7. Consider the following proposition. Proposition 0.1. For all real numbers x and y, if x # y, x > 0, and y > 0, then !+! > 2. Below is an incorrect "proof" of the proposition. Briefly explain why it is incorrect, and rewrite it into a proper proof of the proposition. Proof. Since x and y are positive real numbers, xy is positive and we can multiply both sides of the inequality by xy to obtain + · xy > 2 · xy x² + y? > 2xy. By combining all terms on the left side of the inequality, we see that x? – 2xy + y² > 0 and then by factoring the left side, we obtain (x – y)² > 0. Since x # y, x – y 7 0 and so (x – y)² > 0. This proves that if x # y, x > 0, and y > 0, then + ! > 2.
7. Consider the following proposition. Proposition 0.1. For all real numbers x and y, if x # y, x > 0, and y > 0, then !+! > 2. Below is an incorrect "proof" of the proposition. Briefly explain why it is incorrect, and rewrite it into a proper proof of the proposition. Proof. Since x and y are positive real numbers, xy is positive and we can multiply both sides of the inequality by xy to obtain + · xy > 2 · xy x² + y? > 2xy. By combining all terms on the left side of the inequality, we see that x? – 2xy + y² > 0 and then by factoring the left side, we obtain (x – y)² > 0. Since x # y, x – y 7 0 and so (x – y)² > 0. This proves that if x # y, x > 0, and y > 0, then + ! > 2.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:7. Consider the following proposition.
Proposition 0.1. For all real numbers x and y, if x # y, x > 0, and y > 0, then
!+! > 2.
Below is an incorrect "proof" of the proposition. Briefly explain why it is incorrect,
and rewrite it into a proper proof of the proposition.
Proof. Since x and y are positive real numbers, xy is positive and we can multiply
both sides of the inequality by xy to obtain
+
· xy > 2 · xy
x² + y? > 2xy.
By combining all terms on the left side of the inequality, we see that x² – 2xy +y² > 0
and then by factoring the left side, we obtain (x – y)² > 0. Since x + y, x – y 7 0 and
so (x – y)² > 0. This proves that if x # y, x > 0, and y > 0, then
+! > 2.
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