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
(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.
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 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?
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

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#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.
So

Would I do the mapping

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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.
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



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

I meant to say that I can map all



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).
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.
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 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.
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 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).
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




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#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
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
![\sqrt[r]{x} \sqrt[r]{x}](https://www.thestudentroom.co.uk/latexrender/pictures/94/94ab931c84fc2837dfe49af318c139a7.png)

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#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.
Ah so





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 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.
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 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!
...
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







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 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!
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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