Triangle inequality proof in Spivak's calculus

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Homework Help Overview

The discussion revolves around the proof of the triangle inequality as presented in Spivak's calculus. Participants are examining the logical steps involved in transitioning from squared terms to the absolute value inequality.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the justification for equating expressions in the proof and exploring the implications of inequalities versus equalities. There is also a suggestion to start from fundamental properties of absolute values.

Discussion Status

The discussion is active with participants raising questions about specific steps in the proof and the reasoning behind them. Some have indicated understanding, while others continue to seek clarification on the logical connections made in the proof.

Contextual Notes

There appears to be a focus on the nuances of inequalities and equalities in the context of natural numbers, as well as the implications of assumptions made in the proof.

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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 !
 
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No need to answer, it's understood !
 
You could also start from ##-|a| \leq a \leq |a|##.
 
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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