Bunuel wrote:
Is \((x + y)^2 > x^2 + y^2\)?
(1) \(|x| \leq 0\)
(2) \(\sqrt{y^2} \leq 0\)
Target question: Is (x + y)² > x² + y²?This is a good candidate for
rephrasing the target question.
Take:
(x + y)² > x² + y²Expand and simplify:
x² + 2xy + y² > x² + y²Subtract x² and y² from both sides to get:
2xy > 0Divide both sides by 2 to get:
xy > 0REPHRASED target question: Is xy > 0? Statement 1: |x| ≤ 0 This inequality should seem odd to us, since the absolute value of a number is always GREATER THAN OR EQUAL to zero.
This inequality is saying the absolute value of some number is LESS THAN OR EQUAL to zero.
Since the absolute value of a number can
never be negative, it must us the case that |x| = 0, which means
x = 0.
If
x = 0, then xy = (
0)y = 0, which means we can answer the
REPHRASED target question with certainty.
It is NOT the case that xy > 0Since we can answer the
REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: √(y²) ≤ 0 This inequality should also seem odd to us, because the square root notation instructs us to find the POSITIVE square root of a number, which means √(some number) is always GREATER THAN OR EQUAL to zero.
The given inequality is saying √(some number) is LESS THAN OR EQUAL to zero.
Since √(some number) can
never be negative, it must us the case that √(y²) = 0, which means
y = 0.
If
y = 0, then xy = (x)(
0) = 0, which means we can answer the
REPHRASED target question with certainty.
It is NOT the case that xy > 0Since we can answer the
REPHRASED target question with certainty, statement 2 is SUFFICIENT
Answer:
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