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.
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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 \( n \neq 0 \).
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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.
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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.
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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.
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