Algebraic Number/Examples/Cube Root of (2 plus Root 2)
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Example of Algebraic Number
- $\sqrt [3] {2 + \sqrt 2}$ is an algebraic number.
Proof
Let $x = \sqrt [3] {2 + \sqrt 2}$.
We have that:
\(\ds x^3\) | \(=\) | \(\ds 2 + \sqrt 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3 - 2\) | \(=\) | \(\ds \sqrt 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x^3 - 2}^2\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^6 - 4 x^3 + 4\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^6 - 4 x^3 + 2\) | \(=\) | \(\ds 0\) |
Thus $\sqrt [3] {2 + \sqrt 2}$ is a root of $x^6 - 4 x^3 + 2 = 0$.
Hence the result by definition of algebraic number.
$\blacksquare$
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental