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对齐*：水平间距不一致

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对齐*：水平间距不一致

我是 Latex 的新手，在使用过程中遇到了一些问题align*。在下面的代码中，前两个方程式是水平与后两个方程的间距不同：

\documentclass{article}
\usepackage[margin=1in]{geometry} 
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\dfrac{1}{n}-\dfrac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1, \inf(A)=-1$.\\\\

\textit{Proof}. Suppose for contradiction that $1$ is not an upper bound for $A$, i.e. $\exists n,m \in \mathbb{N} \mid \dfrac{1}{n} - \dfrac{1}{m} > 1$.

\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\text{Because }n(m+1)\geq m+1,\text{ we have}\\
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot\\
\end{align*}

$\newline$Suppose for contradiction that $-1$ is not a lower bound for $A$, i.e. $\ex n,m \in \N \mid$\\$ \dfrac{1}{n} - \dfrac{1}{m} < -1$.
\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\text{Because }m(1+n)\geq 1+n, \text{ we have}\\
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\end{align*}
$\newline\newline$ Claim that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$, or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp$\\$ \left|(1 - \dfrac{1}{m}) - 1\right| < \epsilon $.

\begin{align*}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\end{align*}
Hence, for $M:=\left\lceil\dfrac{1}{\epsilon}\right\rceil+1$, $ m > M \imp \left| 1 - 1 - \dfrac{1}{m}\right|<\epsilon$. Because $\epsilon$ was arbitrarily chosen, it follows that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$.

$\newline\newline$ Claim that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$, or $\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp$\\$ \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon $.

\begin{align*}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \dfrac{1}{\epsilon}\right\rceil+1, n > N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$. Because $\epsilon$ was arbitrarily chosen, it follows that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$.
\end{document}

我不太清楚为什么会发生这种情况。也许这与后两个方程中的方程有关\frac{}{}……如果是这样，我该如何修复间距，以便所有方程都对齐？谢谢。

答案1

你可能正在寻找\intertext：

\documentclass{article}
\usepackage[margin=1in]{geometry} 
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}

For $A:=\left\{\dfrac{1}{n}-\dfrac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1$, $\inf(A)=-1$.

\begin{proof}
Suppose for contradiction that $1$ is not an upper bound for $A$, i.e.\@
$\exists n,m \in \mathbb{N} \mid \dfrac{1}{n} - \dfrac{1}{m} > 1$.

\begin{align*}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
\iff&\frac{m-n}{mn}>1\\
\iff& m-n> mn\\
\iff& m > mn + n\\
\iff& m > n(m+1)\\
\intertext{Because $n(m+1)\geq m+1$, we have}
\iff& m > n(m+1) \geq m+1\\
\iff& m > m+1 &\bot\\
\intertext{\indent Suppose for contradiction that $-1$ is not a lower bound for $A$,
i.e.\@ $\ex n,m \in \N \mid \dfrac{1}{n} - \dfrac{1}{m} < -1$.}
\iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
\iff&\frac{m-n}{mn}<-1\\
\iff& m-n<-mn\\
\iff& m +mn < n\\
\iff& m(1+n)< n\\
\intertext{Because $m(1+n)\geq 1+n$, we have}
\iff& 1+n \leq m(1+n) < n\\
\iff& 1+n < n &\bot
\intertext{\indent Claim that $\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$,
or $\all \epsilon \in (0,\infty) \ex M \in \N \mid m \in \N: m>M \imp
\left|(1 - \dfrac{1}{m}) - 1\right| < \epsilon $.}
\iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
\iff& \left|-\frac{1}{m}\right|<\epsilon\\
\iff& \frac{1}{m}<\epsilon\\
\iff& m > \frac{1}{\epsilon}
\intertext{Hence, for $M:=\left\lceil\dfrac{1}{\epsilon}\right\rceil+1$,
$ m > M \imp \left| 1 - 1 - \dfrac{1}{m}\right|<\epsilon$. Because $\epsilon$ 
was arbitrarily chosen, it follows that 
$\lim\limits_{n=1, m \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = 1$.\endgraf\medskip
\indent Claim that $\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$, 
or $\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp
 \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon $.}
\iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
\iff& \left| \frac{1}{n}\right|<\epsilon\\
\iff& \frac{1}{n} < \epsilon\\
\iff& n > \frac{1}{\epsilon}
\end{align*}
Hence, for $N:=\left\lceil \dfrac{1}{\epsilon}\right\rceil+1$, 
$n > N \imp \left|(\dfrac{1}{n}-1) -(-1)\right| < \epsilon$. 
Because $\epsilon$ was arbitrarily chosen, it follows that 
$\lim\limits_{m=1, n \to \infty} \dfrac{1}{n} - \dfrac{1}{m} = -1$.
\end{proof}

\end{document}

在此处输入图片描述

答案2

编辑：发现错误：前两个环境&的最后一行有太多align

下面是清理后的 MWE（错误仍然存​​在）

  • 不要\dfrac在文本中使用，导致行距过大
  • 不要在文本中使用\\或，你永远不想在文本中手动换行（这是新用户中非常常见的错误）\newline
  • 用于\intertext{...}内部注释align（mathtools提供\shortintertext具有不同间距的注释）

清洁的 MWE

\documentclass{article}
\usepackage[margin=1in]{geometry} 
\usepackage{amsmath,amsthm,amssymb}
\usepackage{amsmath}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ex}{\:\exists\,}
\newcommand{\nex}{\:\nexists\,}
\newcommand{\all}{\:\forall\,}
\newcommand{\imp}{\Longrightarrow}
\begin{document}
For $A:=\left\{\frac{1}{n}-\frac{1}{m}:n,m\in\N\right\}$, $\sup(A) = 1, \inf(A)=-1$.

\begin{proof}
  Suppose for contradiction that $1$ is not an upper bound for $A$,
  i.e.
  $\exists n,m \in \mathbb{N} \mid \frac{1}{n} - \frac{1}{m} > 1$.
  \begin{align*}
    \iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} > 1\\
    \iff&\frac{m-n}{mn}>1\\
    \iff& m-n> mn\\
    \iff& m > mn + n\\
    \iff& m > n(m+1)\\
    \intertext{Because $n(m+1)\geq m+1$, we have}
    \iff& m > n(m+1) \geq m+1\\
    \iff& m > m+1 &\bot
  \end{align*}

  Suppose for contradiction that $-1$ is not a lower bound for $A$,
  i.e. $\ex n,m \in \N \mid \frac{1}{n} - \frac{1}{m} < -1$.
  \begin{align*}
    \iff&\frac{m}{m}\cdot\frac{1}{n} - \frac{n}{n}\cdot\frac{1}{m} < -1\\
    \iff&\frac{m-n}{mn}<-1\\
    \iff& m-n<-mn\\
    \iff& m +mn < n\\
    \iff& m(1+n)< n\\
    \intertext{Because $m(1+n)\geq 1+n$, we have}
    \iff& 1+n \leq m(1+n) < n\\
    \iff& 1+n < n &\bot
  \end{align*}
  Claim that $\lim_{n=1, m \to \infty} \frac{1}{n} -
  \frac{1}{m} = 1$, or $\all \epsilon \in (0,\infty) \ex M \in \N \mid
  m \in \N: m>M \imp$ $ \left|(1 - \frac{1}{m}) - 1\right| < \epsilon
  $.
  \begin{align*}
    \iff& \left| 1 - 1 - \frac{1}{m}\right|<\epsilon\\
    \iff& \left|-\frac{1}{m}\right|<\epsilon\\
    \iff& \frac{1}{m}<\epsilon\\
    \iff& m > \frac{1}{\epsilon}
  \end{align*}
  Hence, for $M:=\left\lceil\frac{1}{\epsilon}\right\rceil+1$,
  $ m > M \imp \left| 1 - 1 - \frac{1}{m}\right|<\epsilon$. Because
  $\epsilon$ was arbitrarily chosen, it follows that
  $\lim_{n=1, m \to \infty} \frac{1}{n} - \frac{1}{m} = 1$.

  Claim that
  $\lim_{m=1, n \to \infty} \frac{1}{n} - \frac{1}{m} = -1$, or
  $\all \epsilon \in (0,\infty) \ex N \in \N \mid n \in \N: n>N \imp$
  $ \left|(\frac{1}{n}-1) -(-1)\right| < \epsilon $.
  \begin{align*}
    \iff& \left| -1 - (-1) + \frac{1}{n}\right|<\epsilon\\
    \iff& \left| \frac{1}{n}\right|<\epsilon\\
    \iff& \frac{1}{n} < \epsilon\\
    \iff& n > \frac{1}{\epsilon}
  \end{align*}
  Hence, for $N:=\left\lceil \frac{1}{\epsilon}\right\rceil+1, n > N
  \imp \left|(\frac{1}{n}-1) -(-1)\right| <
  \epsilon$. Because
  $\epsilon$ was arbitrarily chosen, it follows that
  $\lim\limits_{m=1, n \to \infty} \frac{1}{n} - \frac{1}{m} = -1$.
\end{proof}
\end{document}

这是第一页的图片（清洁后），我明白你所说的不均匀是什么意思

在此处输入图片描述

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