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Triangle inequality proof in Spivak's calculus

  1. Jan 6, 2014 #1
    proof.jpg

    So hi, there's one little thing which I'm not understanding in the proof. After the inequality Spivak considers the two expressions to be equal. Why?!?

    I just don't see why we can't continue with the inequality and when we have factorized the identity to (|a|+|b|)^2 we can just replace (a+b)^2 with (|a+b|)^2 and take the square root of both sides to finally have :

    |a+b| <= |a|+|b|

    Thank you for explaining !
     
  2. jcsd
  3. Jan 6, 2014 #2
    No need to answer, it's understood !
     
  4. Jan 6, 2014 #3

    FeDeX_LaTeX

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    Gold Member

    You could also start from ##-|a| \leq a \leq |a|##.
     
  5. Jan 6, 2014 #4
    Well, I have another question. When Spivak justifies the passage from the squares to |a+b| <= |a|+|b| he says the following : x^2<y^2 supposes that x<y for x,y in N. Now, the only thing bugging me is the following : Why didn't he do the following x^2<=y^2 supposes that x<=y for x,y in N ? Because what he says only justifies the inequality and not the equality ! Like a part is missing ! Am I right ? Thank you!
     
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