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\line{\sevenrm a08.tex[257,phy] \today\hfill}

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Algebraic numbers  are defined as roots of polynomials in one variable with
integer coefficients.

\smallskip\noindent
{\bf Theorem.} The algebraic numbers are closed under the four arithmetic
operations.

\smallskip\noindent
{\bf Proof.} Let $\bar x$, $\bar y$ be roots of polynomials $P(x)$, $Q(y)$
respectively, and let $\bar z=R(\bar x,\bar y)$, for some polynomial~$R$.
We will show that $\bar z$ is algebraic. The polynomial $P(x)↑iQ(y)↑jR(x,y)↑k$
has total degree $pi+qj+rk$, where $p$, $q$, $r$, are the degrees of $P$,
$Q$, $R$ respectively. The polynomials in $x$ and~$y$ of total degree $≤d$
form a vector space of dimension $(d+1)(d+2)/2=O(d↑2)$. 
The number of triples $(i,j,k)$ with $pi+qj+rk≤d$ is $O(d↑3)$ (actually,
about $d↑3/6pqr$), so for large enough~$d$ there is a linear dependency,
giving a nonzero polynomial $f(u,v,w)$ such that $f\bigl(P(x),Q(y),R(x,y)\bigr)
≡0$. Substituting $\bar x$, $\bar y$ for $x$, $y$, we see
$f\bigl(P(\bar x),Q(\bar y),R(\bar x,\bar y)\bigr)=f(0,0,\bar z)$
[RWF: show nontrivial] $=0$, and
$f(0,0,z)$ is a polynomial satisfied by~$\bar z$, which is therefore algebraic.
Letting $R(x,y)$ be $x+y$, $x-y$, or~$xy$ shows closure of the algebraic
numbers under~$+$, $-$, and~$\times$; to show closure under reciprocation,
observe for $x≠0$,
$\sum↓{i=0}↑na↓ix↑i=0$ implies $\sum↓{i=0}↑na↓{n-i}(1/x)↑i=0$.

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\line{\copyright 1984 Robert W. Floyd\hfill}
\line{First draft (not published) March 28, l985.\hfill}

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