Question
In the proof that a over b plus irvity where ABC and d are integers and b and d are non-zero explain why ad + BC over BD is a rational number a by the closure property of quotient of an imaginary number is a rational number b by the closure property ad plus BC over BD reduces to a whole number which is the subset of rational numbers
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GPT-4o mini
To explain why \(\frac{ad + BC}{BD}\) is a rational number, let's break this down step by step.
Given that \(a\), \(b\), \(c\), and \(d\) are integers with \(b\) and \(d\) being non-zero integers, we want to analyze the expression \(\frac{ad + BC}{BD}\).
1. **Understanding the Expression**:
The expression consists of two parts in the numerator:
- \(ad\), which is the product of two integers (hence an integer).
- \(BC\), product of two integers (hence also an integer).
Therefore, the numerator \(ad + BC\) is the sum of two integers, which results in another integer due to the closure property of addition of integers.
2. **Denominator**:
The denominator \(BD\) is the product of two non-zero integers \(b\) and \(d\). The product of non-zero integers is also a non-zero integer.
3. **Forming the Quotient**:
Now, we can observe that \(\frac{ad + BC}{BD}\) forms a fraction where both the numerator and denominator are integers, and the denominator is non-zero.
4. **Conclusion**:
By the definition of rational numbers, a number is considered rational if it can be expressed as \(\frac{m}{n}\) where \(m\) and \(n\) are integers and \(n\) is not zero. Therefore, since both \(ad + BC\) (integer) and \(BD\) (non-zero integer) are integers, \(\frac{ad + BC}{BD}\) is indeed a rational number.
To relate it back to the statements you've provided:
- **A**: "By the closure property of the quotient of an imaginary number is a rational number." This refers more generally to the property that dividing integers that are non-zero gives a rational number, if we were discussing real parts. Imaginary numbers don't directly apply here since \(a\) and \(b\) seem to be considered as integers representing whole parts.
- **B**: "By the closure property, \(ad + BC\) over \(BD\) reduces to a whole number which is a subset of rational numbers." Indeed, if \(ad + BC\) is itself an integer, and since rational numbers include integers, this means the expression remains a rational number.
In simpler terms:
- You're checking if the numerator produces an integer via the sums and products of integers.
- You're confirming the denominator is not zero to maintain rationality when forming the fraction.
This gives you a complete and verified outcome that \(\frac{ad + BC}{BD}\) is rational, verified through its integer construction and avoiding any division by zero.
Given that \(a\), \(b\), \(c\), and \(d\) are integers with \(b\) and \(d\) being non-zero integers, we want to analyze the expression \(\frac{ad + BC}{BD}\).
1. **Understanding the Expression**:
The expression consists of two parts in the numerator:
- \(ad\), which is the product of two integers (hence an integer).
- \(BC\), product of two integers (hence also an integer).
Therefore, the numerator \(ad + BC\) is the sum of two integers, which results in another integer due to the closure property of addition of integers.
2. **Denominator**:
The denominator \(BD\) is the product of two non-zero integers \(b\) and \(d\). The product of non-zero integers is also a non-zero integer.
3. **Forming the Quotient**:
Now, we can observe that \(\frac{ad + BC}{BD}\) forms a fraction where both the numerator and denominator are integers, and the denominator is non-zero.
4. **Conclusion**:
By the definition of rational numbers, a number is considered rational if it can be expressed as \(\frac{m}{n}\) where \(m\) and \(n\) are integers and \(n\) is not zero. Therefore, since both \(ad + BC\) (integer) and \(BD\) (non-zero integer) are integers, \(\frac{ad + BC}{BD}\) is indeed a rational number.
To relate it back to the statements you've provided:
- **A**: "By the closure property of the quotient of an imaginary number is a rational number." This refers more generally to the property that dividing integers that are non-zero gives a rational number, if we were discussing real parts. Imaginary numbers don't directly apply here since \(a\) and \(b\) seem to be considered as integers representing whole parts.
- **B**: "By the closure property, \(ad + BC\) over \(BD\) reduces to a whole number which is a subset of rational numbers." Indeed, if \(ad + BC\) is itself an integer, and since rational numbers include integers, this means the expression remains a rational number.
In simpler terms:
- You're checking if the numerator produces an integer via the sums and products of integers.
- You're confirming the denominator is not zero to maintain rationality when forming the fraction.
This gives you a complete and verified outcome that \(\frac{ad + BC}{BD}\) is rational, verified through its integer construction and avoiding any division by zero.
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