Algebraic numbers proof

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RDKGames

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Any hints on how I can start this?? I know that algebraic numbers are those which satisfy a polynomial equaling 0 but I'm not sure how to use it here at the moment.

ghostwalker

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(Original post by RDKGames)

Any hints on how I can start this?? I know that algebraic numbers are those which satisfy a polynomial equaling 0 but I'm not sure how to use it here at the moment.

Start with the definition of an algebraic number.

So, x is a root of some polynomial p(x)

How might you tweek that polynomial so that root(x) is a root?

RDKGames

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(Original post by ghostwalker)
Start with the definition of an algebraic number.

So, x is a root of some polynomial p(x)

How might you tweek that polynomial so that root(x) is a root?

So $a_1x^n+a_2x^{n-1}+a_3x^{n-2}+...+a_{n-1}x+a_n=0$ is true.

Would I do the mapping $x \mapsto \sqrt{x}$ ? It's what I had initially but it doesn't seem rigorous enough so I dropped it, can't see anything else other than that.

ghostwalker

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(Original post by RDKGames)
So $a_1x^n+a_2x^{n-1}+a_3x^{n-2}+...+a_{n-1}x+a_n=0$ is true.

Would I do the mapping $x \mapsto \sqrt{x}$ ? It's what I had initially but it doesn't seem rigorous enough so I dropped it, can't see anything else other than that.

I suspect you're on the right lines, but I don't know what you mean by do the mapping here.

RDKGames

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(Original post by ghostwalker)
I suspect you're on the right lines, but I don't know what you mean by do the mapping here.

Perhaps I worded it wrong, my mind is going blank today all over :confused:

I meant to say that I can map all

solutions onto their roots, so all x values transfer over to some function of itself, in this case the square root. This gives $a_1(\sqrt{x})^n+a_{2}(\sqrt{x})^ {n-1}+...+a_{n-1}(\sqrt{x})+a_n=0$ which would show that $\sqrt{x}$ is algebraic since it satisfies the polynomial.

ghostwalker

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(Original post by RDKGames)
Perhaps I worded it wrong, my mind is going blank today all over :confused:

I meant to say that I can map all

OK, I follow you, but you've gone the wrong way - you don't know root(x) is a root of that polynomial.

Here's an example:

The simpliest polynomial with 2 as a root is:

The simpliest poly. with root(2) as a root is:

How do you get the latter from the former.

Got to go out now.

math42

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Unless I'm missing something (possible, I'm a bit thick tbh), this seems fairly straightforward.
x^n = (x^(1/2))^(2n).

(Original post by RDKGames)

Any hints on how I can start this?? I know that algebraic numbers are those which satisfy a polynomial equaling 0 but I'm not sure how to use it here at the moment.

RDKGames

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(Original post by ghostwalker)
OK, I follow you, but you've gone the wrong way - you don't know root(x) is a root of that polynomial.

Here's an example:

The simpliest polynomial with 2 as a root is:

The simpliest poly. with root(2) as a root is:

How do you get the latter from the former.

Got to go out now.

Ah so $x \mapsto x^2$ and that gives a new polynomial with $\sqrt{x}$ as the root. How would this be adapted if I wanted to prove that

is also algebraic since for this you'd have to map

onto $\sqrt{x}$ but not all exponents would necessarily be integers if that happened.

(Original post by 13 1 20 8 42)
Unless I'm missing something (possible, I'm a bit thick tbh), this seems fairly straightforward.
x^n = (x^(1/2))^(2n).

I think that's exactly it since from that procedure you get another polynomial with root x as the solution. Can't believe something so simple evaded me so longer than it should've

Thanks.

B_9710

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is algebraic then $\exists P,$ a polynomial with rational coefficients s.t
$\displaystyle a_0+a_1x+\ldots +a_nx^n=0$ .
So if we let

then it is clear that $\sqrt x$ solves
$\displaystyle a_0+a_1y^2+\ldots +a_ny^{2n}=0$ , and this new polynomial clearly has the same coefficients as the first, so they're rational coefficient, so $\sqrt x$ is algebraic. Surely.

ghostwalker

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(Original post by RDKGames)
I think that's exactly it since from that procedure you get another polynomial with root x as the solution. Can't believe something so simple evaded me so longer than it should've

Looks like you've have it cracked now. Sorry my hints weren't hitting the mark :sad:

Obviously, this technique extends to show $\sqrt[r]{x}$ is algebraic $\forall r\in\mathbb{N}$

DFranklin

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#11

(Original post by RDKGames)
Ah so $x \mapsto x^2$ and that gives a new polynomial with $\sqrt{x}$ as the root. How would this be adapted if I wanted to prove that

is also algebraic since for this you'd have to map

onto $\sqrt{x}$ but not all exponents would necessarily be integers if that happened.

As I understand it going the other way constructively (finding the actual poly) is quite hard.

But if x is algebraic then [Q(x):Q] is finite, therefore so is [Q(x^2):Q] so you know a poly exists without haviNg to find it.

RDKGames

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#12

(Original post by DFranklin)
As I understand it going the other way constructively (finding the actual poly) is quite hard.

But if x is algebraic then [Q(x):Q] is finite, therefore so is [Q(x^2):Q] so you know a poly exists without haviNg to find it.

Does this approach fully prove it?? It is not finding any particular polynomials but it turned out more long-winded than I thought it would be. Thanks!

IrrationalRoot

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#13

(Original post by RDKGames)
Does this approach fully prove it?? It is not finding any particular polynomials but it turned out more long-winded than I thought it would be. Thanks!

...

Couldn't this all by replaced by:

Since

is algebraic, by definition there exists a nonzero polynomial

with rational coefficients such that

. Now consider the polynomial

. This is clearly also nonzero with rational coefficients, and $Q(\sqrt x)=P((\sqrt x)^2)=P(x)=0$ , so $\sqrt x$ is a root of

and we're done.

Pretty sure this was what others were getting at.

EDIT: Oh I see you're going the other way here. Well anyway here's a proof of the original way lol.

DFranklin

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#14

(Original post by RDKGames)
Does this approach fully prove it?? It is not finding any particular polynomials but it turned out more long-winded than I thought it would be. Thanks!

Looks OK. My feeling that it was hard came from the more general:

"Given algebraic x, y, with p(x) = 0 and q(y) = 0, find a polynomial r with r(xy) = 0".

I did suspect it wouldn't be quite so bad if you forced x = y.