Question
How does the Closure Property prove that the sum of a rational and irrational number is irrational? (1 point)
• Irrational numbers can be added or subtracted.
• The sum or difference of a rational and irrational number is a rational number.
• The sum of an irrational and rational number can't be predicted.
• The sum or difference of two rational numbers is a rational number
• Irrational numbers can be added or subtracted.
• The sum or difference of a rational and irrational number is a rational number.
• The sum of an irrational and rational number can't be predicted.
• The sum or difference of two rational numbers is a rational number
Answers
GPT-4o mini
To prove that the sum of a rational and an irrational number is irrational using the closure property, the correct statement would be:
• The sum or difference of a rational and irrational number is a rational number.
This statement suggests that if you could somehow assume that the sum of a rational number \( r \) and an irrational number \( i \) is rational, then you would arrive at a contradiction. The closure property states that certain sets of numbers are closed under specific operations (like addition or multiplication).
Here’s the reasoning:
1. Let \( r \) be a rational number.
2. Let \( i \) be an irrational number.
3. Assume that \( r + i \) is rational.
4. If \( r + i = q \) (where \( q \) is rational), then you can rearrange this to \( i = q - r \).
5. Since both \( q \) and \( r \) are rational, the difference \( q - r \) must also be rational (because the set of rational numbers is closed under subtraction).
6. This means that \( i \), which we assumed to be irrational, must also be rational, leading to a contradiction.
Thus, the assumption that the sum of a rational number and an irrational number is rational must be false, which proves that the sum of a rational and an irrational number is indeed irrational.
• The sum or difference of a rational and irrational number is a rational number.
This statement suggests that if you could somehow assume that the sum of a rational number \( r \) and an irrational number \( i \) is rational, then you would arrive at a contradiction. The closure property states that certain sets of numbers are closed under specific operations (like addition or multiplication).
Here’s the reasoning:
1. Let \( r \) be a rational number.
2. Let \( i \) be an irrational number.
3. Assume that \( r + i \) is rational.
4. If \( r + i = q \) (where \( q \) is rational), then you can rearrange this to \( i = q - r \).
5. Since both \( q \) and \( r \) are rational, the difference \( q - r \) must also be rational (because the set of rational numbers is closed under subtraction).
6. This means that \( i \), which we assumed to be irrational, must also be rational, leading to a contradiction.
Thus, the assumption that the sum of a rational number and an irrational number is rational must be false, which proves that the sum of a rational and an irrational number is indeed irrational.