### 数学代写|数论作业代写number theory代考|MATH312

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• Statistical Computing 统计计算
• (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|数论作业代写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|数论作业代写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
$$10^{t} \equiv 1 \quad(\bmod q)$$
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
$$\alpha=a+\frac{b}{10^{t}}+\frac{b}{10^{2 t}}+\cdots=a_{1} a_{2} \cdots a_{s}, b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$ 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.

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

$$4 n\left(s^{2}+s\right)-4\left(r^{2}+r\right)+n-1=0 .$$

$$2 n\left(s^{2}+s\right)-2\left(r^{2}+r\right)+2 k+1=0$$

$$n\left(s^{2}+s\right)-\left(r^{2}+r\right)+2 k+1=0$$

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

$$0.123=0.123000 \cdots=0.122999 \cdots$$

$$\alpha=0 . a_{1} a_{2} \cdots a_{s} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$

$$\alpha=\frac{a}{10^{s}}+\frac{b}{10^{s+t}}+\frac{b}{10^{s+2 t}}+\cdots=\frac{a}{10^{s}}+\frac{b}{10^{s+t}} \frac{1}{1-10^{-t}}$$

$$10^{t} \equiv 1 \quad(\bmod q)$$

$$\alpha=\frac{p r}{10^{t}-1}=a+\frac{b}{10^{t}-1}$$

$$\alpha=a+\frac{b}{10^{t}}+\frac{b}{10^{2 t}}+\cdots=a_{1} a_{2} \cdots a_{s}, b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots b_{t} b_{1} b_{2} \cdots$$

$$10^{m+n} \alpha=\frac{2^{n} 5^{m} p}{q^{\prime}}=\frac{p^{\prime}}{q^{\prime}}$$

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