A question about the definition of $\mathbb{C}$ In the usual definition of the field $\mathbb{C}$, as $\{(a,b):a,b\in\mathbb{R}\}$, the field $\mathbb{R}$ is not exactly a subset of $\mathbb{C}$, but only an isomorphic copy of the subfield $\{(a,0):a\in\mathbb{R}\}$. I am wondering: can we define $\mathbb{C}$ as 
the set 
$$
\mathbb{C}=\mathbb{R}\cup\{(a,b):a,b\in\mathbb{R}\ \text{and}\ b\neq0\}\ ?
$$
In this case, how to define the operations?
$\mathbf{Remark}$: We know that the operations of $\{(a,0)\}$ can be extend to 
$\{(a,b)\}$. Since $\mathbb{R}≅\{(a,0)\}$, can we prove that the operations of $\mathbb{R}$ can be extend to $\mathbb{R}∪\{(a,b):b≠0\}$?
 A: You can instead define $\mathbb{C}$ by
$$\mathbb{C} := \mathbb{R}(i) = \{a + bi \mid a, b \in \mathbb{R}\}$$
where $i^2 = -1$.
Then this is the typical definition of $\mathbb{C}$ and is a field since $i$ is algebraic over $\mathbb{R}$. Furthermore, $\mathbb{R} \subset \mathbb{C}$ by taking $b=0$.
A: You can do it without any difficulty. Consider the map
$$
f\colon\mathbb{R}\cup\underbrace{\{(a,b):a,b\in\mathbb{R},b\ne0\}}_{X}
\to\mathbb{R}\times\mathbb{R}
$$
defined in the obvious way:
$$
f\colon a\in\mathbb{R}\mapsto(a,0),
\qquad
f\colon (a,b)\in X\mapsto(a,b),
$$
and define the operations on $\mathbb{R}\cup X$ by
$$
u+v=f^{-1}(f(u)+f(v)),
\qquad
uv=f^{-1}(f(u)f(v))
$$
However, it is dubious if you gain something with respect to identifying $a$ with $(a,0)$ and setting $i=(0,1)$, so
$$
(a,b)=(a,0)+(b,0)(0,1)=a+bi
$$
A: This is not exactly an answer, but it is too long for a comment.

The traditional construction of $\mathbb{C}$ as an ordered pair of real numbers is the simplest approach possible. But this appears to have to have very non-intuitive definition of multiplication of pairs.
A better approach is the one given by Modern Algebra. Let $\mathbb{R}[x]$ be the set of all polynomials with real coefficients. Let's say that two polynomials $f(x), g(x) \in \mathbb{R}[x]$ are congruent to each other modulo another polynomial $p(x) \in \mathbb{R}[x]$ if both $f(x), g(x)$ leave the same remainder when divided by $p(x)$ or in other words when $p(x) \mid f(x) - g(x)$ and we write $$f(x)\equiv g(x)\pmod{p(x)}$$ Here we assume $p(x)$ to be of positive degree. This relation of congruence is an equivalence relation on $\mathbb{R}[x]$.
Let's fix a polynomial $p(x)$ of positive degree and consider the set of all polynomials congruent to $f(x)$ modulo $p(x)$: $$[f(x)] = \{g(x)\mid g(x)\equiv f(x)\pmod{p(x)}\}$$ Such sets are called equivalence classes and we can define operations of $+$ and $\times$ on these these equivalence classes by $$[f(x)] + [g(x)] = [f(x) + g(x)], [f(x)]\times[g(x)] = [f(x)\times g(x)]$$ And now consider the set of all such equivalence classes $$\mathbb{R}[x]/\langle p(x)\rangle = \{[f(x)]\mid f(x) \in \mathbb{R}\}$$ The magic happens when $p(x)$ is an irreducible polynomial and then it can be proved that the above set forms a field. Moreover this field is isomorphic to $\mathbb{R}$ if $p(x)$ is of degree $1$.
The field of complex numbers is now given by $\mathbb{R}[x]/\langle x^{2} + 1 \rangle$, thus it is the set of equivalence classes modulo the irreducible polynomial $x^{2} + 1$. By convention we denote the equivalence class containing $f(x)$ by the symbol $[r(x)]$ where $r(x)$ is the remainder when $f(x)$ is divided by $p(x)$. So in case of $\mathbb{C} = \mathbb{R}[x]/\langle x^{2} + 1\rangle$ the elements of $\mathbb{C}$ look like $[a + bx]$ (the remainder is always of degree $1$ or less if we divide by $x^{2} + 1$). The sum and product of elements of $\mathbb{C}$ are defined by $$[a + bx] + [c + dx] = [a + c + (b + d)x], [a + bx][c + dx] = [(a + bx)(c + dx)]$$ Let's see what happens when we try to find the remainder when $(a + bx)(c + dx)$ is divided by $x^{2} + 1$.
We have $$(a + bx)(c + dx) = ac + bdx^{2} + (ad + bc)x = (ac - bd) + (ad + bc)x + bd(x^{2} + 1)$$ and thus the remainder is $(ac - bd) + (ad + bc)x$. It follows that $$[a + bx][c + dx] = [(a + bx)(c + dx)] = [ac - bd + (ad + bc)x]$$ This definition of complex numbers can be transformed into the usual one by replacing symbol $[a + bx]$ by $(a, b)$ and we can see now the justification of multiplication rule $$(a, b)(c, d) = (ac - bd, ad + bc)$$ BTW in case you have not guessed the element $[x] \in \mathbb{C}$ is denoted by special symbol $i$ and it has the property $i^{2} + [1] = [0]$.
A: Let $a,b \in \mathbb{C}$. Then there are different possibilities:


*

*$a \in \mathbb{R}, b = (x,y), y \neq 0$: $a + b := (a+x, y)$.

*$a,b \in \mathbb{R}$: $a + b$ is obvious.

*$a = (x_1, y_1), b = (x_2, y_2), y_1 \neq 0 \neq y_2$: $a + b := (x_1 + x_2, y_1 + y_2)$ if $y_1 + y_2 \neq 0$ and $x_1 + x_2$ otherwise.

*$a = (x,y), b \in \mathbb{R}$: As in 1.


You could make the same for multiplication. Anyway: This is stupid and useless.
