holds for all
On the other hand, we have so and you can do the same by writing
The appendix of my calculus book lists Absolute Value Properties:
1. |-a| = |a|
2. |ab| = |a||b|
3. |a/b| = |a|/|b|
4. |a+b| <= |a|+|b|
For 4., it is also mentioned that if 'a' and 'b' differ in sign, then |a+b| is less than |a|+|b|. In "all other cases", |a+b| equals |a|+|b|. I verified that 4. holds true a<0, b<0 (with a few examples).
I questioned about |a-b| and whether it is <,>,=, <=, or greater than or equal to |a|-|b|. I decided to see whether the signs mattered (whether a<0 or a>0 and b<0 and b>0 mattered) and whether the order of "a" and "b" matterred.
With a couple of examples, I found that |a-b| is greater than or equal to |a|-|b|. I found that if 'a' and 'b' differ in signs, then |a-b| > |a|-|b|. However, whether the inequality was "equal" or "greater than" varied with the order in which 'a' and 'b' is subsituted:
When a,b are positive numbers and a<b, the inequality statement is true when |a-b| > |a|-|b|. However, if a and b were reversed in the inequality statement, the statement is true when |b-a| = |b|-|a|.
When a,b are negative numbers and a<b, the inequality statement is true when |a-b| = |a|-|b|. However, if a and b were reversed in the inequality statement, the statement is true when |b-a| > |b|-|a|.
So am I right to conclude that:
(1) |a-b|=>|a|-|b|
(2)
When a,b are positive numbers and a<b, the inequality statement is true when |a-b| > |a|-|b|. However, if a and b were reversed in the inequality statement, the statement is true when |b-a| = |b|-|a|.
When a,b are negative numbers and a<b, the inequality statement is true when |a-b| = |a|-|b|. However, if a and b were reversed in the inequality statement, the statement is true when |b-a| > |b|-|a|.