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, 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, 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, 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.

Answer 1

The correct response to explain why \( \frac{ad + bc}{bd} \) is a rational number is:

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

Here's the reasoning:

1. Closure Property: This property states that the set of integers is closed under addition and multiplication. This means that when you add or multiply integers, the result is also an integer.

2. Since \( a, b, c, \) and \( d \) are integers, we can conclude that:

   • The sum \( ad + bc \) is an integer because it involves integer multiplication and addition.
   • The product \( bd \) is also an integer since it involves integer multiplication.

3. A rational number is defined as a quotient of two integers (where the denominator is not zero). In this case, \( \frac{ad + bc}{bd} \) is a quotient of the integer \( ad + bc \) and the integer \( bd \) (which is non-zero because \( b \) and \( d \) are non-zero).

Therefore, \( \frac{ad + bc}{bd} \) qualifies as a rational number.

Answer 2

so what is the answer