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【离散数学】MIT 6.042J - Fall 2010 - Note for Lecture 2
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\title{\huge \bfseries Mathematics for Computer Science, MIT 6.042J - Fall 2010 - Note for Lecture 2}
\author{\Large Teddy van Jerry}
\date{\today}
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\chead{MIT 6.042J Lecture 2 - Introduction}
\begin{document}
\maketitle
{\Large Lecture 2: Introduction}
\section{Proof by Contradiction}
\subsection{definition}
\begin{definition}
To prove $p$ is True, we assume $p$ is False,
(i.e. $\lnot p$ is True)
then use that thypothesis to derive a false hood or contraction.
\end{definition}
\subsection{Example 1}
$$\sqrt{2} \text{ is irrational.}$$
\begin{proof}
Assume for the purpose of contradiction
that $\sqrt{2}$ is rational.
$$
\begin{aligned}
\Rightarrow & \sqrt{2} = \frac{a}{b}, (a,b)=1 \\
\Rightarrow & 2 = \frac{a^2}{b^2} \\
\Rightarrow & 2b^2 = a^2 \\
\Rightarrow & 2 | a \\
\Rightarrow & 2 | b \\
\Rightarrow & 2 | (a,b) \\
\Rightarrow & (a,b) \neq 1 \\
\Rightarrow & \sqrt{2} \text{ is irrational.}
\end{aligned}
$$
Irrational number was first discovered by ancient Greek people.
But they did not like irrational numbers as
irrational numbers can not be written finitely.
(in ancient Greek, anything infinite was perceived as bad.)
It caused a stir as it proved inconsistent axioms.
They covered it up, and denied the result.
\end{proof}
\subsection{Example 2}
$$90 > 92$$
Proof: by PowerPoint
\begin{proof}
We will take two triangles with total area 90,
and divide them into four triangles with total area greater than 92.
By the `conservation of area' axiom,
this will imply that an area of 90 is larger than an area of 92.
(Picture omitted)
So $90 > 92$.
\end{proof}
It is false because the rectangle is not drawn correctly.
\section{Induction axiom}
\subsection{Definition}
\begin{definition}
Let $P(n)$ be a predicate.
If $P(0)$ is not true and $\forall n \in \mathbb{N}$,
$P(n) \Rightarrow P(n+1)$ is True,
then $\forall n \in \mathbb{N}$, $P(n)$ is true.
\end{definition}
We can view this as a series of dominoes. Each domino knocks over the next.
\subsection{Example 1}
$$\forall n \geqslant 0 \text{, } 1+2+3+\cdots + n = \frac{n(n+1)}{2}$$
or
$$\forall n \geqslant 0 \text{, } \sum_{i = 1}^{n} = \frac{n(n+1)}{2} $$
\begin{proof}
Let $P(n)$ be proposition that $\sum_{i = 1}^{n} = \frac{n(n+1)}{2} $.
Base Case:$$\sum_{i = 1}^{0} = 0 $$
Inductive Step:
(For $n \geqslant 0$, show $P(n) \Rightarrow P(n+1)$ is true.)
Assume $P(n)$ is true for purposes of induction.
(i.e. assume $1+2+3+\cdots + n = \frac{n(n+1)}{2}$)
$$1+2+\cdots + n + (n+1) = \frac{n(n+1)}{2} + (n+1) = \frac{(n+1)(n+2)}{2}$$
\end{proof}
\subsection{Example 2}
$$\forall n \in \mathbb{N}, 3\left|(n^3-n)\right.$$
\begin{proof}
Let $P(n)$ be $3\left|(n^3-n)\right.$.
Base Case: $3|(0-0)$.
Inductive Step: For $n \geqslant 0$, show $P(n) \Rightarrow P(n+1)$ is true.
Examine
$$
\begin{aligned}
(n+1)^3-(n+1) & = n^3+3n^2+3n+1-(n+1) \\
& = n^3+3n^2+2n \\
& = (n^3-n) + 3(n^2+n) \\
\end{aligned}
$$
$$\Rightarrow 3\left|(n+1)^3-(n+1)\right.$$
\end{proof}
\subsection{Example 3}
The induction does not necessarily begin with 0.
We have:
Base Case: $P(n)$ is true.
Inductive Step: $\forall n \geqslant b \text{, } P(n) \Rightarrow P(n+1)$
Conclusion: $\forall n \geqslant b \text{, } P(n)$ is true.
\subsubsection{A false proof}
All horses are the same color.
\begin{proof}
Let $P(n)$ be: In any set of $n$ horses,
the horses are all the same color.
Base Case: One horse is the same color itself.
Inductive Step:
Assume $P(n)$ is true to prove $P(n+1)$ is true.
Consider a set of $n+1$ horses $H_1, H_2, \cdots, H_{n+1}$.
Then $H_1, H_2, \cdots, H_{n}$ are the same color.
And $H_2, H_3, \cdots, H_{n+1}$ are the same color.
Then they share the same color.
\end{proof}
The flaw here: $P(1) \Rightarrow P(2)$ is false as from 1 to $n$
shares no horse with from 2 to $n+1$.
(Or we can say it is false with the dots. The missing link with $n=1$.)
Professor: If you don't check the base case,
you can proof really great stuff.
\subsection{Example 4}
$\forall n, \exists$ way to fill a $2^n \times 2^n$ region with
the L-Shape tiles with a center square missing.
Way 1:
\begin{proof}
Assume $P(n)$ to be $\forall n, \exists$ way to fill a $2^n \times 2^n$ region with
the L-Shape tiles with a \emph{corner} square missing.
Base Case has been checked.
For $n \geqslant 0$, assume $P(n)$ to verify I.H.
so we need to show $P(n+1)$ is true.
Then in the case with $n+1$,
we place the four missing places in the middle,
fill three of them with a L-Shape tile, leaving a missing square.
\end{proof}
Way 2: Assume $P(n)$ to be $\forall n, \exists$ way to fill a $2^n \times 2^n$ region with
the L-Shape tiles with \emph{any} square missing. (Make it stronger.)
\section*{Appendix}
~
This note is written in \LaTeX.
\LaTeX\ code in my blog: \url{https://blog.csdn.net/weixin_50012998/article/details/113527826}\\
The webpage of the video: \url{https://www.bilibili.com/video/BV1zt411M7D2?p=2}
\end{document}
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