- #1

- 424

- 25

Here is an example:

x = a

^{1/2}+ b

^{1/3}

( x - a

^{1/2}) = b

^{1/3}

( x - a

^{1/2})

^{3}= b

I can't figure out how to continue to de-root

*a*.

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- I
- Thread starter swampwiz
- Start date

- #1

- 424

- 25

Here is an example:

x = a

( x - a

( x - a

I can't figure out how to continue to de-root

- #2

- 15,952

- 14,421

In order to get the polynomial, write ##(x-a)p(x)=0## and ##(x-b)q(x)=0.## Then ##(x-(a+b)/2)p(x)q(x)=0## and with ##(a+b)/2## algebraic, we have ##a+b## algebraic, too. But you will need the other roots to set up ##p(x)## and ##q(x)##. That's the only way I see.

- #3

martinbn

Science Advisor

- 2,618

- 994

Expand, solve for root a and square.

Here is an example:

x = a^{1/2}+ b^{1/3}

( x - a^{1/2}) = b^{1/3}

( x - a^{1/2})^{3}= b

I can't figure out how to continue to de-roota.

##

x^3-3x^2\sqrt{a}+3ax-a\sqrt{a} = b

##

##

\sqrt{a}(3x^2+a) = x^3+3ax-b

##

##

a(3x^2+a)^2 = (x^3+3ax-b)^2

##

- #4

- 424

- 25

x = a

- #5

- 15,952

- 14,421

Why?I think I was looking at how to do this for the general case of any number of root terms.

- #6

- 424

- 25

I'd to prove that any value made of arithmetic & root operations can generate a polynomial that has a solution equal to that value - i.e., this would prove that this type of value is an algebraic number.Why?

- #7

- 15,952

- 14,421

This makes no sense. You do not need to know a polynomial for the proof. And without a specific example, you cannot calculate one. You ask for one algorithm that works for any field, for any number of algebraic elements. No way.I'd to prove that any value made of arithmetic & root operations can generate a polynomial that has a solution equal to that value - i.e., this would prove that this type of value is an algebraic number.

- #8

- 424

- 25

If it can be done for any number, then surely there must be some systematic way to do it.This makes no sense. You do not need to know a polynomial for the proof. And without a specific example, you cannot calculate one. You ask for one algorithm that works for any field, for any number of algebraic elements. No way.

- #9

- 15,952

- 14,421

What is any number? Heck, what is even a number? Which fields? Separable extensions? Known Galois group, in case it is Galois? Determine your splitting fields, and calculate all roots. Then apply what I have written in post #2 or proceed as @martinbn suggested in post #3. All you have in the end is Viète.If it can be done for any number, then surely there must be some systematic way to do it.

Your question comes down to: Given any algebraic field extension ##L/K## and an element ##a\in L.## Find its minimal polynomial over ##K##. This means: calculate all powers of ##a## up to ##\dim_K L## and reassemble them with coefficients in ##K## to get ##0.##

That was done in post #3.

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- #10

martinbn

Science Advisor

- 2,618

- 994

There are nonconstructive proves, there doesn't have to be a systematic way to do it.If it can be done for any number, then surely there must be some systematic way to do it.

On the other hand the first result after a search is this one

https://math.stackexchange.com/ques...product-of-two-algebraic-numbers-is-algebraic

- #11

- 424

- 25

So what you are saying is:

In order to get the polynomial, write ##(x-a)p(x)=0## and ##(x-b)q(x)=0.## Then ##(x-(a+b)/2)p(x)q(x)=0## and with ##(a+b)/2## algebraic, we have ##a+b## algebraic, too. But you will need the other roots to set up ##p(x)## and ##q(x)##. That's the only way I see.

Because

0 = ( x - a ) = ( x - b )

then for any polynomials

OK, that's a nice little proof that gets me over the hump!

In order to get the polynomial, write ##(x-a)p(x)=0## and ##(x-b)q(x)=0.## Then ##(x-(a+b)/2)p(x)q(x)=0## and with ##(a+b)/2## algebraic, we have ##a+b## algebraic, too. But you will need the other roots to set up ##p(x)## and ##q(x)##. That's the only way I see.

- #12

- 15,952

- 14,421

???So what you are saying is:

Becausea&bare each algebraic

0 = ( x - a ) = ( x - b )

I mean: because ##\alpha,\beta ## are algebraic, ##K\subseteq K(\alpha ,\beta )## is algebraic. And since ##\alpha +\beta \in K(\alpha ,\beta )## is also algebraic. I would do it in steps: ##K\subseteq K(\alpha )\subseteq K(\alpha ,\beta )##.

then for any polynomialsp( x )&q( x )

OK, that's a nice little proof that gets me over the hump!

- #13

- 424

- 25

I had accidentally posted before finishing the post.???

I mean: because ##\alpha,\beta ## are algebraic, ##K\subseteq K(\alpha ,\beta )## is algebraic. And since ##\alpha +\beta \in K(\alpha ,\beta )## is also algebraic. I would do it in steps: ##K\subseteq K(\alpha )\subseteq K(\alpha ,\beta )##.

0 = ( x - a ) = ( x - b ) = ( x - a ) p( x ) = ( x - b ) q( x ) = ( x - a ) p( x ) q( x ) = ( x - b ) p( x ) q( x )

Summing half of each

0 = ( x - ( a + b ) / 2 ) p( x ) q( x )

thus ( ( a + b ) / 2 ) and as well ( a + b ) must be algebraic.

- #14

- 424

- 25

OK, I think I know what I mean by "an algorithm". If I could prove that any addition, multiplication or root operation having algebraic numbers as arguments results in an algebraic number, then I'm done.

I will now prove why a power, root or product of algebraic numbers is an algebraic number.

GIVEN: a, b -> ALGEBRAIC

THUS: ( a + b ) -> ALGEBRAIC

THUS: ∃ [ f( x ) = 0 = ( x - a ) p( x ) ] &

[ g_{2}( x ) = 0 = ( x + a ) ( x - a ) p( x ) = ( x^{2} - a^{2} ) p( x )

[ g_{3}( x ) = 0 = ( x^{2} + a x + a^{2} ) ( x - a ) p( x ) = ( x^{3} - a^{3} ) p( x )

NOTE: The root operation should be presumed to be a consistent deMoivre root index.

[ g_{1/2}( x ) = 0 = ( x^{1/2} - a^{1/2} ) ( x^{1/2} + a^{1/2} ) p( x ) = ( x^{1/2} - a^{1/2} ) p_{1/2}( x )

[ g_{1/3}( x ) = 0 = ( x^{1/3} - a^{1/3} ) ( x^{2/3} + a^{1/3} x^{1/3} + a^{2/3} ) p( x ) = ( x^{1/3} - a^{1/3} ) p_{1/3}( x )

LET: u = x^{n}

THUS: 0 = ( u - a^{n} ) p( x( u ) )

THUS: a^{n} -> ALGEBRAIC

LET: u = x^{( 1 / n )}

THUS: 0 = ( u - a^{( 1 / n )} ) p_{1/n}( x( u ) )

THUS: a^{( 1 / n )} -> ALGEBRAIC

( a b ) = [ ( a + b )^{2} - ( a^{2} + b^{2} ) ] => SUM OF POWERS OF ALGEBRAICS => SUM OF ALGEBRAICS

THUS: ( a b ) -> ALGEBRAIC

QED

I will now prove why a power, root or product of algebraic numbers is an algebraic number.

GIVEN: a, b -> ALGEBRAIC

THUS: ( a + b ) -> ALGEBRAIC

THUS: ∃ [ f( x ) = 0 = ( x - a ) p( x ) ] &

[ g

[ g

NOTE: The root operation should be presumed to be a consistent deMoivre root index.

[ g

[ g

LET: u = x

THUS: 0 = ( u - a

THUS: a

LET: u = x

THUS: 0 = ( u - a

THUS: a

( a b ) = [ ( a + b )

THUS: ( a b ) -> ALGEBRAIC

QED

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