- #1
tlkieu
- 8
- 1
Just wondering if anyone could confirm if I've headed in the right direction with these
(a) Prove the triangular inequality: |x + y| ≤ |x| + |y|.
(b) Use triangular inequality to prove |x − y| ≥ ||x| − |y||.
(c) Show that if |x − a| < c/2 and |y − b| < c/2 then |(x + y) − (a + b)| < c.
So for (a):
∣x+y∣∣^2 = (x+y)^2
= x^2 + 2xy + y^2
= |x|^2 + 2xy + ∣y∣^2
≤ |x|^2 + 2∣xy∣ + ∣y∣^2
= |x|^2 + 2|x|⋅∣y∣ + ∣y∣^2
= (|x|+∣∣y∣∣)^2
Which shows ∣x+y∣ ≤ |x| + ∣y∣
For (b):
I split it into two proofs
In case (1): If |x| ≥ ∣y∣ we have:
∣|x| − ∣y∣∣ = |x|− ∣y∣, and the proof is finished.
For case (2): If ∣y∣ ≥ |x|:
∣y−x∣ ≥ ∣y∣ − |x| = ∣∣y∣ − |x|∣
That is what I have so far and part (c) I'm not too sure how to approach that one
(a) Prove the triangular inequality: |x + y| ≤ |x| + |y|.
(b) Use triangular inequality to prove |x − y| ≥ ||x| − |y||.
(c) Show that if |x − a| < c/2 and |y − b| < c/2 then |(x + y) − (a + b)| < c.
So for (a):
∣x+y∣∣^2 = (x+y)^2
= x^2 + 2xy + y^2
= |x|^2 + 2xy + ∣y∣^2
≤ |x|^2 + 2∣xy∣ + ∣y∣^2
= |x|^2 + 2|x|⋅∣y∣ + ∣y∣^2
= (|x|+∣∣y∣∣)^2
Which shows ∣x+y∣ ≤ |x| + ∣y∣
For (b):
I split it into two proofs
In case (1): If |x| ≥ ∣y∣ we have:
∣|x| − ∣y∣∣ = |x|− ∣y∣, and the proof is finished.
For case (2): If ∣y∣ ≥ |x|:
∣y−x∣ ≥ ∣y∣ − |x| = ∣∣y∣ − |x|∣
That is what I have so far and part (c) I'm not too sure how to approach that one