What is the question ? Show the first equality ? Treat the cases and .

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- May 24th 2011, 12:59 PM #1

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- May 24th 2011, 01:11 PM #2

- May 24th 2011, 01:12 PM #3

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- May 24th 2011, 06:53 PM #4

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Usually one would consider non-overlapping cases, but I guess the overlapping cases here would work.

Case 1: Suppose x y. What does this imply? If you don't recall what this implies I'll say that addition on**R**satisfy commutation, monotonicity, an equality for inverses, and an equality for neutrals. In other words, for all x, y, z, c on**R**the following properties for + hold:

If x y, (x+c) (y+c) (monotonicity)

(x+y)=(y+x) (commutation)

(x+-(x))=0 (inverse)

x+0=x (neutrality)

Since you started with x y, do you see which property you want to use here first? Do you see what element (function of an element, depending on how you look at it) of**R**you want to select for "c"? If you do, I think you'll infer what I've hinted at. Then check your definition of absolute value. So in this case, abs(x-y) must equal... what? Then replace what abs(x-y) must equal in this case with abs(x-y) in the maximum formula.

Case 2: Suppose x>y. Now I'll point out that for all x, y, c in**R**,

If x>y, then (x+c)>(y+c). Now using a similar technique, along with your definition of absolute value you can show something about the absolute value in this case.

Then you might to want to indicate the cases exhaust**R**and the result consequently follows.

If the above doesn't give you enough clues, I'll happily try and spell out details here.