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数学代写|现代代数代写Modern Algebra代考|Prime Factors and Greatest Common Divisor
In this section, we establish the existence of the greatest common divisor of two integers when at least one of them is nonzero. The Unique Factorization Theorem, also known as the Fundamental Theorem of Arithmetic, is obtained.
Greatest Common Divisor
An integer $d$ is a greatest common divisor of $a$ and $b$ if all these conditions are satisfied:
- $d$ is a positive integer.
- $d \mid a$ and $d \mid b$.
- $c \mid a$ and $c \mid b$ imply $c \mid d$.
The next theorem shows that the greatest common divisor $d$ of $a$ and $b$ exists when at least one of them is not zero. Our proof also shows that $d$ is a linear combination of $a$ and $b$; that is, $d=m a+n b$ for integers $m$ and $n$.
Let $a$ and $b$ be integers, at least one of them not 0 . Then there exists a unique greatest common divisor $d$ of $a$ and $b$. Moreover, $d$ can be written as
$$
d=a m+b n
$$
for integers $m$ and $n$, and $d$ is the smallest positive integer that can be written in this form.
Proof Let $a$ and $b$ be integers, at least one of them not 0 . If $b=0$, then $a \neq 0$, so $|a|>0$. It is easy to see that $d=|a|$ is a greatest common divisor of $a$ and $b$ in this case, and either $d=a \cdot(1)+b \cdot(0)$ or $d=a \cdot(-1)+b \cdot(0)$.
Suppose now that $b \neq 0$. Consider the set $S$ of all integers that can be written in the form $a x+$ by for some integers $x$ and $y$, and let $S^{+}$be the set of all positive integers in $S$. The set $S$ contains $b=a \cdot(0)+b \cdot(1)$ and $-b=a \cdot(0)+b \cdot(-1)$, so $S^{+}$is not empty. By the Well-Ordering Theorem, $S^{+}$has a least element $d$,
$$
d=a m+b n .
$$
We have $d$ positive, and we shall show that $d$ is a greatest common divisor of $a$ and $b$.
By the Division Algorithm, there are integers $q$ and $r$ such that
$$
a=d q+r \text { with } 0 \leq r<d .
$$
From this equation,
$$
\begin{aligned}
r & =a-d q \
& =a-(a m+b n) q \
& =a(1-m q)+b(-n q) .
\end{aligned}
$$
数学代写|现代代数代写Modern Algebra代考|The Euclidean Algorithm
$$
\begin{array}{rlrl}
a & =b q_0+r_1, & 0 \leq r_1<b \
b & =r_1 q_1+r_2, & 0 \leq r_2<r_1 \
r_1 & =r_2 q_2+r_3, & 0 \leq r_3<r_2 \
\vdots & & \vdots \
r_k & =r_{k+1} q_{k+1}+r_{k+2}, & & 0 \leq r_{k+2}<r_{k+1} .
\end{array}
$$
Since the integers $r_1, r_2, \ldots, r_{k+2}$ are decreasing and are all nonnegative, there is a smallest integer $n$ such that $r_{n+1}=0$ :
$$
r_{n-1}=r_n q_n+r_{n+1}, \quad 0=r_{n+1} .
$$
If we put $r_0=b$, this last nonzero remainder $r_n$ is always the greatest common divisor of $a$ and $b$. The proof of this statement is left as an exercise.
As an example, we shall find the greatest common divisor of 1492 and 1776.
Example 1 Performing the arithmetic for the Euclidean Algorithm, we have
$$
\begin{aligned}
1776 & =(1)(1492)+\mathbf{2 8 4} & & \left(q_0=1, r_1=284\right) \
1492 & =(5)(\mathbf{2 8 4})+\mathbf{7 2} & & \left(q_1=5, r_2=72\right) \
\mathbf{2 8 4} & =(3)(\mathbf{7 2})+\mathbf{6 8} & & \left(q_2=3, r_3=68\right) \
\mathbf{7 2} & =(1)(\mathbf{6 8})+\mathbf{4} & & \left(q_3=1, r_4=4\right) \
\mathbf{6 8} & =(\mathbf{4})(17) & & \left(q_4=17, r_5=0\right) .
\end{aligned}
$$
Thus the last nonzero remainder is $r_n=r_4=4$, and $(1776,1492)=4$.
现代代数代考
数学代写|现代代数代写Modern Algebra代考|Prime Factors and Greatest Common Divisor
在本节中，当两个整数中至少有一个非零时，我们建立了它们最大公约数的存在性。得到了唯一因数分解定理，又称算术基本定理。
最大公约数
如果满足所有这些条件，整数$d$是$a$和$b$的最大公约数:
$d$ 是一个正整数。
$d \mid a$ 还有$d \mid b$。
$c \mid a$$c \mid b$暗示$c \mid d$。
下一个定理表明，当$a$和$b$至少有一个不为零时，存在最大公约数$d$。我们的证明还表明$d$是$a$和$b$的线性组合;即对于整数$m$和$n$，为$d=m a+n b$。
设$a$和$b$为整数，其中至少有一个不为0。那么$a$和$b$存在唯一的最大公约数$d$。此外，$d$可以写成
$$
d=a m+b n
$$
对于整数$m$和$n$, $d$是可以写成这种形式的最小的正整数。
证明设$a$和$b$为整数，且至少有一个不为0。如果$b=0$，那么$a \neq 0$，那么$|a|>0$。很容易看出，在这种情况下，$d=|a|$是$a$和$b$的最大公约数，$d=a \cdot(1)+b \cdot(0)$或$d=a \cdot(-1)+b \cdot(0)$也是如此。
现在假设$b \neq 0$。考虑所有整数的集合$S$，这些整数可以写成$a x+$的形式，对于某些整数$x$和$y$，设$S^{+}$为$S$中所有正整数的集合。集合$S$包含$b=a \cdot(0)+b \cdot(1)$和$-b=a \cdot(0)+b \cdot(-1)$，因此$S^{+}$不是空的。根据良序定理，$S^{+}$有一个最小元素$d$，
$$
d=a m+b n .
$$
我们有$d$是正数，我们将证明$d$是$a$和$b$的最大公约数。
通过除法算法，存在整数$q$和$r$，使得
$$
a=d q+r \text { with } 0 \leq r<d .
$$
由这个方程，
$$
\begin{aligned}
r & =a-d q \
& =a-(a m+b n) q \
& =a(1-m q)+b(-n q) .
\end{aligned}
$$
数学代写|现代代数代写Modern Algebra代考|The Euclidean Algorithm
$$
\begin{array}{rlrl}
a & =b q_0+r_1, & 0 \leq r_1<b \
b & =r_1 q_1+r_2, & 0 \leq r_2<r_1 \
r_1 & =r_2 q_2+r_3, & 0 \leq r_3<r_2 \
\vdots & & \vdots \
r_k & =r_{k+1} q_{k+1}+r_{k+2}, & & 0 \leq r_{k+2}<r_{k+1} .
\end{array}
$$
由于整数$r_1, r_2, \ldots, r_{k+2}$都是递减的且都是非负的，因此存在一个最小整数$n$，使得$r_{n+1}=0$:
$$
r_{n-1}=r_n q_n+r_{n+1}, \quad 0=r_{n+1} .
$$
如果我们代入$r_0=b$，最后一个非零余数$r_n$总是$a$和$b$的最大公约数。这句话的证明留作练习。
作为一个例子，我们将找出1492和1776的最大公约数。
例1执行欧几里得算法的算术，我们有
$$
\begin{aligned}
1776 & =(1)(1492)+\mathbf{2 8 4} & & \left(q_0=1, r_1=284\right) \
1492 & =(5)(\mathbf{2 8 4})+\mathbf{7 2} & & \left(q_1=5, r_2=72\right) \
\mathbf{2 8 4} & =(3)(\mathbf{7 2})+\mathbf{6 8} & & \left(q_2=3, r_3=68\right) \
\mathbf{7 2} & =(1)(\mathbf{6 8})+\mathbf{4} & & \left(q_3=1, r_4=4\right) \
\mathbf{6 8} & =(\mathbf{4})(17) & & \left(q_4=17, r_5=0\right) .
\end{aligned}
$$
因此最后的非零余数是$r_n=r_4=4$和$(1776,1492)=4$。
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