I have the following triangle inequality proof, I just dont understand any part of it so if able please explain it in more detail.
I have the following triangle inequality proof, I just dont understand any part of it so if able please explain it in more detail.
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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I have the following triangle inequality proof, I just dont understand any part of it so if able please explain it in more detail.
![Theorem
(The Triangle Inequality) For every two real numbers x and y,
[x+y] ≤ [x] + [y].
Proof Since |x + y = |x] + [y] if either x or y is 0, we can assume that .x and y are nonzero. We
proceed by cases.
Case 1. x > 0 and y > 0. Then x + y > 0 and
Case 2. x < 0 and y < 0. Since x + y <0,
|x + y = x + y = [x] + [y].
|x + y = −(x + y) = (−x) + (−y) = |x] + [yl.
Case 3. One of x and y is positive and the other is negative. Assume, without loss of
generality, that x > 0 and y < 0. We consider two subcases.
Subcase 3.1.x+y ≥ 0. Then
|x] + |y| = x + (−y) = x−y>x+y= |x + yl.
Subcase 3.2. x+y < 0. Here,
|x| + y =x + (−y) = x − y > −x − y = −(x + y) = |x +y].
Therefore, x + y ≤ [x] + [y] for every two real numbers.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42044228-cff6-4391-9596-005bad4f5c7f%2Fd38dd518-e8e5-4cfd-bb0c-e44dca23b1f1%2Fo7035k_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem
(The Triangle Inequality) For every two real numbers x and y,
[x+y] ≤ [x] + [y].
Proof Since |x + y = |x] + [y] if either x or y is 0, we can assume that .x and y are nonzero. We
proceed by cases.
Case 1. x > 0 and y > 0. Then x + y > 0 and
Case 2. x < 0 and y < 0. Since x + y <0,
|x + y = x + y = [x] + [y].
|x + y = −(x + y) = (−x) + (−y) = |x] + [yl.
Case 3. One of x and y is positive and the other is negative. Assume, without loss of
generality, that x > 0 and y < 0. We consider two subcases.
Subcase 3.1.x+y ≥ 0. Then
|x] + |y| = x + (−y) = x−y>x+y= |x + yl.
Subcase 3.2. x+y < 0. Here,
|x| + y =x + (−y) = x − y > −x − y = −(x + y) = |x +y].
Therefore, x + y ≤ [x] + [y] for every two real numbers.
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