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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.

Ask by Murray Mitchell. in the United States
Jan 24,2025

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\( \frac{a}{b} < \frac{e}{f} \)

Solución

To compare \( \frac{a}{b} \) and \( \frac{e}{f} \), we will analyze the given conditions step by step. ### Known Conditions: 1. \( \frac{a}{b^{\prime}} \) and \( \frac{c}{d^{\prime}} \) are rational numbers. 2. \( \frac{f}{f} \) is a rational number (which equals 1). 3. \( \frac{d}{b} < \frac{c}{d^{\prime}} \) 4. \( \frac{c}{d} < \frac{e}{f^{\prime}} \) ### Step-by-Step Analysis: 1. **Understanding the first inequality:** \[ \frac{d}{b} < \frac{c}{d^{\prime}} \] This implies that \( d \cdot d^{\prime} < c \cdot b \). Rearranging gives us: \[ c \cdot b > d \cdot d^{\prime} \] 2. **Understanding the second inequality:** \[ \frac{c}{d} < \frac{e}{f^{\prime}} \] This implies that \( c \cdot f^{\prime} < e \cdot d \). Rearranging gives us: \[ e \cdot d > c \cdot f^{\prime} \] ### Comparing \( \frac{a}{b} \) and \( \frac{e}{f} \): To compare \( \frac{a}{b} \) and \( \frac{e}{f} \), we need to relate \( a \) and \( e \) through the inequalities we have. From the first inequality, we have: \[ c \cdot b > d \cdot d^{\prime} \] This suggests that \( c \) is relatively larger than \( \frac{d \cdot d^{\prime}}{b} \). From the second inequality, we have: \[ e \cdot d > c \cdot f^{\prime} \] This suggests that \( e \) is relatively larger than \( \frac{c \cdot f^{\prime}}{d} \). ### Conclusion: To compare \( \frac{a}{b} \) and \( \frac{e}{f} \), we can infer that since \( c \) is larger than a certain ratio involving \( d \) and \( d^{\prime} \), and \( e \) is larger than a ratio involving \( c \) and \( f^{\prime} \), we can conclude that: - If \( a \) is related to \( c \) in a way that maintains the inequalities, we can say that \( \frac{a}{b} \) is likely to be less than \( \frac{e}{f} \) based on the transitive nature of inequalities. Thus, we can conclude that: \[ \frac{a}{b} < \frac{e}{f} \] This conclusion is based on the relationships established by the inequalities provided.

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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!

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