To explain why \(\frac{ad + bc}{bd}\) is a rational number when \(a\), \(b\), \(c\), and \(d\) are integers with \(b\) and \(d\) being non-zero:
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Definition of Rational Numbers: A number is considered rational if it can be expressed as a quotient of two integers, where the denominator is not zero.
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Numerator \(ad + bc\): Here \(ad + bc\) is the result of the addition of two products, \(ad\) and \(bc\). Since \(a\), \(b\), \(c\), and \(d\) are integers, both \(ad\) and \(bc\) are also integers (the product of integers is an integer). Thus, when we add these two integers, \(ad + bc\) is itself an integer.
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Denominator \(bd\): Since both \(b\) and \(d\) are given to be non-zero integers, their product \(bd\) is also a non-zero integer. This is important because a rational number cannot have a zero denominator.
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Combining the Numerator and Denominator: Now, we have \(\frac{ad + bc}{bd}\). The numerator \(ad + bc\) is an integer, and the denominator \(bd\) is a non-zero integer. Therefore, we are looking at a quotient of two integers.
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Conclusion: Since the quotient of an integer (the numerator) by a non-zero integer (the denominator) satisfies the definition of a rational number, we conclude that \(\frac{ad + bc}{bd}\) is indeed a rational number.
Your original statement seemed to mix concepts, particularly the mention of imaginary numbers. The reasoning based on closure properties of integers regarding addition and multiplication leading to integers is the correct approach here. Thus, the final understanding is that \(\frac{ad + bc}{bd}\) is a rational number because it is a quotient of two integers, where the denominator is non-zero.