0
$\begingroup$

V is a vector space where $$V = \{\mathrm{rotations}\} = \{\theta : θ ~ \text{is a real number and} ~ 0 ≤ θ < 2π\}$$

Addition is defined by $$θ_1 + θ_2 := (θ_1 + θ_2) ~ \mathrm{mod} ~ 2π$$

Scaling by real numbers is defined by $$rθ = rθ ~ \mathrm{mod} ~ 2π$$

My question as to do with the axioms of Additive Associativity and Additive Inverses.

AA is defined as $$(u+v)+w=u+(v+w) \tag{$u,v,w ∈ V$}$$ and AI is defined as: there exists a vector w such that w=-v where $$v+w=0_v \tag{$v,w ∈ V$}$$

With regards to AA, I am unsure of how mod2π would effect the addition. So if u=θ1, v=θ2 and w=θ3, then (u+v)+w would be $$((θ_1+θ_2) ~ \mathrm{mod} ~ 2π)+θ_3) ~ \mathrm{mod} ~2π$$ right? How can I prove that is the same thing as $$(θ_1+(θ_2+θ_3) ~ \mathrm{mod} ~2π) ~ \mathrm{mod} ~2π$$ wihtout losing generality? Can the mod be pulled out of the expression and done afterwards since θ is a real number?

Secondly, for AI I assume that $$w=-v$$ would not mean literally negative v, but rather the inverse that would provide the zero vector since there are no negative elements in V. For example, if $$v=3π/2$$ then $$w=π/2$$ then $$v+w=0$$ as defined by the addition of vectors in this space. Am I right in assuming this?

Thanks for all your help!

$\endgroup$
  • $\begingroup$ Welcome to MSE. Please use MathJax to format both, your questions and your answers :) $\endgroup$ – mrtaurho Feb 10 at 20:00
  • $\begingroup$ What is the field your vector space is meant to be defined on? $\endgroup$ – celtschk Feb 10 at 20:24
  • $\begingroup$ I believe that the vector field would be R^1 since the space is that of rotations. $\endgroup$ – Matt Feb 10 at 20:34
  • 1
    $\begingroup$ You are right about additive inverses. $-v$ really means $2\pi - v$. $\endgroup$ – Nick Feb 10 at 21:16
  • $\begingroup$ @Matt: I meant the field of scalars; field here being the algebraical structure.For example, it might be the real numbers or the complex numbers. $\endgroup$ – celtschk Feb 10 at 21:53
0
$\begingroup$

Yes. The "mods" can be pulled out of the parentheses. Here's why:

The expression $(\theta_1 + \theta_2) ~ \mathrm{mod} ~ 2\pi$ really means any number of the form $(\theta_1 + \theta_2) + 2k\pi$ where $k$ is an integer. Similarly, the entire expression $((\theta_1 + \theta_2) ~ \mathrm{mod} ~ 2\pi + \theta_3) ~ \mathrm{mod} ~ 2\pi$ is a number of the form

$$ ((\theta_1 + \theta_2) + 2k\pi + \theta_3) + 2m\pi = (\theta_1+\theta_2)+\theta_3 + 2(k+m)\pi = ((\theta_1+\theta_2)+\theta_3) ~ \mathrm{mod} ~ 2\pi $$

And similarly for $\theta_1 + (\theta_2 + \theta_3)$.

$\endgroup$
  • $\begingroup$ Awesome, so I was on the right track then! Thank you for both of your answers! I will now close the thread. $\endgroup$ – Matt Feb 10 at 21:29
0
$\begingroup$

See both of u/Nick 's comments for the answer to this question. "Mods" can be pulled out of the expression and the additive inverse would be 2$\pi$ - v.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.