Simple Proof From Loomis and Sternberg's Calculus

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SUMMARY

The forum discussion centers on proving the statement "if ab = 0, then either a = 0 or b = 0" using concepts from Loomis and Sternberg's Advanced Calculus. Participants emphasize the importance of understanding axioms of real numbers and vector spaces, particularly the closure of non-zero reals under multiplication. They suggest using contraposition and case analysis to tackle the proof, highlighting that simple proofs often require subtle reasoning. Ultimately, the discussion concludes that the proof is achievable by demonstrating both cases: when a = 0 and when a ≠ 0.

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  • Understanding of real number axioms and properties
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  • #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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