rename Chinese Remainder Theorem to Sunzi's Theorem

Author: rzach

content/incompleteness/representability-in-q/beta-function.tex

@@ -42,7 +42,7 @@
 minimization---however, we're allowed to use addition, multiplication,
 and~$\Char{=}$. There are various ways to prove this lemma, but one of
 the cleanest is still G\"odel's original method, which used a
-number-theoretic fact called the Chinese Remainder theorem.
+number-theoretic fact called Sunzi's Theorem (traditionally, the ``Chinese Remainder Theorem'').
 
 \begin{defn}
 Two natural numbers $a$ and $b$ are \emph{relatively prime} iff their
@@ -55,7 +55,7 @@
 same remainder when divided by~$c$.
 \end{defn}
 
-Here is the \emph{Chinese Remainder theorem}:
+Here is Sunzi's Theorem:
 \begin{thm}
 Suppose $x_0$, \dots,~$x_n$ are (pairwise) relatively prime. Let
 $y_0$, \dots,~$y_n$ be any numbers. Then there is a number $z$ such that
@@ -67,7 +67,7 @@
 \end{align*}
 \end{thm}
 
-Here is how we will use the Chinese Remainder theorem: if $x_0$,
+Here is how we will use Sunzi's Theorem: if $x_0$,
 \dots,~$x_n$ are bigger than $y_0$, \dots,~$y_n$ respectively, then we
 can take $z$ to code the sequence $\tuple{y_0, \dots,y_n}$. To
 recover~$y_i$, we need only divide $z$ by~$x_i$ and take the
@@ -150,9 +150,8 @@
 \]
 and let $d_1 = \fact{j}$. By \olref{rel-prime} above, we know that
 $1+d_1$, $1+2 d_1$, \dots, $1+(n+1) d_1$ are relatively prime, and
-by~\olref{less} that all are greater than $a_0$, \dots,~$a_n$. By the
-Chinese Remainder theorem there is a value~$d_0$ such that for
-each~$i$,
+by~\olref{less} that all are greater than $a_0$, \dots,~$a_n$. By
+Sunzi's Theorem there is a value~$d_0$ such that for each~$i$,
 \[
 d_0 \equiv a_i \mod (1+(i+1)d_1)
 \]