \in \mathbb{Q} \} \). Prove that the sets \( A \) and \( B \) have no elements in common.** **Explanation:** The problem involves showing that two sets constructed by adding rational numbers to given irrational numbers are disjoint, under the condition that the difference between the irrational numbers is also irrational. Here’s the detailed reasoning: 1. **Definitions of Sets:** - Set \( A = \{ x + r : r \in \mathbb{Q} \} \): This represents all numbers formed by adding any rational number \( r \) to the irrational number \( x \). - Set \( B = \{ y + r : r \in \mathbb{Q} \} \): This represents all numbers formed by adding any rational number \( r \) to the irrational number \( y \). 2. **Condition:** - \( x - y \) is irrational. 3. **Proof Strategy:** - Assume for contradiction that there exists some element common to both sets \( A \) and \( B \). - Then there exist rational numbers \( r_1 \) and \( r_2 \) such that \( x + r_1 = y + r_2 \). - Rearranging gives \( x - y = r_2 - r_1 \), which implies that \( x - y \) is rational, a contradiction. Therefore, the sets \( A \) and \( B \) have no elements in common.
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