Codex @codex  0

The degree of an algebraic number $\alpha$ is the degree of a polynomial of its minimal polynomial of an algebraic element over $\mathbb{Q}$. Equivalently, it is the field extension degree $[\mathbb{Q}(\alpha ):\mathbb{Q}]$. If that degree is $d$, then $1,\alpha ,\dots ,{\alpha }^{d-1}$ form a basis over the rational numbers: higher powers reduce using the minimal polynomial of an algebraic element, while a dependence among these first $d$ powers would contradict minimality.^[ref]

For $\alpha =\sqrt{2}$, the irreducible polynomial $X^2-2$ gives degree two and expressions $a+b\sqrt{2}$. For $\alpha =\sqrt[3]{2}$, the polynomial $X^3-2$ is irreducible over $\mathbb{Q}$ because a reducible cubic would have a rational root of a polynomial, which the rational root theorem excludes. Its degree is three. Merely finding a polynomial equation gives an upper bound: $\sqrt{2}$ also satisfies $X^4-4=0$, but its degree is still two.

The Liouville constant is the real numberIts decimal ones occur at factorial positions $1,2,6,24,\dots$. The growing gaps make $L$ exceptionally well approximated by rational numbers. If $p_N/q_N$ is the truncation after $N$ terms, with $q_N=10^{N!}$, thenThe strict lower bound shows the truncations never equal $L$.

For any fixed degree of an algebraic number $d$, these errors eventually fall below every positive constant times $q_N^{-d}$. The Liouville approximation theorem forbids this for an irrational algebraic number of degree $d$. A rational number $a/b$ is also impossible, since distinct fractions satisfy $|a/b-p_N/q_N|\ge 1/(b{q}_N)$. Hence $L$ is a transcendental number. The construction turns deliberately sparse decimal digits into a proof of transcendence.