From 8bf848e97c48a4685d5b013d3be49883793dc35c Mon Sep 17 00:00:00 2001 From: jxu <7989982+jxu@users.noreply.github.com> Date: Sat, 14 Jan 2023 23:11:44 -0500 Subject: [PATCH 1/3] Replace p with m notation p_i should only be used for primes, while CRT works for any coprime m_i (esp. case of prime powers p^e) --- src/algebra/chinese-remainder-theorem.md | 20 ++++++++++---------- 1 file changed, 10 insertions(+), 10 deletions(-) diff --git a/src/algebra/chinese-remainder-theorem.md b/src/algebra/chinese-remainder-theorem.md index cc198dbda..2c0c25b71 100644 --- a/src/algebra/chinese-remainder-theorem.md +++ b/src/algebra/chinese-remainder-theorem.md @@ -10,36 +10,36 @@ The Chinese Remainder Theorem (which will be referred to as CRT in the rest of t ## Formulation -Let $p = p_1 \cdot p_2 \cdots p_k$, where $p_i$ are pairwise relatively prime. In addition to $p_i$, we are also given a set of congruence equations +Let $m = m_1 \cdot m_2 \cdots m_k$, where $m_i$ are pairwise relatively prime. In addition to $m_i$, we are also given a set of congruence equations $$\begin{align} - a &\equiv a_1 \pmod{p_1} \\\\ - a &\equiv a_2 \pmod{p_2} \\\\ + a &\equiv a_1 \pmod{m_1} \\\\ + a &\equiv a_2 \pmod{m_2} \\\\ &\ldots \\\\ - a &\equiv a_k \pmod{p_k} + a &\equiv a_k \pmod{m_k} \end{align}$$ -where $a_i$ are some given constants. The original form of CRT then states that the given set of congruence equations always has *one and exactly one* solution modulo $p$. +where $a_i$ are some given constants. The original form of CRT then states that the given set of congruence equations always has *one and exactly one* solution modulo $m$. ### Corollary A consequence of the CRT is that the equation -$$x \equiv a \pmod{p}$$ +$$x \equiv a \pmod{m}$$ is equivalent to the system of equations $$\begin{align} - x &\equiv a_1 \pmod{p_1} \\\\ + x &\equiv a_1 \pmod{m_1} \\\\ &\ldots \\\\ - x &\equiv a_k \pmod{p_k} + x &\equiv a_k \pmod{m_k} \end{align}$$ -(As above, assume that $p = p_1 p_2 \cdots p_k$ and $p_i$ are pairwise relatively prime). +(As above, assume that $m = m_1 m_2 \cdots m_k$ and $m_i$ are pairwise relatively prime). ## Garner's Algorithm -Another consequence of the CRT is that we can represent big numbers using an array of small integers. For example, let $p$ be the product of the first $1000$ primes. From calculations we can see that $p$ has around $3000$ digits. +Another consequence of the CRT is that we can represent big numbers using an array of small integers. For example, let $p$ be the product of the first $1000$ primes. From calculations we can see that $P$ has around $3000$ digits. Any number $a$ less than $p$ can be represented as an array $a_1, \ldots, a_k$, where $a_i \equiv a \pmod{p_i}$. But to do this we obviously need to know how to get back the number $a$ from its representation. In this section, we discuss Garner's Algorithm, which can be used for this purpose. We seek a representation on the form From 0b180538c5f06623474d697943ec45cf6ff70c12 Mon Sep 17 00:00:00 2001 From: jxu <7989982+jxu@users.noreply.github.com> Date: Sun, 15 Jan 2023 09:16:21 -0500 Subject: [PATCH 2/3] Restore Garner's Algorithm notation to lowercase --- src/algebra/chinese-remainder-theorem.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/algebra/chinese-remainder-theorem.md b/src/algebra/chinese-remainder-theorem.md index 2c0c25b71..23fa3c655 100644 --- a/src/algebra/chinese-remainder-theorem.md +++ b/src/algebra/chinese-remainder-theorem.md @@ -39,7 +39,7 @@ $$\begin{align} ## Garner's Algorithm -Another consequence of the CRT is that we can represent big numbers using an array of small integers. For example, let $p$ be the product of the first $1000$ primes. From calculations we can see that $P$ has around $3000$ digits. +Another consequence of the CRT is that we can represent big numbers using an array of small integers. For example, let $p$ be the product of the first $1000$ primes. From calculations we can see that $p$ has around $3000$ digits. Any number $a$ less than $p$ can be represented as an array $a_1, \ldots, a_k$, where $a_i \equiv a \pmod{p_i}$. But to do this we obviously need to know how to get back the number $a$ from its representation. In this section, we discuss Garner's Algorithm, which can be used for this purpose. We seek a representation on the form From 1ecb5fac1db6e5e3268959e413d50b0b4f823934 Mon Sep 17 00:00:00 2001 From: Oleksandr Kulkov Date: Sun, 15 Jan 2023 16:21:46 +0100 Subject: [PATCH 3/3] relatively prime -> coprime --- src/algebra/chinese-remainder-theorem.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/src/algebra/chinese-remainder-theorem.md b/src/algebra/chinese-remainder-theorem.md index 23fa3c655..82f3e5eb8 100644 --- a/src/algebra/chinese-remainder-theorem.md +++ b/src/algebra/chinese-remainder-theorem.md @@ -10,7 +10,7 @@ The Chinese Remainder Theorem (which will be referred to as CRT in the rest of t ## Formulation -Let $m = m_1 \cdot m_2 \cdots m_k$, where $m_i$ are pairwise relatively prime. In addition to $m_i$, we are also given a set of congruence equations +Let $m = m_1 \cdot m_2 \cdots m_k$, where $m_i$ are pairwise coprime. In addition to $m_i$, we are also given a set of congruence equations $$\begin{align} a &\equiv a_1 \pmod{m_1} \\\\ @@ -35,7 +35,7 @@ $$\begin{align} x &\equiv a_k \pmod{m_k} \end{align}$$ -(As above, assume that $m = m_1 m_2 \cdots m_k$ and $m_i$ are pairwise relatively prime). +(As above, assume that $m = m_1 m_2 \cdots m_k$ and $m_i$ are pairwise coprime). ## Garner's Algorithm pFad - Phonifier reborn

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