Name for the reals augmented with an $x$ such that $x^2 = x$ 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 $e^{a+bx} = e^a + (e^{a+b} - e^a)x$.
 A: 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.
A: 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.
A: 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 $\infty^2=\infty$.
