Algebraic Number/Examples/Cube Root of 2 plus Root 3
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Example of Algebraic Number
- $\sqrt [3] 2 + \sqrt 3$ is an algebraic number.
Proof
Let $x = \sqrt [3] 2 + \sqrt 3$.
We have that:
\(\ds x - \sqrt 3\) | \(=\) | \(\ds \sqrt [3] 2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x - \sqrt 3}^3\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x - \sqrt 3} \paren {x - \sqrt 3}^2\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x - \sqrt 3} \paren {x^2 - 2 x \sqrt 3 + 3}\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3 - 3 \sqrt 3 x^2 + 9 x - 3 \sqrt 3\) | \(=\) | \(\ds 2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^3 + 9 x - 2\) | \(=\) | \(\ds 3 \sqrt 3 x^2 + 3 \sqrt 3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 3 \sqrt 3 \paren {x^2 + 1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x^3 + 9 x - 2}^2\) | \(=\) | \(\ds 3^2 {\sqrt 3}^2 \paren {x^2 + 1}^2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^6 + 18 x^4 - 4 x^3 + 81 x^2 - 36 x + 4\) | \(=\) | \(\ds 27 \paren {x^4 + 2 x^2 + 1}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 27 x^4 + 54 x^2 + 27\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23\) | \(=\) | \(\ds 0\) |
Thus $\sqrt [3] 2 + \sqrt 3$ is a root of a non-zero polynomial with rational coefficients, namely $x^6 - 9 x^4 - 4 x^3 + 27 x^2 - 36 x - 23 = 0$.
Hence the result by definition of algebraic number.
$\blacksquare$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $158 \ \text{(a)}$