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# 数学代写|实分析代写Real Analysis代考|Order properties of Q

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## 数学代写|实分析代写Real Analysis代考|Order properties of Q

On the set $\mathbb{Q}$, a linear order relation $<$ is defined by ” $aa$ ( $b$ is greater than $c$ The law of trichotomy states that a rational number $a$ is one of following : $a<0, a=0,00$.

A rational number $a$ is said to be positive if $a>0$ and is said tc negative if $a<0$.

1. If $a, b, c \in \mathbb{Q}$ and $a<c, c<b$ both hold, we write $a<c<b$. We that $c$ lies between $a$ and $b$.
2. The ficld $\mathbb{Q}$ together with the order relation defined on $\mathbb{Q}$ satisf. O1-O4 becomes an ordered field.

If $x$ and $y$ be any two rational numbers and $x<y$, there exists a rational number’ $r$ such that $x<r<y$. That is, between any two rational numbers there exists a rational number.
\begin{aligned} & x<y \Rightarrow x+y<y+y \text {, by } O 3 \ & \Rightarrow \frac{1}{2}(x+y)<\frac{1}{2}(2 y) \text {, by } O 4 \ & \text { i.e., } \quad \frac{1}{2}(x+y)<y \text {. } \ & \text { Again, } x<y \Rightarrow x+x<x+y \text {, by } O 3 \ & \Rightarrow \quad \frac{1}{2}(2 x)<\frac{1}{2}(x+y) \text {, by } O 4 \ & \text { i.e., } \quad x<\frac{1}{2}(x+y) \text {. } \ & \end{aligned}
Therefore we have $x<\frac{1}{2}(x+y)<y$. Then $r=\frac{1}{2}(x+y)$.
We observe that between two rational numbers $x$ and $y$ (where $x<y$ ) there exists another rational number $\frac{1}{2}(x+y)$. Again between $x$ and $\frac{1}{2}(x+y)$ ( since $x<\frac{1}{2}(x+y)$ ) there exists another rational number and the process can be continued indefinitely.

We say that between any two rational numbers $x$ and $y$ (where $x<y$ ) there exist infinitely many rational numbers. This is expressed by saying that the set $\mathbb{Q}$ is dense and this property of $\mathbb{Q}$ is called the density property of $\mathbb{Q}$.

Because of this density property of $\mathbb{Q}$, between any two rational numbers $x$ and $y$ we can interpolate infinitely many rational numbers.

## 数学代写|实分析代写Real Analysis代考|Geometrical representation of rational numbers

Rational numbers can be represented by points on a straight line. Let $X^{\prime} X$ be a directed line. We take a point $O$ on the line. $O$ divides the line into two parts. The part to the right of $O$ is called the positive side and the part to the left of $O$ is called the negative side.

Let us take a point $A$ to the right of $O$. Let $O$ represent the rational number zero and $A$ represent the rational number one. Taking the distance $O A$ as the unit distance on some chosen scale, each rational number can be represented by a unique point on the line. First of all, the positive integers $2,3, \cdots$ are represented by the points $A_2, A_3, \cdots$ lying to the right of $O$ where $O A_2=2 O A, O A_3=3 O A, \cdots$ and the negative integers $-1,-2, \cdots$ are represented by the points $A_1^{\prime}, A_2^{\prime}, \cdots$ lying to the left $O$ such that $O A_1^{\prime}=O A, O A_2^{\prime}=2 O A, \cdots$

To represent a positive rational number $r$ of the form $\frac{p}{q}$ where $p, q$ are positive integers, we measure $p$ times the distance $O A$ to the right of $O$ and get a point $B$ and then measure the $q$ th part of the distance $O B$ to the right of $O$ to the get the point $P . P$ represents the rational number $r$. If $r$ be a negative rational number $(-s)$ then the point $P^{\prime}$ to the left of $O$ (where $O P^{\prime}=O P$ and $P$ represents $s$ ) represents $r$.

Thus every rational number can be made to correspond to a point on the line. If a point that corresponds to a rational number be called a rational point then we observe that between any two rational points there lie infinitely many rational points. If all the rational numbers be plotted as points on the line it appears that the whole line is covered by rational points i.e., the whole line is composed of only rational points.

## 数学代写|实分析代写Real Analysis代考|Order properties of Q

\begin{aligned} & x<y \Rightarrow x+y<y+y \text {, by } O 3 \ & \Rightarrow \frac{1}{2}(x+y)<\frac{1}{2}(2 y) \text {, by } O 4 \ & \text { i.e., } \quad \frac{1}{2}(x+y)<y \text {. } \ & \text { Again, } x<y \Rightarrow x+x<x+y \text {, by } O 3 \ & \Rightarrow \quad \frac{1}{2}(2 x)<\frac{1}{2}(x+y) \text {, by } O 4 \ & \text { i.e., } \quad x<\frac{1}{2}(x+y) \text {. } \ & \end{aligned}

## MATLAB代写

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