# 数学代写|伽罗瓦理论代写Galois Theory代考|MATH121

## 数学代写|伽罗瓦理论代写Galois Theory代考|Approximate Constructions

For the technical drawing expert we emphasise that we are discussing exact constructions. There are many approximate constructions for trisecting the angle, for instance, but no exact methods. Dudley (1987) is a fascinating collection of approximate methods that were thought by their inventors to be exact. Figure 10 is a typical example. To trisect angle BOA, draw line BE parallel to OA. Mark off AC and CD equal to OA, draw arc DE with centre $\mathrm{C}$ and radius CD. Drop a perpendicular EF to OD and draw arc FT centre $\mathrm{O}$ radius $\mathrm{OF}$ to meet BE at T. Then angle TOA approximately trisects angle BOA. See Exercise 7.10.

The Greeks were well aware that by going outside the Platonic constraints, all three classical problems can be solved. Archimedes and others knew that angles can be trisected using a marked ruler, as in Figure 11. The ruler has marked on it two points distance $r$ apart. Given $\angle \mathrm{AOB}=\theta$ draw a circle centre $\mathrm{O}$ with radius $r$, cutting $\mathrm{OA}$ at $\mathrm{X}, \mathrm{OB}$ at Y. Place the ruler with its edge through $\mathrm{X}$ and one mark on the line $\mathrm{OY}$ at $\mathrm{D}$; slide it until the other marked point lies on the circle at $\mathrm{E}$. Then $\angle \mathrm{EDO}=\theta / 3$. For a proof, see Exercise 7.4. Exercise $7.14$ shows how to duplicate the cube using a marked ruler.

Setting your compass up against the ruler so that the pivot point and the pencil effectively constitute such marks also provides a trisection, but again this goes beyond the precise concept of a ‘ruler-and-compass construction’. Many other uses of ‘exotic’ instruments are catalogued in Dudley (1987), which examines the history of trisection attempts. Euclid may have limited himself to an unmarked ruler (plus compass) because it made his axiomatic treatment more convincing. It is not entirely clear what conditions should apply to a marked ruler-the distance between the marks causes difficulties. Presumably it ought to be constructible, for example.

The Greeks solved all three problems using conic sections, or more recondite curves such as the conchoid of Nichomedes or the quadratrix (Klein 1962, Coolidge 1963). Archimedes tackled the problem of squaring the circle in a characteristically ingenious manner, and proved a result which would now be written
$$3 \frac{10}{71}<\pi<3 \frac{1}{7}$$
This was a remarkable achievement with the limited techniques available, and refinements of his method can approximate $\pi$ to any required degree of precision.

## 数学代写|伽罗瓦理论代写Galois Theory代考|Constructions in C

We begin by formalising the notion of a ruler-and-compass construction. Assume that initially we are given two distinct points in the plane. Equivalently, by Euclid’s Axiom 1, we can begin with the line segment that joins them. These points let us choose an origin and set a scale. So we can identify the Euclidean plane $\mathbb{R}^2$ with $\mathbb{C}$, and assume that these two points are 0 and 1.

Euclid dealt with finite line segments (his condition (1) above) but could make them as long as he pleased by extending the line (condition (2)). We find it more convenient to work with infinitely long lines (modelling an infinitely long ruler), which in effect combines Euclid’s conditions into just one: the possibility of drawing the (infinitely long) line that passes through two given distinct points. From now on, ‘line’ is always used in this sense.
If $z_1, z_2 \in \mathbb{C}$ and $0 \leq r \in \mathbb{R}$, define
\begin{aligned} L\left(z_1, z_2\right) &=\text { the line joining } z_1 \text { to } z_2 \quad\left(z_1 \neq z_2\right) \ C\left(z_1, r\right) &=\text { the circle centre } z_1 \text { with radius } r>0 \end{aligned}
We now define constructible points, lines, and circles recursively:
Definition 7.2. For each $n \in \mathbb{N}$ define sets $\mathcal{P}n, \mathcal{L}_n$, and $\mathcal{C}_n$ of $n$-constructible points, lines, and circles, by: \begin{aligned} \mathcal{P}_0=&{0,1} \ \mathcal{L}_0=& \emptyset \ \mathcal{C}_0=& \emptyset \ \mathcal{L}{n+1}=&\left{L\left(z_1, z_2\right): z_1, z_2 \in \mathcal{P}n\right} \ \mathcal{C}{n+1}=&\left{C\left(z_1,\left|z_2-z_3\right|\right): z_1, z_2, z_3 \in \mathcal{P}n\right} \ \mathcal{P}{n+1}=&\left{z \in \mathbb{C}: z \text { lies on two distinct lines in } \mathcal{L}{n+1}\right} \cup \ &\left{z \in \mathbb{C}: z \text { lies on a line in } \mathcal{L}{n+1} \text { and a circle in } \mathcal{C}{n+1}\right} \cup \ &\left{z \in \mathbb{C}: z \text { lies on two distinct circles in } \mathcal{C}{n+1}\right} \end{aligned}

# 伽罗瓦理论代考

## 数学代写|伽罗瓦理论代写Galois Theory代考|Approximate Constructions

$$3 \frac{10}{71}<\pi<3 \frac{1}{7}$$

## 数学代写|伽罗瓦理论代写Galois Theory代考|Constructions in C

$L\left(z_1, z_2\right)=$ the line joining $z_1$ to $z_2 \quad\left(z_1 \neq z_2\right) C\left(z_1, r\right) \quad=$ the circle centre $z_1$ with radius $r>0$

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