
\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}
我正在努力使
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}.
疊加層的第三張幻燈片為紅色,第四張幻燈片為常規黑色,但在第四張幻燈片中它完全消失了。為什麼是這樣?
答案1
-命令\only
將列印以下內容僅有的在那張投影片上,您可以定義。因此,您\only<3>
將僅在覆蓋層 3 上列印該句子。因此你的問題是看不到文字。
如果您想在覆蓋 3 上進行一些特殊準備,但想在其他覆蓋上正常顯示,請使用該\alt
命令。如果採用一組覆蓋(在您的情況<3>
和第一對大括號中,您可以定義該覆蓋中應該發生的情況。在任何其他覆蓋範圍上,使用第二對大括號的內容。
請參閱投影機手冊對於更多指令,如\uncover
, \invisible
, \visible
, ...
\color{red}
在您的情況下,呼叫覆蓋 3\color
就足夠了。著色將在周圍環境中自動結束(此處:itemize
)。
(為了證明,我添加了第五個覆蓋層,它\color{green}
不會影響環境之外的任何內容。如果您想要對命令進行更多控制,您當然可以使用\textcolor{}
或複製並貼上文本,如下所示\alt<3>{\color{red} text}{pure uncolored text}
:)
微量元素:
\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}
結果:
答案2
您可能想使用 \alert 指令
\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}