using the primitive element of biquadratic field
with rational coefficients.
If is cubed (http://planetmath.org/CubeOfANumber), the result no terms with :
Thus, if we subtract from this the product , the term vanishes:
Dividing this equation by () yields
Similarly, we have
The (2) and (3) may be interpreted as such polynomials as intended.
Multiplying the equations (2) and (3) we obtain a corresponding for the square root of which also lies in the quartic field :
For example, in the special case we have
Remark. The sum (1) of two square roots of positive squarefree integers is always irrational, since in the contrary case, the equation (3) would say that would be rational; this has been proven impossible here (http://planetmath.org/SquareRootOf2IsIrrationalProof).
|Title||using the primitive element of biquadratic field|
|Date of creation||2013-03-22 17:54:25|
|Last modified on||2013-03-22 17:54:25|
|Last modified by||pahio (2872)|
|Synonym||expressing two square roots with their sum|
|Synonym||irrational sum of square roots|