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Real Analysis problem (easy), Triangle inequality

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    > a[1], a[2], a[3], .. , a[n] are arbitrary real numbers, prove that;

    abs(sum(a, i = 1 .. n)) <= sum(abs(a), i = 1 .. n)

    2. Relevant equations



    3. The attempt at a solution

    I have uploaded my attempt as a pdf file, since I'm not too familiar with the practice of writing it with correct notation on this great site :) i apologize. I am a 17-year old HS student, so i apologize if my attempt is way off or seems a bit thick .. This analysis is all new to me :)
     

    Attached Files:

  2. jcsd
  3. Oct 2, 2011 #2
    You have the right approach, but be a little careful about your logic. You're using "OK, I want to prove statement A. But statement A implies statement B. Statement B is true, so statement statement A is true. " This is bad logic; let's say I want to prove 1=0. I add 5 to both sides to get 6=5 and I multiply both sides by 0 to get 0=0. This is true so 6=5! See what I mean? xP But it is kinda easy to do that--especially in these cases. xP

    Anyway, in your case what you want to do is prove that it's true for n=1. (It's obviously true. o.o) Then, assume it's true for n=k. Add [itex]|a_{k+1}|[/itex] to both sides and use the triangle inequality on the left. Hence, the inequality is true for all natural numbers.

    Remember, induction is proving that IF n=k is true, THEN n=k+1 is true, which is what I did above. (Proving that n=1 is true, of course. xD)
     
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