I am looking to write a proof (without breaking it into cases) for the following:Prove that |x+y| ≤ |x| + |y| using the following assumptions:-|x| ≤ x ≤ |x|if c ≤ r, then c ≥ -rif r ≥ -c, and r ≤ c, then -c ≤ r ≤ c
Before I go on, I have to apologize. I was unable to come up with a proof of my own (I kept getting stuck), so I searched the internet (this property is famously known as the "Triangle Inequality", and has applications in number theory, calculus, physics, and linear algebra) and found two different proofs that appeared side-by-side on numerous sites. I have copied and pasted these proofs into this post BUT I must warn you that I myself do not believe either of them. Though these proofs seem unanimously accepted by the people of the internet, they do not seem correct to me (I will explain why later). I am doing my best to answer your question, but just know that.
The red question marks indicate areas where the proof fails for me. If you can find an explanation for these areas, let me know.
Thank you for your help. Looking at that, however, I realized that I gave you the wrong assumptions. The first one (-|x| ≤ x ≤ |x|) is correct, but the next two assumptions are:
ReplyDeleteFor all real numbers r and c with c ≥ 0, if -c ≤ r ≤ c, then |r| ≤ c.
For all real numbers r and c with c ≥ 0, if |r|≤ c, then -c ≤ r ≤ c.
Maybe that will help? I have stared at this conjecture for a long time and am still coming up empty. Thanks again for your help.
Because neither of the proofs I found use those two assumptions, changing them does not change the proofs.
ReplyDeleteThe second proof I posted would be perfectly acceptable if we knew that xy ≤ |x||y|.
Actually, looking at the first proof again, the assumption "if -c ≤ r ≤ c, then |r| ≤ c" would give justification for the conclusion and therefore make the whole proof valid.
ReplyDeleteHere's how:
-|x|-|y| ≤ x + y ≤ |x| + |y| (which I got from adding the x and y inequalities) can be changed to -(|x| + |y|) ≤ x + y ≤ |x| + |y|, which implies that |x + y| ≤ |x| + |y| by letting r = x + y and letting c = |x| + |y| and using |r| ≤ c.
Let me know if that makes sense. I will rewrite the first proof correctly and edit the actual post when I have time.
Beautiful and Strange, where the hell have you been?
ReplyDeleteWe have missed you dearly on Figment.
Come back this instant!
From,
Ariel Magic-esi
(We will stalk you...)
This is my first time go to see at here and i am really pleassant to read all at alone place.
ReplyDeletemy web site :: www.tamporno.net