Are set operations same as lambdas? I'm teaching myself algebra (programmer by profession), and I wanted to clear a constant doubt that I'm having.
My doubt arose from the self-study of algebraic structures- Groups, Fields and Rings and these SE hits:
What are the differences between rings, groups, and fields?
Can anyone explain the difference between a ring, a group, and a field in a way so that your average 15 year old can understand?
Let's use the partial definition of a group as:

A group is a set of elements $E$ with an operation $\circ{}$ to combine the elements of $E$....

Most of the answers above use addition, subtraction, multiplication, and division as the operation $\circ$ to explain these concepts. Is there some special reason behind this other than illustration purposes? My programmer brain wonders why can't I use any lambda with binary arity instead, of course, as long as it follows the definitional requirements of the particular structure? Say, lambda x,y: math.log(x,y).
That leads to the last related question- an integer ring $Z_m$ consists of:


*

*The set $Z_m$ = $\{0, 1, ..., m-1\}$

*Two operations $+$, and $\times$ such that ....


By now, I know that the inverse of operation $+$ exists, but not necessarily for $\times$ for an integer ring. Then, can I assume that there are rings out there that could be working with different sets and operations that are not even remotely close to the arithmetic operations of addition and multiplication?
 A: Ultimately, I would say that it depends on what you would like to do with your binary operation $\circ.$ Given a set $S$ with a binary operation $\circ,$ you have a magma. Going one step further, you could make sure that $\circ$ is associative, and you'd wind up with a semigroup. If you desire an identity element $e_S,$ then you are dealing with a monoid. Last but not least, requiring your elements to have inverses under $\circ$ will give you a group.
Given that $S$ is a commutative group under $\circ,$ an additional binary operation $*$ will make $S$ into a ring if
1.) $S$ is a monoid under $*$ and
2.) the operation $*$ is distributive, i.e., $r * (s \circ t) = (r * s) \circ (r * t)$ and $(s \circ t) * r = (s * r) \circ (t * r).$
Consequently, the answer to your last question is probably no because the operation $\circ$ acts in an analogous manner to addition of integers, and the operation $*$ acts similarly to multiplication of $n \times n$ matrices.
A: It depends on what you mean by 'remotely close', but you could consider the following: Take any set $X$ (if a ring needs to have a multiplicative identity, $X$ may not be empty). Now on the power set $\mathcal P (X)$ define addition of $A,B\subseteq X$ via $A\Delta B := (A\setminus B) \cup (B\setminus A)$ and multiplication as $A\cap B$. Then $\mathcal P(X)$ is a ring.
Rings of this type  are called Boolean rings, since they arise when you study Boolean algebras.
