Assignment 6
Author:
Frank Yang
Last Updated:
11년 전
License:
Other (as stated in the work)
Abstract:
Assignment for Math Structure
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\title{Assignment 6}
\author{Frank Yang}
\begin{document}
\maketitle
\section{Homework list}
\begin{itemize}
\item Section 3.3: 2, 12, 24
\item Section 4.1: 26ac
\item Section 4.2: 6ab, 12cdef
\end{itemize}
\section{Solution}
\begin{enumerate}
\item[2]
We give a proof by induction.
Let $S(n)=1+5+9+...+(4n-3)$, where n is a positive integer.
We want to prove that for every n, $S(n)=2n^2-n$.
Basis step:
$S(1)=2 \times 1^2-1=1$, which is same with sum of 1.
Inductive step:
Assume $S(k)=2k^2-k$. We want to show $S(k+1)=2(k+1)^2-k$.
$$S(k+1)=1+5+9+...+(4k-3)+(4(k+1)-3)=S(k)+4(k+1)-3$$
$$S(k+1)=2k^2-k+4(k+1)-3=2k^2+4k+2-1-k$$
$$2k^2+4k+2-1-k=2(k^2+2k+1)-1-k=2(k+1)^2-(k+1)$$
So, we have shown that if $S(k)=2k^2-k$, then $S(k+1)=2(k+1)^2-k$. Since the statement is also true for the basis case, $S(n)=2n^2-n$ for every positive integer n.
\item [12]
We give a proof by induction.
Let $b(n)=1/3+1/15+...+1/(4n^2-1)$, where n is a positive integer.
(a)$b_{1}=1/3,b_{2}=2/5,b_{3}=3/7,b_{4}=4/9, b_{5}=5/11$
(b)$b_{n}=n/(2n+1)$
(c)We give a proof by induction.
We want to prove that for every n, $b_{n}=n/(2n+1)$.
Basis step:
$b_{1}=1/(2\times 1 +1)=1/3$, which is the same with sum of $1/(4\times 1^2-1)=1/3$.
Inductive step:
Assume $b_{k}=k/(2k+1)$. We want to show $b_{k+1}=(k+1)/(2(k+1)+1)=(k+1)/(2k+3)$.
$$b_{k+1}=b_{k}+1/(4(k+1)^2-1)=k/(2k+1)+1/(4(k+1)^2-1)$$
$$=k/(2k+1)+1/(4k^2+8k+3)=k/(2k+1)+1/((2k+1)(2k+3))$$
$$=k(2k+3)/((2k+1)(2k+3))+1/((2k+1)(2k+3)=(2k^2+3k+1)/((2k+1)(2k+3))$$
$$=(k+1)(2k+1)/((2k+1)(2k+3))$$
$$=(k+1)/(2k+3)$$
We have shown that if $b_{k}=k/(2k+1)$, then $b_{k+1}=(k+1)/(2(k+1)+1)=(k+1)/(2k+3)$. Since the statement is also true for the basis case, $b_{n}=n/(2n+1)$ for every positive integer n.
(d) Based on our previous proof, the statement is equivalent as $b_{\infty}$ converges. Since $\displaystyle \lim_{n \to \infty}b_{n}=\displaystyle \lim_{n \to \infty}n/(2n+1)=1/2$. The sum of this series converges to $1/2$.
\item[24]
We give a proof by induction. Let $f_{n}$ be the $n$th Fibonacci, where n is a integer $\geq 0$. We want to show that for every n, $f_{n}<2^n$.
Basis step:
$f_{0}=0$, which is less then $2^0=1$.
$f_{1}=1$, which is less then $2^1=2$.
Inductive step:
Assume $f_{k}<2^k$ and $f_{k+1}<2^{k+1}$. We want to show $f_{k+2}<2^{k+2}$.
$$f_{k+2}=f_{k}+f_{k+1}<2^k+2^{k+1}<2^{k+1}+2^{k+1}=2^{k+2}$$
We have shown that if $f_{k}<2^k$ and $f_{k+1}<2^{k+1}$, then$f_{k+2}<2^{k+2}$. Since the statement is also true for the two basis cases, $f_{n}<2^n$ for all n.
\item[26a]
We give a proof by induction.
We want to prove for all $I$, $A \cap (\displaystyle \bigcup_{i\in I} A_{i}) = \displaystyle \bigcup_{i\in I}(A \cap A_{i}) $.
Basis Step:
Assume there are only two sets in $A_{i}$ family, $A_{1}$ and $A_{2}$.
From basic set identities, we know that $A \cap (A_{1} \cup A_{2}) = (A \cap A_{1}) \cup (A \cap A_{2})$. Thus, the statement is true for basis case.
Inductive step:
Assume for any integer $k>2$, $A \cap (\displaystyle \bigcup_{i\in k} A_{i}) = \displaystyle \bigcup_{i\in k}(A \cap A_{i})$. We are going to show that $A \cap (\displaystyle \bigcup_{i\in k+1} A_{i}) = \displaystyle \bigcup_{i\in k+1}(A \cap A_{i})$.
$$A \cap (\displaystyle \bigcup_{i\in k+1} A_{i}) = A \cap (\displaystyle \bigcup_{i\in k} A_{i} \cup A_{k+1})$$
Since $\displaystyle \bigcup_{i\in k} A_{i}$ can be treated as a single set,
$$A \cap (\displaystyle \bigcup_{i\in k} A_{i} \cup A_{k+1})=(A \cap \displaystyle \bigcup_{i\in k} A_{i}) \cup (A \cap A_{k+1})$$
$$=\displaystyle \bigcup_{i\in k}(A \cap A_{i}) \cup (A \cap A_{k+1})$$
$$=\displaystyle \bigcup_{i\in k+1}(A \cap A_{i})$$
\item[26c]
We give a proof by induction.
We want to prove for all $I$, $(\displaystyle \bigcap_{i\in I} A_{i})^{c} = \displaystyle \bigcup_{i\in I} A_{i}^{c}$.
Basis Step:
Assume there are only two sets in $A_{i}$ family, $A_{1}$ and $A_{2}$.
From basic set identities, we know that $ (A_{1} \cap A_{2})^c = A_{1}^c \cup A_{2}^c$. Thus, the statement is true for basis case.
Inductive step:
Assume for any integer $k>2$, $(\displaystyle \bigcap_{i\in k} A_{i})^{c} = \displaystyle \bigcup_{i\in k} A_{i}^{c}$.
We are going to show that $(\displaystyle \bigcap_{i\in k+1} A_{i})^{c} = \displaystyle \bigcup_{i\in k+1} A_{i}^{c}$.
$$(\displaystyle \bigcap_{i\in k+1} A_{i})^{c}=(\displaystyle \bigcap_{i\in k} A_{i} \cap A_{k+1})^{c}$$
Since $\displaystyle \bigcap_{i\in k} A_{i}$ can be treated as a single set,
$$(\displaystyle \bigcap_{i\in k} A_{i} \cap A_{k+1})^{c}=(\displaystyle \bigcap_{i\in k} A_{i})^{c} \cup A_{k+1}^{c}$$
$$=(\displaystyle\bigcup_{i\in k} A_{i}^{c}) \cup A_{k+1}^{c}$$
$$=\displaystyle \bigcup_{i\in k+1} A_{i}^{c}$$
\item[6]
(a) not reflexive, symmetric, not antisymmetric, not transitive
(b) reflexive, symmetric, not antisymmetric, not transitive
\item[12]
(c) Assume $R$ and $S$ are symmetric and odered pair $(a,b)$ is in $R \cap S$. Then, $(a,b)$ is in $R$ and in $S$. Since both relations are symmetric, $(b,a)$ is also in $R$ and in $S$. So $(b,a)$ is in $R \cap S$, which shows $R \cap S$ is symmetric.
(d) Assume $R$ and $S$ are symmetric and odered pair $(a,b)$ is in $R \cup S$. Then, $(a,b)$ is in $R$ or in $S$. Since both relations are symmetric, $(b,a)$ is also in $R$ or in $S$. So $(b,a)$ is in $R \cup S$, which shows $R \cup S$ is symmetric.
(e) Assume $R$ and $S$ are transitive and odered pairs $(a,b)$, $(b,c)$ are in $R \cap S$. Then, $(a,b)$, $(b,c)$ are in $R$ and in $S$. Since both relations are transitive, $(a,c)$ is also in $R$ and in $S$. So $(a,c)$ is in $R \cap S$, which shows $R \cap S$ is transitive.
(e) Assume $R$ and $S$ are transitive and odered pairs $(a,b)$, $(b,c)$ are in $R \cup S$. Then, $(a,b)$, $(b,c)$ are in $R$ or in $S$. Since both relations are transitive, $(a,c)$ is in $R$ or in $S$. So $(a,c)$ is in $R \cup S$, which shows $R \cup S$ is transitive.
\end{enumerate}
\end{document}