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.
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ghostwalker
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(Original post by 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.
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?
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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.
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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.
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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 x>0 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.
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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 x>0 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.
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:
x-2=0

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

x^2-2=0

How do you get the latter from the former.

Got to go out now.
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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)
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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.
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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:
x-2=0

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

x^2-2=0

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 x^2 is also algebraic since for this you'd have to map x 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.
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B_9710
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If  x 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  x=y^2 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.
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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}
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DFranklin
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(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 x^2 is also algebraic since for this you'd have to map x 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.
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RDKGames
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(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!

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IrrationalRoot
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(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 x is algebraic, by definition there exists a nonzero polynomial P(t) with rational coefficients such that P(x)=0. Now consider the polynomial Q(t)=P(t^2). 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 Q(t) 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.
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DFranklin
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(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.
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