## 数学代写|数论作业代写number theory代考|MATH3170

2022年12月23日

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## 数学代写|数论作业代写number theory代考|EXISTENCE OF TRANSCENDENTAL NUMBERS

The first question we need to address about transcendental numbers is whether or not there are any! It is clear that algebraic numbers exist: for a start, all rational numbers are algebraic, and we have also given a few examples of irrational algebraic numbers. However, it is conceivable that every complex number could be a root of a rational polynomial, in which case transcendental numbers would not exist.

Notice, by the way, that we have so far only seen algebraic numbers of degree up to 4 . It is not at all clear that algebraic numbers of arbitrarily high degree exist. If, for example, we were to consider polynomials with real (rather than rational) coefficients, then there would be no irreducible polynomials of degree greater than 2 . The situation in this case would therefore be very simple: all real numbers would be algebraic (over $\mathbb{R}$ ) of degree 1 , and all nonreal complex numbers would be algebraic (over $\mathbb{R}$ ) of degree 2 . Among the complex numbers there would be no algebraic numbers of higher degree, and no transcendental numbers.

The existence of transcendental numbers was first proved by Joseph Liouville, who attempted to show that $e$ is not an algebraic number. He failed in this aim but achieved enough to allow him in 1844 (and again, using different techniques, in 1851) to give specific examples of transcendental numbers. A completely different proof was given three decades later by Georg Cantor: a proof which is perhaps simpler, though, as it does not provide any specific examples of transcendentals, possibly somehow beside the point as far as number theory is concerned. We shall begin with Cantor’s proof.

Cantor proved the existence of transcendental numbers simply by showing that there are, in a sense, more complex numbers than algebraic numbers. Specifically, the set of complex numbers is uncountable – this follows immediately from the uncountability of the reals, proved by Cantor in 1874 – while, as we shall now show, the set of (complex) algebraic numbers is countable.
First, a slightly informal proof. Recall that an algebraic number is, (almost) by definition, a root of a non-zero polynomial with integral coefficients. Define the height of any such polynomial to be the maximum of the absolute values of its coefficients: that is, if $f(z)=a_n z^n+a_{n-1} z^{n-1}+\cdots+a_1 z+a_0$ with all $a_k$ integers and $a_n \neq 0$, then
$$H(f)=\max \left(\left|a_n\right|,\left|a_{n-1}\right|, \ldots,\left|a_1\right|,\left|a_0\right|\right) .$$
The height of any polynomial $f \neq 0$ is a positive integer. Let $m$ be a positive integer and consider all polynomials $f$ with $\operatorname{deg} f+H(f)=m$. Any such polynomial has degree less than $m$ and therefore at most $m$ non-zero coefficients; each of these coefficients is an integer from $-m$ to $m$. Therefore, the number of $f$ satisfying $\operatorname{deg} f+H(f)=m$ is at most $(2 m+1)^m$ and is hence finite. So we can construct a list of all non-zero polynomials over $\mathbb{Z}$ by writing down those with degree plus height equal to 1 , followed by those with degree plus height equal to 2 , and so on. We can now write down all the roots of the first polynomial in our list, followed by all the roots of the second, and so forth; deleting any repetitions in this list, we have listed all (real and complex) algebraic numbers once each. Since the set of algebraic numbers can be arranged in a list it is countable, and therefore cannot include all complex numbers. Thus transcendental numbers exist.

## 数学代写|数论作业代写number theory代考|APPROXIMATION OF REAL NUMBERS BY RATIONALS

Instead of relying on Cantor’s countability argument we can go back to Liouville’s earlier proofs. These make it possible to explicitly construct transcendental numbers and are therefore, from the number-theoretic point of view, more interesting than Cantor’s proof. They also avoid the intuitionist objections mentioned above – though perhaps not completely so, as they do make use of the Mean Value Theorem from elementary calculus.

Liouville’s methods derive from an investigation of
Joseph Liouville
$(1809-1882)$
the problem of approximating real numbers by rationals. Let $\alpha \in \mathbb{R}$; we wish to ask how closely $\alpha$ can be approximated by rational numbers $p / q$. That is, we want to know how small
$$\left|\alpha-\frac{p}{q}\right|$$
can be made by a suitable choice of the rational $p / q$. Unfortunately, this problem is too easy to be of any interest: as the rationals are dense in $\mathbb{R}$, the difference (3.4), for any $\alpha$, can be made as small as desired by chonsing a large value of $q$ and an appropriate $p$. Specifically, if we want the difference to be smaller than a positive number $\varepsilon$, we choose $q>1 / 2 \varepsilon$ and let $p$ be the closest integer to $q \alpha$. Then
$$|q \alpha-p| \leq \frac{1}{2} \quad \Rightarrow \quad\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q}<\varepsilon .$$
This observation, though not very interesting in itself, may suggest a more significant approach, namely, to insist that the closeness of approximation should depend on the denominator of the approximating fraction. In other words, we shall be interested in a fairly weak approximation if it is given by a fraction with very small denominator, whereas if the denominator is large we shall expect the approximation to be exceptionally close. One way to achieve this is to try to solve an inequality such as
$$\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^2},$$
where $\alpha$ is a given real number and we seek rational $p / q$. In this case, if we are forced to choose a large value of $q$, we do at least know that the approximation is much closer than we had previously with
$$\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q} .$$

# 数论作业代写

## 数学代写|数论作业代写number theory代考|EXISTENCE OF TRANSCENDENTAL NUMBERS

H(F)=最大限度(|一种n|,|一种n−1|,…,|一种1|,|一种0|).

## 数学代写|数论作业代写number theory代考|APPROXIMATION OF REAL NUMBERS BY RATIONALS

Liouville 的方法源自对 Joseph Liouville的调查 $(1809-1882)$

$$\left|\alpha-\frac{p}{q}\right|$$

$$|q \alpha-p| \leq \frac{1}{2} \Rightarrow\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q}<\varepsilon .$$

$$\left|\alpha-\frac{p}{q}\right|<\frac{1}{q^2},$$

$$\left|\alpha-\frac{p}{q}\right| \leq \frac{1}{2 q}$$

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