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Triangle Inequality

  1. Jun 20, 2009 #1
    http://math.ucsd.edu/~wgarner/math4c/derivations/other/triangleinequal_files/eq0007S.gif [Broken]

    Why did they introduce the <= sign?

    I cannot think of any numbers that would violote the =. So why introduce the <?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 20, 2009 #2
    Because if "a" or "b" (not both) is negative, then the answer would be less than the formula with the absolute value of "a" or "b".
     
  4. Jun 20, 2009 #3
    That's not true, plugin a=-3 and b=2. In fact, try any set of numbers and you will see.
     
  5. Jun 20, 2009 #4

    HallsofIvy

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    Okay, I will: [itex]|-3+2|^2= 1^2= 1. |-3|+ |2|= 3+2= 5[/itex] and [itex]5^2= 25[/itex]. 1 is definitely less than 25!

    Now, what do YOU get? (Or did you do |a+b|2 and (a+b)2 rather than |a+b|2 and (|a|+ |b|)^2?)
     
  6. Jun 20, 2009 #5

    Fredrik

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    I think (or at least I hope) that the OP is referring to the last line (not the one before). The last one should start with a =.
     
  7. Jun 20, 2009 #6

    HallsofIvy

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    That is not my interpretation of that particular way of writing mathematics.

    If it were [itex]|a+b|^2= (a+ b)(a+ b)= a^2+ 2ab+ b^2\le |a|^2+ 2|a||b|+ |b|^2= (|a|+ |b|)^2[/itex], in one line, then, yes, the last two are equal. But my understanding of
    [tex]\begin{array}{cc}|a+ b|^2&= (a+b)(a+b)\\ &= a^2+ 2ab+ b^2\\ &\le |a|^2+ 2|a||b|+ |b|^2 \\ &\le (|a|+ |b|)^2[/tex]
    is that the left side, here [itex]|a+ b|^2[/itex], is "copied" down the left. That is, it is
    [tex]\begin{array}{cc}|a+ b|^2&= (a+b)(a+b)\\|a+ b|^2&= a^2+ 2ab+ b^2\\|a+ b|^2&\le |a|^2+ 2|a||b|+ |b|^2 \\|a+ b|^2&\le (|a|+ |b|)^2[/tex]
     
  8. Jun 20, 2009 #7
    Your interpretation is correct (but I know you don't need me to tell you that). Most authors write this way, and it even saves ink!
     
  9. Jun 20, 2009 #8
    The last one should be with <=. I understand why it is there; I was just pissed because I would have never thought to put the <= after putting up absolute values around 2ab.

    Thanks for the help.

    HallsOfIvy, I do like your interpretation of the math better, more lucid. The GIF I posted is hotlinked from some website I found. Unfortunately, the book I got this problem (spivak) from uses the same notation as the GIF image.
     
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