# Confirming Axioms of Vector Spaces that rely on modular arithmetic

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?

• What is the field your vector space is meant to be defined on? – celtschk Feb 10 at 20:24
• I believe that the vector field would be R^1 since the space is that of rotations. – Matt Feb 10 at 20:34
• You are right about additive inverses. $-v$ really means $2\pi - v$. – Nick Feb 10 at 21:16
• @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. – celtschk Feb 10 at 21:53

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