Appendix : The set of algebraic numbers

The set is closed

If $\alpha,\beta\in \overline{\mathbb{Q}}$, then $\alpha+\beta,\alpha-\beta,\alpha\beta$ and $\alpha/\beta\in \overline{\mathbb{Q}}$.

[ proof ]
Let minimal polynomials of $\alpha,\beta\,(\neq 0)$ over $\mathbb{Q}$ be $A(X),B(X)$ respectively, where

$$\begin{eqnarray*} A(X)=X^m+a_{m\!-\!1}X^{m\!-\!1}+\cdots+a_1 X+a_0\\ (a_0(\ne... ...ts+b_1 X+b_0\\ (b_0(\neq\!0),b_1,\cdots,b_{n-1}\in\mathbb{Q}). \end{eqnarray*}$$

And let

$$\begin{eqnarray*} &&\{\alpha^p\beta^q\}=\{\lambda_j\}_{j=1}^{N},\\ &&\text{wh... ...=0,1,\cdots,m-1,\\ q=0,1,\cdots,n-1,\\ N=mn. \end{cases}\\ \end{eqnarray*}$$

For $A(\alpha)=0,B(\beta)=0$,

$$\begin{eqnarray*} &&\alpha^m=-a_{m\!-\!1}\alpha^{m\!-\!1}-\cdots-a_1 \alpha-a_0,\\ &&\beta^n=-b_{n-1}\beta^{n-1}-\cdots-b_1 \beta-b_0. \end{eqnarray*}$$

So $(\alpha+\beta)\lambda_k$ is a linear combination of $\{\lambda_j\}$ over $\mathbb{Q}$, that is

$$\forall \lambda_k,\; \exists C_{k1},\cdots,C_{kN}\in\mathb... ...lpha+\beta)\lambda_k=C_{k1}\lambda_1+\cdots+C_{kN}\lambda_{N}.$$

Hence by moving the term $(\alpha+\beta)\lambda_k$ to the other side,

$$\left\{\begin{array}{r@{\,}r@{\,}r@{\,}r@{}r} (C_{11}\!-\!... ..._{NN}\!-\!(\alpha\!+\!\beta))\lambda_N&=0. \end{array}\right.$$

For $\lambda_k\neq 0$,

$$\left\vert\begin{array}{cccc} C_{11}\!-\!(\alpha\!+\!\beta... ...ots &C_{NN}\!-\!(\alpha\!+\!\beta) \end{array}\right\vert=0.$$

Therefore $\alpha+\beta\in \overline{\mathbb{Q}}$. Similarly $\alpha\beta\in \overline{\mathbb{Q}}$. And for

$$\begin{eqnarray*} &&(-1)^n(-\beta)^n+(-1)^{n-1}b_{n-1}(-\beta)^{n-1} +\cdots+(... ...}\right)^{n-1} +\cdots+b_{n-1}\left(\frac{1}{\beta}\right)+1=0, \end{eqnarray*}$$

then $-\beta,1/\beta\in \overline{\mathbb{Q}}$, that is, $\alpha-\beta,\alpha/\beta\in \overline{\mathbb{Q}}$. [ Q.E.D. ]

The proof shows that with respect to an algebraic degree,

• if $\deg\beta=n$, then $$\deg(-\beta)=\deg(1/\beta)=n$$,
• if $\deg\alpha=m,\deg\beta=n$, then $\deg(\alpha+\beta),\deg(\alpha\beta)\leqq mn$.

The set can not be expanded

An algebraic number over $\overline{\mathbb{Q}}$ is the one over $\mathbb{Q}$.

[ proof ]
Let $\xi$ be an algebraic number of degree 2 over $\overline{\mathbb{Q}}$, then

$$\begin{eqnarray*} &&\exists\alpha,\beta\in\overline{\mathbb{Q}},\;\xi^2+\alpha\xi+\beta=0. \end{eqnarray*}$$

And let

$$\begin{eqnarray*} &&\{\alpha^p\beta^q\}\cup\{\xi\alpha^p\beta^q\} =\{\lambda_j... ... p=0,1,\cdots,m-1,\\ q=0,1,\cdots,n-1,\\ N=2mn. \end{cases}\end{eqnarray*}$$

For

$$\begin{eqnarray*} &&\xi^2\alpha^p\beta^q =(-\alpha\xi-\beta)\alpha^p\beta^q =... ...pha-a_0,\\ &&\beta^n=-b_{n-1}\beta^{n-1}-\cdots-b_1 \beta-b_0, \end{eqnarray*}$$

$\xi\lambda_k$ is a linear combination of $\{\lambda_j\}$ over $\mathbb{Q}$, that is

$$\forall \lambda_k,\; \exists C_{k1},\cdots,C_{kN}\in\mathbb{Q},\; \xi\lambda_k=C_{k1}\lambda_1+\cdots+C_{kN}\lambda_{N}.$$

Hence by moving the term $\xi\lambda_k$ to the other side,

$$\left\{\begin{array}{r@{\,}r@{\,}r@{\,}r@{}r} (C_{11}\!-\!... ...a_2&+\cdots&+(C_{NN}\!-\!\xi)\lambda_N&=0. \end{array}\right.$$

For $\lambda_k\neq 0$,

$$\left\vert\begin{array}{cccc} C_{11}\!-\!\xi&C_{12}&\cdots... ...N1}&C_{N2} &\cdots &C_{NN}\!-\!\xi \end{array}\right\vert=0.$$

Therefore $\xi\in \overline{\mathbb{Q}}$.

When $\deg \xi>2$ over $\overline{\mathbb{Q}}$, $\xi\in \overline{\mathbb{Q}}$ is proved similarly. If

$$\begin{eqnarray*} &&\exists \alpha_l\in\overline{\mathbb{Q}},\, \xi^n+\alpha_{... ...lpha_l-1,\\ N=n\deg\alpha_{n-1}\cdots\deg\alpha_0, \end{cases}\end{eqnarray*}$$

then $\xi\lambda_k$ is a linear combination of $\{\lambda_j\}$ over $\mathbb{Q}$ and

$$\exists C\in M_N(\mathbb{Q}),\,\det(C-\xi E)=0.$$

Therefore $\xi\in \overline{\mathbb{Q}}$. [ Q.E.D. ]

e and pi are transcendental pi is transcendental