Simple Proof From Loomis and Sternberg's Calculus

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    Calculus Proof
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Homework Help Overview

The discussion revolves around proving a fundamental property of multiplication in real numbers, specifically that if the product of two elements is zero, then at least one of the elements must also be zero. This is being explored in the context of Loomis and Sternberg's Advanced Calculus, which suggests a focus on foundational mathematical principles.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss various approaches to the proof, including the use of axioms of real numbers, contraposition, and case analysis. Questions arise about the validity of certain steps and the implications of the axioms being used.

Discussion Status

Participants are actively engaging with the problem, offering hints and discussing the nuances of the proof. Some have provided guidance on using case analysis, while others express uncertainty about the implications of their reasoning. There is a recognition of the difficulty of seemingly simple proofs, and participants are exploring different interpretations of the problem.

Contextual Notes

There is mention of constraints related to the axioms of real numbers and vector spaces, as well as the challenge of working with different mathematical structures. Some participants express concern about the complexity of the material being studied and its appropriateness for their current understanding.

  • #31
Claim: x \vec{0} = \vec{0}

Let x be some arbitrary scalar in \mathbb{R}.

x \vec{0} = x ( \vec{0} + \vec{0}) \iff

x \vec{0} = x \vec{0} + x \vec{0} \iff

x \vec{0} - x \vec{0} = x \vec{0} + x \vec{0} - x \vec{0} \iff

\vec{0} = x \vec{0}

Because we had supposed x was arbitrary, meaning that we have not assumed anything about x beyond the fact that it is merely in \mathbb{R}, then this statement must be true of all real numbers.
 
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  • #32
yeah! very nice. ok, I am now fully convinced that ##a\mathbf{v}=\mathbf{0}## implies either ##a=0## or ##\mathbf{v}=\mathbf{0}## :) It takes quite a lot of work to prove things that seem fairly intuitive. But it's kind of satisfying too, right? I'm more physics than maths, so I don't have much experience doing proofs. But I would like to learn a bit more 'proper mathematics'. In fact, I was reading about groups this week.
 

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