# Algebraic numbers proof

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#1

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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4 years ago
#2
(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.

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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#3
(Original post by ghostwalker)

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

How might you tweek that polynomial so that root(x) is a root?
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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4 years ago
#4
(Original post by 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.
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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#5
(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

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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4 years ago
#6
(Original post by 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.
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.
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4 years ago
#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 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.
0
#8
(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 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.

(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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4 years ago
#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.
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4 years ago
#10
(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
Obviously, this technique extends to show is algebraic
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4 years ago
#11
(Original post by 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.
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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#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!

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4 years ago
#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 , 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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4 years ago
#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.
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