\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.5in} \pagestyle{fancy} \lhead{\color{blue} Lecture Notes} \chead{\color{black} \large The Set of All Integers} \rhead{ \footnotesize page \ \thepage} \cfoot{} \lfoot{\footnotesize \copyright $\;$ Hidegkuti, Williams 2018} \rfoot{\footnotesize Last revised: August 1, 2018} \textwidth 7.6in \textheight 9.7in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \begin{document} %TCIMACRO{% %\TeXButton{parskip 0.06 stretch 1.12}{\setlength{\parskip}{0.06in} %\setstretch{1.12}}}% %BeginExpansion \setlength{\parskip}{0.06in} \setstretch{1.12}% %EndExpansion Recall that the set of all natural numbers (also called counting numbers) is $% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion =\left\{ 1,\,2,\,3,\,4,\,\ldots \right\} $. \ This set was historically the first set that people used. \ In this course, it will be a recurring theme that a mathematical system or set would be enlarged. \ This was the case with the natural numbers. \ Why would mathematicians of past centuries feel the need to step beyond the natural numbers? \ One reason is closure. Recall the meaning of closure. \ The set $% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ is closed under addition. \ In other words, the sum of \textit{any} two natural numbers is also a natural number. \ $% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ is also closed under multiplication. \ However, we do not have closure under subtraction and division. \ We \textit{can} find a subtraction, say $% 3-12$, or a division, $10\div 7$ that do not result in natural number. \ If we stay within the set of natural numbers, this means that the results for $% 3-12$ or $10\div 7$ do not exist. \ It is a common theme in mathematics to work towards closure. \ \vspace{0.04in}\vspace{0.04in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt}}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion The set of all natural numbers, $% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion =\left\{ 1,\,2,\,3,\,4,\,\ldots \right\} $ is closed under addition and multiplication, but not under subtraction and division.% %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \vspace{0.05in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt}}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \textbf{Definition: }\ The set of all integers, denoted by $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $, is the set \FRAME{dtbpF}{2.904in}{0.3009in}{0pt}{}{}{pic1integers.bmp}{% \special{language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file "F";width 2.904in;height 0.3009in;depth 0pt;original-width 7.2809in;original-height 0.7809in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic1integers.bmp';file-properties "XNPEU";}}% %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \vspace{0.05in} %TCIMACRO{% %\TeXButton{\begin{minipage}{4.5in}}{\begin{minipage}{4.5in} %\begin{flushleft}}}% %BeginExpansion \begin{minipage}{4.5in} \begin{flushleft}% %EndExpansion The set of integers completely contains the set of natural numbers. \ In other words, the set of all natural numbers is a subset of the set of all integers, $% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion \subseteq %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $. \ We can also imagine that we started with the natural numbers and added zero and the opposite of each natural number to form the set of all integers.% \vspace{0.4in} %TCIMACRO{% %\TeXButton{\end{minipage}}{\end{flushleft} %\end{minipage}}}% %BeginExpansion \end{flushleft} \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{2.5in}}{\begin{minipage}{2.5in} %\setlength{\leftskip}{25pt}}}% %BeginExpansion \begin{minipage}{2.5in} \setlength{\leftskip}{25pt}% %EndExpansion \FRAME{itbpF}{2.9239in}{1.8109in}{0in}{}{}{pic2integers.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.9239in;height 1.8109in;depth 0in;original-width 3.0727in;original-height 1.8931in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic2integers.bmp';file-properties "XNPEU";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.05in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} % %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt} %\begin{flushleft} %\setlength{\leftskip}{20pt} %\setlength{\hangindent}{58pt}}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt} \begin{flushleft} \setlength{\leftskip}{20pt} \setlength{\hangindent}{58pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \textbf{Definition: }\ The \textbf{opposite} of $3$ is written as $-3$. \ For any number, the sum of the number and its opposite is zero. \ Another expression for the opposite is the \textbf{additive inverse}.% %TCIMACRO{% %\TeXButton{end of box}{\end{flushleft} %\end{boxedminipage}}}% %BeginExpansion \end{flushleft} \end{boxedminipage}% %EndExpansion \vspace{0.05in} The opposite of $3$ is $-3$. \ The opposite of $-3$ is $3$. \ The opposite of zero is zero itself.\vspace{0.05in} The negative sign already has a meaning, that of subtraction. \ We now are facing an ambiguity that is often the source of confusion. \ Does a negative sign denote the opposite of a number, or does it denote subtraction? \ This is a question that we often need to ask ourselves. While the answer always clearly exists, it very much depends on the context. \ For example, the negative sign in $-3$ clearly denotes that we are talking about the opposite of $3$ or negative $3$. \ However, if we place a number in front of it, like in $8-3$, the same sign here denotes subtraction. \ And what about $8\left( -3\right) $? \ Now the parentheses tells us that the negative sign does not denote subtraction, rather it describes the number after it as negative. \begin{center} \begin{tabular}{ccccc} $-3$ & ~~~~~~~~~~~ & $8-3$ & ~~~~~~~~~~~ & $8\left( -3\right) $ \\ the opposite of $3$ & & subtraction & & the opposite of $3$% \end{tabular} \end{center} \pagebreak We often depict integers with a number line. \ \FRAME{dtbpF}{4.0923in}{% 0.6261in}{0pt}{}{}{pic3integers.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 4.0923in;height 0.6261in;depth 0pt;original-width 7.3025in;original-height 1.094in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic3integers.bmp';file-properties "XNPEU";}}% \vspace{0.05in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt}}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \textbf{Definition: }\ \ (Ordering on $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $) \ Between two integers, the one on the right is greater. %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \vspace{0.05in} Another way of envisioning this is to think of a positive number as money and a negative number as debt, and ask: who is richer? \ $2<10$ was obvious, and also that $-5<3$, but now we see that between $-100$ and $-2$, $-2$ is greater. \ After all, the person who has no money and only $2$ dollars of debt is better off than another person who has no money and a $100$ dollars of debt. \ And so $-100<-2$. \ We can swap inequalites, as long as the smaller part of the inequality sign points to the smaller number.\vspace{0.06in} \begin{tabular}{lllllll} $-100<-2$ & ~~~ & read: $-100$ is less than $-2$ & ~~~~~~~~~~~ & $-100\leq -2 $ & ~~~ & read: $-100$ is less than or equal to $-2$ \\ $-2>-100$ & & read: \ $-2$ is greater than $-100$ & & $-2\geq -100$ & & read: \ $-2$ is greater than or equal to $-100$% \end{tabular} \ Wether we plot them on a number line or think money and debt, we will agree that $-1000\,000$ \ (negative one million) is less than $5$. \ But what \ if we wanted to compare the size of numbers, ignoring their signs? \ Suppose we want to say that a million dollar debt is a lot of debt. \ In this case, we use the concept of the absolute value of a number.\vspace{0.05in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\begin{flushleft} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt} %\setlength{\hangindent}{58pt}}}% %BeginExpansion \begin{boxedminipage}{7.6in} \begin{flushleft} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt} \setlength{\hangindent}{58pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \textbf{Definition: }\ \ The \textbf{absolute value of a number} is its distance from zero on the number line. \ We denote the absolute value of a number $x$ by $\left\vert \,x\,\right\vert $. %TCIMACRO{% %\TeXButton{end of box}{\end{flushleft} %\end{boxedminipage}}}% %BeginExpansion \end{flushleft} \end{boxedminipage}% %EndExpansion \vspace{0.05in} Distances can never be negative. \ $-5$ is $5$ units away from zero on the number line. \ So is $5$, only it is in the other direction. \ So, the absolute value of $5$ and $-5$ are both $5$. \ \vspace{0.05in} \textbf{Example 1.} \ Compute each of the following.\vspace{0.05in} %TCIMACRO{\TeXButton{par 58}{\setlength{\parindent}{58pt}}}% %BeginExpansion \setlength{\parindent}{58pt}% %EndExpansion a) \ $\left\vert -2\right\vert $ \qquad \qquad b) \ $\left\vert \,2\,\right\vert $ \qquad \qquad c) \ $\left\vert \,0\,\right\vert $ \qquad \qquad d) \ $-\left\vert -5\right\vert $ \vspace{0.05in}\vspace{0.05in} %TCIMACRO{% %\TeXButton{par 0 }{\setlength{\parindent}{0pt} %}}% %BeginExpansion \setlength{\parindent}{0pt} % %EndExpansion \textbf{Solution:} \ a) \ The number $-2$ is $2$ units away from zero on the number line. \ Thus $\left\vert -2\right\vert =2$. %TCIMACRO{% %\TeXButton{par 49 }{\setlength{\parindent}{49pt} % %}}% %BeginExpansion \setlength{\parindent}{49pt} % %EndExpansion b) \ The number $2$ is $2$ units away from zero on the number line. \ Thus $% \left\vert \,2\,\right\vert =2$. c) \ The distance between zero and zero on the number line is zero. \ Thus $% \left\vert \,0\,\right\vert =0$. %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion d) \ This is an example for two negatives not making a positive. \ The way we can read this as: \textit{the opposite of the absolute value of negative five}. \ The absolute value of $-5$ is $5.$ \ The opposite of that is $-5$. \ \newline Using notation, $-\left\vert -5\right\vert =-5$. %TCIMACRO{% %\TeXButton{par 0 hang 108}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{108pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{108pt}% %EndExpansion \textbf{Addition of Integers:} \ Again, think money and debt. \ Positive numbers represent money, negative numbers represent debt. \ Adding zero to anything will leave the other number unchanged.\vspace{0.07in} \pagebreak \textbf{Example 2.} \ Compute each of the following sums. %TCIMACRO{\TeXButton{par 58}{\setlength{\parindent}{58pt}}}% %BeginExpansion \setlength{\parindent}{58pt}% %EndExpansion a) \ $-4+7$ \qquad\ \ \ \ b) \ $\,-3+\left( -8\right) \,$ \qquad\ \ \ \ c) \ $\,3+\left( -14\right) $ \qquad\ \ \ \ d) \ $-7+2$ \qquad\ \ \ \ e) \ $-2+0$ %TCIMACRO{% %\TeXButton{par 0 hang 66}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{66pt}% %EndExpansion \textbf{Solution:} \ a) \ We think of $-4+7$ as follows. We start with a person who has no money and is in debt by $4$ dollars. \ To this, we add $7$ dollars. \ So the person pays off all that $4$ dollar debt and is still left with $3$ dollars. So $-4+7=3$. %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion b) \ We think of $-3+\left( -8\right) $ as follows. We start with a person who has no money and is in debt by $3$ dollars. \ To this, we add another debt of $8$ dollars. \ So this person is still now in debt by $11$ dollars. So $-3+\left( -8\right) =-11$. %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion c) \ We think of $3+\left( -14\right) $ as follows. We start with a person who has $3$ dollars. \ To this, we add a debt of $14$ dollars. \ So the person pays off all the debt he can - that is $3$ dollars and is still in debt by $11$ dollars. So $3+\left( -14\right) =-11$. %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion d) \ We think of $-7+2$ as follows. We start with a person who has no money and is in debt by $7$ dollars. \ Then this person gets $2$ dollars. \ So the person pays off all the debt she can. After she pays off $2$ dollars of debt, she has no money and is still left with $5$ dollars of debt. So $% -7+2=-5$. %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion e) \ Adding zero to any number leaves the other number unchanged. \ Therefore, $-2+0=-2$.\vspace{0.05in} %TCIMACRO{% %\TeXButton{par 0 hang 0}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion We already know how to add two positive numbers, and we know that the sum is positive. \ Adding two negative numbers is similar, we are summing debts. \ So we know that the sum of two negative numbers is also negative. \ If we add a negative and a positive number, the result may be positive or negative, depending on which number's size (or absolute value) is greater.% \vspace{0.05in} Adding zero to any number leaves that number unchanged. \ In other words, for any integer $x$, $x+0=x$ and $0+x=x$. \ In the language of algebra, we refer to a number that has no effect in an operation as an identity or identity element.\vspace{0.05in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt} %\begin{flushleft} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt} %\setlength{\hangindent}{63pt} %}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt} \begin{flushleft} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt} \setlength{\hangindent}{63pt} % %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \textbf{Definition: }\ \ \ When added to any integer, zero has no effect. \ Because of this behavior, we call zero \newline an \textbf{additive identity}. %TCIMACRO{% %\TeXButton{end of box}{\end{flushleft} %\end{boxedminipage}}}% %BeginExpansion \end{flushleft} \end{boxedminipage}% %EndExpansion \vspace{0.05in} %TCIMACRO{\TeXButton{\begin{minipage}{0.8in}}{\begin{minipage}{0.8in}}}% %BeginExpansion \begin{minipage}{0.8in}% %EndExpansion \FRAME{itbpF}{0.7325in}{0.678in}{0in}{}{}{question.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.7325in;height 0.678in;depth 0in;original-width 1.0533in;original-height 0.9729in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'question.bmp';file-properties "XNPEU";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{6in}}{\begin{minipage}{6in} %\begin{flushleft} %\setlength{\leftskip}{25pt} %\setlength{\rightskip}{25pt}}}% %BeginExpansion \begin{minipage}{6in} \begin{flushleft} \setlength{\leftskip}{25pt} \setlength{\rightskip}{25pt}% %EndExpansion Discussion: \ Based on its behavior, can you find a multiplicative identity within the set of all integers?\vspace{0.1in} %TCIMACRO{% %\TeXButton{\end{minipage}}{\end{flushleft} %\end{minipage}}}% %BeginExpansion \end{flushleft} \end{minipage}% %EndExpansion Now that we can add integers, we need to return to absolute values. \ The absolute value sign is also a grouping symbol that overwrites order of operations. \ So if there is a sum (or any other expression) within the absolute value sign, we need to perform those until we are left with just a number within the absolute value sign. \ Then we take the absolute value of that number. \textbf{Example 3.} \ Compute each of the following. %TCIMACRO{\TeXButton{par 58}{\setlength{\parindent}{58pt}}}% %BeginExpansion \setlength{\parindent}{58pt}% %EndExpansion a) \ $\left\vert -9+4\right\vert $ \qquad \qquad b) \ $\,\left\vert -9\right\vert +\left\vert \,4\,\right\vert \,$ \qquad \qquad c) \ $% \left\vert \,8\,\right\vert +\left\vert -7\right\vert $ \qquad \qquad d) \ $% \left\vert \,8+\left( -7\right) \right\vert $ %TCIMACRO{% %\TeXButton{par 0 hang 66}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{66pt}% %EndExpansion \textbf{Solution:} \ a) \ The absolute value sign is also a pair of parentheses. \ We perform the addition $-9+4$\ inside, and get $-5$. \ Then we take the absolute value of $-5$.\newline $\left\vert -9+4\right\vert =\left\vert -5\right\vert =5$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion b) \ In this example we take the absolute values and then add.\newline $\left\vert -9\right\vert +\left\vert \,4\,\right\vert =9+4=13$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion c) \ We take the absolute values and then add.\newline $\left\vert \,8\,\right\vert +\left\vert -7\right\vert =8+7=15$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion d) \ We first perform the addition inside and then take the absolute value.% \newline $\left\vert \,8+\left( -7\right) \right\vert =\left\vert \,1\,\right\vert =1$ %TCIMACRO{% %\TeXButton{par 0 hang 0}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion \pagebreak \textbf{Subtraction of Integers:} \ the following statement is always true, and is often extremely useful.\vspace{0.05in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt}}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion %TCIMACRO{\TeXButton{\large}{\large}}% %BeginExpansion \large% %EndExpansion \textbf{To subtract is to add the opposite.}% %TCIMACRO{\TeXButton{\normalsize}{\normalsize}}% %BeginExpansion \normalsize% %EndExpansion %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \vspace{0.05in} Of course, we don't always use this fact. \ In the subtraction $10-3$, we would only complicate things by applying this fact. \ It would still get us the right result. \ Instead of subtracting positive $3$, we add its opposite, negative $3$.% \begin{equation*} 10-3=7\text{ \ \ \ \ \ and also, \ \ \ }10+\left( -3\right) =7 \end{equation*}% Consider the subtraction $100-\left( -20\right) $. \ We are asked to subtract negative $20$. \ To subtract is to add the opposite. \ So, instead of subtracting negative $20$, we will add its opposite, positive $20$.% \begin{equation*} 100-\left( -20\right) =100+20=120 \end{equation*} \textbf{Example 4.} \ Compute each of the following. %TCIMACRO{\TeXButton{par 58}{\setlength{\parindent}{58pt}}}% %BeginExpansion \setlength{\parindent}{58pt}% %EndExpansion a) \ $-7-8$ \qquad \qquad b) \ $\,-9-\left( -5\right) \,$ \qquad \qquad c) \ $1-7$ \qquad \qquad d) \ $6-\left( -3\right) $ %TCIMACRO{% %\TeXButton{par 0 hang 66}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{66pt}% %EndExpansion \textbf{Solution:}\ \ a) \ First, the negative sign in front of the $7$ cannot denote subtraction. \ We are asked to subtract positive $8$ from negative $7$. \ To subtract is to add the opposite. \ Instead of subtracting positive $8$, we will add its opposite, negative $8$.\newline $-7-8=-7+\left( -8\right) =-15$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion b) \ To subtract is to add the opposite. \ Instead of subtracting negative $5 $, we will add its opposite, positive $5$.\newline $-9-\left( -5\right) =-9+5=-4$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion c) \ To subtract is to add the opposite. \ Instead of subtracting positive $7 $, we will add its opposite, negative $7$.\newline $1-7=1+\left( -7\right) =-6$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion d) \ To subtract is to add the opposite. \ Instead of subtracting negative $3 $, we will add its opposite, positive $3$.\newline $6-\left( -3\right) =6+3=9$ %TCIMACRO{% %\TeXButton{par 0 hang 0}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion Why is $100-\left( -20\right) =100+20$ ? Even if we understand how to compute this, it would be nice to understand why this is correct. \ So here is one way to think about this.\vspace{0.05in} %TCIMACRO{% %\TeXButton{\begin{minipage}{4.5in}}{\begin{minipage}{4.5in} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt} %\begin{flushleft}}}% %BeginExpansion \begin{minipage}{4.5in} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt} \begin{flushleft}% %EndExpansion Imagine that we have both a bank account a credit card with a bank. \ Suppose that at the moment, we have $150$ dollars in the bank but we also owe $50$ dollars to the bank on the credit card. \ So our net worth is $100$ dollars. %TCIMACRO{% %\TeXButton{\end{minipage}}{\end{flushleft} %\end{minipage}}}% %BeginExpansion \end{flushleft} \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3in}}{\begin{minipage}{3in} %\setlength{\leftskip}{20pt} % %\begin{flushright}}}% %BeginExpansion \begin{minipage}{3in} \setlength{\leftskip}{20pt} \begin{flushright}% %EndExpansion \begin{tabular}{|c|c|c|} \hline Money & Debt on & Total \\ in bank & credit card & Net worth \\ \hline $150$ & $50$ & $100$ \\ \hline \end{tabular}% \vspace{0.2in}% %TCIMACRO{% %\TeXButton{\end{minipage}}{\end{flushright} %\end{minipage}}}% %BeginExpansion \end{flushright} \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{4.5in}}{\begin{minipage}{4.5in} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt} %\begin{flushleft}}}% %BeginExpansion \begin{minipage}{4.5in} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt} \begin{flushleft}% %EndExpansion Suppose now that we have collected enough bonus points on the credit card to earn rewards. \ So the bank reduces our credit card debt by $20$ dollars. \ (i.e. subtracts $20$ debt, i.e. subtracts negative $20$). \ We still have our $150$ in cash, but now our debt is reduced to $30$ dollars. \ So our net worth is now $120$ dollars. \ That is $20$ dollars more than before. %TCIMACRO{% %\TeXButton{\end{minipage}}{\end{flushleft} %\end{minipage}}}% %BeginExpansion \end{flushleft} \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3in}}{\begin{minipage}{3in} %\setlength{\leftskip}{20pt} %\begin{flushright}}}% %BeginExpansion \begin{minipage}{3in} \setlength{\leftskip}{20pt} \begin{flushright}% %EndExpansion \begin{tabular}{c|c|c|c|} \cline{2-4} & \multicolumn{1}{|c|}{Money} & Debt on & Total \\ & \multicolumn{1}{|c|}{in bank} & credit card & Net worth \\ \hline \multicolumn{1}{|c|}{before} & \multicolumn{1}{|c|}{$150$} & $50$ & $100$ \\ \hline \multicolumn{1}{|c|}{after} & \multicolumn{1}{|c|}{$150$} & $30$ & $120$ \\ \hline \end{tabular}% \vspace{0.2in}% %TCIMACRO{% %\TeXButton{\end{minipage}}{\end{flushright} %\end{minipage}}}% %BeginExpansion \end{flushright} \end{minipage}% %EndExpansion After all, reducing our debt by $20$ dollars is almost the same as if someone gave us $20$ dollars so that we can pay off some of our debts.% \vspace{0.05in} \pagebreak \textbf{Multiplication of Integers:} \ Multiply the absolute values. \ If two integers have the same sign, their product is positive. \ If two integers have different signs, their product is negative. If any of the factors is zero, the product is zero. \textbf{Example 5.} \ Compute each of the following. %TCIMACRO{\TeXButton{par 58}{\setlength{\parindent}{58pt}}}% %BeginExpansion \setlength{\parindent}{58pt}% %EndExpansion a) \ $-3\cdot 5$ \qquad \qquad b) \ $\,-4\left( -5\right) \,$ \qquad \qquad c) \ $10\left( -2\right) $ \qquad \qquad d) \ $0\left( -3\right) $ \qquad \qquad e) \ $-1\left( 8\right) $ %TCIMACRO{% %\TeXButton{par 0 hang 66}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{66pt}% %EndExpansion \textbf{Solution:}\ \ a) \ The product of a negative and a positive number is negative.\newline $-3\cdot 5=-15$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion b) \ The product of two negative numbers is positive.\newline $-4\left( -5\right) =20$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion c) \ The product of a positive and a negative number is negative.\newline $10\left( -2\right) =-20$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion d) \ If any of the factors is zero, the product is zero.\newline $0\left( -3\right) =0$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion e) \ The product of a negative and a positive number is negative.\newline $-1\left( 8\right) =-8$ %TCIMACRO{% %\TeXButton{par 0 hang 0}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion Notice that if we multiply any integer by $-1$, the result is the opposite of that integer. \ This will be very useful later.\vspace{0.1in} Why do these rules work this way? \ Here is one possible explanation. \ Multiplication is defined as repeated addition.\ For example, $4\cdot 7$ means that we add $7$ to itself, $4$ times.% \begin{equation*} 4\cdot 7=7+7+7+7=28 \end{equation*}% Consider now $4\cdot \left( -7\right) $. \ This means that we add $-7$ to itself, $4$ times \begin{equation*} 4\cdot \left( -7\right) =-7+\left( -7\right) +\left( -7\right) +\left( -7\right) =-28 \end{equation*}% The logic becomes a bit tortured, but it also works with the first factor being negative. \ Consider now the product $-5\cdot 8$. \ We can interpret the first negative sign as repeated subtraction. \ So, we are subtracting $8$ repeatedly, $5$ times. If we feel that we don't have anything to subtract the first $8$ from, we can fix that by inserting a zero. \ We know that adding zero will not change any value.% \begin{eqnarray*} -5\cdot 8 &=&-8-8-8-8-8\text{ } \\ &=&0-8-8-8-8-8\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ to subtract is to add the opposite} \\ &=&0+\left( -8\right) +\left( -8\right) +\left( -8\right) +\left( -8\right) +\left( -8\right) \\ &=&-40 \end{eqnarray*}% The most interesting case is probably when we are multiplying two negative numbers. \ Consider the product $-4\cdot \left( -10\right) $. \ The first negative sign is interpreted as repeated subtraction, the second one is that we are subtracting negative numbers. \ So we are subtracting negative $10$ repeatedly, $4$ times. \ If we need something to subtract the first\ negative $10$ from, we will just insert a zero at the beginning. \begin{eqnarray*} -4\cdot \left( -10\right) &=&0-\left( -10\right) -\left( -10\right) -\left( -10\right) -\left( -10\right) \text{ \ \ \ \ \ \ \ \ to subtract is to add the opposite} \\ &=&0+10+10+10+10=40 \end{eqnarray*} \pagebreak \textbf{Division of integers:} \ We will deal with zero later. \ For the quotient of any two non-zero integers, the rules are very simple and similar to those of multiplication. \ Divide the absolute values. \ If the the integers have the same sign, the quotient is positive. \ If they have different signs, the quotient is negative. \textbf{Example 6.} \ Compute each of the following. %TCIMACRO{\TeXButton{par 58}{\setlength{\parindent}{58pt}}}% %BeginExpansion \setlength{\parindent}{58pt}% %EndExpansion a) \ $14\div \left( -2\right) $ \qquad \qquad b) \ $\,-24\div \left( -6\right) \,$ \qquad \qquad c) \ $-10\div 5$ %TCIMACRO{% %\TeXButton{par 0 hang 66}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{66pt}% %EndExpansion \textbf{Solution:}\ \ a) \ Divide the absolute values. \ The quotient of a positive and a negative number is negative.\newline $14\div \left( -2\right) =-7$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion b) \ Divide the absolute values. \ The quotient of two negative numbers is positive.\newline $-24\div \left( -6\right) =4$ %TCIMACRO{% %\TeXButton{par 49 hang 66}{\setlength{\parindent}{49pt} %\setlength{\hangindent}{66pt}}}% %BeginExpansion \setlength{\parindent}{49pt} \setlength{\hangindent}{66pt}% %EndExpansion c) \ Divide the absolute values. \ The quotient of a negative and a positive number is negative.\newline $-10\div 5=-2\vspace{0.07in}$ %TCIMACRO{% %\TeXButton{par 0 hang 0}{\setlength{\parindent}{0pt} %\setlength{\hangindent}{0pt}}}% %BeginExpansion \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt}% %EndExpansion Division is often denoted with a horizontal bar. \ The same computations can also be written as$\vspace{0.07in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{14}{-2}=-7$ \ and \ $\dfrac{-24}{-6}=4$\ \ and\ \ $\dfrac{-10}{5}=-2\vspace{0.07in}$ Why do these rules work this way? \ Division is defined in terms of multiplication backward. \ In other words,$\vspace{0.07in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{20}{4}=5$ \ is true because \ $% 4\cdot 5=20\vspace{0.07in}$ Let us apply this idea. \ \ What is the result of $14\div \left( -2\right) $% ? \begin{equation*} \dfrac{14}{-2}=\fbox{?}\text{ \ would be true because \ }-2\cdot \fbox{?}=14 \end{equation*}% Since $-2\cdot 7$ would result in $-14,$ we can only choose $-7$ to make the multiplication backward work. \ $-2\left( -7\right) =14$, therefore $\dfrac{% 14}{-2}=-7$.% \begin{equation*} \text{Similarly, \ }\dfrac{-24}{-6}=\fbox{?}\text{ \ would be true because \ }-6\cdot \fbox{?}=-24 \end{equation*}% We need to multiply $-6$ by a positive number to get a negative product. \ Only positive $4$ will work, and so $\dfrac{-24}{-6}=4$.\vspace{0.05in} \textbf{Division by Zero:} \ Now that we understand that division is defined in terms of multiplication backward, we can easily deal with zero. \ The expressions~~$\dfrac{0}{3}~~$and$~~\dfrac{3}{0}~~$look very similar, and yet they are very different. \begin{equation*} \dfrac{0}{3}=\fbox{?}\text{ \ would be true because \ }3\cdot \fbox{?}=0 \end{equation*}% In this case, we can only use zero to make the multiplication backward work. \ Let us investigate the other case.% \begin{equation*} \dfrac{3}{0}=\fbox{?}\text{ \ would be true because \ }0\cdot \fbox{?}=3 \end{equation*}% Now we are in trouble. \ If we multiply any number by zero, the product is zero. \ Therefore, we can not meaningfully complete this division, and so we say that $\dfrac{3}{0}$ is undefined. \ In written notation, $\ \dfrac{3}{0}=% \func{undefined}$. We have established that we cannot divide a non-zero number by zero. \ What about $\dfrac{0}{0}$? \ \begin{equation*} \dfrac{0}{0}=\fbox{?}\text{ \ would be true because \ }0\cdot \fbox{?}=0 \end{equation*}% Now the problem is that \textit{every} number would work, because any number times zero is zero. \ Mathematicians prefer one clean answer as a result of an operation. \ We do not like an operation that results in several numbers, let alone every number! \ So, one fundamental rule of mathematics is that division by zero is not allowed. \ \ Indeed, division by zero is not just an error: it is one of the worst errors.\vspace{0.05in}\vspace{0.05in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt} %}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt} % %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \textbf{The first commandment of mathemtics: \ }% %TCIMACRO{\TeXButton{\color{red}}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion %TCIMACRO{\TeXButton{\large}{\large}}% %BeginExpansion \large% %EndExpansion \textit{Thou shall not divide by zero. Ever...}% %TCIMACRO{\TeXButton{\large}{\large}}% %BeginExpansion \large% %EndExpansion %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \vspace{0.05in}\vspace{0.05in} \textbf{Changes in Notation} \ \vspace{0.1in} With the introduction of negative numbers, our notation will have to be modified. \ It is a widely accepted convention that if there are several signs (operations or negative) between two numbers, a pair of parentheses must separate them. \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{tabular}{clclclc} $-2+-6$ & ~~~~~~~~~~~ & $-5--3$ & ~~~~~~~~~~~ & $-3\cdot -4$ & ~~~~~~~~~~~ & $-30\div -5$ \\ \multicolumn{1}{l}{$+-$ is not allowed} & & \multicolumn{1}{l}{$--$ \ is not allowed} & & \multicolumn{1}{l}{$\cdot ~-$ \ is not allowed} & & \multicolumn{1}{l}{$\div -$ \ is not allowed}% \end{tabular}% \vspace{0.05in} For this reason, until a few decades ago, we used to put a pair of parentheses around \textit{every negative number}.\vspace{0.05in} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{tabular}{lllllll} $\left( -2\right) +\left( -6\right) $ & ~~~~~~~~~~~~~~~~~~~~ & $\left( -5\right) -\left( -3\right) $ & ~~~~~~~~~~~~~~~~~~~~~ & $\left( -3\right) \cdot \left( -4\right) $ & ~~~~~~~~~~~~~~~~~~~~~~ & $\left( -30\right) \div \left( -5\right) $ \\ \multicolumn{1}{c}{old style} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{ old style} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{old style} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{old style}% \end{tabular}% \vspace{0.05in} The development of mathematical notation is an ongoing process. \ The most important goal in notation is clarity. \ As long as clarity is not jeopardized, mathematicians are in the habit of omitting things. \ A few decades ago we stopped putting the parentheses around the first negative number in the line or inside a parentheses, because there was no risk that we would read the sign incorrectly as subtraction. \ Also, there is rarely an operation sign in front of the first number. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{tabular}{cllllll} $-2+\left( -6\right) $ & ~~~~~~~~~~~~~~~~~~~~ & $-5-\left( -3\right) $ & ~~~~~~~~~~~~~~~~~~~~~ & $-3\cdot \left( -4\right) $ & ~~~~~~~~~~~~~~~~~~~~~ & $-30\div \left( -5\right) $ \\ \multicolumn{1}{l}{more modern} & & more modern & & more modern & & more modern% \end{tabular} \vspace{0.05in} In the case of multiplication, we can omit one more thing. \ Recall that multiplication is the default operation; if we see two numbers with \textit{% nothing} between them, that indicates multiplication. \ For example, there is no operation sign or parentheses in $2x$ or $ab$ and yet it is clear that the operation is multiplication. \ Now that most negative numbers must be placed in parentheses, we can often omit the dot indicating multiplication. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{tabular}{clclclc} $-2+\left( -6\right) $ & ~~~~~~~ & $-5-\left( -3\right) $ & ~~~~~~~~ & $% -3\left( -4\right) $ & ~~~~~~~~~~ & $-30\div \left( -5\right) $ \\ \multicolumn{1}{l}{cannot omit anything} & & \multicolumn{1}{l}{cannot omit anything} & & \multicolumn{1}{l}{we can omit the dot} & & \multicolumn{1}{l}{cannot omit anything}% \end{tabular} The most common modern style is minimalistic, ommitting as much as possible, as long as confusion is avoided. This can lead to apparent irregularities in notation. \ For example, our notation will be $2\left( -3\right) $, \ but when we swap the two factors, it will be $-3\cdot 2$.\vspace{0.07in}\vspace{% 0.07in} \textbf{Our Larger Number Sysytem}\vspace{0.07in} %TCIMACRO{% %\TeXButton{box start - 7.6in}{\begin{boxedminipage}{7.6in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt} %\setlength{\leftskip}{20pt} %\setlength{\rightskip}{20pt}}}% %BeginExpansion \begin{boxedminipage}{7.6in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\leftskip}{20pt} \setlength{\rightskip}{20pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion %TCIMACRO{\TeXButton{\large}{\large}}% %BeginExpansion \large% %EndExpansion \textbf{The set of all integers is closed under addition, subtraction, and multiplication. \ It is not closed under division.}% %TCIMACRO{\TeXButton{\normalsize}{\normalsize}}% %BeginExpansion \normalsize% %EndExpansion %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \vspace{0.05in} As a matter of fact, the set of all integers ($% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $) is the smallest set that contains the set of all natural numbers ($% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $) and is closed under subtraction. \ Another way to state this is that $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $ is the closure of $% %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ under subtraction.\vspace{0.05in} In the future, we will further enlarge our number system to obtain closure under division. \pagebreak \FRAME{itbpF}{0.5561in}{0.576in}{0.2508in}{}{}{work.jpg}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.5561in;height 0.576in;depth 0.2508in;original-width 4.6977in;original-height 4.875in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'work.jpg';file-properties "XNPEU";}} \ \ {\Large \ Practice Problems} \begin{enumerate} \item Label each of the following statements as true or false. a) \ $-3\in %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c) \ $% %TCIMACRO{\U{2124} }% %BeginExpansion \mathbb{Z} %EndExpansion \subseteq %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e) \ $-\left\vert -2\right\vert =-2$ b) \ $-3\not\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $-3\geq -3$ \ \ \ \ \ \ \ \ \ \ \ \ \ f) \ For every integer $x$, \ $\left\vert \,x\,\right\vert \geq x$ \item Label each of the following statements as true or false.% %TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}% %BeginExpansion \begin{multicols}{2}% %EndExpansion a) \ Every integer is a natural number. b) \ Every natural number is an integer. c) \ $3<-5$ \ or \ $-2>-8$ d) \ $3<-5$ \ and \ $-2>-8$ e) \ Zero is also called the additive identity. f) \ $5\leq 5$ and $-\left\vert -8\right\vert =-8$ g) \ $-2<-2$ \ or \ $\left\vert -2\right\vert >\left\vert -10\right\vert $ %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item Place an inequality sign between the given numbers to make the statement true. a) \ $5~~~~~-7$ \ \ \ \ \ \ \ \ \ \ \ b) \ $-12~~~~~-4$ \ \ \ \ \ \ \ \ \ c) \ $0~~~~~-8$ \ \ \ \ \ \ \ \ \ \ d) \ $-1~~~~~-4$ \ \ \ \ \ \ \ \ \ \ e) \ $% -7~~~~~-7$ \item Simplify each of the following. a) \ $\left\vert -5\right\vert $ \ \ \ \ \ \ \ b) \ $\left\vert \,5\,\right\vert $ \ \ \ \ \ \ \ c) \ $-\left\vert \,5\,\right\vert $ \ \ \ \ \ \ d) \ $-\left\vert -5\right\vert $ \ \ \ \ \ \ e) \ $\left\vert \,0\,\right\vert $ \ \ \ \ \ \ \ f) \ $\left\vert -12+9\right\vert $ \ \ \ \ \ \ g) \ $\left\vert -12\right\vert +\left\vert \,9\,\right\vert $ \item Perform the indicated operations.% %TCIMACRO{\TeXButton{5col begin}{\begin{multicols}{5}}}% %BeginExpansion \begin{multicols}{5}% %EndExpansion a) \ $-2+7\vspace{0.05in}$ b) \ $-7-\left( -4\right) \vspace{0.05in}$ c) \ $12\div \left( -2\right) \vspace{0.05in}$ d) \ $5\left( -3\right) \vspace{0.05in}$ e) \ $-8\cdot 0\vspace{0.05in}$ f) \ $-3-\left( -10\right) \vspace{0.05in}$ g) \ $-20\div 0$ $\vspace{0.05in}$ h) \ $-12\div 3\vspace{0.05in}$ i) \ $-4\cdot 7\vspace{0.05in}$ j) \ $-6-\left\vert -7\right\vert \vspace{0.05in}$ k) \ $-3\div \left( -3\right) \vspace{0.05in}$ l) \ $\left\vert \,9\,\right\vert +\left( -1\right) \vspace{0.05in}$ m) \ $-3-0\vspace{0.05in}$ n) \ $0\left( -4\right) \vspace{0.05in}$ o) \ $\dfrac{-5}{0}\vspace{0.05in}$ p) \ $0\div \left( -1\right) \vspace{0.05in}$ q) \ $9+\left\vert -1\right\vert \vspace{0.05in}$ r) \ $\left\vert 9+\left( -1\right) \right\vert \vspace{0.05in}$ %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \end{enumerate} \pagebreak \FRAME{itbpF}{0.7524in}{0.7524in}{0.2214in}{}{}{answers.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.7524in;height 0.7524in;depth 0.2214in;original-width 8.6455in;original-height 8.6455in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'answers.jpg';file-properties "XNPEU";}} \ \ \ {\large Answers for Practice Problems} \begin{enumerate} \item a) \ true \ \ \ b) \ true \ \ \ c) false \ \ \ \ d) \ true \ \ \ \ e) \ true \ \ f) \ true \item a) \ false \ \ \ b) \ true \ \ \ c) true \ \ \ \ d) \ false \ \ \ e) \ true \ \ \ \ f) \ true \ \ g) \ false \item a) \ $5>-7$ \ or \ $5\geq -7$ \ \ \ \ \ b) \ $-12<-4$ \ or \ $-12\leq -4$ \ \ \ \ \ c) \ $0>-8$ \ or \ $0\geq -8$ \ \ \ \ d) \ $-1>-4$ \ or \ $% -1\geq -4$\vspace{0.05in}\ \newline e) \ $-7\geq -7$ \ or \ $-7\leq -7$ \item a) \ $5$ \ \ \ \ \ b) \ $5$ \ \ \ \ c) \ $-5$ \ \ \ \ d) \ $-5$ \ \ \ \ e) \ $0$ \ \ \ \ f) \ $3$ \ \ \ \ g) \ $21$ \item a) \ $5$ \ \ \ \ b) \ $-3$ \ \ \ \ c) \ $-6$ \ \ \ \ d) \ $-15$ \ \ \ e) \ $0$ \ \ \ f) \ $7$ \ \ \ \ g) \ undefined \ \ \ h) \ $-4$ \ \ \ \ i) \ $% -28$ \ \ \ \ j) \ $-13$ \ \ \ \ k) \ $1$ \vspace{0.05in}\ \ \ \ \ l) \ $8$ \ \ \ \ \ m) \ $-3$ \ \ \ \ n) \ $0$ \ \ \ \ o) \ undefined \ \ \ \ p) \ $0$ \ \ \ q) \ $10$ \ \ \ \ r) \ $8$\vspace{0.4in}\vspace{6in} \end{enumerate} %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}