if |x-y| < c, then |x| < |y| + c?

Answer 1

Answer is yes but I didn't follow 1st poster's reasoning.
This proof requires the identity that |a+b| <= |a| + |b| for all a,b.
(also, c > 0 as 1st poster showed)

Proof:
If x - y >= 0, then,
x - y < c
x < y + c
|x| < |y + c| <= |y| + |c| = |y| + c

if x - y < 0 then,
- (x - y) < c
-x + y < c
-x < -y + c
|-x| < |-y + c|
|x| < |-y + c| <= |-y| + |c| = |y| + c

Answer 2

yes.

|x-y| will always be positive.
therefore, if |x-y|
therefore, |x| will always be less than |y| + c.

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doug - your example: x = 2 and y = -3.

it is true that |x-y|=5.
but, c therefore must be greater than 5.
2 (|x|) is still less than 3(|y|)+c.

Answer 3

No. Let x = 2 and y = -3. Then |x-y| = 5 but |x| - |y| = -1

In general, |x ± y| =/= |x| ± |y|


Doug