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Chih-Hung Chang edited untitled.tex
over 8 years ago
Commit id: ba9569eb20bc0fa5984a8999a7df334b1e732d9c
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\newtheorem{theorem}{Theorem}
這篇短文,想藉由一次小考的題目來點出大家在寫證明題時常見的錯誤;由於每個人的寫法不盡相同,這邊所提的只是一個大方向,請大家自行判斷文章所指出的,是否就是自己曾經犯過的錯誤。
\section{1. 是非題}
\begin{enumerate}
% ----------------------------------------------------------------
\section{是非題}
是非題答題方式很簡單:對的給證明,錯的給反例;但也是最難寫的題型,因為判斷對錯本身就不是一件容易的事。撇除這個部份不談,是非題比較容易犯的,嚴格來說不是錯,而是寫法上失焦。我們來看底下這個例子。
\begin{itemize}
\item If $x$ and $y$ are integers of the same parity, then $xy$ and $(x+y)^2$ are of the same parity. (Two integers are of the same parity if they are both odd or both even.)
\end{itemize}
% ----------------------------------------------------------------
\item Let $A$ and $B$ be two sets. If $A \setminus B = B \setminus A$, then $A \setminus B = \varnothing$.
% ----------------------------------------------------------------
\item Let $n \in \mathbb{Z}$. If $n^3 + n$ is even, then $n$ is even. \section{運算元混淆}
% ---------------------------------------------------------------- \begin{itemize}
\item
For every positive irrational number $b$, there is an irrational number $a$ such that $0 < a < b$. Let $A$ and $B$ be two sets. If $A \setminus B = B \setminus A$, then $A \setminus B = \varnothing$.
\end{itemize}
% ----------------------------------------------------------------
\item There are infinitely many primes. \section{說明不夠詳細}
% ---------------------------------------------------------------- \begin{itemize}
\item For every positive
integer $n$ irrational number $b$, there
exists is an
odd integer $m$ irrational number $a$ such that
$2^{2n} + m$ is $0 < a
perfect square. < b$.
\end{itemize}
% ----------------------------------------------------------------
\item $(1 + \frac{1}{n})^n < n$ for every integer $n \geq 3$. \section{倒因為果}
% ---------------------------------------------------------------- \begin{itemize}
\item $n^3 + 1 > n^2 + n$ for every integer $n \geq 2$.
% ----------------------------------------------------------------
\item Suppose that $m, n, r \in \mathbb{N}$. If $r$ divides $mn$, then either $r$ divides $m$ or $r$ divides $n$.
% ----------------------------------------------------------------
\item Let $n \in \mathbb{N}$ be a composite number.
\begin{enumerate}
\item Suppose that $n = a b$ for some $a, b \in \mathbb{N}$. Then either $a \leq \sqrt{n}$ or $b \leq \sqrt{n}$.
\item There exists a prime factor $p$ of $n$ such that $p \leq \sqrt{n}$.
\end{enumerate}
\end{enumerate} \end{itemize}