roots of polynomial with real coeff (1 Viewer)

[Sirius Black]

If (2-sqrt3) is a root for a polynomial with real coefficients, is 2+sqrt3 also a root
I thought this rule only applied to complex root and its conjugate right? but the solution says it it the way i stated above, which means (2+sqrt3) is a conj. of
(2-sqrt3)?!
Could anyone help?

 

[Antwan23q]

well if the coefficients are integers, then it has to be the conjugate, i think.

 

[acmilan]

I thought only complex numbers could have conjugates? The conjugate of a real number, in this case 2-sqrt3, is that same real number. Well its what I thought anyways.

 

[gman03]

well if the coefficients are integers, then it has to be the conjugate, i think.

I think it works for rational coefficient as well....

I know two types of conjugate - complex and the surd

 

[Antwan23q]

well i dont think they call it a conjugate. i thought it was that as well, but I had to do a question and all i had to go on was that a polynomial had 3 degrees, monic, coefficents were integers and u had a root like that, and ud to find the equation. and they only way u could is by saying that it was that the conjugate was the other root

 

[richz]

its just a rule learn it

 

[who_loves_maths]

Originally Posted by Sirius Black
If (2-sqrt3) is a root for a polynomial with real coefficients, is 2+sqrt3 also a root
I thought this rule only applied to complex root and its conjugate right? but the solution says it it the way i stated above, which means (2+sqrt3) is a conj. of
(2-sqrt3)?!
Could anyone help?

Originally Posted by acmilan
I thought only complex numbers could have conjugates? The conjugate of a real number, in this case 2-sqrt3, is that same real number. Well its what I thought anyways.

hi guys,

the rule that if (2-sqrt3) is a root then so is (2+sqrt3), is applicable to only polynomials with RATIONAL coefficients, just like how the complex conjugate root theorem applies to polynomials with REAL coefficients.
(the reason isn't actually very complicated, it's simply a manifestation of the properties of numbers fields and number sets. the fact that the real numbers are a subset of the complex numbers, and that the rational numbers are a subset of the real numbers is the underlying symmetry behind this.)

as with the meaning and definition of what a 'conjugate' is, the thing to remember is that in the 4u course, we use the term conjugate to apply to complex numbers - but that is a gross oversimplification of the correct name complex conjugates.
so with surds, you have real conjugates.

in general, conjugates are simply numbers over a certain number field that do not exist in sub-number fields or subsets of that fields, and which when added or multiplied generate numbers that are exclusively within a smaller sub-field or subset. {eg. complex conjugates give you real numbers, and real conjugates give you rational numbers ---> (2-sqrt3 + 2+sqrt3) = 4, and, (2-sqrt3)(2+sqrt3) = 1}

it's because the surds and the 'i's exhibit similar (or mathematically equivalent) properties that they both apply to polynomials with rational coefficients.

of course it's best to, like what xrtzx said in his/her post, just learn it as a rule.

P.S. you can actually prove the conjugate result for the surds in general over a polynomial with rational coefficients in the same way you would prove the complex conjugate theorem for the complex numbers.

 

[who_loves_maths]

i just want to add something to my last post for clarity:

when i mentioned real conjugates and how the "real conjugate theorem" exists for polynomials with rational coefficients, it's important to realise that the 'real' numbers i was referring to are in the form of surds. (exact forms, or exact values).

that is, the theorem does not work for real numbers such as 'pi' or 'e', but this is NOT due to the incompleteness of the theorem, in fact there is good reason why real numbers that cannot be expressed in exact form (ie. root extractions with the four operations) are not a part of this theorem/result - it is because those numbers are transcendental.
and the thing with ALL transcendental numbers is that they cannot be roots of any polynomial of ANY degree with rational coefficients.
hence, the 'real conjugate root theorem' for polynomials with rational coefficients is a complete theorem on its own - since the roots, should they be real, are never transcendental numbers like 'pi' or 'e', and are ALWAYS in surd form.

from this, of course, follows the result that ALL polynomials of ODD degree with rational coefficients must have at least ONE rational root.
[this is equivalent to the result that polynomials of odd degree with real coefficients must have at least one real root.]

hope i cleared a few things in my last post here

 

[Sirius Black]

I thought only complex numbers could have conjugates? The conjugate of a real number, in this case 2-sqrt3, is that same real number. Well its what I thought anyways.

that's what i thought too-real number has a conj of itself :uhhuh:

 

[Sirius Black]

i just want to add something to my last post for clarity:

when i mentioned real conjugates and how the "real conjugate theorem" exists for polynomials with rational coefficients, it's important to realise that the 'real' numbers i was referring to are in the form of surds. (exact forms, or exact values).

that is, the theorem does not work for real numbers such as 'pi' or 'e', but this is NOT due to the incompleteness of the theorem, in fact there is good reason why real numbers that cannot be expressed in exact form (ie. root extractions with the four operations) are not a part of this theorem/result - it is because those numbers are transcendental.
and the thing with ALL transcendental numbers is that they cannot be roots of any polynomial of ANY degree with rational coefficients.
hence, the 'real conjugate root theorem' for polynomials with rational coefficients is a complete theorem on its own - since the roots, should they be real, are never transcendental numbers like 'pi' or 'e', and are ALWAYS in surd form.

from this, of course, follows the result that ALL polynomials of ODD degree with rational coefficients must have at least ONE rational root.
[this is equivalent to the result that polynomials of odd degree with real coefficients must have at least one real root.]

hope i cleared a few things in my last post here

wow cool i got it now....thanx a lot!