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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数学代写|离散数学代写discrete math代考|Direct Inference (Modus Ponens) and Proofs
We concluded our last section with a proof that the sum of even numbers is even. There were several crucial ingredients in that proof. One consisted of introducing symbols for members of the universe of integers numbers. In other words, rather than saying “suppose we have two integers,” we introduced symbols for the two members of our universe we assumed we had. How did we know to use algebraic symbols? As usual, there is more than one possible answer to this question. In this case, our intuition was probably based on thinking about what an even number is, and realizing that the definition itself is essentially symbolic. (You may argue that an even number is just twice another number, and you would be right. Apparently there are no symbols are in that definition. But they really are there; they are the phrases “even number” and “another number.” Since we all know algebra is easier with symbolic variables rather than words, we should recognize that it makes sense to use algebraic notation.) Thus this decision was based on experience, not logic.
Next we assumed the two integers we were talking about were even. We then used the definition of even numbers, and our previous parenthetic comment suggests that it was natural for us to use the definition symbolically. The definition tells us that if $m$ is an even number, then there exists another integer $i$ such that $m=2 i$. We combined this with the assumption that $m$ is even to conclude that in fact there does exist an integer $i$ such that $m=2 i$. This is an example of using the principle of direct inference (called modus ponens in Latin, though we won’t worry about Latin names in this course.) The principle says that from $p$ and $p \Rightarrow q$ we may conclude $q$. This principle is a cornerstone of logical arguments. If you think about the truth table for $p \Rightarrow q$, you can view this rule as describing one row of this truth table. There are quite a few such rules (called rules of inference) that people commonly use in proofs without explicitly stating them. We will try to tease out which rules we were using in the proof before beginning a formal study of rules of inference.
We then used algebra to show that because $m=2 i$ and $n=2 j$, there exists a $k$ such that $m+n=2 k$ (our $k$ was $i+j$ ). Next we used the definition of even number again. We then used a rule of inference which says that if, by assuming $p$, we may prove $q$, then the statement $p \Rightarrow q$ is true. We finally reached a grand conclusion that for all pairs of integers, if they are even, then their sum is even. Here is another rule of inference, one of the more difficult to describe. We introduced the variables $m$ and $n$. We used only well-known consequences of the fact that they were in the universe of integers in our proof. Thus we felt justified in asserting that what we concluded about $m$ and $n$ is true for any pair of integers. We might say that we were treating $m$ and $n$ as generic members of our universe. Thus our rule of inference says that if we can prove a statement for a generic member of our universe, then we can conclude it is true for every member of our universe. Perhaps the reason this rule is hard to put into words is that it is not simply a description of a truth table, but is a principle that we use in order to prove universally quantified statements.
数学代写|离散数学代写discrete math代考|Rules of inference
We have seen the ingredients of a typical proof. What do we mean by a proof in general? A proof of a statement is a convincing argument that the statement is true. To be more precise about it, we can agree that a proof consists of a sequence of statements, each of which is either a hypothesis (to be described in more detail in our description of rules of inference below), a generally accepted fact, or the result of one of the following rules of inference for compound statements.
1 From an example $x$ that does not satisfy $p(x)$, we may conclude $\neg p(x)$.
2 From $p(x)$ and $q(x)$, we may conclude $p(x) \wedge q(x)$.
3 From either $p(x)$ or $q(x)$, we may conclude $p(x) \vee q(x)$.
4 From either $q(x)$ or $\neg p(x)$ we may conclude $p(x) \Rightarrow q(x)$.
5 From $p(x) \Rightarrow q(x)$ and $q(x) \Rightarrow p(x)$ we may conclude $p(x) \Leftrightarrow q(x)$.
6 From $p(x)$ and $p(x) \Rightarrow q(x)$ we may conclude $q(x)$.
7 From $p(x) \Rightarrow q(x)$ and $q(x) \Rightarrow r(x)$ we may conclude $p(x) \Rightarrow r(x)$.
8 If we can derive $q(x)$ from the hypothesis that $x$ satisfies $p(x)$, then we may conclude $p(x) \Rightarrow q(x) .$
9 If we can derive $p(x)$ from the hypothesis that $x$ is a (generic) member of our universe $U$, we may conclude $\forall x \in U(p(x))$.
10 From an example of an $x \in U$ satisfying $p(x)$ we may conclude $\exists x \in U(p(x))$.
数学代写|离散数学代写DISCRETE MATH代考|Contrapositive rule of inference
In Exercise 3.3-2 we saw that if we know that $\neg q \rightarrow \neg p$, then we can conclude that $p \rightarrow q$. To see what that is good for, consider the following example.
Prove that if $n$ is a positive integer with $n^{2}>100$, then $n>10$.
Proof: Suppose $n$ is not greater than 10. (Now we use the rule of algebra for inequalities which says that if $x \leq y$ and $c \geq 0$, then $c x \leq c y$.) Then since $1 \leq n \leq 10$,
$$
n \cdot n \leq n \cdot 10 \leq 10 \cdot 10=100 .
$$
Thus $n^{2}$ is not greater than 100 . Therefore, if $n^{2}>100, n$ must be greater than $10 .$
We could give an intuitive explanation of the reasoning (and doing so would be a good exercise), but we don’t need to because we can see we are just using the result of Exercise 2 .
We adopt the result of Exercise 2 as a rule of inference, called the contrapositive rule of inference.
11 From $\neg q \Rightarrow \neg p$ we may conclude $p \Rightarrow q$.
In our proof of the Chinese Remainder Theorem, Theorem 2.3.6, we wanted to prove that for a certain function $f$ that if $x$ and $y$ were different integers between 0 and $m n-1$, then $f(x) \neq f(y)$. To prove this we assumed that in fact $f(x)=f(y)$ and proved that $x$ and $y$ were not different integers between 0 and $m n-1$. Had we known the principle of contrapositive inference, we could have concluded then and there that $f$ was one-to-one. Instead, we used the more common principle of proof by contradiction, discussed in detail below, to complete our proof. If you look back at the proof, you will see that we might have been able to shorten it by a sentence by using contrapositive inference.
A quick look at the truth tables for $p \Rightarrow q$ and $q \Rightarrow p$ should convince us that these two statements are not equivalent. $q \Rightarrow p$ is called the converse of $p \Rightarrow q$. It is surprising how often people, even professional mathematicians, absent-mindedly prove the converse of a statement when they mean to prove the statement itself. Try not to join this crowd!
离散数学代写
数学代写|离散数学代写DISCRETE MATH代考|DIRECT INFERENCE 米这d在s磷这n和ns和证明
我们在上一节结束时证明了偶数之和是偶数。在这个证明中,有几个关键因素。其中之一是为整数宇宙的成员引入符号。换句话说,我们不是说 “假设我们有两个整数”,而是为我们假设的宇宙中的两个成员引入符号。我们怎么知道要使用代数符号呢?像往常一样,这个问题有一个以上的可能答案。在这种情况下,我们的直觉可能是基于对什么是偶数的思考,并意识到这个定义本身本质上就是符号。 你可能会说,偶数只是另一个数字的两倍,你会是对的。显然,该定义中没有任何符号。但它们确实存在;它们是 “偶数 “和 “另一个数 “这两个短语。既然我们都知道代数用符号变量比用文字更容易,我们应该认识到使用代数符号是有意义的)。因此,这个决定是基于经验,而不是逻辑。
接下来我们假设我们讨论的两个整数是偶数。然后我们使用了偶数的定义,我们之前的括号注释表明我们很自然地使用符号的定义。定义告诉我们,如果米是偶数,则存在另一个整数一世这样米=2一世. 我们将其与以下假设相结合米甚至得出结论,事实上确实存在一个整数一世这样米=2一世. 这是一个使用直接推理原理的例子C一种ll和d米这d在sp这n和ns一世n大号一种吨一世n,吨H这在GH在和在这n′吨在这rr是一种b这在吨大号一种吨一世nn一种米和s一世n吨H一世sC这在rs和.原理说从p和p⇒q我们可以得出结论q. 这个原则是逻辑论证的基石。如果你考虑真值表p⇒q,您可以将此规则视为描述此真值表的一行。有很多这样的规则C一种ll和dr在l和s这F一世nF和r和nC和人们通常在没有明确说明的情况下在证明中使用。在开始正式研究推理规则之前,我们将尝试梳理出我们在证明中使用了哪些规则。
然后我们用代数来证明,因为米=2一世和n=2j,存在一个ķ这样米+n=2ķ 这在r$ķ$在一种s$一世+j$. 接下来我们再次使用偶数的定义。然后我们使用了一个推理规则,它说如果,通过假设p, 我们可以证明q, 那么语句p⇒q是真的。我们终于得出了一个宏大的结论,对于所有整数对,如果它们是偶数,那么它们的和就是偶数。这是另一个推理规则,其中一个更难以描述。我们引入了变量米和n. 我们只使用了众所周知的结果,即它们在我们的证明中存在于整数的宇宙中。因此,我们认为有理由断言我们得出的结论是米和n对任何整数对都成立。我们可以说我们正在治疗米和n作为我们宇宙的一般成员。因此,我们的推理规则说,如果我们可以证明对我们宇宙的一般成员的陈述,那么我们可以得出结论,它对我们宇宙的每个成员都是正确的。也许这条规则很难用语言表达的原因是它不仅仅是对真值表的描述,而是我们用来证明普遍量化的陈述的原则。
数学代写|离散数学代写DISCRETE MATH代考|RULES OF INFERENCE
我们已经看到了典型证明的成分。我们所说的一般证明是什么意思?陈述的证明是陈述真实的令人信服的论据。更准确地说,我们可以同意一个证明由一系列陈述组成,每个陈述要么是一个假设吨这b和d和sCr一世b和d一世n米这r和d和吨一种一世l一世n这在rd和sCr一世p吨一世这n这Fr在l和s这F一世nF和r和nC和b和l这在,普遍接受的事实,或以下复合语句的推理规则之一的结果。
1 From an example $x$ that does not satisfy $p(x)$, we may conclude $\neg p(x)$.
2 From $p(x)$ and $q(x)$, we may conclude $p(x) \wedge q(x)$.
3 From either $p(x)$ or $q(x)$, we may conclude $p(x) \vee q(x)$.
4 From either $q(x)$ or $\neg p(x)$ we may conclude $p(x) \Rightarrow q(x)$.
5 From $p(x) \Rightarrow q(x)$ and $q(x) \Rightarrow p(x)$ we may conclude $p(x) \Leftrightarrow q(x)$.
6 From $p(x)$ and $p(x) \Rightarrow q(x)$ we may conclude $q(x)$.
7 From $p(x) \Rightarrow q(x)$ and $q(x) \Rightarrow r(x)$ we may conclude $p(x) \Rightarrow r(x)$.
8 If we can derive $q(x)$ from the hypothesis that $x$ satisfies $p(x)$, then we may conclude $p(x) \Rightarrow q(x) .$
9 If we can derive $p(x)$ from the hypothesis that $x$ is a (generic) member of our universe $U$, we may conclude $\forall x \in U(p(x))$.
10 From an example of an $x \in U$ satisfying $p(x)$ we may conclude $\exists x \in U(p(x))$.
数学代写|离散数学代写DISCRETE MATH代考|CONTRAPOSITIVE RULE OF INFERENCE
在练习 3.3-2 中,我们看到如果我们知道¬q→¬p,那么我们可以得出结论p→q. 要了解这有什么好处,请考虑以下示例。
证明如果n是一个正整数n2>100, 然后n>10.
证明:假设n不大于 10。ñ这在在和在s和吨H和r在l和这F一种lG和br一种F这r一世n和q在一种l一世吨一世和s在H一世CHs一种是s吨H一种吨一世F$X≤是$一种nd$C≥0$,吨H和n$CX≤C是$.那么自从1≤n≤10,
n⋅n≤n⋅10≤10⋅10=100.
因此n2不大于 100 。因此,如果n2>100,n必须大于10.
我们可以对推理给出一个直观的解释一种ndd这一世nGs这在这在ldb和一种G这这d和X和rC一世s和,但我们不需要,因为我们可以看到我们只是在使用练习 2 的结果。
我们采用练习 2 的结果作为推理规则,称为对立推理规则。
11 从¬q⇒¬p我们可以得出结论p⇒q.
在我们证明中国剩余定理的定理 2.3.6 中,我们想证明对于某个函数F如果X和是是 0 到 0 之间的不同整数米n−1, 然后F(X)≠F(是). 为了证明这一点,我们假设事实上F(X)=F(是)并证明了X和是不是 0 和 0 之间的不同整数米n−1. 如果我们知道对立推理的原理,我们就可以在当时和那里得出结论:F是一对一的。相反,我们使用了更常见的反证法原理来完成我们的证明,下面将详细讨论。如果你回头看这个证明,你会发现我们可能已经能够通过使用对立推理将它缩短一个句子。
快速浏览真值表p⇒q和q⇒p应该让我们相信这两个陈述是不等价的。q⇒p被称为逆p⇒q. 令人惊讶的是,即使是专业数学家,当他们想要证明一个陈述本身时,他们常常心不在焉地证明一个陈述的反面。尽量不要加入这个人群!
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。