Absolute value theorem that I can't convince myself of

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Discussion Overview

The discussion revolves around the absolute value inequality for real numbers, specifically the triangle inequality: \(\left|a+b\right| \leq \left|a\right| + \left|b\right|\). Participants explore whether this inequality is proven, an axiom, or something else, and provide reasoning to support its validity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions the status of the triangle inequality, asking if it is proven or an axiom.
  • Another participant provides a link to the Wikipedia page on the triangle inequality, suggesting it is a well-established concept.
  • A different participant emphasizes the importance of the inequality and encourages others to convince themselves of its truth.
  • Further, a participant outlines a reasoning process through various cases based on the signs of \(a\) and \(b\) to demonstrate the validity of the inequality.

Areas of Agreement / Disagreement

There is no explicit consensus on the status of the triangle inequality as proven or axiomatic, and the discussion includes multiple perspectives on understanding its validity.

Contextual Notes

The reasoning provided relies on specific cases based on the signs of \(a\) and \(b\), which may not cover all scenarios or assumptions involved in the inequality.

JennyInTheSky
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While reading my text, I came across an inequality that I couldn't convince myself of...

For real numbers a,b: \left|a+b|<= |a|+|b|. Is this something proven? Or is it an axiom or something?
 
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It's a pretty important inequality. I highly suggest you convince yourself of its truth :D
 
To convince yourself of its truth consider what the effect of the signs of a and b have on the inequality.

Case 1( a and b are positive):
|a+b| = a + b = |a|+|b|

Case 2 (a is positive and b is non-positive):
Let b = -y then a and y are positive. If a-y is positive:
|a+b|=|a-y| = a-y \leq a \leq |a| + |b|
If a-y is non-positive, then y-a is positive and:
|a+b|=|a-y| = y-a \leq y = |b| \leq |a| + |b|

Case 3 (a and b are negative):
Let a = -x, b = -y:
|a+b| = |-(x+y)| = x+y = |a|+|b|
 

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