Abstract Algebra Homework #1
Forfatter
Chas Lewis
Sidst opdateret
9 år siden
Licens
LaTeX Project Public License 1.3c
Resumé
First abstract algebra home work.
First abstract algebra home work.
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\begin{document}
\setcounter{section}{1} % Homework Number
\begin{center}
{\Large MAT 3530 Homework \#1} \\
due 28.\ August 2015 \\
Chas Lewis
\end{center}~\\
\begin{question}
The \emph{power set} of a finite set $A$ is the set $\mathcal{P}(A)$ of all subsets of $A$.
Determine a formula for $\#\mathcal{P}(A)$ in terms of $\#A$. Prove your formula is correct.
\end{question}
\begin{proof}
If the power set of $\mathcal{P}$(A) contains the totality of all elements in A ie a$\in$A and A$\in$${P}$(A), then there also exists a relation between the cardinality of ${P}$ and the cardinality of A. We can show this by example A=\{1,2\} therefore ${P}$(A)=[\{0\},\{1\}] $\cup$ ${P}$(2)=[\{2\}]. So the cardinality of ${P}$(1) equals the cardinality of A, therefore we can show the cardinality of a added to the cardinality of ${P}$(1) equals the cardinality of ${P}$(A).
\end{proof}
\begin{question}
Let $f: X \to Y$ be a function, and let $A, B \subset X$.
\begin{enumerate}
\item Prove $f(A \cup B) = f(A) \cup f(B)$.\\
We first need to show f(A$\cup$B) $\subset$ f(A) $\cup$f(B) \
Let's say y$\in$f(A$\cup$B), and there is x$\in$f(A) \
or x$\in$B so f(x)$\in$f(B) \
in either case y=f(x)$\in$f(A)$\cup$f(B) \
secondly we need to show f(A)$\cup$f(B) $\subset$ f(A$\cup$B) \
y$\in$f(A)$\cup$f(B), so y$\in$f(A) or y$\in$f(B) \
if y$\in$f(A), then there is a$\in$A so f(a)=y such that y=f(A)$\in$f(A$\cup$B) since a$\in$f(A$\cup$B)\
also if y=f(B), then y$\in$f(A$\cup$B) and the equality is complete.\
\item Prove $f(A \cap B) \subset f(A) \cap f(B)$. \
Say y$\in$f(A$\cap$B) and x$\in$A$\cap$B \
so I can then say that x$\in$A, y$\in$f(A) and x$\in$b and y$\in$f(B). \
So y$\in$$\subset$f(A)$\cap$f(B).
\item Find a counterexample to $f(A \cap B) = f(A) \cap f(B)$.\
\end{enumerate}
\end{question}
\begin{proof}
Let's say A = \{1,2\} and b = \{1,3\}. Now we multiply A, which yields A*A=\{1,2,2,4\} and B*A=\{1,3,3,9\}. Now, f(A$\cap$B)=\{1,2\} but f(A)$\cap$f(B)=\{1,0\}.
\end{proof}
\begin{question}
Define the relation $\sim$ on $\QQ$ by $r \sim s$ if and only if $r - s \in \ZZ$. Prove $\sim$ is an equivalence relation.
\end{question}
\begin{proof}
We know that $\mathbb{Z}$$\subset$$\mathbb{Q}$ and if c$\in$r-s then c$\in$$\mathbb{Q}$ and c$\in$$\mathbb{Z}$ so r$\sim$s, r$\sim$c, and s$\sim$c so it is transitive. also, r-r$\in\ZZ,\QQ$ and s-s$\in\ZZ,\QQ$ so it is reflexive. Finally since r-s$\in\ZZ$ then s-r$\in\ZZ$ because r$\sim$s so it is symmetric.
\end{proof}
\begin{question}
Let $f: X \to Y$ be a function. Prove that the relation $u \sim v$ if and only if $f(u) = f(v)$ is an equivalence relation.
\end{question}
\begin{proof}
Say u$\in$x and v$\in$y if f(u)=f(v) and v$\in$f(v) and u$\in$f(u) then f(v)$\sim$f(u) and this show reflexive as well as symmetric. If c$\in$y then c$\in$f(v),f(u) thus c$\sim$u,c$\sim$v, and of course u$\sim$v which is transitive.
\end{proof}
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