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$\tilde{\Bbb{C}}$ was defined in the following manner

$\tilde{\Bbb{C}} = \Bbb{R} \cdot 1 + \Bbb{R} \cdot e$

with $1 \cdot 1 = 1, 1 \cdot e = e \cdot 1, e \cdot e = 1$

Could you elaborate more on this please? Or simply what is its name?

george
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2 Answers2

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A way to think about this set is $\mathbb{R}[\sqrt{1}]$ (in contrast to $\mathbb{C}=\mathbb{R}[\sqrt{-1}]$. $e$ here is a formal symbol whose square is 1, just like $i$ is a formal symbol whose square is $-1$. A little more formally is that this is isomorphic to $\mathbb{R}[x]/(x^2-1)$, where $(x^2-1)$ is the ideal generated by $x^2-1$.

Stella Biderman
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  • Yeah, just realized that. Edited, along with a comparison of this field and $\mathbb{C}$ – Stella Biderman Apr 03 '14 at 15:23
  • What is the difference between $\mathbb{R}[\sqrt{1}]$ and $\Bbb{R}$ – george Apr 03 '14 at 16:21
  • That notation is used to draw attention to the fact that $e$ is a formal symbol which acts like a square root of 1 that doesn't lie within $\mathbb{R}$. This is a notation that I don't think is formally correct but I have encountered in number theory – Stella Biderman Apr 03 '14 at 18:29
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There are three ways of extending multiplication to the plane which are useful in elementary geometry. One writes a planar vector formally as $a+be$ for some new element $e$. The arithmetic is then determined by knowledge of $e^2$. The best known is the case of the complex numbers ($e^2=-1$) (related to eulidean geometry). The case $e^2=0$ leads to the so-called dual numbers. Finally, $e^2=1$ is appropriate for hyperbolic geometry. A suitable reference is Yaglom "Complex numbers in geometry". Interesting exercise: compute the exponential function $\exp(a+b e)$ using the power series expansion of the exponential function in the other two cases.

jena
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