using the primitive element of biquadratic field
Let m and n be two distinct squarefree integers ≠1. We want to express their square roots as polynomials of
α:=√m+√n | (1) |
with rational coefficients.
If α is cubed (http://planetmath.org/CubeOfANumber), the result no terms with √mn:
α3 | =(√m)3+3(√m)2√n+3√m(√n)2+(√n)3 | ||
=m√m+3m√n+3n√m+n√n | |||
=(m+3n)√m+(3m+n)√n |
Thus, if we subtract from this the product (3m+n)α, the √n term vanishes:
α3-(3m+n)α=(-2m+2n)√m |
Dividing this equation by -2m+2n (≠0) yields
√m=α3-(3m+n)α2(-m+n). | (2) |
Similarly, we have
√n=α3-(m+3n)α2(m-n). | (3) |
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 mn which also lies in the quartic field ℚ(√m,√n)=ℚ(√m+√n):
√mn=α6-4(m+n)α4+(3m2+10mn+3n2)α24(-m2+2mn-n2) |
For example, in the special case m:=2,n:=3 we have
√2=α3-9α2,√3=-α3-11α2,√6=-α6+20α4-99α24. |
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 √n would be rational; this has been proven impossible here (http://planetmath.org/SquareRootOf2IsIrrationalProof).
Title | using the primitive element of biquadratic field |
---|---|
Canonical name | UsingThePrimitiveElementOfBiquadraticField |
Date of creation | 2013-03-22 17:54:25 |
Last modified on | 2013-03-22 17:54:25 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 12 |
Author | pahio (2872) |
Entry type | Application |
Classification | msc 11R16 |
Synonym | expressing two square roots with their sum |
Synonym | irrational sum of square roots |
Related topic | BinomialTheorem |