From 7e9f75ebd83e968e7cd4ddfb09fcf3869b5657e2 Mon Sep 17 00:00:00 2001 From: Antti H S Laaksonen Date: Thu, 20 Apr 2017 22:54:08 +0300 Subject: [PATCH] Improve grammar --- chapter21.tex | 27 +++++++++++++-------------- 1 file changed, 13 insertions(+), 14 deletions(-) diff --git a/chapter21.tex b/chapter21.tex index 7f2d221..bdc56ac 100644 --- a/chapter21.tex +++ b/chapter21.tex @@ -37,11 +37,11 @@ are integers, if not otherwise stated. \index{factor} \index{divisor} -A number $a$ is a \key{factor} or \key{divisor} of a number $b$ +A number $a$ is called a \key{factor} or a \key{divisor} of a number $b$ if $a$ divides $b$. If $a$ is a factor of $b$, we write $a \mid b$, and otherwise we write $a \nmid b$. -For example, the factors of the number 24 are +For example, the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. \index{prime} @@ -49,15 +49,14 @@ For example, the factors of the number 24 are A number $n>1$ is a \key{prime} if its only positive factors are 1 and $n$. -For example, the numbers 7, 19 and 41 are primes. -The number 35 is not a prime, because it can be -divided into the factors $5 \cdot 7 = 35$. +For example, 7, 19 and 41 are primes, +but 35 is not a prime, because $5 \cdot 7 = 35$. For each number $n>1$, there is a unique \key{prime factorization} \[ n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k},\] -where $p_1,p_2,\ldots,p_k$ are primes and +where $p_1,p_2,\ldots,p_k$ are distinct primes and $\alpha_1,\alpha_2,\ldots,\alpha_k$ are positive numbers. -For example, the prime factorization for the number 84 is +For example, the prime factorization for 84 is \[84 = 2^2 \cdot 3^1 \cdot 7^1.\] The \key{number of factors} of a number $n$ is @@ -66,7 +65,7 @@ because for each prime $p_i$, there are $\alpha_i+1$ ways to choose how many times it appears in the factor. For example, the number of factors -of the number 84 is +of 84 is $\tau(84)=3 \cdot 2 \cdot 2 = 12$. The factors are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84. @@ -74,14 +73,14 @@ The factors are The \key{sum of factors} of $n$ is \[\sigma(n)=\prod_{i=1}^k (1+p_i+\ldots+p_i^{\alpha_i}) = \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1},\] where the latter formula is based on the geometric progression formula. -For example, the sum of factors of the number 84 is +For example, the sum of factors of 84 is \[\sigma(84)=\frac{2^3-1}{2-1} \cdot \frac{3^2-1}{3-1} \cdot \frac{7^2-1}{7-1} = 7 \cdot 4 \cdot 8 = 224.\] The \key{product of factors} of $n$ is \[\mu(n)=n^{\tau(n)/2},\] because we can form $\tau(n)/2$ pairs from the factors, each with product $n$. -For example, the factors of the number 84 +For example, the factors of 84 produce the pairs $1 \cdot 84$, $2 \cdot 42$, $3 \cdot 28$, etc., and the product of the factors is $\mu(84)=84^6=351298031616$. @@ -91,7 +90,7 @@ and the product of the factors is $\mu(84)=84^6=351298031616$. A number $n$ is \key{perfect} if $n=\sigma(n)-n$, i.e., $n$ equals the sum of its factors between $1$ and $n-1$. -For example, the number 28 is perfect, +For example, 28 is a perfect number, because $28=1+2+4+7+14$. \subsubsection{Number of primes} @@ -471,7 +470,7 @@ because $36 \bmod 17 = 2$ and $6^{-1} \bmod 17 = 3$. However, a modular inverse does not always exist. For example, if $x=2$ and $m=4$, the equation \[ x x^{-1} \bmod m = 1 \] -cannot be solved, because all multiples of the number 2 +cannot be solved, because all multiples of 2 are even and the remainder can never be 1 when $m=4$. It turns out that the value of $x^{-1} \bmod m$ can be calculated exactly when $x$ and $m$ are coprime. @@ -487,7 +486,7 @@ x^{-1} = x^{m-2}. \] For example, if $x=6$ and $m=17$, then \[x^{-1}=6^{17-2} \bmod 17 = 3.\] -Using this formula, we can calculate the modular inverse +Using this formula, we can calculate modular inverses efficiently using the modular exponentation algorithm. The above formula can be derived using Euler's theorem. @@ -684,7 +683,7 @@ For example, $(3,4,5)$ is a Pythagorean triple. If $(a,b,c)$ is a Pythagorean triple, all triples of the form $(ka,kb,kc)$ are also Pythagorean triples where $k>1$. -A Pythagorean triple is \key{primitive} if +A Pythagorean triple is \emph{primitive} if $a$, $b$ and $c$ are coprime, and all Pythagorean triples can be constructed from primitive triples using a multiplier $k$.