Simple Proof From Loomis and Sternberg's Calculus

  • Thread starter Thread starter Bashyboy
  • Start date Start date
  • Tags Tags
    Calculus Proof
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
31 replies · 4K views
Claim: [itex]x \vec{0} = \vec{0}[/itex]

Let [itex]x[/itex] be some arbitrary scalar in [itex]\mathbb{R}[/itex].

[itex]x \vec{0} = x ( \vec{0} + \vec{0}) \iff[/itex]

[itex]x \vec{0} = x \vec{0} + x \vec{0} \iff[/itex]

[itex]x \vec{0} - x \vec{0} = x \vec{0} + x \vec{0} - x \vec{0} \iff[/itex]

[itex]\vec{0} = x \vec{0}[/itex]

Because we had supposed x was arbitrary, meaning that we have not assumed anything about x beyond the fact that it is merely in [itex]\mathbb{R}[/itex], then this statement must be true of all real numbers.
 
Last edited:
Physics news on Phys.org
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.