The Field of Real and Complex Numbers

# 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)
\begin{align} \quad a * a^{-1} = a * \frac{1}{a} = 1 \end{align}
(2)
\begin{align} \quad a^{-1} * a = \frac{1}{a} * a = 1 \end{align}

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)
\begin{align} \quad z^{-1} = \frac{1}{a + bi} \end{align}

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)
\begin{align} \quad z^{-1} = \frac{1}{a + b\sqrt{-1}} \cdot \frac{a - b \sqrt{-1}}{a - b \sqrt{-1}} = \frac{a - bi}{a^2 + b^2} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i \end{align}

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)
\begin{align} \quad y * z = (a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i \end{align}

And:

(6)
\begin{align} \quad z * y = (c + di)(a + bi) = ca + cbi + dai + dbi^2 = (ca - db) + (cb + da)i = (ac - bd) + (ad + bc)i \end{align}

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