|a+b| = |a| + |b| implies a and b parallel?

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Discussion Overview

The discussion revolves around the theorem concerning vectors in \(\mathbb{R}^k\) that states if \(|a| + |b| = |a + b|\), then \(|a|\) and \(|b|\) are parallel to each other in the same direction. Participants are exploring the validity of this theorem, attempting to prove or disprove it, and discussing related mathematical concepts.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the theorem and requests proof or disproof, indicating they have proven the converse.
  • Another participant suggests a method to prove the theorem using the inner product and provides a calculation involving \(|a+b|^2\) and the expansion of the inner product.
  • A participant follows up with a derivation that leads to the conclusion that the angle \(\theta\) between vectors \(a\) and \(b\) is zero, implying they are parallel.
  • A repeated inquiry about the theorem is made, along with a counterexample suggesting that if \(a\) and \(b\) are parallel, the equality \(|a+b| = |a| + |b|\) does not hold under certain conditions.

Areas of Agreement / Disagreement

Participants are engaged in a debate regarding the theorem, with some supporting its validity through mathematical reasoning while others question its truth by providing counterexamples. No consensus is reached on the theorem's validity.

Contextual Notes

The discussion includes assumptions about the properties of vectors and their magnitudes, as well as the implications of the inner product. There are unresolved aspects regarding the conditions under which the theorem holds true.

julypraise
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Is the following theorem true:

Theorem: Suppose [itex]a, \, b \in \mathbb{R}^k[/itex]. If [itex]|a| + |b| = |a + b|[/itex], then [itex]|a|[/itex] and [itex]|b|[/itex] are parallel to each other in the same direction.

I proved the converse, but I couldn't prove the theorem above. Please post the proof or the disproof of it, or a link of them. Thanks.
 
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Hi julypraise! :smile:

The trick to proving this is to calculate the inproduct

[tex]<a+b,a+b>[/tex]

in two different ways. The first way involves

[tex]<a+b,a+b>=|a+b|^2=(|a|+|b|)^2[/tex]

The other way starts as

[tex]<a+b,a+b>=|a|^2+|b|^2+2<a,b>[/tex]

Can you finish it?
 
Thank you so much, micromass.

So, by using the trick, I derived that

[itex]2|a||b| = 2<a, \, b>[/itex], and by the definition of inproduct

[itex]|a||b| = |a||b|cos\theta[/itex]

and therefore, [itex]\theta = 0[/itex]

where [itex]\theta[/itex] is the angle between.
 
julypraise said:
Is the following theorem true:

Theorem: Suppose [itex]a, \, b \in \mathbb{R}^k[/itex]. If [itex]|a| + |b| = |a + b|[/itex], then [itex]|a|[/itex] and [itex]|b|[/itex] are parallel to each other in the same direction.

I proved the converse, but I couldn't prove the theorem above. Please post the proof or the disproof of it, or a link of them. Thanks.

If `a` and `b` are parallel then ∃c∈ℝ s.t., a=cb. So then, ||a+b||=||cb+b||=||b||(|c+1|)≠||a||+||b||. Since ||a||=||cb||=(|c|)||b||.
 
Last edited:

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