The Field of Rational Functions

We will now look at yet another special field. Consider the set of rational functions denoted $R(x)$ whose elements are functions in the form $\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are real valued polynomials. This set under the operations of standard addition and standard multiplication forms a field, which we will prove.

Let $f_1, f_2, f_3 \in R(x)$ such that $f_1 = \frac{p_1}{q_1}$, $f_2 = \frac{p_2}{q_2}$ and $f_3 = \frac{p_3}{q_3}$.

• FA1: $f_1 + f_2 = \frac{p_1}{q_1} + \frac{p_2}{q_2} = \frac{p_1q_2 + p_2q_1}{q_1q_2}$ and so $f_1 + f_2 \in R(x)$.

• FA2: $f_1 + f_2 = \frac{p_1q_2 + p_2q_1}{q_1q_2} = \frac{p_2q_1 + p_1q_2}{q_2q_1} = \frac{p_2}{q_2} + \frac{p_1}{q_1} = f_2 + f_1$.

• FA3: $f_1 + (f_2 + f_3) = \frac{p_1}{q_1} + \frac{p_2q_3 + p_3q_2}{q_2q_3} + \frac{p_1q_2q_3 + q_1p_2q_3 + q_1p_3q_2}{q_1q_2q_3} = \frac{p_1q_2 + p_2q_1}{q_1q_2} + \frac{p_3}{q_3} = (f_1 + f_2) + f_3$

• FA4: The additive identity is the rational function $f_0 = 0$, that is $f_1 + f_0 = \frac{p_1}{q_1} + 0 = f_1$.

• FA5: The additive inverse of the rational function $f_1$ is $-f_1$, that is $f_1 + (-f_1) = \frac{p_1}{q_1} - \frac{p_1}{q_1} = 0$.

• FM1: $f_1 \cdot f_2 = \frac{p_1}{q_1} \cdot \frac{p_2}{q_2} = \frac{p_1p_2}{q_1q_2}$ and so $f_1 \cdot f_2 \in R(x)$.

• FM2: $f_1 \cdot f_2 = \frac{p_1p_2}{q_1q_2} = \frac{p_2p_1}{q_2q_1} = f_2 \cdot f_1$.

• FM3: $f_1 \cdot (f_2 \cdot f_3) = \frac{p_1}{q_1} \cdot \left [ \frac{p_2p_3}{q_2q_3}\right ] = \frac{p_1[p_2p_3]}{q_1[q_2q_3]} = \frac{[p_1p_2]p_3}{[q_1q_2]q_3} = \left [ \frac{p_1p_2}{q_1q_2} \right ] \cdot \frac{p_3}{q_3} = (f_1 \cdot f_2) \cdot f_3$.

• FM4: The multiplicative identity is the rational function $f_I = 1$, that is $f_1 \cdot f_I = \frac{p_1}{q_1} \cdot 1 = f_1$.

• FM5: The multiplicative inverse of the rational function $f_1$ is $f_1^{-1} := \frac{q_1}{p_1}$, that is $f_1 \cdot f_2 = \frac{p_1}{q_1} \cdot \frac{q_1}{p_1} = \frac{p_1q_1}{p_1q_1} = 1 = f_I$.

• FD: $f_1 (f_2 + f_3) = \frac{p_1}{q_1} \cdot \left ( \frac{p_2q_3 + p_2q_3}{q_2q_3}\right) = \frac{p_1p_2q_3 + p_1p_2q_3}{q_1q_2q_3} = \frac{p_1q_2 + p_2q_1}{q_1q_2} + \frac{p_1q_3 + p_3q_1}{q_1q_3} = f_1 \cdot f_2 + f_1 \cdot f_3$.

Since $R(x)$ satisfies all eleven field axioms, then the set $R(x)$ paired with the operations of addition and multiplication form a field.