seanieg89
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- 2007
Re: HSC 2013 4U Marathon
The main question is why a given algebraic number must be the root of a unique minimal (with respect to degree) monic irreducible polynomial with rational coeffs? (note we need uniqueness as well as existence here, think about why.)
If you resolve this question, it is clear that the minimal polynomial of this particular root must have degree at least 2 (else it would be rational). And therefore the polynomial you have constructed is the "minimal polynomial" of p+sqrt(q). The rest of the argument is fine.
My argument is less general but works well here. (And is what I hoped a student would find by thinking of the complex conjugate root theorem and trying to mimic it.)
At the beginning I should probably have proven that two numbers of that form are equal iff their rational and surdic parts are equal, which is easy to do but am tired and lazy now. Other than that it is a pretty rigorous argument.
Its pretty good, certainly everything you have said is true and this method extends nicely to more complicated situations.Can you please criticise this proof? I am starting to have doubts about it.
===
Start with the polynomial,
We need to make the lowest reducible rational polynomial that has that as a root, so we simply:
The above polynomial has the root
It is the simplest most reducible rational polynomial that has p + root(q) as a root.
Now consider:
Where a_1, a_2 ... are all rational polynomial functions of lowest reducibility.
If is a root then among the series a_1, a_2... there must exist a_k for some k such that is a root.
Since a_k is irreducible and rational, then that polynomial must be:
And that polynomial also has the root
Therefore that is also a root of P(x).
Is there a gap or something in this? Or is it all good?
The main question is why a given algebraic number must be the root of a unique minimal (with respect to degree) monic irreducible polynomial with rational coeffs? (note we need uniqueness as well as existence here, think about why.)
If you resolve this question, it is clear that the minimal polynomial of this particular root must have degree at least 2 (else it would be rational). And therefore the polynomial you have constructed is the "minimal polynomial" of p+sqrt(q). The rest of the argument is fine.
My argument is less general but works well here. (And is what I hoped a student would find by thinking of the complex conjugate root theorem and trying to mimic it.)
At the beginning I should probably have proven that two numbers of that form are equal iff their rational and surdic parts are equal, which is easy to do but am tired and lazy now. Other than that it is a pretty rigorous argument.