Question
In the proof that ab+cd
, where a, b, c, and d are integers and b and d are non-zero, explain why ad+bcbd
is a rational number.(1 point)
Responses
By the Closure Property, ad+bcbd
reduces to a whole number, which is a subset of rational numbers.
By the Closure Property, the fraction with numerator A d plus b c and denominator b d reduces to a whole number, which is a subset of rational numbers.
By the Closure Property, ad+bc
and bd
are both integers, and so ad+bcbd
is a quotient of two integers.
By the Closure Property, A d plus b c and b d are both integers, and so the fraction with numerator A d plus b c and denominator b d is a quotient of two integers.
By the Closure Property, a quotient of imaginary numbers is a rational number.
By the Closure Property, a quotient of imaginary numbers is a rational number.
By the Closure Property, ad+bc
and bd
are both quotients of integers, and so ad+bcbd
is a quotient of two integers.
By the Closure Property, A d plus b c and b d are both quotients of integers, and so the fraction with numerator A d plus b c and denominator b d is a quotient of two integers.
, where a, b, c, and d are integers and b and d are non-zero, explain why ad+bcbd
is a rational number.(1 point)
Responses
By the Closure Property, ad+bcbd
reduces to a whole number, which is a subset of rational numbers.
By the Closure Property, the fraction with numerator A d plus b c and denominator b d reduces to a whole number, which is a subset of rational numbers.
By the Closure Property, ad+bc
and bd
are both integers, and so ad+bcbd
is a quotient of two integers.
By the Closure Property, A d plus b c and b d are both integers, and so the fraction with numerator A d plus b c and denominator b d is a quotient of two integers.
By the Closure Property, a quotient of imaginary numbers is a rational number.
By the Closure Property, a quotient of imaginary numbers is a rational number.
By the Closure Property, ad+bc
and bd
are both quotients of integers, and so ad+bcbd
is a quotient of two integers.
By the Closure Property, A d plus b c and b d are both quotients of integers, and so the fraction with numerator A d plus b c and denominator b d is a quotient of two integers.
Answers
GPT-4o mini
To determine why \( \frac{ad + bc}{bd} \) is a rational number, we can use the Closure Property of integers and rational numbers. The correct response among the given options would be:
**By the Closure Property, \( ad + bc \) and \( bd \) are both integers, and so \( \frac{ad + bc}{bd} \) is a quotient of two integers.**
**Explanation:**
1. **Integer Operations**: Since \( a, b, c, \) and \( d \) are integers and \( b \) and \( d \) are non-zero, the products \( ad \) and \( bc \) are also integers (as the product of integers is an integer).
2. **Addition of Integers**: The sum \( ad + bc \) is an integer (as the sum of integers is also an integer).
3. **Denominator**: The product \( bd \) is an integer, and because \( b \) and \( d \) are non-zero, \( bd \) is also non-zero.
4. **Quotient of Integers**: Therefore, \( \frac{ad + bc}{bd} \) is a quotient of two integers (with the denominator being non-zero), which means it is a rational number by definition.
Hence, the process demonstrates that \( \frac{ad + bc}{bd} \) is indeed a rational number.
**By the Closure Property, \( ad + bc \) and \( bd \) are both integers, and so \( \frac{ad + bc}{bd} \) is a quotient of two integers.**
**Explanation:**
1. **Integer Operations**: Since \( a, b, c, \) and \( d \) are integers and \( b \) and \( d \) are non-zero, the products \( ad \) and \( bc \) are also integers (as the product of integers is an integer).
2. **Addition of Integers**: The sum \( ad + bc \) is an integer (as the sum of integers is also an integer).
3. **Denominator**: The product \( bd \) is an integer, and because \( b \) and \( d \) are non-zero, \( bd \) is also non-zero.
4. **Quotient of Integers**: Therefore, \( \frac{ad + bc}{bd} \) is a quotient of two integers (with the denominator being non-zero), which means it is a rational number by definition.
Hence, the process demonstrates that \( \frac{ad + bc}{bd} \) is indeed a rational number.