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!
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 $\theta_1 + (\theta_2 + \theta_3)$.