\documentclass{beamer}
\mode<presentation>
\usepackage{amsfonts,amsmath,amssymb,graphicx,xcolor}
\newtheorem{thm}{Theorem}
\begin{document}
\title{...}
\begin{frame}
\titlepage
\end{frame}
\begin{frame}
\begin{thm}
$\sqrt{2}$ is irrational.
\end{thm}\pause
\begin{proof}
The proof is by contradiction.\pause
\begin{itemize}
\item\only<3>{\textcolor{red}{Suppose, for a contradiction, that $\sqrt{2}$ is rational.
That is, there are coprime integers $a$ and $b$ such that $\sqrt{2}=\frac{a}{b}.$}}\pause
\item\only<4>{\textcolor{red}{$\sqrt{2}$}}
\end{itemize}
\end{proof}
\end{frame}
\end{document}
我正在努力
Suppose, for a contradiction, that $\sqrt{2}$ is rational.
That is, there are coprime integers $a$ and $b$ such that \sqrt{2}=\frac{a}{b}.
覆盖图的第三张幻灯片中是红色,第四张幻灯片中是普通黑色,然而在第四张幻灯片中,黑色完全消失了。这是为什么呢?
答案1
-命令\only
将打印以下内容仅有的在该幻灯片上,您定义。因此,您\only<3>
将只在覆盖层 3 上打印该句子。它在第 4 层和任何后续层上都是不可见的。因此,您看不到文本的问题就出现了。
如果您希望在覆盖层 3 上进行一些特殊准备,但在其他覆盖层上保持正常外观,请使用该\alt
命令。如果采用覆盖层集(在您的案例中<3>
以及在第一对括号中,您可以定义该覆盖层中应该发生什么。在任何其他覆盖层上,将使用第二对括号的内容。
查看投影机手册用于进一步的命令,如\uncover
,,,...\invisible
\visible
就您而言,调用\color{red}
覆盖 3就足够了\color
。作为命令,后面的所有内容都将为红色。着色将在周围环境处自动结束(此处:)itemize
。
(我添加了第五个覆盖,以证明它\color{green}
不会影响环境之外的任何东西。如果您想更好地控制命令,您当然可以使用\textcolor{}
或复制和粘贴文本,如下所示\alt<3>{\color{red} text}{pure uncolored text}
:)
梅威瑟:
\documentclass{beamer}
\mode<presentation>
\usepackage{amsfonts,amsmath,amssymb,graphicx,xcolor}
\newtheorem{thm}{Theorem}
\begin{document}
\title{...}
\begin{frame}
\titlepage
\end{frame}
\begin{frame}
\begin{thm}
$\sqrt{2}$ is irrational.
\end{thm}\pause
\begin{proof}
The proof is by contradiction.\pause
\begin{itemize}
\item\alt<3>{\color{red}}{} Suppose, for a contradiction, that
$\sqrt{2}$ is rational. That is, there are coprime integers
$a$ and $b$ such that $\sqrt{2}=\frac{a}{b}.$\pause
\item\alt<4>{\color{red}}{\color{green}}$\sqrt{2}$
\end{itemize}
\only<5>{Easy Peasy!}
\end{proof}
\end{frame}
\end{document}
结果:
答案2
您可能需要使用 \alert 命令
\documentclass{beamer}
\mode<presentation>
\usepackage{amsfonts,amsmath,amssymb,graphicx,xcolor}
\newtheorem{thm}{Theorem}
\begin{document}
\title{...}
\begin{frame}
\titlepage
\end{frame}
\begin{frame}
\begin{thm}
$\sqrt{2}$ is irrational.
\end{thm}\pause
\begin{proof}
The proof is by contradiction.\pause
\begin{itemize}
\item\alert<3>{Suppose, for a contradiction, that $\sqrt{2}$ is rational. That is, there are coprime integers $a$ and $b$ such that $\sqrt{2}=\frac{a}{b}.$}\pause
\item\alert<4>{$\sqrt{2}$}
\end{itemize}
\end{proof}
\end{frame}
\end{document}