## 数学代写|离散数学作业代写discrete mathematics代考|MATH200

2023年3月27日

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## 数学代写|离散数学作业代写discrete mathematics代考|Proofs by Contraposition and Proofs by Contradiction

When we cannot easily employ a direct proof, we make use of an indirect proof. Indirect proofs do not start with the premises and end with the conclusion. There are two general types of indirect proofs, namely, proofs by contraposition and proofs by contradiction.
A proof by contraposition is based on the law of contrapositive, that is, the conditional statement $p \rightarrow q$ is equivalent to its contrapositive $\bar{q} \rightarrow \bar{p}$. In other words, in a proof by contraposition of $p \rightarrow q$, we take $\bar{q}$ as a premise, and we show that $\bar{p}$ must follow.
Example 4.7
Prove the following statements using a proof by contraposition.
(a) If a real number is irrational, then its square root is irrational.
(b) If $r=m n$, where $m$ and $n$ are positive integers, then $m \leq \sqrt{r}$ or $n \leq \sqrt{r}$.
Solution
(a) By letting $x$ be an arbitrary real number, we need to prove that if $x$ is irrational, then $\sqrt{x}$ is irrational. Using a proof by contraposition, we want to prove that if $\sqrt{x}$ is not irrational, then $x$ is not irrational, or equivalently if $\sqrt{x}$ is rational, then $x$ is rational. If $\sqrt{x}$ is rational, then $\sqrt{x}=\frac{m}{n}$ for some integers $m$ and $n \neq 0$. As a result, we have $x=\frac{m m^2}{n^2}$, which is the quotient of integers. Hence $x$ is rational. We just showed the negation of the hypothesis of the original conditional statement is true.

(b) Using a proof by contraposition, we want to prove that if $m \leq \sqrt{r}$ or $n \leq \sqrt{r}$ is false, then $r=m n$ is false, or equivalently if both $m>\sqrt{r}$ and $n>\sqrt{r}$ are true, then $r=m n$ is false. However, if $m>\sqrt{r}$ and $n>\sqrt{r}$, then $m n>r$. This shows that $m n \neq r$, which contradicts the premise $m n=r$. We just showed the negation of the hypothesis of the original conditional statement is true.
When a conditional statement is true but the conclusion is false, then the hypothesis is false. In other words, if the implication $\bar{p} \rightarrow q$ is true, but $q$ is false, then $\bar{p}$ is false or, equivalently, $p$ is true. Note that a contradiction is a proposition of the form $r \wedge \bar{r}$, where $r$ may be any proposition, thus it is always false regardless of the truth value of $r$. Therefore if we show $\bar{p} \rightarrow(r \wedge \bar{r})$ is true, then $p$ is true. The method of proof by contradiction, which is an indirect proof, states that if the supposition that statement $p$ is false leads logically to a contradiction, then $p$ is true.

## 数学代写|离散数学作业代写discrete mathematics代考|Proof by Cases and Proofs by Exhaustion

Sometimes we need to partition the proof into several disjoint parts whose union is the complete theorem and then prove each part individually. Suppose we must prove $p \rightarrow q$ and that $p$ is equivalent to $p_1 \vee p_2 \vee \ldots \vee p_n$ (where $p_1, p_2, \ldots, p_n$ are the cases). To prove a conditional statement of the form $\left(p_1 \vee p_2 \vee \ldots \vee p_n\right) \rightarrow q$, we prove $\left(p_1 \rightarrow q\right) \wedge\left(p_2 \rightarrow q\right) \wedge \ldots \wedge\left(p_n \rightarrow q\right)$, as the two statements are equivalent. Such a proof is called a proof by cases, as we have
$$(p \rightarrow q) \leftrightarrow\left(\left(p_1 \vee p_2 \vee \ldots \vee p_n\right) \rightarrow q\right) \leftrightarrow\left(\left(p_1 \rightarrow q\right) \wedge\left(p_2 \rightarrow q\right) \wedge \ldots \wedge\left(p_n \rightarrow q\right)\right)$$
Example 4.9
Assuming $k$ is a positive integer, show that $m=k^3-k$ is an even integer.
Solution
Using a proof by cases, we consider two mutually exclusive cases for $k$, that is, $k$ is even, and $k$ is odd, as every positive integer falls into one of these two mutually exclusive cases. Assuming $k$ is even, then for some integer $n$, we have $k=2 n \rightarrow m=k^3-k=k(k+1)(k-1)=2 n(2 n+1)(2 n-1)$, that is, $m$ is even, as it is a multiple of 2 . Assuming $k$ is odd, then for some integer $n$, we have $k=2 n+1 \rightarrow m=k^3-k=(k-1) k(k+1)=2 n(2 n+1)(2 n+2)$, that is, $m$ is even, as it is a multiple of 2 .
A proof by cases must check all possible cases that arise in a theorem. However, when each case involves checking an example, such a proof is called a proof by exhaustion or an exhaustive proof. Note that when the number of cases is infinitely many or just very large, then neither proof by cases nor proof by exhaustion is possible or even feasible.

# 离散数学代写

## 数学代写|离散数学作业代写discrete mathematics代考|Proofs by Contraposition and Proofs by Contradiction

(a) 如果一个实数是无理数，那么它的平方根也是无理 数。
(b) 如果 $r=m n$ ，在哪里 $m$ 和 $n$ 是正整数，那么 $m \leq \sqrt{r}$ 或者 $n \leq \sqrt{r}$.

(a) 通过让 $x$ 是任意实数，我们需要证明如果 $x$ 是无理数， 那么 $\sqrt{x}$ 是不合理的。使用反证法，我们想证明如果 $\sqrt{x}$ 不是无理数，那么 $x$ 不是不合理的，或者等效地如果 $\sqrt{x}$ 是有理数，那么 $x$ 是理性的。如果 $\sqrt{x}$ 是有理数，那么 $\sqrt{x}=\frac{m}{n}$ 对于一些整数 $m$ 和 $n \neq 0$. 结果，我们有 $x=\frac{m m^2}{n^2}$ ，这是整数的商。因此 $x$ 是理性的。我们只是 证明了原始条件语句假设的否定为真。
(b) 使用反证法，我们想证明如果 $m \leq \sqrt{r}$ 或者 $n \leq \sqrt{r}$ 是假的，那么 $r=m n$ 是假的，或者如果两者都是 $m>\sqrt{r}$ 和 $n>\sqrt{r}$ 是真的，那么 $r=m n$ 是假的。然 而，如果 $m>\sqrt{r}$ 和 $n>\sqrt{r}$ ，然后 $m n>r$. 这表明 $m n \neq r$, 这与前提相矛盾 $m n=r$. 我们只是证明了原 始条件语句假设的否定为真。

## 数学代写|离散数学作业代写discrete mathematics代考|Proof by Cases and Proofs by Exhaustion

$$(p \rightarrow q) \leftrightarrow\left(\left(p_1 \vee p_2 \vee \ldots \vee p_n\right) \rightarrow q\right) \leftrightarrow\left(\left(p_1 \rightarrow q\right)\right.$$

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## 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。