If 1/x < 1/y, then
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x < y
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x > y
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x < y if xy < 0
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x < y if xy > 0
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x > y if xy < 0
Reveal answer
Fill a bubble to check yourself
C
Correct answer
Explanation
If xy > 0 (both positive or both negative), then taking reciprocals preserves the inequality: 1/x < 1/y implies x > y. Wait, let me reconsider: if x and y are both positive and 1/x < 1/y, then x > y. If x and y are both negative and 1/x < 1/y, then x > y still holds. But the question asks when x < y. Actually, option C says x < y if xy < 0. If xy < 0, one is positive, one negative. Suppose x is positive, y is negative: then 1/x > 0 > 1/y, so 1/x > 1/y, contradicting 1/x < 1/y. Suppose x is negative, y is positive: then 1/x < 0 < 1/y, so 1/x < 1/y holds, and x < y (negative < positive). So option C is correct.