Algebraic Number/Examples/Root 3 plus Root 2

Example of Algebraic Number

$\sqrt 3 + \sqrt 2$ is an algebraic number.

Let $x = \sqrt 3 + \sqrt 2$.

We have that:

$\ds x - \sqrt 2$

$=$

$\ds \sqrt 3$

$\ds \leadsto \ \ $

$\ds \paren {x - \sqrt 2}^2$

$=$

$\ds 3$

$\ds \leadsto \ \ $

$\ds x^2 - 2 \sqrt 2 x + 2$

$=$

$\ds 3$

$\ds \leadsto \ \ $

$\ds x^2 - 1$

$=$

$\ds 2 \sqrt 2 x$

$\ds \leadsto \ \ $

$\ds \paren {x^2 - 1}^2$

$=$

$\ds 8 x^2$

$\ds \leadsto \ \ $

$\ds x^4 - 2 x^2 + 1$

$=$

$\ds 8 x^2$

$\ds \leadsto \ \ $

$\ds x^4 - 10 x^2 + 1$

$=$

$\ds 0$

Thus $\sqrt 3 + \sqrt 2$ is a root of $x^4 - 10 x^2 + 1 = 0$.

Hence the result by definition of algebraic number.

$\blacksquare$

1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $47 \ \text {(a)}$