Algebraic Number/Examples/Root 2

Example of Algebraic Number

$\sqrt 2$ is an algebraic number.

Proof

We have that:

\(\ds x^2 - 2\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds \pm \sqrt 2\)

Quadratic Formula

Thus $\sqrt 2$ is a root of $x^2 - 2$.

Hence the result by definition of algebraic number.

$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $76$
• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): algebraic number