30. Higher Order Thinking Suppose \( \frac{a}{b^{\prime}} \frac{c}{d^{\prime}} \) and \( \frac{f}{f} \) represent three rational numbers. If \( \frac{d}{b} \) is less than \( \frac{c}{d^{\prime}} \) and \( \frac{c}{d} \) is less than \( \frac{e}{f^{\prime}} \) compare \( \frac{a}{b} \) and \( \frac{e}{f} \) Explain.
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The Deep Dive
To tackle this comparison, we start by noting the relationships given in the problem. Since \( \frac{d}{b} < \frac{c}{d^{\prime}} \) implies that \( d \) is less than some fraction of \( c \), and \( \frac{c}{d} < \frac{e}{f^{\prime}} \) indicates that \( c \) is less than some fraction of \( e \), we gather that as long as \( b \) and \( d^{\prime} \) are positive, it affirms their relative sizes. This means, ultimately, the ratios \( \frac{a}{b} \) and \( \frac{e}{f} \) hinge upon how the other fractions stack against each other. However, since \( \frac{c}{d^{\prime}} \) and \( \frac{e}{f^{\prime}} \) are determined to be greater than the fractions of \( b \) and \( d \) respectively, we can derive that the ratio \( \frac{a}{b} \) must be lower than \( \frac{e}{f} \), as they are inherently compared through the other inequalities provided. In simpler terms, if you imagine yourself on a race track, \( a \), \( b \), and their ratios are at the starting line against \( e \) and \( f \), it’s clear that \( e \) is at an advantage since it's linked to a higher competition benchmark. So clearly, \( \frac{a}{b} < \frac{e}{f} \). In summary, through the relationships and inequalities given, we can conclude that \( \frac{a}{b} \) is less than \( \frac{e}{f} \), showcasing a fun play with logic and fractions!