如果你也在 怎样代写离散数学discrete math这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。离散数学discrete math是研究可以被认为是 “离散”（类似于离散变量，与自然数集有偏射）而不是 “连续”（类似于连续函数）的数学结构。离散数学研究的对象包括整数、图形和逻辑中的语句。相比之下，离散数学不包括 “连续数学 “中的课题，如实数、微积分或欧几里得几何。离散对象通常可以用整数来列举；更正式地说，离散数学被定性为处理可数集的数学分支（有限集或与自然数具有相同心数的集）。然而，”离散数学 “这一术语并没有确切的定义。
离散数学discrete math的研究在二十世纪后半叶有所增加，部分原因是数字计算机的发展，它以 “离散 “的步骤操作，并以 “离散 “的比特存储数据。离散数学的概念和符号在研究和描述计算机科学分支的对象和问题时非常有用，如计算机算法、编程语言、密码学、自动定理证明和软件开发。反过来说，计算机实现在将离散数学的思想应用于现实世界的问题中也很重要，例如在运筹学中。
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我们提供的离散数学discrete math及其相关学科的代写，服务范围广, 其中包括但不限于:
数学代写|离散数学代写discrete math代考|Inverses mod p
In the last section we explored multiplication in $Z_{n}$. We saw in the special case with $n=12$ and $a=4$ that if we used multiplication by $a$ in $Z_{n}$ to encrypt a message, then our receiver would need to be able to solve, for $x$, the equation $a \cdot{ }{n} x=b$ in order to decode a received message $b$. In the end of section exercises there are exercises that show that with some values of $n$, equations of the form $a \cdot{ }{n} x=b$ have a unique solution, while for other values of $n$ we could have equations with no solutions, or equations with more than one solution. Notice that if $n=m k$, then the equation $m \cdot n x=0$ will always have at least the two solutions, $k$ and 0 . Thus if we want the equation $a \cdot{ }{n} x=b$ to have a unique solution in $Z{n}$ for every $a \neq 0$ and every $b$ in $Z_{n}$, then $n$ must be a prime number.
If you experimented with the prime numbers 5,7 , and 11 in the last set of problems (and if you didn’t, now would be a great time to do so), you probably recognized that what is relevant for solving the equation $a \cdot{ }{n} x=b$ is the question of whether $a$ has a multiplicative inverse in $Z{n}$, that is, whether there is another number $a^{\prime}$ such that $a^{\prime}{ }{n} a=1$. If $a$ does have the inverse $a^{\prime}$, then the unique solution to the equation $a \cdot{ }{n} x=b$ is $a^{\prime}{ }{n} b$. Once we realize the importance of inverses, we find it relatively easy to make computations that convince us of that every nonzero element in $Z{5}, Z_{7}$, and $Z_{11}$ does have a multiplicative inverse, so every equation of the form $a \cdot n x=b$ (with $a \neq 0$ ) does have a unique solution. The evidence we have for $p=5,7$, and 11 suggests that whenever $p$ is a prime then every nonzero element of $Z_{p}$ has an inverse in $Z_{p}$, and therefore every equation of the form $a \cdot p=b$ (with $a \neq 0$ ) has a unique solution. This leads us to focus on trying to show that for each prime $p$, each element of $Z_{p}$ has a multiplicative inverse.
Thus we are interested in showing that for each nonzero $a$ in $Z_{p}$, the equation
$$
a \cdot p x=1
$$
has a solution. What does this equation mean, though? We can re-express it as
$$
a \cdot p x-{ }_{p} 1=0
$$
or
$$
(a x-1) \bmod p=0 .
$$
This means that $a x-1$ is a multiple of $p$, so that there is some integer $y$ such that $a x-1=p y$, or
$$
a x-p y=1
$$
数学代写|离散数学代写discrete math代考|Greatest Common Divisors (GCD)
2.2-1 Suppose $m$ is not a prime, and $a$ and $x$ are integers such that $a \cdot m x=1$ in $Z_{m}$. What equation involving $m$ in the integers does this give us?
2.2-2 Suppose that $a$ and $m$ are integers such that $a x-m y=1$, for some integers $x$ and $y$. What does that tell us about being able to find a (multiplicative) inverse for $a$ $(\bmod m)$ ?
2.2-3 If $a x-m y=1$ for integers $x$ and $y$, can $a$ and $m$ have any common divisors other than 1 and $-1$ ?
In Exercise 2.2-1 we saw that if the equation
$$
a \cdot m x=1
$$
has a solution, then there is a number $y$ such that
$$
a x-m y=1
$$
In Exercise 2.2-2, we saw that if $x$ and $y$ are integers such that $a x-m y=1$, then $a x-1$ is a multiple of $m$, and so $x$ is a multiplicative inverse of $a$ in $Z_{m}$. Thus we have actually proved the following:
数学代写|离散数学代写DISCRETE MATH代考|The GCD Algorithm
Our algorithm for Exercise 2.2-4 is based on Lemma 2.2.3 and the observation that if $k=n q$, for any $q$, then $n=\operatorname{gcd}(k, n)$. It is convenient to assume that $n \leq k$, so if this is not the case, we exchange $k$ and $n$. We first write $k=n q+r$ in the usual way. If $r=0$, then we return $n$ as the greatest common divisor. Otherwise,we apply our algorithm to find the greatest common divisor of $n$ and $r$. Finally, we return the result as the greatest common divisor of $k$ and $n$.
As an example, consider finding
$$
\operatorname{gcd}(24,14)
$$
We can write
$$
24=1(14)+10 .
$$
In this case $k=24, n=14, q=1$ and $r=10$. Thus we can apply Lemma $2.2 .3$ and conclude that
$$
\operatorname{gcd}(24,14)=\operatorname{gcd}(14,10) \text {. }
$$
We therefore continue our computation of $\operatorname{gcd}(14,10)$, by writing $14=10 \cdot+4$, and have that
$$
\operatorname{gcd}(14,10)=\operatorname{gcd}(10,4)
$$
Now,
$$
10=4 \cdot 2+2
$$
and so
$$
\operatorname{gcd}(10,4)=\operatorname{gcd}(4,2)
$$
Now
$$
4=2 \cdot 2+0,
$$
so that 2 is a divisor of four and is thus the greatest common divisor of 2 and 4 . Since 2 is a divisor of 4 , this last equation forms our “base case”. Thus we have that
$$
\operatorname{gcd}(24,14)=\operatorname{gcd}(4,2)=2
$$
离散数学代写
数学代写|离散数学代写DISCRETE MATH代考|INVERSES MOD P
在上一节中，我们探讨了乘法从n. 我们在特殊情况下看到n=12和一种=4如果我们使用乘法一种在从n加密消息，那么我们的接收者需要能够解决，因为$x$, the equation $a \cdot{ }{n} x=b$ in order to decode a received message $b$. In the end of section exercises there are exercises that show that with some values of $n$, equations of the form $a \cdot{ }{n} x=b$ have a unique solution, while for other values of $n$ we could have equations with no solutions, or equations with more than one solution. Notice that if $n=m k$, then the equation $m \cdot n x=0$ will always have at least the two solutions, $k$ and 0 . Thus if we want the equation $a \cdot{ }{n} x=b$ to have a unique solution in $Z{n}$ for every $a \neq 0$ and every $b$ in $Z_{n}$, t 必须是质数。
如果您在最后一组问题中尝试了素数 5,7 (and if you didn’t, now would be a great time to do so), you probably recognized that what is relevant for solving the equation $a \cdot{ }{n} x=b$ is the question of whether $a$ has a multiplicative inverse in $Z{n}$, that is, whether there is another number $a^{\prime}$ such that $a^{\prime}{ }{n} a=1$. If $a$ does have the inverse $a^{\prime}$, then the unique solution to the equation $a \cdot{ }{n} x=b$ is $a^{\prime}{ }{n} b$. Once we realize the importance of inverses, we find it relatively easy to make computations that convince us of that every nonzero element in $Z{5}, Z_{7}$, and $Z_{11}$ does have a multiplicative inverse, so every equation of the form $a \cdot n x=b$ (with $a \neq 0$ ) does have a unique solution. The evidence we have for $p=5,7$, and 11 suggests that whenever $p$ is a prime then every nonzero element of $Z_{p}$ has an inverse in $Z_{p}$, and therefore every equation of the form $a \cdot p=b$ (with $a \neq 0$ ) has a unique solution. This leads us to focus on trying to show that for each prime $p$, each element of $Z_{p}$ has a multiplicative inverse.
Thus we are interested in showing that for each nonzero $a$ in $Z_{p}$, the equation
$$
a \cdot p x=1
$$
has a solution. What does this equation mean, though? We can re-express it as
$$
a \cdot p x-{ }_{p} 1=0
$$
or
$$
(a x-1) \bmod p=0 .
$$
This means that $a x-1$ is a multiple of $p$, so that there is some integer $y$ such that $a x-1=p y$, or
$$
a x-p y=1
$$
数学代写|离散数学代写DISCRETE MATH代考|GREATEST COMMON DIVISORS GCD
2.2-1 假设米不是素数，并且一种和X是整数，使得一种⋅米X=1在从米. 涉及什么方程米在整数中，这给了我们吗？
2.2-2 假设一种和米是整数，使得一种X−米是=1, 对于一些整数X和是. 这告诉我们能够找到一个米在l吨一世pl一世C一种吨一世在和逆为一种 (反对米)?
2.2-3 如果一种X−米是=1对于整数X和是， 能够一种和米有除 1 以外的任何公约数和−1?
在练习 2.2-1 中，我们看到如果方程
一种⋅米X=1
有解，则有数是这样
一种X−米是=1
在练习 2.2-2 中，我们看到如果X和是是整数，使得一种X−米是=1， 然后一种X−1是的倍数米， 所以X是乘法逆一种在从米. 因此，我们实际上证明了以下内容：
数学代写|离散数学代写DISCRETE MATH代考|THE GCD ALGORITHM
我们的练习 2.2-4 的算法是基于引理 2.2.3 和观察，如果ķ=nq, 对于任何q， 然后n=gcd(ķ,n). 可以方便地假设n≤ķ，所以如果不是这样，我们交换ķ和n. 我们先写ķ=nq+r以通常的方式。如果r=0，然后我们返回n作为最大公约数。否则，我们应用我们的算法找到最大公约数n和r. 最后，我们将结果作为最大公约数返回ķ和n.
例如，考虑寻找
$$
\operatorname{gcd}(24,14)
$$
We can write
$$
24=1(14)+10 .
$$
In this case $k=24, n=14, q=1$ and $r=10$. Thus we can apply Lemma $2.2 .3$ and conclude that
$$
\operatorname{gcd}(24,14)=\operatorname{gcd}(14,10) \text {. }
$$
We therefore continue our computation of $\operatorname{gcd}(14,10)$, by writing $14=10 \cdot+4$, and have that
$$
\operatorname{gcd}(14,10)=\operatorname{gcd}(10,4)
$$
Now,
$$
10=4 \cdot 2+2
$$
and so
$$
\operatorname{gcd}(10,4)=\operatorname{gcd}(4,2)
$$
Now
$$
4=2 \cdot 2+0,
$$
so that 2 is a divisor of four and is thus the greatest common divisor of 2 and 4 . Since 2 is a divisor of 4 , this last equation forms our “base case”. Thus we have that
$$
\operatorname{gcd}(24,14)=\operatorname{gcd}(4,2)=2
$$
数学代写|离散数学代写discrete math代考 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
微观经济学代写
微观经济学是主流经济学的一个分支，研究个人和企业在做出有关稀缺资源分配的决策时的行为以及这些个人和企业之间的相互作用。my-assignmentexpert™ 为您的留学生涯保驾护航 在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的数学Mathematics代写服务。我们的专家在图论代写Graph Theory代写方面经验极为丰富，各种图论代写Graph Theory相关的作业也就用不着 说。
线性代数代写
线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。
博弈论代写
现代博弈论始于约翰-冯-诺伊曼（John von Neumann）提出的两人零和博弈中的混合策略均衡的观点及其证明。冯-诺依曼的原始证明使用了关于连续映射到紧凑凸集的布劳威尔定点定理，这成为博弈论和数学经济学的标准方法。在他的论文之后，1944年，他与奥斯卡-莫根斯特恩（Oskar Morgenstern）共同撰写了《游戏和经济行为理论》一书，该书考虑了几个参与者的合作游戏。这本书的第二版提供了预期效用的公理理论，使数理统计学家和经济学家能够处理不确定性下的决策。
微积分代写
微积分，最初被称为无穷小微积分或 “无穷小的微积分”，是对连续变化的数学研究，就像几何学是对形状的研究，而代数是对算术运算的概括研究一样。
它有两个主要分支，微分和积分；微分涉及瞬时变化率和曲线的斜率，而积分涉及数量的累积，以及曲线下或曲线之间的面积。这两个分支通过微积分的基本定理相互联系，它们利用了无限序列和无限级数收敛到一个明确定义的极限的基本概念 。
计量经济学代写
什么是计量经济学？
计量经济学是统计学和数学模型的定量应用，使用数据来发展理论或测试经济学中的现有假设，并根据历史数据预测未来趋势。它对现实世界的数据进行统计试验，然后将结果与被测试的理论进行比较和对比。
根据你是对测试现有理论感兴趣，还是对利用现有数据在这些观察的基础上提出新的假设感兴趣，计量经济学可以细分为两大类：理论和应用。那些经常从事这种实践的人通常被称为计量经济学家。
Matlab代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习和应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。