How do Boolean rings work? In a Boolean ring, every element equals to its square. But isn't this property only satisfied by $1$ and $0$? I don't understand how this ring works.
I also have a similar question about dual numbers. If $\varepsilon^2=0$, then $\varepsilon=0$, but I know it is not that simple. So, are we extending real numbers? For example, for i to make sense, we extended reals to complex numbers. Maybe we extended real numbers such that there exists a number apart from $0$ and its square is $0$?
 A: 
In a Boolean ring, every element equals to its square. But isn't this property only satisfied by $1$ and $0$? I don't understand how this ring works.

No. First consider $\mathbb{Z}_2=\{0,1\}$, which is a ring with addition and multiplication $\text{mod }2$ and with that it is a boolean ring. Not very interesting though. Now consider $R=\mathbb{Z}_2\times\cdots\times\mathbb{Z}_2$, the product of $n$ copies of $\mathbb{Z}_2$, with pointwise addition and multiplication. Since any element $r\in R$ is of the form $r=(r_1,\ldots,r_n)$, where $r_i=0$ or $r_i=1$, then $r^2=r$. So it is a boolean ring, but clearly it has more elements than $0$ and $1$ only, in fact it has $2^n$ elements. So for example, if $n=3$, then $(0,1,0)$ is an element that is equal to its own square, but it is neither zero $(0,0,0)$ nor one $(1,1,1)$.
Note that every finite boolean ring is of the form above.

I also have a similar question about dual numbers. If $\varepsilon^2=0$, then $\varepsilon=0$, but I know it is not that simple.

No, this is not true at all: $\varepsilon^2=0$ doesn't imply $\varepsilon=0$. In general, rings with the property "$ab=0$ implies $a=0$ or $b=0$" are known as integral domains. The ring of dual numbers is not an integral domain. And there are many important non-integral domains out there.

So, are we extending real numbers? For example, for it to make sense, we extended reals to complex numbers.

Yes, we do extend real numbers. But not into complex numbers. Complex numbers and dual numbers are two very different ways of extending real numbers.

Maybe we extended real numbers such that there exists a number apart from $0$ and its square is $0$?

Yes, exactly.
