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If you add an $x$ such that $x^2=-1$ to the reals, you get the complex numbers. If you add an $x$ such that $x^2=0$ to the reals, you get the dual numbers. If you add an $x$ such that $x^2=+1$ to the reals, you get the split complex numbers.

If you add an $x$ such that $x^2=x$ to the reals, you get... what? What's the name for that number system? You know, the one where $e^{a+bx} = e^a + (e^{a+b} - e^a)x$.

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    $\begingroup$ I believe this will be isomorphic to $\Bbb{R}\times \Bbb{R}$. Which incidentally is the same as the split complex numbers. $\endgroup$
    – jgon
    Commented May 22, 2015 at 5:06
  • $\begingroup$ I don't understand (sorry if it's just me), but don't $0, 1$ already satisfy that? $\endgroup$
    – MT_
    Commented May 22, 2015 at 5:08
  • $\begingroup$ @Soke, I was assuming we were taking $\Bbb{R}[x]/(x^2-x)$. $\endgroup$
    – jgon
    Commented May 22, 2015 at 5:08
  • $\begingroup$ @jgon Oh okay, the material is probably just going over my head. $\endgroup$
    – MT_
    Commented May 22, 2015 at 5:10
  • $\begingroup$ @Soke 0 and 1 do satisfy $v^2 = v$, but the question is about what you get when you add a third solution. $\endgroup$ Commented May 22, 2015 at 5:32

3 Answers 3

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If $x^2=x$, then

$$ (2x-1)^2=4x^2-4x+1=4x-4x+1=1 $$

So you actually have the split-complex numbers in disguise.

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  • $\begingroup$ Doesn't this just show that we can embed the split complex numbers into the number system of interest? It doesn't show they're isomorphic. $\endgroup$ Commented May 22, 2015 at 5:11
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    $\begingroup$ @goblin: They're both $2$-dimensional... $\endgroup$
    – Micah
    Commented May 22, 2015 at 5:11
  • $\begingroup$ It's pretty easy to find an inverse to show the embedding is onto and $1-1$ $\endgroup$ Commented May 22, 2015 at 5:14
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    $\begingroup$ @Strilanc: Yes, exactly. In general, if you try to extend the real numbers by appending something that satisfies a quadratic relation, it always will end up being a disguised version of one of the three systems in your question. $\endgroup$
    – Micah
    Commented May 22, 2015 at 5:16
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    $\begingroup$ indeed, adding $x$ so that $x^2+bx+c=0$, the resulting ring must be isomorphic to one of the three listed above, depending on whether $b^2-4c$ is negative, zero, or positive. $\endgroup$ Commented May 22, 2015 at 5:19
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Similarly to Micah's answer, but showing an ismorphism to $\Bbb{R}\times \Bbb{R}$, note that

\begin{align} (a(1-x)+bx)(c(1-x)+dx) &= ac(1^2-2x+x^2)+(ad+bc)(x-x^2)+bdx^2 \\ &= ac(1-2x+x)+(ad+bc)(x-x)+bdx \\ &=ac(1-x)+bdx. \end{align}

Therefore the mapping to $\Bbb{R}\times \Bbb{R}$: $\varphi(a(1-x)+bx)=(a,b)$ preserves multiplication. Showing that it preserves addition is easy. Similarly the kernel must be trivial since if $\varphi(a(1-x)+bx)=0$, we have $a=b=0$, so $a(1-x)+bx=0$. Finally, this homomorphism is trivially surjective. Note that the mapping is well defined since $1-x$ and $x$ form a basis.

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Such objects are called idempotents. Split-complex numbers have two idempotents, $1/2+j/2$ and $1/2-j/2$.

Triplex numbers also have idempotents, for instance, $\frac{1+i+j}2$ and $\frac{2-i-j}2$ (totally 8).

Also, consider the one-point or two-point compactifications of real numbers (the later is called "extended real line", $\overline{\mathbb{R}}$). In both, there is one idempotent, $\infty$, because $\infty^2=\infty$.

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