Rational Number is Algebraic of Degree 1

Theorem

Let $r \in \Q$ be a rational number.

Then $r$ is an algebraic number of degree $1$.

Proof

Let $r$ be expressed in the form:

$r = \dfrac p q$

By Rational Number is Algebraic, $r$ can be expressed as the root of the linear function:

$q x - p = 0$

A polynomial of degree zero is a constant polynomial, and has no roots.

Hence the result by definition of degree of algebraic number.

$\blacksquare$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental