Question

How does the closure property prove that the sum of a rational and irrational number is irrational

Answer 1

To understand how the closure property can help demonstrate that the sum of a rational number and an irrational number is irrational, we need to first clarify the definitions of rational and irrational numbers as well as the closure property.

### Definitions:
- **Rational Number**: A number that can be expressed as the quotient of two integers \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
- **Irrational Number**: A number that cannot be expressed as a quotient of two integers. Examples include numbers like \(\sqrt{2}\) and \(\pi\).

### Closure Property:
The closure property of a set states that performing a specified operation (like addition or multiplication) on members of that set will yield a result that is also a member of that set. The set of rational numbers is closed under addition, meaning that the sum of two rational numbers is always a rational number.

### Proof that the Sum of a Rational and an Irrational Number is Irrational:

Let’s denote:
- \( r \) as a rational number,
- \( i \) as an irrational number.

We want to prove that the sum \( s = r + i \) is irrational.

1. **Assuming for contradiction**: Let's suppose \( s \) (the sum \( r + i \)) is rational. If \( s \) is rational, by definition of rational numbers, it can be written as \( s = \frac{m}{n} \) where \( m \) and \( n \) are integers, and \( n \neq 0 \).

2. **Rearranging the equation**: From our assumption \( s = r + i \), we can rearrange this to find \( i \):
\[
i = s - r
\]
Given that both \( s \) and \( r \) are assumed to be rational (since we assumed \( s \) is rational and \( r \) is defined as rational), the difference of two rational numbers \( s \) and \( r \) is also rational. This means \( i \) would also be rational.

3. **Contradiction**: However, this is a contradiction because we initially defined \( i \) as an irrational number. So our assumption that \( s = r + i \) is rational must be false.

4. **Conclusion**: Therefore, the sum \( s = r + i \) must be irrational.

### Summary
The sum of a rational number and an irrational number is irrational because assuming the sum could be rational leads to a contradiction. The closure property of the rational numbers (which states that the addition of two rationals results in a rational) plays a crucial role in this proof, showing that if the sum were rational, then both summands would also need to be rational, contradicting the definition of the irrational number.