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\newcommand{\br}{\mathbb R} % this is the symbol for the real numbers. mathbb is the "blackboard bold" font
\newcommand{\bz}{\mathbb Z} % integers
\newcommand{\bq}{\mathbb Q} % rationals
\newcommand{\bn}{\mathbb N} % natural numbers
\newcommand{\co}{\colon\thinspace} % this is a good way to write functions, i.e instead of $f: X \to Y$, you would write $f\co X \to Y$. It looks nicer.
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\begin{document}
% This is for the title
{\Large {\bf Homework 3}} \hfill MATH 301 \\
{Solutions to graded problem} \\[-2mm]
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\bigskip
%\textbf{Reading.} Reach Chapter 1 in the course textbook (\emph{Abstract Algebra} by Thomas W. Judson.
\setcounter{Exercise}2
\bigskip
\begin{ExGraded}
Let \( n \in \bn \).
Prove that given any \( m \in \bz \) there exists a unique element \( a \in \{0, 1, 2, \ldots, n-1 \} \) such that \( m \equiv a \pmod n \).
\end{ExGraded}
\begin{Solution}
Let \( m \in \bz \).
By the division algorithm, there exists unique \( q, a \in \bz \) such that \( m = qn + a \) and \( a \in \{0, 1, \ldots, n-1\} \).
Rearranging the above equality, we have that \( m-a = qn \), and hence \( n \mid m-a \).
This implies that \( m \equiv a \pmod n \) and \( a \in \{0, 1, \ldots, n-1\} \), as desired.
It is left to show that \( a \) is unique: let \( a' \in \{0, 1, \ldots, n-1\} \) such that \( m \equiv a' \pmod n \).
Similar to the above (but in reverse), there exists \( q' \in \bz \) such that \( m = q'n + a' \).
As \( 0 \leq a' < n \), the uniqueness component of the division algorithm implies that \( a' = a \).
Therefore, there exists a unique \( a \in \{0, \ldots, n-1\} \) such that \( m \equiv a \pmod n \).
\end{Solution}
\setcounter{Exercise}5
\medskip
\begin{ExChallenge}
Let \( m,n \in \bn \) be relatively prime, and let \( a, b \in \bz \).
Prove that there exists \( x \in \bz \) such that
\begin{align*}
x &\equiv a \pmod m \\
x &\equiv b \pmod n .
\end{align*}
(Hint: Start by writing 1 as a linear combination of \( m \) and \( n \).)
\end{ExChallenge}
\begin{Solution}
As \( m \) and \( n \) are relatively prime, there exists \( s, t \in \bz \) such that \( 1 = ms + nt \).
From this, we see that \( 1- ms = nt \) and \( 1 - nt = ms \), and hence, \( ms \equiv 1 \pmod n \) and \( nt \equiv 1 \pmod m \).
Setting \( x = bms + ant \), we have \[ x \equiv 0 + ant \equiv a(1) \equiv a \pmod m \] and \[ x \equiv bms + 0 \equiv b(1) \equiv b \pmod n. \]
Therefore, \( x \) is as desired.
\end{Solution}
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