如何在方程式中添加等效符号

Question

我正在尝试改变

对此

这是我的代码：

    \begin{align*}
\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}} 
     & = (\overline{\underbrace{zz\cdots z}_{n}})^2  \\
\dfrac{x(10^{2n}-1)}{9} - \dfrac{y(10^n-1)}{9} 
     & = \dfrac{z^2(10^n-1)^2}{9^2}\\9x(10^n+1) - 9y 
     & = z^2(10^n-1)\\9x \cdot 10^n + (9x-9y) 
     & = z^2 \cdot 10^n - z^2 
     \end{align*}

Answer 1

或许是这个。

\documentclass{article}
\usepackage{amsmath}
\begin{document}
    \begin{alignat*}{4}
&\qquad&
\smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}} & = (\overline{\underbrace{zz\cdots z}_{n}})^2  \\
\Leftarrow\kern-4pt\Rightarrow&&
\dfrac{x(10^{2n}-1)}{9} - \dfrac{y(10^n-1)}{9} 
     & = \dfrac{z^2(10^n-1)^2}{9^2}\\
\Leftarrow\kern-4pt\Rightarrow&&
9x(10^n+1) - 9y 
     & = z^2(10^n-1)\\
\Leftarrow\kern-4pt\Rightarrow&&
9x \cdot 10^n + (9x-9y) 
     & = z^2 \cdot 10^n - z^2 
     \end{alignat*}
\end{document}

正如 Zarko 所说，还提供了一个双头双线的amspackage宏。但是，它比我上面提供的自编宏要窄得多。\Leftrightarrow\Leftarrow\kern-4pt\Rightarrow

Answer 2

您还可以\ArrowBetweenLines尝试mathtools：

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}

\begin{document}

 \begin{alignat*}{2}
 &  &  \smash[b]{\underbrace{\overline{xx\cdots x}}_\text{$2n$}}-\smash[b]{\underbrace{\overline{yy\cdots y}}_\text{$n$}}
     & = (\overline{\underbrace{zz\cdots z}_{n}})^2 \\
\ArrowBetweenLines
 &  & \dfrac{x(10^{2n}-1)}{9} - \dfrac{y(10^n-1)}{9} & = \dfrac{z^2(10^n-1)^2}{9^2}\\%
\ArrowBetweenLines
 &  & 9x(10^n+1) - 9y   & = z^2(10^n-1)\\
\ArrowBetweenLines
 &  & 9x \cdot 10^n + (9x-9y) & = z^2 \cdot 10^n - z^2
 \end{alignat*}

\end{document}