https://file.aiquickdraw.com/mathbot/file/1725927556822neyei1pp.jpg

The image presents a proof by contradiction that explores the relationship between the sum of a rational number \( y \) and an irrational number \( x \). Let's analyze the reasoning provided.

### Problem Analysis

1. **Given Information:**
   - \( y = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
   - \( x \) is an irrational number.

2. **Goal:** 
   - To show that the sum \( x + y \) (where \( y = \frac{a}{b} \)) is irrational, by assuming the contrary (that the sum is rational) and deriving a contradiction.

### Proof Outline

1. **Assumption (Contradiction Hypothesis):**
   - Assume that \( x + y = \frac{m}{n} \), where \( m \) and \( n \) are integers and \( n \neq 0 \). This means that \( x + y \) is rational.

2. **Steps to Derive Contradiction:**
   - Substitute \( y = \frac{a}{b} \) into the equation: 
     \[
     x + \frac{a}{b} = \frac{m}{n}
     \]
   - Isolate \( x \):
     \[
     x = \frac{m}{n} - \frac{a}{b}
     \]
   - Combine the fractions:
     \[
     x = \frac{mb - an}{nb}
     \]
   - Notice that both \( mb - an \) and \( nb \) are integers (since they are products and differences of integers). Thus, \( x \) is expressed as the ratio of two integers, implying that \( x \) is rational.

3. **Contradiction:**
   - This contradicts the original assumption that \( x \) is irrational. Hence, the assumption that \( x + y \) is rational must be false.

### Conclusion:
- The sum of a rational number \( y \) and an irrational number \( x \) is irrational.

### Final Question from the Image:
- The question asks: "Based on what we established about the classification of \( x \) and using the closure of integers, what does the equation tell you about the type of number \( x \) must be for the sum to be rational? What conclusion can you now make about the result of adding a rational and an irrational number?"

**Answer:**
For the sum to be rational, both components must be rational. Since we know \( x \) is irrational, the sum \( x + y \) cannot be rational. Therefore, adding a rational number to an irrational number always results in an irrational number.

---

Would you like further details on this topic or have any specific questions? Here are some related questions you might find interesting:

1. What is the definition of a rational number?
2. Can the product of a rational and an irrational number be rational?
3. How can we prove that the square root of 2 is irrational?
4. What are some common misconceptions about rational and irrational numbers?
5. How does the concept of closure under addition apply to rational and irrational numbers?

**Tip:** Remember that a proof by contradiction is a powerful technique to demonstrate that something cannot be true by assuming the opposite and finding a logical inconsistency.

What can we conclude about the result of adding a rational number to an irrational number based on the proof by contradiction?

Explore a proof by contradiction that demonstrates why the sum of a rational number and an irrational number is always irrational. This problem covers concepts in rational numbers, irrational numbers, and algebraic manipulation, suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation