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By admin 数学代写 2022年2月23日

This section will give a proof of a remarkable result.
I think its proof, based on the pigeon hole principle, is even more interesting than the result obtained.
Dirichlet proved it in the 1830 ‘s. Jacobi used similar ideas around the same time in studying elliptic functions.
The theorem involves the sum of integer multiples of numbers whose ratio is irrational. If $a / b$ is irrational,
then it is not possible that $m a+n b=0, m, m \in \mathbb{Z}$ because if this were so, you would have $\frac{-m}{n}=\frac{b}{a}$
and so the ratio of $a, b$ is rational after all. Even though you cannot get 0 (which you can get if the ratio of $a$ and $b$ is rational)
you can get such an integer combination arbitrarily small. Dirichlet did this in the 1830 ‘s long before the real numbers
were constructed from the rational numbers in 1858 and published in 1872 .

定理 1. If $a, b$ are real numbers and a/b is not rational, then for every $\varepsilon>0$
there exist integers $m, n$ such that $|m a+n b|<\varepsilon$.

证明 . Proof: Let $P_{N}$ denote all combinations of the form $m a+n b$ where $m, n$ are integers and $|m|,|n| \leq N$. Thus there are $(2 N+1)^{2}$ of these integer combinations and all of them are contained in the interval $I \equiv[-N(|a|+|b|), N(|a||+| b \mid)]$. Now pick an integer $M$ such that
$$
(2 N)^{2}2$. Now partition the interval $I$ into $M$ equal intervals. If $l$ is the length of one of these intervals, then
$$
\begin{aligned}
l M &=2 N(|a|+|b|) \
l &=\frac{2 N(|a|+|b|)}{M}<\frac{2 N(|a|+|b|)}{(2 N)^{2}}=\frac{2(|a|+|b|)}{4} \frac{1}{N} \equiv \frac{C}{N}
\end{aligned}
$$
Now as mentioned, all of the points of $P_{N}$ are contained in $I$ and there are more of these points than there are intervals in the partition of $I$ which has only $M$ intervals.

Therefore, some interval contains two points of $P_{N} \cdot 3^{3}$ But each interval has length no more than $C / N$ and so there exist $k, \hat{k}, l, \hat{l}$ integers such that
$$
|k a+l b-(\hat{k} a+\hat{l} b)| \equiv|m a+n b|<\frac{C}{N} $$ Now let $\varepsilon>0$ be given. Choose $N$ large enough that $C / N<\varepsilon$. Then the above inequality holds for some integers $m, n$.

微积分note Integer Multiples of Irrational Numbers

本节将证明一个显着的结果。
我认为它的证明，基于鸽子洞原理，比得到的结果更有趣。
Dirichlet 在 1830 年代证明了这一点。 Jacobi 大约在同一时间使用了类似的想法来研究椭圆函数。
该定理涉及比率为无理数的整数倍数之和。如果 $a / b$ 是非理性的，
那么 $m a+nb=0, m, m \in \mathbb{Z}$ 是不可能的，因为如果是这样，你将有 $\frac{-m}{n}=\frac{b} {a}$
所以 $a, b$ 的比例毕竟是合理的。即使你不能得到 0（如果 $a$ 和 $b$ 的比率是合理的，你可以得到）
你可以得到这样一个任意小的整数组合。狄利克雷在 1830 年代早在实数出现之前就这样做了
于 1858 年由有理数构成，并于 1872 年出版。

定理 2. 如果 $a, b$ 是实数并且 a/b 不是有理数，那么对于每个 $\varepsilon>0$
存在整数 $m, n$ 使得 $|m a+n b|<\varepsilon$。

\begin{证明}
证明：令$P_{N}$ 表示$m a+n b$ 形式的所有组合，其中$m, n$ 是整数，$|m|,|n| \leq N$。因此有 $(2 N+1)^{2}$ 这些整数组合，它们都包含在区间 $I \equiv[-N(|a|+|b|), N(|a| |+| b \mid)]$。现在选择一个整数 $M$ 使得
$$
(2 N)^{2}2$。现在将区间 $I$ 划分为 $M$ 相等的区间。如果 $l$ 是这些间隔之一的长度，则
$$
\开始{对齐}
l M &=2 N(|a|+|b|) \
l &=\frac{2 N(|a|+|b|)}{M}<\frac{2 N(|a|+|b|)}{(2 N)^{2}}=\frac {2(|a|+|b|)}{4} \frac{1}{N} \equiv \frac{C}{N}
\end{对齐}
$$
现在如前所述，$P_{N}$ 的所有点都包含在 $I$ 中，并且这些点的数量比 $I$ 分区中只有 $M$ 间隔的间隔要多。

因此，某个区间包含两个点 $P_{N} \cdot 3^{3}$ 但每个区间的长度不超过 $C / N$ ，因此存在 $k, \hat{k}, l, \ hat{l}$ 个整数，使得
$$
|k a+l b-(\hat{k} a+\hat{l} b)| \equiv|m a+n b|<\frac{C}{N} $$ 现在让$\varepsilon>0$ 给出。选择足够大的$N$，使$C / N<\varepsilon$。那么上述不等式对某些整数 $m, n$ 成立。
\end{证明}