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Mathematics LibreTexts

3.2: Identity for Addition

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    13975
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    %\documentclass{book} %\setlength{\textwidth}{4.5in} %\setlength{\topmargin}{0.05in} %\setlength{\textheight}{8.3in} %\setlength{\parskip}{10pt} %\setlength{\parindent}{0.0in} %\setlength{\oddsidemargin}{-0.01in} %\setlength{\evensidemargin}{-0.01in} %\pagestyle{headings} %\begin{document} %\noindent %\large %%\begin{flushleft} %%\setcounter{page}{82} %%\setcounter{chapter}{4} %%\pagenumbering{arabic}%\pagestyle{empty} % \chapter{Additional Properties of Real Numbers} \copyright \ H. Feiner 2011 \section{Identity for Addition}%section 1 %\setlength{\unitlength}{10pt} %\begin{picture}(0,8)(-10,-5) %\put(10,0){$2\ +\ 3\ =\ 3\ +\ 2$} %\put(5,-4){Position 1} %\put(7,-6){Position 2} %\put(13,-4){Position 1} %\put(17,-6){Position 2} %\put(9.2,-3){\vector(1,2){1.2}} %\put(11.2,-5){\vector(1,4){1.2}} %\put(16.2,-3.2){\vector(-1,4){0.7}} %\put(20.2,-5){\vector(-1,2){2.2}} %\put(10.3,1){\line(0,1){2.2}} %\put(10.3,3.2){\line(1,0){7.2}} %\put(17.5,3.2){\vector(0,-1){2.2}} %\put(12.6,1){\line(0,1){1.2}} %\put(12.6,2.2){\line(1,0){2.5}} % \put(15.1,2.2){\vector(0,-1){1.2}} %\end{picture}\\ $0$ is the identity element for addition because $0+5=5$ AND $5+0=5$. In general missing where $a$ is any real number. ********************************************************* Is $0$ the identity element for subtraction? Is $5-0=5$ a true statement? Yes, but $0-5\not=5$. \\[10pt]Therefore $0$ is not the identity element for subtraction. $0$ has an interesting property for multiplication. $0\cdot a=0$ and $a\cdot 0=0$ for all real numbers $a$. *********************************************************\\[20pt] \section{Identity for Multiplication}%section 2 %\setlength{\unitlength}{10pt} %\begin{picture}(0,8)(-10,-5) %\put(10,0){$2\ +\ 3\ =\ 3\ +\ 2$} %\put(5,-4){Position 1} %\put(7,-6){Position 2} %\put(13,-4){Position 1} %\put(17,-6){Position 2} %\put(9.2,-3){\vector(1,2){1.2}} %\put(11.2,-5){\vector(1,4){1.2}} %\put(16.2,-3.2){\vector(-1,4){0.7}} %\put(20.2,-5){\vector(-1,2){2.2}} %\put(10.3,1){\line(0,1){2.2}} %\put(10.3,3.2){\line(1,0){7.2}} %\put(17.5,3.2){\vector(0,-1){2.2}} %\put(12.6,1){\line(0,1){1.2}} %\put(12.6,2.2){\line(1,0){2.5}} % \put(15.1,2.2){\vector(0,-1){1.2}} %\end{picture}\\ $1$ is the identity element for multiplication because $1\cdot 5=5$ AND $5\cdot 1=5$. In general missing where $a$ is any real number. ********************************************************* Is $1$ the identity element for division? Is $5\div 1=5$ a true statement? Yes, but $1\div 5\not=5$.\\[10pt] Therefore $1$ is not the identity element for division.\\ $0$ has an interesting property for division. $a\div0$ is not defined for all real numbers $a$. Here is one explanation: Consider $12\div 3=\displaystyle \frac{12}{3}=4$. Rewrite as $(4)(3)=12$. In general the division-to-multiplication process is $a\div b=\displaystyle \frac{a}{b}=c$ with $(b)(c)=a$.\\[10pt] Now let's apply the division-to-multiplication process when the denominator is $0$. $12\div 0=\displaystyle \frac{12}{0}=x$ is rewritten $(0)(x)=12$. But $0x=0$, not $12$, above. No product $0\cdot x$ results in the dividend (numerator) $12$. This means {\bf division by }$0$ is undefined. Note that $0$ is defined, although you may wish it were undefined if you earn $0$ as your score on your test.\\ In general $a\div 0=\displaystyle \frac{a}{0}=c$ because $(0)(c)=a$ is impossible ($a\not = 0$). ********************************************************* \section{Opposites (Additive Inverses) and Reciprocals (Multiplicative Inverses)}%section 3 What number added to $5$ results in the identity for addition?\\[-0.1in] $5+(-5)=0$\\ In general, $a+(-a)=0$. $-5$ above is called the opposite (or additive inverse) of $5$. Every real number has a unique opposite. $0$ is its own opposite.\\ \setlength{\unitlength}{10pt} \begin{picture}(0,5)(0,-3) \put(0,0){\vector(1,0){35}} \multiput(1,-0.5)(2,0){17}{\line(0,1){1}} \put(0.5,-1.3){$-8$} \put(2.5,-1.3){$-7$} \put(4.5,-1.3){$-6$} \put(6.5,-1.3){$-5$} \put(8.5,-1.3){$-4$} \put(10.5,-1.3){$-3$} \put(12.5,-1.3){$-2$} \put(14.5,-1.3){$-1$} \put(16.7,-1.3){$0$} \put(18.7,-1.3){$1$} \put(20.7,-1.3){$2$} \put(22.7,-1.3){$3$} \put(24.7,-1.3){$4$} \put(26.7,-1.3){$5$} \put(28.7,-1.3){$6$} \put(30.7,-1.3){$7$} \put(32.7,-1.3){$8$} \put(7.0,0){\circle*{0.6}} \put(27.0,0){\circle*{0.6}} \put(16.5,1){\vector(-1,0){9.5}} \put(17.5,1){\vector(1,0){9.5}} \put(9.5,2){\scriptsize Same distance, \hspace{0.5in} opposite sides of $0$} \end{picture} Example 1: What is the opposite of $-3$? {\bf Solution:} The opposite of $-3$ is $3$. Thus $-(-3)=3$.\\[10pt] What number multiplied by $5$ results in the identity for multiplication? $5\cdot \displaystyle \frac{1}{5}=1$\\ $\displaystyle \frac{1}{5}$ is called the reciprocal (or multiplicative inverse) of $5$. Every real number has a unique opposite except for $0$ because division by $0$ is undefined. In general, $a\cdot \displaystyle \frac{1}{a}=1$ ($a\not =0$). ********************************************************* \section{Absolute Value}%section 4 The absolute value of a number is the measure of its distance from $0$ (to the left or to the right). \setlength{\unitlength}{10pt} \begin{picture}(0,7)(0,-3) \put(0,0){\vector(1,0){35}} \multiput(1,-0.5)(2,0){17}{\line(0,1){1}} \put(0.5,-1.3){$-8$} \put(2.5,-1.3){$-7$} \put(4.5,-1.3){$-6$} \put(6.5,-1.3){$-5$} \put(8.5,-1.3){$-4$} \put(10.5,-1.3){$-3$} \put(12.5,-1.3){$-2$} \put(14.5,-1.3){$-1$} \put(16.7,-1.3){$0$} \put(18.7,-1.3){$1$} \put(20.7,-1.3){$2$} \put(22.7,-1.3){$3$} \put(24.7,-1.3){$4$} \put(26.7,-1.3){$5$} \put(28.7,-1.3){$6$} \put(30.7,-1.3){$7$} \put(32.7,-1.3){$8$} \put(5.0,0){\circle*{0.6}} \put(23.0,0){\circle*{0.6}} \put(16.5,1){\vector(-1,0){11.5}} \put(17.5,1){\vector(1,0){5.5}} \put(18.0,2){\scriptsize $3$ is $3$ units away from $0$.} \put(5.5,2){\scriptsize $-6$ is $6$ units away from $0$.} \end{picture} The absolute value of $-6$, written $|-6|$, is $6$. The absolute value of $3$, written $|3|$, is $3$. You were probably taught to copy the number between the absolute value bars and erase the negative sign, if any. That probably served its purpose, but it is insufficient. The formal definition of absolute value is $$\hbox{Definition }\hspace{0.5in}|x|=\left\{\begin{array}{rcl} x&\hbox{if}&0\le x\\ -x&\hbox{if}&x< 0\\ \end{array}\right. \hspace{1.5in}$$\\ {\bf Example 2:} Use the definition of absolute value to find $|-6|$. {\bf Solution:} $|-6|=\left\{\begin{array}{rcl} {\scriptsize \hbox{(omit the upper path)}}\\[5pt] -(-6)=6&\hbox{since}&-6\le 0\\ \end{array}\right.$ \\[10pt] {\bf Example 3:} Use the definition of absolute value to find $|3|$. {\bf Solution:} $|3|=\left\{\begin{array}{rcl} 3&\hbox{since}&0\le 3\\ {\scriptsize \hbox{(omit the lower path)}}\\[5pt] \end{array}\right.$\\[10pt] {\bf Example 4:} Evaluate $-|x|$ if $x=-6$ {\bf Solution:} $\begin{array}{rcl lll} -|x|&=&-|-6|&\hbox{Substitute $-6$ for $x$.}\\ &=&-6 &\hbox{since $|-6|=6$} \end{array}$ Do not confuse $-(-6)=6$ with $-|-6|=-6$. *********************************************************\\[-50pt] \section{Inequality Symbol}%section 5 There can only be one true symbol $<, =, >$ relating a pair of real numbers (trichotomy). $a<b$,>b$. Place $a$ and $b$ on a number line. The number to the right is the larger one. For example $-6<-1$ or $-1>-6$. $-6\le -1$ or $-1\ge -6$ would also be correct. \setlength{\unitlength}{10pt} \begin{picture}(0,5)(0,-2) \put(0,0){\vector(1,0){35}} \multiput(1,-0.5)(2,0){17}{\line(0,1){1}} \put(0.5,-1.3){$-8$} \put(2.5,-1.3){$-7$} \put(4.5,-1.3){$-6$} \put(6.5,-1.3){$-5$} \put(8.5,-1.3){$-4$} \put(10.5,-1.3){$-3$} \put(12.5,-1.3){$-2$} \put(14.5,-1.3){$-1$} \put(16.7,-1.3){$0$} \put(18.7,-1.3){$1$} \put(20.7,-1.3){$2$} \put(22.7,-1.3){$3$} \put(24.7,-1.3){$4$} \put(26.7,-1.3){$5$} \put(28.7,-1.3){$6$} \put(30.7,-1.3){$7$} \put(32.7,-1.3){$8$} \put(5.0,0){\circle*{0.6}} \put(15.0,0){\circle*{0.6}} %\put(23.0,0){\circle*{0.6}} %\put(31.0,0){\circle*{0.6}} %\put(16.5,1){\vector(-1,0){11.5}} %\put(17.5,1){\vector(1,0){5.5}} %\put(17.5,2){$3$ is $3$ units away from $0$.} %\put(4.5,2){$-6$ is $6$ units away from $0$.} \end{picture}$ $\\[10pt] {\bf Example 5:} Insert $<$, $=$, or $>$ between $-6 \ _{\_\_\_\_} \ -3$. {\bf Solution:} $-6<-3$ because $-3$ is to the right of $-6$. Do you have trouble remembering which side of $<$ is the larger number? \setlength{\unitlength}{10pt} \begin{picture}(0,9)(6,-7) \put(10,0){\circle{2}} \put(10,-4){\circle{4}} \put(10,-2){\line(0,1){1}} \put(9,-5.4){\line(-1,-3){1}} \put(11,-5.4){\line(1,-3){1}} \put(8.4,-3.0){\line(-2,1){1}} \put(11.8,-3.0){\line(2,1){1}} \put(9.4,-4.5){$-6$} \put(17,-5.1){\oval(4,2)[b]} \put(14.6,-5.05){\line(1,-1){1.0}} \put(14.6,-5.1){\line(1,-1){1.0}} \put(14.6,-5.15){\line(1,-1){1.0}} \put(14.6,-5.05){\line(1,1){1.0}} \put(14.6,-5.1){\line(1,1){1.0}} \put(14.6,-5.15){\line(1,1){1.0}} \put(15.6,-4.15){\line(1,0){5.5}} \put(21.1,-4.15){\line(0,-1){1.1}} \put(21.1,-5.25){\line(-1,0){2.2}} \put(17.1,-3.8){knife} \put(17.9,-5.05){handle} \put(15.5,-5.8){blade} \put(14.1,-8.05){point} \put(14.5,-7.25){\vector(0,1){1.8}} \put(28.5,-3.05){\parbox[t]{1.8in}{The stronger (bigger) number holds the handle of the knife, while the smaller (weaker) number gets the point of the blade.}} \end{picture}$ $\\[-17pt] \begin{picture}(0,9)(-9,-14) \put(10,0){\circle{2}} \put(10,-4){\circle{4}} \put(10,-2){\line(0,1){1}} \put(9,-5.4){\line(-1,-3){1}} \put(11,-5.4){\line(1,-3){1}} \put(8.4,-3.0){\line(-2,1){1}} \put(11.8,-3.0){\line(2,1){1}} \put(9.4,-4.5){$-3$} \end{picture}$ $\\[-50pt] You may have been taught that the bigger fish (on the right) eats the smaller fish(on the left.)\\ Use of $\not=$:\\ If $a\not=b$ then $a<b$>b$ (one of these is true). Thus $-6\not=3$ means $3< -6$ (false) or $3> -6$ (true).\\ Use of $\le$: \\ $a\le b$ is true if either $a<b$>$. Thus $x>\$20,000$ or $\$20,000<x$.\\[10pt]>x$. \section{Rounding} Rounding is a means for estimating, approximating.\\ {\bf Example 12:} Round $14,371$ to the nearest hundred. {\bf Solution:}\\[40pt] \begin{picture}(0,0)(8,-6) \put(10,0){$1\ 4\ 3\ 7\ 1$} \put(12.5,-1.8){\circle{1.1}} \put(10,-2){$1\ 4\ 3\ 7\ 1$\ \hbox{Circle the hundred digit ``3"}} \put(10,-4){$1\ 4\ 3\ 0\ 0$\ \hbox{Substitute 0 for digits to the right of the circle.}} \put(12.5,-3.8){\circle{1.1}} \put(14.0,-5.8){The first dropped digit, 7, was $\ge 5$,} \put(14.0,-7.1){so the circled digit is incremented.} \end{picture}\\[-5pt] $14,371$ rounded to the nearest hundred is $14,400$\\[-20pt] \setlength{\unitlength}{10pt} \begin{picture}(0,5)(0,-3) \put(0,0){\vector(1,0){25}} \multiput(1,-0.5)(10,0){3}{\line(0,1){1}} \put(0.5,-1.3){$14,300$} \put(9.7,-1.3){$14,350$} \put(15.2,-1.3){$14,371$} \put(19.9,-1.3){$14,400$} \put(17.8,0){\circle*{0.6}} \end{picture} Note that $14,371$ is in the second half of the number line segments above (to the right of $14,350$). The number is closer to $14,400$ than $14,300$.\\[10pt] {\bf Example 13:}\\ Round $14,371$ to the nearest thousand.\\ {\bf Solution:}\\[50pt] \begin{picture}(0,0)(8,-6) \put(10,0){$1\ 4\ 3\ 7\ 1$} \put(11.5,-1.8){\circle{1.1}} \put(10,-2){$1\ 4\ 3\ 7\ 1$\ \hbox{Circle the thousand digit ``4"}} \put(10,-4){$1\ 4\ 0\ 0\ 0$\ \hbox{Substitute 0 for digits to the right of the circle.}} \put(11.5,-3.8){\circle{1.1}} \put(14.0,-5.8){The first dropped digit, 3 , was $< 5$, } \put(14.0,-7.1){so the circled digit remains unchanged.} \end{picture}\\[-1pt] $14,371$ rounded to the nearest thousand is $14,000$\\[-20pt] \setlength{\unitlength}{10pt} \begin{picture}(0,5)(0,-3) \put(0,0){\vector(1,0){25}} \multiput(1,-0.5)(10,0){3}{\line(0,1){1}} \put(0.3,-1.3){$14,000$} \put(9.9,-1.3){$14,500$} \put(5.3,-1.3){$14,371$} \put(19.7,-1.3){$15,000$} \put(07.0,0){\circle*{0.6}} \end{picture} Note that $14,371$ is in the first half of the number line segments above. The number is closer to $14,000$ than $15,000$. Note: If the number to be rounded (say $14,500$) falls exactly between the extremes ($14,000$ and $15,000$) always round up to the extreme right ($15,000$). %\end{document}%$$$$$ \section{Exercise 3} Justify the following statements or state why the statement is false. \begin{enumerate} \item %%% 1 $5+(-5)=0$ \item %%% 2 $\left(\displaystyle \frac{4}{9}\right)\left(\displaystyle \frac{9}{4}\right)=1$ \item %%% 3 $(-2.8)(1)=-2.8$ \item %%% 4 $(-2.8)(0)=0$ \item %%% 5 $(-2.8)+(0)=-2.8$ \item %%% 6 $6(7\cdot 8)=(7\cdot 8)6$ \item %%% 7 Every real number has a reciprocal. \item %%% 8 The sum of any number $x$ and the absolute value of its opposite is $2x$. \item %%% 9 Find two numbers equal to their reciprocals. \item %%% 10 The sum of a number and its opposite is 0. \item %%% 11 The product of a number and its reciprocal is 1. \item %%% 12 $0$ is the identity for subtraction because $x-0=x$. \item %%% 13 $1$ is the identity for division because $\displaystyle \frac{x}{1}=x$. \item %%% 14 Find two numbers such that $|x|=-6$. \item %%% 15 Find two numbers such that $|x|=6$. \item %%% 16 Find two numbers such that $|x|>6$. \item %%% 17 Is $-1\le -5$ true or false?. \item %%% 18 There are at most 270 calories in a consumer product. What is the highest number of calories in the product? (Use a whole number.) \item %%% 19 According to kidshealth.org, most full-term babies weigh at least 6 lbs 2 oz. According to this data, what is the highest weight a full-term baby can weigh? (Assume ``most babies" in the statement is replaced by ``all babies".) (See \\(www.ehow.com/facts$\_$5349623$\_$normal-birth-weight-length.html) %and 9 lbs. 2 oz. \item %%% 20 $x\ge x$ is a true statement. Correct? \item %%% 21 Round 23.4567 to the nearest hundredth. \end{enumerate} \fbox{\huge STOP!} The solutions follow: Keep them covered up till you have worked out each of the problems above. \begin{enumerate} \item %%% 1 $5+(-5)=0$\\[10pt] {\bf Solution:}\\[10pt] $5+(-5)=0$ because $5$ and $-5$ are opposites. \item %%% 2 $\left(\displaystyle \frac{4}{9}\right)\left(\displaystyle \frac{9}{4}\right)=1$\\[10pt] {\bf Solution:}\\[10pt] $\left(\displaystyle \frac{4}{9}\right)$and $\left(\displaystyle \frac{9}{4}\right)$ are reciprocals. \item %%% 3 $(-2.8)(1)=-2.8$\\[10pt] {\bf Solution:}\\ [10pt] $(-2.8)(1)=-2.8$ because $1$ is the identity for addition. \item %%% 4 $(-2.8)(0)=0$ \\[10pt] {\bf Solution:}\\ [10pt] $(-2.8)(0)=0$ Any number times 0 is 0 (Multiplication property of $0$). \item %%% 5 $(-2.8)+(0)=-2.8$\\[10pt] {\bf Solution:}\\ [10pt] $(-2.8)+(0)=-2.8$ The sum of a number and 0 is that number. $0$ is the identity for addition. \item %%% 6 $6(7\cdot 8)=(7\cdot 8)6$\\[10pt] {\bf Solution:}\\ [10pt] $6(7\cdot 8)=(7\cdot 8)6$ Multiplication is commutative. $7\cdot 8$ and $6$ have commuted to each other's position. \item %%% 7 Every real number has a reciprocal.\\[10pt] {\bf Solution:}\\ [10pt] Every real number has a reciprocal. This statement is false. $0$ does not have a reciprocal $\displaystyle \frac{1}{0}$ because division by $0$ is undefined. Do not say 0 is undefined. If you score 0 on a test you are very well aware of its meaning. \item %%% 8 The sum of any number $x$ and the absolute value of its opposite is $2x$.\\[10pt] {\bf Solution:}\\ [10pt] The statement is false if $x\ge 0$ as illustrated by the following:\\ Let $x=7$. The opposite of $x$ is $-7$. The sum of $x=7$ and its opposite is \\ $7+(-7)=0$, \\ not $2x=(2)(7)=14$. The statement is true if $x<0$:\\ Let $x=-3$. The opposite of $x$ is $-(-3)=3$. The sum of $|x|=|-3|=3$ and its opposite is $-x=-(-3)=3$ is $3+(3)=2(3)=6$. \item %%% 9 Find two numbers equal to their reciprocals.\\[10pt] {\bf Solution:} If $x=1$ its reciprocal is $\displaystyle \frac{1}{1}=1$. If $x=-1$ its reciprocal is $\displaystyle \frac{1}{-1}=-1$. \item %%% 10 The sum of a number and its opposite is 0.\\[10pt] {\bf Solution:} \\[10pt] The sum of a number and its opposite is 0. $x+(-x)=0$. Let $x=4$ then $-x=-4$ is its opposite. The sum is $x+(-x)=4+(-4)=0$. Let $x=-3$ then $-x=-(-3)=3$ is its opposite. The sum is $x+(-x)=-3+(3)=0$. \item %%% 11 The product of a number and its reciprocal is 1.\\[10pt] {\bf Solution:}\\[10pt] The product of a number and its reciprocal is 1 if $x\not= 0$. $x\left(\displaystyle \frac{1}{x}\right)=1$.\\ If $x=0$, the statement is false because $0$ does not have a reciprocal. \item %%% 12 $0$ is the identity for subtraction because $x-0=x$.\\[10pt] {\bf Solution:}\\[10pt] $0$ is the identity for subtraction because $x-0=x$. The statement is false because $0-x=-x$, not $x$. Let $x=3$. $3-0=3$ but $0-3=-3\not = 3$. \item %%% 13 $1$ is the identity for division because $\displaystyle \frac{x}{1}=x$.\\[10pt] {\bf Solution:}\\[10pt] $1$ is not the identity for division because even though \\ $\displaystyle \frac{x}{1}=x$, $\displaystyle \frac{1}{x}\not= x$. \item %%% 14 Find two numbers such that $|x|=-6$.\\[10pt] {\bf Solution:}\\[10pt] Find two numbers such that $|x|=-6$. You cannot find any numbers because absolute value is never negative. This problem has no solution. \item %%% 15 Find two numbers such that $|x|=6$.\\[10pt] {\bf Solution:}\\[10pt] Find two numbers such that $|x|=6$.\\ \setlength{\unitlength}{10pt} \begin{picture}(0,5)(0,-3) \put(0,0){\vector(1,0){35}} \multiput(1,-0.5)(2,0){17}{\line(0,1){1}} \put(0.5,-1.3){$-8$} \put(2.5,-1.3){$-7$} \put(4.5,-1.3){$-6$} \put(6.5,-1.3){$-5$} \put(8.5,-1.3){$-4$} \put(10.5,-1.3){$-3$} \put(12.5,-1.3){$-2$} \put(14.5,-1.3){$-1$} \put(16.7,-1.3){$0$} \put(18.7,-1.3){$1$} \put(20.7,-1.3){$2$} \put(22.7,-1.3){$3$} \put(24.7,-1.3){$4$} \put(26.7,-1.3){$5$} \put(28.7,-1.3){$6$} \put(30.7,-1.3){$7$} \put(26.7,-1.3){$8$} \put(5.0,0){\circle*{0.6}} %\put(15.0,0){\circle*{0.6}} %\put(23.0,0){\circle*{0.6}} \put(29.0,0){\circle*{0.6}} %\put(16.5,1){\vector(-1,0){11.5}} %\put(17.5,1){\vector(1,0){5.5}} %\put(17.5,2){$3$ is $3$ units away from $0$.} %\put(4.5,2){$-6$ is $6$ units away from $0$.} \end{picture}\\ The numbers are $x=-6$ and $x=6$. \item %%% 16 Find two numbers such that $|x|>6$.\\[10pt] {\bf Solution:}\\[10pt] Find two numbers such that $|x|>6$. There are infinitely many solutions, like $x=7$, $x=6,543$, $x=6.03$, $x=-6.5$, $x=-27$.\\ \setlength{\unitlength}{10pt} \begin{picture}(0,5)(0,-3) \put(0,0){\vector(1,0){35}} \multiput(1,-0.5)(2,0){17}{\line(0,1){1}} \put(0.5,-1.3){$-8$} \put(2.5,-1.3){$-7$} \put(4.5,-1.3){$-6$} \put(6.5,-1.3){$-5$} \put(8.5,-1.3){$-4$} \put(10.5,-1.3){$-3$} \put(12.5,-1.3){$-2$} \put(14.5,-1.3){$-1$} \put(16.7,-1.3){$0$} \put(18.7,-1.3){$1$} \put(20.7,-1.3){$2$} \put(22.7,-1.3){$3$} \put(24.7,-1.3){$4$} \put(26.7,-1.3){$5$} \put(28.7,-1.3){$6$} \put(30.7,-1.3){$7$} \put(32.7,-1.3){$8$} \put(5.0,0){\circle{0.6}} \put(28.9,0.05){\line(1,0){5}} \put(28.9,0.1){\line(1,0){5}} \put(28.9,0.15){\line(1,0){5}} \put(28.9,0.2){\line(1,0){5}} \put(28.9,0.25){\line(1,0){5}} \put(4.7,-0.05){\line(-1,0){5}} \put(4.7,-0.1){\line(-1,0){5}} \put(4.7,-0.15){\line(-1,0){5}} \put(4.7,-0.2){\line(-1,0){5}} \put(4.7,-0.25){\line(-1,0){5}} %\put(23.0,0){\circle*{0.6}} \put(29.0,0){\circle{0.6}} %\put(16.5,1){\vector(-1,0){11.5}} %\put(17.5,1){\vector(1,0){5.5}} %\put(17.5,2){$3$ is $3$ units away from $0$.} %\put(4.5,2){$-6$ is $6$ units away from $0$.} \end{picture} \item %%% 17 Is $-1\le -5$ true or false?.\\[10pt] {\bf Solution:}\\[10pt] $-1\le -5$ is false because %$-5$ is not to the right of $-1$. \\ $-1< -5$ is false and $-1= -5$ is false.\\ \setlength{\unitlength}{10pt} \begin{picture}(0,5)(0,-3) \put(0,0){\vector(1,0){35}} \multiput(1,-0.5)(2,0){17}{\line(0,1){1}} \put(0.5,-1.3){$-8$} \put(2.5,-1.3){$-7$} \put(4.5,-1.3){$-6$} \put(6.5,-1.3){$-5$} \put(8.5,-1.3){$-4$} \put(10.5,-1.3){$-3$} \put(12.5,-1.3){$-2$} \put(14.5,-1.3){$-1$} \put(16.7,-1.3){$0$} \put(18.7,-1.3){$1$} \put(20.7,-1.3){$2$} \put(22.7,-1.3){$3$} \put(24.7,-1.3){$4$} \put(26.7,-1.3){$5$} \put(28.7,-1.3){$6$} \put(30.7,-1.3){$7$} \put(32.7,-1.3){$8$} \put(7.0,0){\circle*{0.6}} %\put(28.9,0.1){\line(1,0){5}} %\put(28.9,0.15){\line(1,0){5}} %\put(28.9,0.2){\line(1,0){5}} %\put(28.9,0.25){\line(1,0){5}} %\put(4.7,-0.05){\line(-1,0){5}} %\put(4.7,-0.1){\line(-1,0){5}} %\put(4.7,-0.15){\line(-1,0){5}} %\put(4.7,-0.2){\line(-1,0){5}} %\put(4.7,-0.25){\line(-1,0){5}} %\put(23.0,0){\circle*{0.6}} \put(15.0,0){\circle*{0.6}} %\put(16.5,1){\vector(-1,0){11.5}} %\put(17.5,1){\vector(1,0){5.5}} %\put(17.5,2){$3$ is $3$ units away from $0$.} %\put(4.5,2){$-6$ is $6$ units away from $0$.} \end{picture} \item %%% 18 There are at most $270$ calories in a consumer product. What is the highest number of calories in the product? (Use a whole number.)\\[10pt] {\bf Solution:}\\[10pt] %There are at most 270 calories in a product. What is the highest number of calories in the product? (Use a whole number.) \\ The highest number is $270$ calories. ``at most" means ``up to" and including. \item %%% 19 Most full-term babies weigh at least $6$ lbs $2$ oz. According to this data, what is the highest weight a full-term baby can weigh?\\[10pt] {\bf Solution:}\\[10pt] ``at least" means ``that much or more". There is no largest number. The problem has no solution according to the given data. Of course there is a largest number ($9$ lbs. $2$ oz. according to the website), but it is not given in the statement of the problem. \item %%% 20 $x\ge x$ is a true statement. Correct?\\[10pt] {\bf Solution:}\\[10pt] $x\ge x$ is indeed a true statement because $x=x$ for all real numbers. \item 4%%% 21 Round $23.4567$ to the nearest hundredth.\\[10pt] {\bf Solution:}\\[70pt] \begin{picture}(0,0)(8,-6) \put(10,0){$2\ 3.\ 4\ 5\ 6\ 7$} \put(14,-1.8){\circle{1.1}} \put(10,-2){$2\ 3.\ 4\ 5\ 6\ 7$\ \ \ \hbox{Circle the hundredth digit ``5"}} \put(10,-4){$2\ 3.\ 4\ 5\ 0\ 0$\ \ \ \hbox{Substitute 0 for digits to the right of the circle.}} \put(14,-3.8){\circle{1.1}} \put(10.0,-5.8){$2\ 3.\ 4\ 6$\ \ \ \ \ The first dropped digit, 6, is $\ge 5$.} \put(16.3,-7.2){The circled digit is incremented.} \end{picture}\\[15pt] $23.4567$ rounded to the nearest hundredth is $23.46$\\[-20pt] \setlength{\unitlength}{12pt} \begin{picture}(0,5)(0,-3) \put(0,0.5){\vector(1,0){25}} \multiput(1,-0.0)(10,0){3}{\line(0,1){1}} \put(0.5,-1.3){$23.45$} \put(9.9,-1.3){$23.455$} \put(13.3,1.3){$23.4567$} \put(19.7,-1.3){$23.46$} \put(14.0,0.5){\circle*{0.6}} \end{picture} Note that $23.4567$ is in the second half of the number line segments above. The number is closer to $123.46$ than $23.45$. \end{enumerate} \section{Supplemental Homework} %%\end{flushleft} %\end{document}