Question
How does the Closure Property prove that the sum of a rational and irrational number is irrational?(1 point) Responses The sum or difference of a rational and irrational number is a rational number. The sum or difference of a rational and irrational number is a rational number. The sum or difference of two rational numbers is a rational number. The sum or difference of two rational numbers is a rational number. Irrational numbers can be added or subtracted. Irrational numbers can be added or subtracted. The sum of an irrational and rational number can’t be predicted.
Answers
Answered by
GPT-4o mini
To prove that the sum of a rational and an irrational number is irrational, we rely on the closure property of rational numbers.
1. **Closure Property of Rational Numbers**: This property states that the sum or difference (and also the product) of two rational numbers is always a rational number.
2. **Definitions**:
- A **rational number** is any number that can be expressed as the quotient of two integers (where the denominator is not zero).
- An **irrational number** is any real number that cannot be expressed as the quotient of two integers.
3. **Proof by Contradiction**:
- Let's assume we have a rational number \( r \) and an irrational number \( i \).
- If we add them, \( s = r + i \).
- Suppose, for contradiction, that \( s \) is rational.
- Since \( r \) is rational and, under the closure property, the sum or difference of two rational numbers is rational, if \( s \) is rational, we can rearrange to express \( i = s - r \).
- Here, \( s - r \) (which is a rational number minus a rational number) must also be rational.
4. **Contradiction**: This implies that we could express the irrational number \( i \) as a rational number, which contradicts the definition of an irrational number.
Thus, we conclude that the sum of a rational number and an irrational number must be irrational.
**Correct Response**: The sum of a rational and an irrational number cannot be predicted to be rational; it is irrational. The other responses are incorrect because they either contradict the established properties or state the opposite of what is true.
1. **Closure Property of Rational Numbers**: This property states that the sum or difference (and also the product) of two rational numbers is always a rational number.
2. **Definitions**:
- A **rational number** is any number that can be expressed as the quotient of two integers (where the denominator is not zero).
- An **irrational number** is any real number that cannot be expressed as the quotient of two integers.
3. **Proof by Contradiction**:
- Let's assume we have a rational number \( r \) and an irrational number \( i \).
- If we add them, \( s = r + i \).
- Suppose, for contradiction, that \( s \) is rational.
- Since \( r \) is rational and, under the closure property, the sum or difference of two rational numbers is rational, if \( s \) is rational, we can rearrange to express \( i = s - r \).
- Here, \( s - r \) (which is a rational number minus a rational number) must also be rational.
4. **Contradiction**: This implies that we could express the irrational number \( i \) as a rational number, which contradicts the definition of an irrational number.
Thus, we conclude that the sum of a rational number and an irrational number must be irrational.
**Correct Response**: The sum of a rational and an irrational number cannot be predicted to be rational; it is irrational. The other responses are incorrect because they either contradict the established properties or state the opposite of what is true.
There are no human answers yet. A form for humans to post answers is coming very soon!