Proof check? Absolute value inequalities. Real Analysis

illjay7005

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Sep 15, 2009
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Prove that |a|?b iff-b?a?b

Part 1.

absval a <=b
so 0<=b-absval a
so 0<=b-a or 0<=b-(-a)
so a<=b or -b<=a
hence -b<=a<=b

Part 2.

just reverse of part 1

Just want to make sure I am correct. Thanks!!!
 
illjay7005 said:
Prove that |a|?b iff-b?a?b

Part 1.

absval a <=b
so 0<=b-absval a
so 0<=b-a or 0<=b-(-a)
so a<=b or -b<=a
hence -b<=a<=b

Part 2.

just reverse of part 1

Just want to make sure I am correct. Thanks!!!

Correct .... according to me.

However, in a test or homework problem, write out the second part fully to get full credit.
 
Make clear cases... what allows you to do that absolute value trick is the "trichotomy ordering property" of the real numbers.

There are two cases: \(\displaystyle a \ge 0\) or \(\displaystyle a < 0\).
Using the definition of the \(\displaystyle |\cdot|\) function:
Case 1: \(\displaystyle a \ge 0 \iff |a| = a\)
Insert proof here
Case 2: \(\displaystyle a < 0 \iff |a| = -a\)
Insert proof here
 
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