Suppose that and y are real numbers. Show that and max{x,y} min{x,y} = = x+y+|x−y| 2 x + y - x - y 2
Suppose that and y are real numbers. Show that and max{x,y} min{x,y} = = x+y+|x−y| 2 x + y - x - y 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question

Transcribed Image Text:Exercise 1.3.2. Here is a paraphrase of the exercise.
Suppose that
and y are real numbers. Show that
and
max{x,y} =
min{x,y}
=
x + y + x - y
2
x + y − |x − y\¸
-
2
Remarks
If Exercise 1.2.10 had max and min in place of sup and inf, then the solution would be
easy. The main difficulty in the exercise is that the suprema and the infima may not belong to
the respective sets. Nonetheless, there must be elements in the sets within an arbitrarily
small positive & of the suprema and the infima. The key to the proof is that establishing
equality up to an arbitrarily small error is good enough.
Exercise 1.3.2 is a standard application of the method of proof by cases. The trichotomy
property of Definition 1.1.1 says that either a <y, or x = y, or > y. The upshot is that
the "lattice" operations of maximum and minimum can be obtained from the operations of
addition, multiplication, and absolute value.
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