Algebraic Number/Examples/Root 2

Example of Algebraic Number

$\sqrt 2$ is an algebraic number.

We have that:

$\ds x^2 - 2$

$=$

$\ds 0$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \pm \sqrt 2$

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

Hence the result by definition of algebraic number.

$\blacksquare$

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
2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): algebraic number