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数学竞赛代写|Stanford大学数学夏令营SUMaC代写

Definition 2.7.10 A set $S \subseteq \mathbb{R}$ is dense in $\mathbb{R}$ if whenever $a<b, S \cap(a, b) \neq \emptyset$
Thus the above theorem says $Q$ is “dense” in $\mathbb{R}$.
You probably saw the process of division in elementary school. Even though you saw it at a young age it is very profound and quite difficult to understand. Suppose you want to do the following problem $\frac{79}{22}$. What did you do? You likely did a process of long division which gave the following result. $\frac{79}{22}=3$ with remainder 13 . This meant $79=3(22)+13$.You were given two numbers, 79 and 22 and you wrote the first ass some multiple of the second added to a third number which was smaller than the second number. Can this always be done? The answer is in the next theorem and depends here on the Archimedean property of the real numbers.
Theorem 2.7.11 Suppose $0<a$ and let $b \geq 0$. Then there exists a unique integer $p$ and real number $r$ such that $0 \leq r<a$ and $b=p a+r$.
Proof: Let $S \equiv{n \in \mathbb{N}: a n>b}$. By the Archimedean property this set is nonempty. Let $p+1$ be the smallest element of $S$. Then $p a \leq b$ because $p+1$ is the smallest in $S$. Therefore, $r \equiv b-p a \geq 0$. If $r \geq a$ then $b-p a \geq a$ and so $b \geq(p+1) a$ contradicting $p+1 \in S$. Therefore, $r<a$ as desired.
To verify uniqueness of $p$ and $r$, suppose $p_{1}$ and $r_{1}, i=1,2$, both work and $r_{2}>r_{1}$. Then a little algebra shows $p_{1}-p_{2}=\frac{m_{2}-r_{1}}{a} \in(0,1)$. Thus $p_{1}-p_{2}$ is an integer between 0 and 1 , contradicting Theorem $2.7 .8$.The case that $r_{1}>r_{2}$ cannot occur either by similar reasoning. Thus $r_{1}=r_{2}$ and it follows that $p_{1}=p_{2}$.
This theorem is called the Euclidean algorithm when $a$ and $b$ are integers. In this case, you would have $r$ is an integer because it equals an integer.
$2.8$ Arithmetic of Integers
Here we consider some very important algebraic notions including the Euclidean algorithm just mentioned and issues related to whether two numbers are relatively prime, prime numbers and so forth. The following definition describes what is meant by a prime number and also what is meant by the word “divides”.Definition 2.8.1 The number a divides the number $b$ if, in Theorem 2.7.11. $r=0$. That is, there is zero remainder. The notation for this is $a \mid b$, read a divides $b$ and $a$ is called a factor of $b$. A prime number is one which has the property that the only numbers which divide it are itself and 1 and it is at least 2. The greatest common divisor of two positive integers $m, n$ is that number $p$ which has the property that $p$ divides both $m$ and $n$ and also if $q$ divides both $m$ and $n$, then $q$ divides $p$. Two integers are relatively prime if their greatest common divisor is one. The greatest common divisor of $m$ and $n$ is denoted as $(m, n)$.
There is a phenomenal and amazing theorem which relates the greatest common divisor to the smallest number in a certain set. Suppose $m, n$ are two positive integers. Then if $x, y$ are integers, so is $x m+y n$. Consider all integers which are of this form. Some are positive such as $1 m+1 n$ and some are not. The set $S$ in the following theorem consists of exactly those integers of this form which are positive. Then the greatest common divisor of $m$ and $n$ will be the smallest number in $S$. This is what the following theorem says.
定义 2.7.10 一个集合 $S \subseteq \mathbb{R}$ 在 $\mathbb{R}$ 中是稠密的,如果每当 $ab}$。根据阿基米德性质,这个集合是非空的。令 $p+1$ 为 $S$ 的最小元素。然后 $p a \leq b$ 因为 $p+1$ 是 $S$ 中最小的。因此,$r \equiv b-p a \geq 0$。如果 $r \geq a$ 则 $b-p a \geq a$ 等 $b \geq(p+1) a$ 与 $p+1 \in S$ 矛盾。因此,$rr_{1}$。然后一个小代数显示 $p_{1}-p_{2}=\frac{m_{2}-r_{1}}{a} \in(0,1)$。因此 $p_{1}-p_{2}$ 是一个介于 0 和 1 之间的整数,这与定理 $2.7 .8$ 相矛盾。$r_{1}>r_{2}$ 的情况也不能通过类似的推理发生。因此 $r_{1}=r_{2}$ 并且它遵循 $p_{1}=p_{2}$。 当$a$ 和$b$ 是整数时,这个定理称为欧几里得算法。在这种情况下,您将 $r$ 是一个整数,因为它等于一个整数。 $2.8$ 整数算术 这里我们考虑一些非常重要的代数概念,包括刚才提到的欧几里得算法以及两个数是否互质、素数等问题。下面的定义描述了质数的含义以及“除法”一词的含义。定义 2.8.1 如果在定理 2.7.11 中,数 a 除数 $b$。 $r=0$。也就是说,余数为零。这个符号是$a \mid b$,读a 除$b$,$a$ 称为$b$ 的因数。素数是一个具有这样的性质的数,即唯一能除它的数是它自己和 1,并且它至少是 2。两个正整数 $m,n$ 的最大公约数是具有这个性质的数 $p$ $p$ 可以同时除以 $m$ 和 $n$,并且如果 $q$ 可以同时除以 $m$ 和 $n$,那么 $q$ 也可以除以 $p$。两个整数互质,如果它们的最大公约数是一。 $m$ 和$n$ 的最大公约数记为$(m, n)$。 有一个惊人的定理将最大公约数与某个集合中的最小数联系起来。假设 $m, n$ 是两个正整数。那么如果 $x, y$ 是整数,那么 $x m+y n$ 也是。考虑所有具有这种形式的整数。有些是积极的,例如 $1 m+1 n$,有些则不是。以下定理中的集合 $S$ 恰好由这种形式的正整数组成。那么 $m$ 和 $n$ 的最大公约数将是 $S$ 中的最小数。这就是下面的定理所说的。

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