# Algebraic numbers proof

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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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#2

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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.

**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.

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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(Original post by

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?

**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?

Would I do the mapping ? 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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#4

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So is true.

Would I do the mapping ? 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.

**RDKGames**)So is true.

Would I do the mapping ? 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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(Original post by

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

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

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 which would show that is algebraic since it satisfies the polynomial.

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#6

(Original post by

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

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 which would show that is algebraic since it satisfies the polynomial.

**RDKGames**)Perhaps I worded it wrong, my mind is going blank today all over

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 which would show that is algebraic since it satisfies the 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.**

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#7

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

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.

x^n = (x^(1/2))^(2n).

**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.

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(Original post by

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.

**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.**
(Original post by

Unless I'm missing something (possible, I'm a bit thick tbh), this seems fairly straightforward.

x^n = (x^(1/2))^(2n).

**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).

Thanks.

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#9

If is algebraic then a polynomial with rational coefficients s.t

.

So if we let then it is clear that solves

, and this new polynomial clearly has the same coefficients as the first, so they're rational coefficient, so is algebraic. Surely.

.

So if we let then it is clear that solves

, and this new polynomial clearly has the same coefficients as the first, so they're rational coefficient, so is algebraic. Surely.

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#10

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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

**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

Obviously, this technique extends to show is algebraic

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#11

(Original post by

Ah so and that gives a new polynomial with 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 but not all exponents would necessarily be integers if that happened.

**RDKGames**)Ah so and that gives a new polynomial with 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 but not all exponents would necessarily be integers if that happened.

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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(Original post by

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.

**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.

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#13

(Original post by

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!

...

**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!

...

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 , so 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.

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#14

(Original post by

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!

**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!

"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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