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Let $a,b,c$ be odd primes. In particular, $ab, ac, bc$ are all odd numbers. We can use this to our advantage, since then $\sqrt[ac]{x} : \Bbb{R} \to \Bbb{R}$ is well-defined and a bijection.

Let $V = \Bbb{R}^3$ as sets and define scalar multiplication on $V$ by $\lambda \in \Bbb{R}, \ v=(x,y,z) \in V \implies \lambda\cdot v = (\lambda^{bc} x, \lambda^{ac} y, \lambda^{ab} z)$. Define addition of vectors by: $$ u,v \in V, u = (x,y,z), v = (x', y', z') \implies \\ u + v = ((\sqrt[bc]{x} + \sqrt[bc]{x'})^{bc}, (\sqrt[ac]{y} + \sqrt[ac]{y'})^{ac}, (\sqrt[ab]{z} + \sqrt[ab]{z'}))^{ab}) $$

What vector space is this?

Does it actually form a ring too?

share|improve this question
    
Oh, I should have actually written down what I was thinking before objecting. Sorry. –  Karl Kronenfeld Nov 15 '13 at 3:02
    
Re your scalar multiplication, I don't think it is distributive. $(\lambda + \mu) \cdot v = ([\lambda + \mu]^{bc}x$, .. etc. ) $\ne \lambda \cdot v + \mu \cdot v$. –  Betty Mock Nov 15 '13 at 4:32

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