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It's very standard to view rotations about the origin in $\mathbb{R}^2$ as a group $\mathbb{SO}(2)$, with the zero rotation as an identity and composition of rotations as addition. This can also be viewed as the interval $[0,2\pi)$ with $0$ as the additive identity and addition defined mod $2\pi$.

We can lift this to a Rng structure by defining multiplication $\hat\times:[0,2\pi)\times[0,2\pi)\rightarrow[0,2\pi)$ as $$a\hat\times b=\frac{ab}{2\pi},$$ so each rotation is treated as a fraction of a complete rotation of $2\pi$. Since $a,b<2\pi$ we have that $\frac{ab}{2\pi}\leq\min\{a,b\},$ with equality holding iff $a=0$ or $b=0$. This is a rng and not a ring because the multiplicative identity we would like, $2\pi$, cannot be allowed in the set on consequence of non-uniqueness for the additive identity.

Is this a well known construction, or obviously trivial as a geometric structure for some reason I'm missing?

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    $\begingroup$ It seems multiplication does not distribute over addition: $\pi\hat{\times}(\pi+\pi)=\pi\hat{\times} 0=0$, but $\pi\hat{\times}\pi+\pi\hat{\times}\pi=\pi/2+\pi/2=\pi$. $\endgroup$
    – Julian Rosen
    Commented Dec 3, 2017 at 22:57
  • $\begingroup$ @JulianRosen Ah, thank you. This may be what I was missing. On further consideration, this lack of well-behaved interplay will likely make this a useless construction. If you'd like to post your comment as an answer I'll gladly accept. $\endgroup$
    – Alec Rhea
    Commented Dec 3, 2017 at 22:58
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    $\begingroup$ Can we please stop using all those stupid "rng", "rig" and similar puns? How can it be good mathematical notation when you never know if the author made a mistype or is discussing some special structure? $\endgroup$
    – Anton Fetisov
    Commented Dec 4, 2017 at 0:48
  • $\begingroup$ How would you call them then? $\endgroup$
    – fosco
    Commented Dec 5, 2017 at 8:50
  • $\begingroup$ @Fosco Loregian: I call them ring, and I call a ring with one a ring with one. Different areas of mathematics have different generic examples, for some having a multiplicative identity is always satisfied, for some this is a strong condition. So every talk dealing with rings should start with "A ring is a commutative ring with one" or whatever conventions you prefer. $\endgroup$
    – Jan-Christoph Schlage-Puchta
    Commented Dec 6, 2017 at 11:31

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The multiplication does not give a Rng structure because it does not distribute over addition: $$ \pi\hat{\times}\big(\pi+\pi)=\pi \hat{\times} 0=0\neq \pi = \frac{\pi}{2}+\frac{\pi}{2}=\pi\hat{\times}\pi +\pi\hat{\times}\pi. $$

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