In the proof that ab+cd

Question

, 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, 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.

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.

Answer 1

To explain why \( \frac{ad + bc}{bd} \) is a rational number, we can focus on the integer properties of the components involved.

Correct Response: "By the Closure Property, \( ad + bc \) and \( bd \) are both integers, and so \( \frac{ad + bc}{bd} \) is a quotient of two integers."

Explanation:

• \( a \), \( b \), \( c \), and \( d \) are integers by the problem statement.
• The multiplication of integers \( a \) and \( d \) gives \( ad \), which is an integer.
• Similarly, the multiplication of integers \( b \) and \( c \) gives \( bc \), which is also an integer.
• Adding these two integer results, \( ad + bc \), yields another integer.
• \( b \) and \( d \) are non-zero integers, so their product \( bd \) is also a non-zero integer.
• Therefore, \( \frac{ad + bc}{bd} \) is a quotient of two integers (with the denominator being non-zero), which means it is a rational number.

This demonstrates that the expression is indeed rational based on the properties of integers and the definition of rational numbers.