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Difficulty: HardQuestion Stem Simplification and Target Rephrasing

For all real numbers aa and bb with a≠ba \neq b, the Data Sufficiency Yes/No target question "Is a2−b2(a−b)2>1\frac{a^2 - b^2}{(a - b)^2} > 1?" is algebraically equivalent to the simplified target question "Is a>ba > b?"

Answer: Answer

Answer

The statement is False because rephrasing the target inequality yields 2ba−b>0\frac{2b}{a-b} > 0, which requires analyzing the signs of both bb and a−ba-b, rather than evaluating whether a>ba > b alone.
The statement is False. Correct simplification of a2−b2(a−b)2>1\frac{a^2 - b^2}{(a - b)^2} > 1 leads to 2ba−b>0\frac{2b}{a-b} > 0. This inequality requires 2b2b and a−ba-b to have identical signs, which holds either when b>0b > 0 and a>ba > b or when b<0b < 0 and a<ba < b. Because a>ba > b can be true while b<0b < 0 (making the original inequality false), the proposed rephrasing is invalid.

Step-by-Step Solution

1
Factor the algebraic expression in the numerator of the target question.
The numerator a2−b2a^2 - b^2 factors as (a−b)(a+b)(a-b)(a+b), yielding (a−b)(a+b)(a−b)2>1\frac{(a-b)(a+b)}{(a-b)^2} > 1.
Factoring allows simplification of common terms between the numerator and denominator.
2
Simplify the fraction by canceling common non-zero terms.
Since a≠ba \neq b, a−b≠0a - b \neq 0, so a+ba−b>1\frac{a+b}{a-b} > 1.
Canceling (a−b)(a-b) is valid as long as a≠ba \neq b.
3
Compare the fraction to zero by subtracting 1 from both sides.
\frac{a+b}{a-b} - 1 > 0 \implies \frac{(a+b) - (a-b)}{a-b} > 0 \implies \frac{2b}{a-b} > 0.
Subtracting 1 avoids multiplying by a variable expression (a−b)(a-b) whose sign is unknown.
4
Determine the conditions under which 2ba−b>0\frac{2b}{a-b} > 0.
The quotient is positive when 2b2b and a−ba-b have the same sign: Case 1 (b>0b > 0 and a>ba > b) OR Case 2 (b<0b < 0 and a<ba < b).
A quotient is strictly positive if and only if its numerator and denominator share the same sign.
5
Test whether "Is a>ba > b?" is equivalent to the derived condition using a counterexample.
If a=2a = 2 and b=−1b = -1, then a>ba > b is true (2>−12 > -1). However, 2(−1)2−(−1)=−23≯0\frac{2(-1)}{2 - (-1)} = -\frac{2}{3} \ngtr 0.
Finding a scenario where a>ba > b is true but the original inequality fails proves the two target questions are not algebraically equivalent.

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Question Stem Simplification and Target Rephrasing
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Question Stem Simplification and Target Rephrasing

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