Proof check? Absolute value inequalities. Real Analysis

[illjay7005]

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!!!

 

[Deleted member 4993]

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.

 

[illjay7005]

Thank you!!

 

[daon]

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