The Complex Numbers Form a Field Containing the Real Numbers: i Squared Is -1 and Not Real, the Canonical Form a + bi and Its Arithmetic, and Numerals

In the setting of The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness, let $\mathbb{C}$, its sum $+$ and product $\cdot$, and $i$ be as in The Complex Numbers: Pairs of Real Numbers with Their Addition and Multiplication, and the Imaginary Unit §complex, The Complex Numbers: Pairs of Real Numbers with Their Addition and Multiplication, and the Imaginary Unit §operations and The Complex Numbers: Pairs of Real Numbers with Their Addition and Multiplication, and the Imaginary Unit §imaginary-unit. Negatives and reciprocals in $\mathbb{C}$ are as in Commutative Rings, Fields and Ordered Fields: Standard Notation §rings and Commutative Rings, Fields and Ordered Fields: Standard Notation §fields, which apply by the clause field.

¶ $\mathbb{C}$, with $+$, $\cdot$, $(0,0)$ and $(1,0)$, is a field. The negative of $(a,b)\in\mathbb{C}$ is $-(a,b)=(-a,-b)$; and if $(a,b)\neq(0,0)$, then $a^{2}+b^{2}\neq0$ and

$$(a,b)^{-1}=\Big(\frac{a}{a^{2}+b^{2}},\,\frac{-b}{a^{2}+b^{2}}\Big).$$

¶ The map $\iota:\mathbb{R}\to\mathbb{C}$, $a\mapsto(a,0)$, is injective, $\iota(0)=(0,0)$ and $\iota(1)=(1,0)$, and for all $a,b\in\mathbb{R}$

$$\iota(a+b)=\iota(a)+\iota(b),\qquad \iota(ab)=\iota(a)\cdot\iota(b),\qquad \iota(-a)=-\iota(a);$$

if moreover $a\neq0$, then $\iota(a)\neq(0,0)$ and $\iota(a^{-1})=\iota(a)^{-1}$.

¶ $i\cdot i=-\iota(1)$.

¶ $i\neq\iota(a)$ for every $a\in\mathbb{R}$.

¶ For all $a,b\in\mathbb{R}$, $(a,b)=\iota(a)+\iota(b)\cdot i$. Hence for every $z\in\mathbb{C}$ there is exactly one pair $(a,b)$ of real numbers with $z=\iota(a)+\iota(b)\cdot i$.

¶ For all $a,b,c,d\in\mathbb{R}$,

$$\big(\iota(a)+\iota(b)\cdot i\big)+\big(\iota(c)+\iota(d)\cdot i\big)=\iota(a+c)+\iota(b+d)\cdot i,\qquad \big(\iota(a)+\iota(b)\cdot i\big)\cdot\big(\iota(c)+\iota(d)\cdot i\big)=\iota(ac-bd)+\iota(ad+bc)\cdot i.$$

¶ For every $n\in\mathbb{N}_{0}$, let $n_{\mathbb{R}}$ and $n_{\mathbb{C}}$ be the images of $n$ in $\mathbb{R}$ and in $\mathbb{C}$, the latter with $+$, $\cdot$, $(0,0)$ and $(1,0)$ as in the clause field. Then $n_{\mathbb{C}}=\iota(n_{\mathbb{R}})$.