### 数学代写|数论代写Number theory代考|MTH2106

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数论代写Number theory代考|IRRATIONAL SURDS

The following result is well known, and was, essentially, proved by Pythagoras or one of his followers.
Theorem 1.1. $\sqrt{2}$ is irrational.
Proof by contradiction. Suppose that $\sqrt{2}=p / q$, where $p$ and $q$ are integers with no common factor, and with $q \neq 0$. Squaring both sides and multiplying by $q^{2}$, we have $p^{2}=2 q^{2}$. Thus $p^{2}$ is even and so $p$ is even, say $p=2 r$. Substituting for $p$ gives $q^{2}=2 r^{2}$ and so $q$ is even. Thus $p$ and $q$ have a common factor of 2 , and this contradicts our initial assumption. Therefore, $\sqrt{2}$ is irrational.

Plato records that his teacher Theodorus proved the irrationality of $\sqrt{n}$ for $n$ up to 17 . Historians of mathematics have wondered why he stopped just here; the question is made harder by the fact that we don’t know exactly how Theodorus’ proof ran. The following proof of the irrationality of $\sqrt{n}$ for certain values of $n$ suggests a possible reason for stopping just before $n=17$.
First, if $n=4 k$, then the irrationality of $\sqrt{n}$ is equivalent to that of $\sqrt{k}$; and if $n=4 k+2$, then the method used above for $n=2$ can be employed with only minor changes. So we concentrate on odd values of $n$. If $n$ is odd

and $\sqrt{n}=p / q$, then $n q^{2}=p^{2}$ and $p$ and $q$ must both be odd; substituting $p=2 r+1$ and $q=2 s+1$ and rearranging yields
$$4 n\left(s^{2}+s\right)-4\left(r^{2}+r\right)+n-1=0 .$$
Consider the case $n=4 k+3$. Cancelling 2 from the above equation gives
$$2 n\left(s^{2}+s\right)-2\left(r^{2}+r\right)+2 k+1=0$$
which is clearly impossible as the left-hand side is odd. This method does not work directly for $n=4 k+1$, so we consider as a subsidiary case $n=8 k+5$. Substituting as above and cancelling 4 we obtain
$$n\left(s^{2}+s\right)-\left(r^{2}+r\right)+2 k+1=0$$
but as $r^{2}+r$ and $s^{2}+s$ are both even, this is again impossible.
The remaining possibility is that $n=8 k+1$; but it appears that this case has to be split up into still further subcases, and the proof becomes much more complicated (try it!), so we shall stop here. Therefore, we have proved the following.

## 数学代写|数论代写Number theory代考|IRRATIONAL DECIMALS

The following well-known result characterises rational numbers in terms of their decimals. Note that the eventually periodic decimal expansions include the finite expansions, for instance, $0.123=0.123000 \cdots=0.122999 \cdots$.

Theorem 1.7. Rationality of decimals. A real number $\alpha$ is rational if and only if it has an eventually periodic decimal expansion.

Proof. Firstly, suppose that $\alpha$ has an eventually periodic expansion. Without loss of generality we may assume that $0<\alpha<1$, say
$$\alpha=0 . a_{1} a_{2} \cdots a_{s} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$
Let $a$ and $b$ be the non-negative integers with digits $a_{1} a_{2} \cdots a_{s}$ and $b_{1} b_{2} \cdots b_{t}$ respectively; then
$$\alpha=\frac{a}{10^{s}}+\frac{b}{10^{s+t}}+\frac{b}{10^{s+2 t}}+\cdots=\frac{a}{10^{s}}+\frac{b}{10^{s+t}} \frac{1}{1-10^{-t}},$$
which is rational. Conversely, suppose that $\alpha=p / q$ is rational, and initially assume that neither 2 nor 5 is a factor of $q$. Choose $t=\phi(q)$, where $\phi$ is Euler’s function: see definition $1.6$ in the appendix to this chapter. By Euler’s Theorem we have
$$10^{t} \equiv 1(\bmod q)$$
and so $q$ is a factor of $10^{t}-1$, say $10^{t}-1=q r$. Hence we can write
$$\alpha=\frac{p r}{10^{t}-1}=a+\frac{b}{10^{t}-1}$$
here we have used the division algorithm to guarantee that $0 \leq b<10^{t}-1$. We can thus write $b$ as a number of $t$ digits, say $b=b_{1} b_{2} \cdots b_{t}$; it is possible that $b_{1}$ is zero. Similarly, write $a=a_{1} a_{2} \cdots a_{s}$. Then
$$\alpha=a+\frac{b}{10^{t}}+\frac{b}{10^{2 t}}+\cdots=a_{1} a_{2} \cdots a_{s} \cdot b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$

and we see that $\alpha$ has an eventually periodic decimal expansion. To complete the proof we must also consider the case when $q$ has 2 or 5 as a factor. Let $q=2^{m} 5^{n} q^{\prime}$, where neither 2 nor 5 is a factor of $q^{\prime}$; then
$$10^{m+n} \alpha=\frac{2^{n} 5^{m} p}{q^{\prime}}=\frac{p^{\prime}}{q^{\prime}}$$
say; by the previous argument, the decimal expansion of $10^{m+n} \alpha$ is eventually periodic. The expansion of $\alpha$ contains exactly the same digits (with the decimal point shifted $m+n$ places), so it too is eventually periodic.

Alternative proof (sketch). To show that every rational number $\alpha=p / q$ has an eventually periodic decimal expansion, suppose without loss of generality that $0 \leq \alpha<1$, and consider how to compute the expansion
$$\alpha=0 . a_{1} a_{2} a_{3} \cdots$$
by division. We divide $10 p$ by $q$; the quotient is $a_{1}$ and the remainder, say, $p_{1}$. Dividing $10 p_{1}$ by $q$ gives quotient $a_{2}$ and remainder $p_{2}$, and so on. Since division by $q$ gives only a finite number of possible remainders, the remainder $p_{k}$ must at some stage be the same as a previous remainder $p_{j}$. From this point on the whole procedure repeats and we have $a_{k}=a_{j}, p_{k+1}=p_{j+1}$, $a_{k+1}=a_{j+1}$ and so on. Exercise. Write out this proof in more detail.

## 数学代写|数论代写Number theory代考|IRRATIONALITY OF THE EXPONENTIAL CONSTANT

Once we get beyond radical expressions and decimals, irrationality proofs, for the most part, become significantly harder. A notable exception is the irrationality of the exponential constant $e$. Apart from the intrinsic interest of the result, its proof provides our first glimpse of an idea which will recur again and again in irrationality arguments, and which we shall employ extensively in Chapters 2 and $5 .$

Proof. Assume that $e=p / q$ is rational. That is,
$$\frac{p}{q}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots$$
and for any positive integer $n$, we have
$$\frac{p n !}{q}=n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\cdots+1+R$$
where $R$ (which depends on $n$ ) is given by
$$R=\frac{n !}{(n+1) !}+\frac{n !}{(n+2) !}+\cdots$$
We can estimate $R$ in terms of a geometric series:
$$R=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots<\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots=\frac{1}{n} .$$
In particular, choose $n=q$. Then
$$R=\frac{p n !}{q}-\left(n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\cdots+1\right)$$
is clearly an integer; but using (1.1), we have $0<R<1$. This is impossible, and so $e$ is irrational.

Observe that this proof relies essentially on an infinite series for $e$, and therefore has to involve concepts of calculus. In some sense this may be surprising, as number theory is usually thought of as studying discrete systems while calculus is the science of the continuous; in another sense there should be no surprise, as it is not even possible to define the number $e$ without recourse to calculus techniques. Whether it is in fact a surprise or not, we shall find that many of our future proofs will be expressed in terms of calculus.

## 数学代写|数论代写Number theory代考|IRRATIONAL SURDS

4n(s2+s)−4(r2+r)+n−1=0.

2n(s2+s)−2(r2+r)+2ķ+1=0

n(s2+s)−(r2+r)+2ķ+1=0

## 数学代写|数论代写Number theory代考|IRRATIONAL DECIMALS

10吨≡1(反对q)

10米+n一个=2n5米pq′=p′q′

## 数学代写|数论代写Number theory代考|IRRATIONALITY OF THE EXPONENTIAL CONSTANT

pq=1+11!+12!+13!+⋯

pn!q=n!+n!1!+n!2!+⋯+1+R

R=n!(n+1)!+n!(n+2)!+⋯

R=1n+1+1(n+1)(n+2)+⋯<1n+1+1(n+1)2+⋯=1n.

R=pn!q−(n!+n!1!+n!2!+⋯+1)

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