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
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ISBN:9780470458365
Author:Erwin Kreyszig
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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.
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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