# The Field of Real and Complex Numbers

Recall from the Fields page that a field is set $F$ with two binary operations $+$ and $*$ that satisfy the ring axioms and the following additional axioms:

- For every $a \in F \setminus \{ 0 \}$ there exists an element $a^{-1} \in F$ such that $a * a^{-1} = 1$ and $a^{-1} * a = 1$ (The existence of inverses under $*$).

- For all $a, b \in F$ we have that $a * b = b * a$ (Commutativity of $*$).

On The Ring of Real and Complex Numbers page we saw that the set of real numbers $\mathbb{R}$ and the set of complex numbers $\mathbb{C}$ under the operations of standard addition $+$ and multiplication $*$ form rings. We will now verify that further $(\mathbb{R}, +, *)$ and $(\mathbb{C}, +, *)$ are fields.

## The Field of Real Numbers

For every element $a \in \mathbb{R} \setminus \{ 0 \}$ we note that $a^{-1} = \frac{1}{a} \in \mathbb{R}$ and:

(1)So for every nonzero real number $a$ the multiplicative inverse of $a$ is the real number $\frac{1}{a}$.

Furthermore, we already know from our experience with the real numbers that multiplication is commutative, so $(\mathbb{R}, +, *)$ is a field.

## The Field of Complex Numbers

Let $z \in \mathbb{C}$. Then $z = a + bi$ where $a, b \in \mathbb{R}$ and $i^2 = -1$. We want to find $z^{-1} \in \mathbb{C}$ such that $z * z^{-1} = 1$ and $z^{-1} * z = 1$. First consider:

(3)Clearly $z^{-1}$ satisfies $z * z^{-1} = 1$ and $z^{-1} * z = 1$, but is $z^{-1} \in \mathbb{C}$? The answer is yes. If we rationalize the denominator we see that:

(4)Now let $y, z \in \mathbb{C}$ where $z = a + bi$ and $y = c + di$ for $a, b, c, d \in \mathbb{R}$. Then:

(5)And:

(6)Therefore $y * z = z * y$. Hence $(\mathbb{C}, +, *)$ is a field.