1.2: Operations with Real Numbers

Last updated
Save as PDF

Page ID: 6227

Anonymous
LibreTexts

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

Learning Objectives

Review the properties of real numbers.
Simplify expressions involving grouping symbols and exponents.
Simplify using the correct order of operations.

Working with Real Numbers

In this section, we continue to review the properties of real numbers and their operations. The result of adding real numbers is called the sum⁵³ and the result of subtracting is called the difference⁵⁴. Given any real numbers a, b, and c, we have the following properties of addition:

Table $\PageIndex{1}$
Additive Identity Property:⁵⁵	a+0=0+a=a
Additive Inverse Property:⁵⁶	a+(−a)=(−a)+a=0
Associative Property:⁵⁷	(a+b)+c=a+(b+c)
Commutative Property:⁵⁸	a+b=b+a

It is important to note that addition is commutative and subtraction is not. In other words, the order in which we add does not matter and will yield the same result. However, this is not true of subtraction.

$5+10=10+5$ $5−10≠10−5$

$15=15$ $−5≠5$

We use these properties, along with the double-negative property for real numbers, to perform more involved sequential operations. To simplify things, make it a general rule to first replace all sequential operations with either addition or subtraction and then perform each operation in order from left to right.

Example $\PageIndex{1}$:

Simplify: $−10−(−10)+(−5)$.

Solution

Replace the sequential operations and then perform them from left to right.

$−10−(−10)+(−5)=−10+10−5$ $\color{Cerulean}{Replace −(−) with addition (+)}$.

$\color{Cerulean}{Replace +(−) with subtraction (-).}$

$=0−5$

$=−5$

Answer

$−5$

Adding or subtracting fractions requires a common denominator⁵⁹. Assume the common denominator c is a nonzero integer and we have

$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$ and $\frac{a}{c}−\frac{b}{c}=\frac{a−b}{c}$

Example $\PageIndex{2}$:

Simplify: $\frac{2}{9}−\frac{1}{15}+\frac{8}{45}$.

Solution

First determine the least common multiple (LCM) of $9, 15, and 45$. The least common multiple of all the denominators is called the least common denominator⁶⁰ (LCD). We begin by listing the multiples of each given denominator:

$\{9,18,27,36,45,54,63,72,81,90,\dots\}$ $\color{Cerulean}{Multiples of 9}$

$\{15,30,45,60,75,90,\dots\}$ $\color{Cerulean}{Multiples of 15}$

$\{45,90,135\dots\}$ $\color{Cerulean}{Multiples of \: 45}$

Here we see that the LCM$(9, 15, 45) = 45$. Multiply the numerator and the denominator of each fraction by values that result in equivalent fractions with the determined common denominator.

$\frac{2}{9}−\frac{1}{15}+\frac{8}{45}=\frac{2}{9}⋅\color{Cerulean}{\frac{5}{5}}$$−\frac{1}{15}⋅\color{Cerulean}{\frac{3}{3}}$$+\frac{8}{45}$

$=\frac{10}{45}−\frac{3}{45}+\frac{8}{45}$

Once we have equivalent fractions, with a common denominator, we can perform the operations on the numerators and write the result over the common denominator.

$=\frac{10−3+8}{45}$

$=\frac{15}{45}$

And then reduce if necessary,

$=\frac{15\color{Cerulean}{÷15}}{45\color{Cerulean}{÷15}}$

$=\frac{1}{3}$

Answer

$\frac{1}{3}$

Finding the LCM using lists of multiples, as described in the previous example, is often very cumbersome. For example, try making a list of multiples for $12$ and $81$. We can streamline the process of finding the LCM by using prime factors.

$12=2^{2}⋅3$

$81=3^{4}$

The least common multiple is the product of each prime factor raised to the highest power. In this case,

$LCM(12,81)=2^{2}⋅3^{4}=324$

Often we will find the need to translate English sentences involving addition and subtraction to mathematical statements. Below are some common translations.

$n+2 \color{Cerulean}{The\: sum\: of\: a\: number\: and\: 2.}$

$2−n \color{Cerulean}{The\: difference\: of\: 2\: and\: a\: number.}$

$n−2 \color{Cerulean}{Here\: 2\: is\: subtracted\: from\: a\: number.}$

Example $\PageIndex{3}$:

What is $8$ subtracted from the sum of $3$ and $\frac{1}{2}$?

Solution

We know that subtraction is not commutative; therefore, we must take care to subtract in the correct order. First, add $3$ and $\frac{1}{2}$ and then subtract $8$ as follows:

Perform the indicated operations.

$(3+\frac{1}{2})−8=(\frac{3}{1}⋅\color{Cerulean}{\frac{2}{2}}$$+\frac{1}{2})−8$

$=(\frac{6+1}{2})−8$

$=\frac{7}{2}−\frac{8}{1}⋅\color{Cerulean}{\frac{2}{2}}$

$=\frac{7−16}{2}$

$=−\frac{9}{2}$

Answer

$−\frac{9}{2}$

The result of multiplying real numbers is called the product⁶¹ and the result of dividing is called the quotient⁶². Given any real numbers a, b, and c, we have the following properties of multiplication:

Table $\PageIndex{2}$
Zero Factor Property:⁶³	a⋅0=0⋅a=0
Multiplicative Identity Property:⁶⁴	a⋅1=1⋅a=a
Associative Property:⁶⁵	(a⋅b)⋅c=a⋅(b⋅c)
Commutative Property:⁶⁶	a⋅b=b⋅a

It is important to note that multiplication is commutative and division is not. In other words, the order in which we multiply does not matter and will yield the same result. However, this is not true of division.

$5⋅10=10⋅5$ $5÷10≠10÷5$

$50=50$ $0.5≠2$

We will use these properties to perform sequential operations involving multiplication and division. Recall that the product of a positive number and a negative number is negative. Also, the product of two negative numbers is positive.

Example $\PageIndex{4}$:

Multiply: 5(−3)(−2)(−4).

Solution

Multiply two numbers at a time as follows:

Answer

$−120$

Because multiplication is commutative, the order in which we multiply does not affect the final answer. However, when sequential operations involve multiplication and division, order does matter; hence we must work the operations from left to right to obtain a correct result.

Example $\PageIndex{5}$:

Simplify: 10÷(−2)(−5).

Solution

Perform the division first; otherwise the result will be incorrect.

Notice that the order in which we multiply and divide does affect the result. Therefore, it is important to perform the operations of multiplication and division as they appear from left to right.

Answer

$25$

The product of two fractions is the fraction formed by the product of the numerators and the product of the denominators. In other words, to multiply fractions, multiply the numerators and multiply the denominators:

$\frac{a}{b}⋅\frac{c}{d}=\frac{ac}{bd}$

Example $\PageIndex{6}$:

Multiply $−\frac{4}{5}⋅\frac{25}{12}$.

Solution

Multiply the numerators and multiply the denominators. Reduce by dividing out any common factors.

Screenshot (133).png — Figure $\PageIndex{4}$

Answer:

$−\frac{5}{3}$

Two real numbers whose product is $1$ are called reciprocals⁶⁷. Therefore, $\frac{a}{b}$ and $\frac{b}{a}$ are reciprocals because $\frac{a}{b}⋅\frac{b}{a}=\frac{ab}{ab}=1$. For example,

$\frac{2}{3}⋅\frac{3}{2}=\frac{6}{6}=1$

Because their product is $1, \frac{2}{3}$ and $\frac{3}{2}$ are reciprocals. Some other reciprocals are listed below:

$\frac{5}{8}$ and $\frac{8}{5}$ $7$ and $\frac{1}{7}$ $−\frac{4}{5}$ and $−\frac{5}{4}$

This definition is important because dividing fractions requires that you multiply the dividend by the reciprocal of the divisor.

$\frac{a}{b}÷\color{Cerulean}{\frac{c}{d}}$ $=\frac { \frac { a } { b } } { \frac { c } { d } } \cdot \color{OliveGreen}{\frac { \frac { d } { c } } { \frac { d } { c } }}$ $=\frac { \frac { a } { b } \cdot \frac { d } { c } } { 1 } = \frac { a } { b } \cdot \color{Cerulean}{\frac { d } { c }}$

In general,

$\frac { a } { b } \div \color{Cerulean}{\frac { c } { d }}$ $= \frac { a } { b } \cdot \color{Cerulean}{\frac { d } { c }}$ $= \frac { a d } { b c }$

Example $\PageIndex{7}$:

Simplify: $\frac{5}{4}÷\frac{3}{5}⋅\frac{1}{2}$.

Solution

Perform the multiplication and division from left to right.

$\frac{5}{4}÷\color{Cerulean}{\frac{3}{5}}$ $⋅\frac{1}{2}=\frac{5}{4}⋅\color{Cerulean}{\frac{5}{3}}$ $⋅\frac{1}{2}$

$=\frac{5⋅5⋅1}{4⋅3⋅2}$

$=\frac{25}{24}$

In algebra, it is often preferable to work with improper fractions. In this case, we leave the answer expressed as an improper fraction.

Answer

$\frac{25}{24}$

Exercise $\PageIndex{1}$

Simplify: $\frac{1}{2}⋅\frac{3}{4}÷\frac{1}{8}$.

Answer: $3$ www.youtube.com/v/4zV-fyepZkk

Grouping Symbols and Exponents

In a computation where more than one operation is involved, grouping symbols help tell us which operations to perform first. The grouping symbols⁶⁸ commonly used in algebra are:

$( )$ $\color{Cerulean}{Parentheses}$

$[ ]$ $\color{Cerulean}{Brackets}$

$\{ \}$ $\color{Cerulean}{Braces}$

$-$ $\color{Cerulean}{Fraction\: bar}$

All of the above grouping symbols, as well as absolute value, have the same order of precedence. Perform operations inside the innermost grouping symbol or absolute value first.

Example $\PageIndex{8}$:

Simplify: $2−(\frac{4}{5}−\frac{2}{15})$.

Solution

Perform the operations within the parentheses first.

$2−(\frac{4}{5}−\frac{2}{15})= 2−(\frac{4}{5}⋅\color{Cerulean}{\frac{3}{3}}$$−\frac{2}{15})$

$=2−(\frac{12}{15}−\frac{2}{15})$

$=2−(\frac{10}{15})$

$=\frac{2}{1}⋅\color{Cerulean}{\frac{3}{3}}$$−\frac{2}{3}$

$=\frac{6-2}{3}$

$=\frac{4}{3}$

Answer:

$\frac{4}{3}$

Example $\PageIndex{9}$:

Simplify: $\frac { 5 - | 4 - ( - 3 ) | } { | - 3 | - ( 5 - 7 ) }$.

Solution

The fraction bar groups the numerator and denominator. Hence, they should be simplified separately.

$\frac { 5 - | 4 - ( - 3 ) | } { | - 3 | - ( 5 - 7 ) } = \frac { 5 - | 4 + 3 | } { | - 3 | - ( - 2 ) }$

$= \frac { 5 - | 7 | } { | - 3 | + 2 }$

$= \frac { 5 - 7 } { 3 + 2 }$

$= \frac { - 2 } { 5 }$

$= - \frac { 2 } { 5 }$

Answer:

$−\frac{2}{5}$

If a number is repeated as a factor numerous times, then we can write the product in a more compact form using exponential notation⁶⁹. For example,

$5⋅5⋅5⋅5=54$

The base⁷⁰ is the factor and the positive integer exponent⁷¹ indicates the number of times the base is repeated as a factor. In the above example, the base is $5$ and the exponent is $4$. Exponents are sometimes indicated with the caret (^) symbol found on the keyboard, $5^4 = 5*5*5*5$. In general, if a is the base that is repeated as a factor n times, then

When the exponent is $2$ we call the result a square⁷², and when the exponent is $3$ we call the result a cube⁷³. For example,

$5^{2}=5⋅5=25$ $\color{Cerulean}{"5\: squared”}$

$5^{3}=5⋅5⋅5=125$ $\color{Cerulean}{“5\: cubed”}$

If the exponent is greater than $3$, then the notation $a^{n}$ is read, “a raised to the nth power.” The base can be any real number,

$(2.5)^{2}=(2.5)(2.5)=6.25$

$(−\frac{2}{3})^{3}=(−\frac{2}{3})(−\frac{2}{3})(−\frac{2}{3})=−\frac{8}{27}$

$(−2)^4=(−2)(−2)(−2)(−2)=16$

$−2^{4}=−1⋅2⋅2⋅2⋅2=−16$

Notice that the result of a negative base with an even exponent is positive. The result of a negative base with an odd exponent is negative. These facts are often confused when negative numbers are involved. Study the following four examples carefully:

Table $\PageIndex{3}$
*The base is $(−3)$.*	*The base is $3$.*
$(−3)^{4}=(−3)(−3)(−3)(−3)=+81$ $(−3)^{3}=(−3)(−3)(−3)=−27$	$−3^{4}=−1⋅3⋅3⋅3⋅3=−81$ $−3^{3}=−1⋅3⋅3⋅3=−27$

The parentheses indicate that the negative number is to be used as the base.

Example $\PageIndex{10}$:

Calculate:

$(−\frac{1}{3})^{3}$
$(−\frac{1}{3})^{4}$

Solution

Here $−\frac{1}{3}$ is the base for both problems.

1.Use the base as a factor three times.

$(−\frac{1}{3})^{3}=(−\frac{1}{3})(−\frac{1}{3})(−\frac{1}{3})$

$=−\frac{1}{27}$

2.Use the base as a factor four times.

\((−\frac{1}{3})^{4}=(−\frac{1}{3})(−\frac{1}{3})(−\frac{1}{3})(−\frac{1}{3})

$=+\frac{1}{81}$

Answers:

−$\frac{12}{7}$
$\frac{1}{81}$

Exercise $\PageIndex{2}$

Simplify:

$−2^{4}$
$(−2)^{4}$

Answer

1. −16

2. 16

www.youtube.com/v/o3X52psRJtg

Order of Operations

When several operations are to be applied within a calculation, we must follow a specific order to ensure a single correct result.

Perform all calculations within the innermost parentheses or grouping symbol first.
Evaluate all exponents.
Apply multiplication and division from left to right.
Perform all remaining addition and subtraction operations last from left to right.

Note that multiplication and division should be worked from left to right. Because of this, it is often reasonable to perform division before multiplication.

Example $\PageIndex{11}$:

Simplify: $5^{3} − 24 ÷ 6 ⋅ \frac{1}{2} + 2.$

Solution

First, evaluate $5^{3}$ and then perform multiplication and division as they appear from left to right.

\begin{aligned} 5 ^ { 3 } - 24 \div 6 \cdot \frac { 1 } { 2 } + 2 & = 5 ^ { 3 } - 24 \div 6 \cdot \frac { 1 } { 2 } + 2 \\ & = 125 - 24 \div 6 \cdot \frac { 1 } { 2 } + 2 \\ & = 125 - 4 \cdot \frac { 1 } { 2 } + 2 \\ & = 125 - 2 + 2 \\ & = 123 + 2 \\ & = 125 \end{aligned}

Multiplying first would have led to an incorrect result.

Answer:

$125$

Example $\PageIndex{12}$:

Simplify: $- 10 - 5 ^ { 2 } + ( - 3 ) ^ { 4 }$.

Solution

Take care to correctly identify the base when squaring.

$\begin{aligned} - 10 - 5 ^ { 2 } + ( - 3 ) ^ { 4 } & = - 10 - 25 + 81 \\ & = - 35 + 81 \\ & = 46 \end{aligned}$

Answer:

$46$

We are less likely to make a mistake if we work one operation at a time. Some problems may involve an absolute value, in which case we assign it the same order of precedence as parentheses.

Example $\PageIndex{13}$:

Simplify: $7 - 5 \left| - 2 ^ { 2 } + ( - 3 ) ^ { 2 }\right.$.

Solution

Begin by performing the operations within the absolute value first.

$\begin{aligned} 7 - 5 \left| - 2 ^ { 2 } + ( - 3 ) ^ { 2 } \right| & = 7 - 5 | - 4 + 9 | \\ & = 7 - 5 | 5 | \\ & = 7 - 5 \cdot 5 \\ & = 7 - 25 \\ & = - 18 \end{aligned}$

Subtracting $7−5$ first will lead to incorrect results.

Answer:

$−18$

Exercise $\PageIndex{3}$

Simplify: $- 6 ^ { 2 } - \left[ - 15 - ( - 2 ) ^ { 3 } \right] - ( - 2 ) ^ { 4 }$.

Answer

$-45$

www.youtube.com/v/DnaviQzLPA0

Key Takeaways

Addition is commutative and subtraction is not. Furthermore, multiplication is commutative and division is not.
Adding or subtracting fractions requires a common denominator; multiplying or dividing fractions does not.
Grouping symbols indicate which operations to perform first. We usually group mathematical operations with parentheses, brackets, braces, and the fraction bar. We also group operations within absolute values. All groupings have the same order of precedence: the operations within the innermost grouping are performed first.
When using exponential notation $a^{n}$, the base a is used as a factor n times. Parentheses indicate that a negative number is to be used as the base. For example, $(−5)^{2}$ is positive and $−5^{2}$ is negative.
To ensure a single correct result when applying operations within a calculation, follow the order of operations. First, perform operations in the innermost parentheses or groupings. Next, simplify all exponents. Perform multiplication and division operations from left to right. Finally, perform addition and subtraction operations from left to right.

Exercise $\PageIndex{4}$

Perform the operations. Reduce all fractions to lowest terms.

$33−(−15)+(−8)$
$−10−9+(−6)$
$−23+(−7)−(−10)$
$−1−(−1)−1$
$\frac{1}{2}+\frac{1}{3}−\frac{1}{6}$
$−\frac{1}{5}+\frac{1}{2}−\frac{1}{10}$
$\frac{2}{3}−(−\frac{1}{4})−\frac{1}{6}$
$−\frac{3}{2}−(−\frac{2}{9})−\frac{5}{6}$
$\frac{3}{4}−(−\frac{1}{2})−\frac{5}{8}$
$−\frac{1}{5}−\frac{3}{2}−(−\frac{7}{10})$
Subtract $3$ from $10$.
Subtract $−2$ from $16$.
Subtract $−\frac{5}{6}$ from $4$.
Subtract $−\frac{1}{2}$ from $\frac{3}{2}$.
Calculate the sum of $−10$ and $25$.
Calculate the sum of $−30$ and $−20$.
Find the difference of $10$ and $5$.
Find the difference of $−17$ and $−3$.

Answer

1. $40$

3. $−20$

5. $\frac{2}{3}$

7. $\frac{3}{4}$

9. $\frac{5}{8}$

11. $7$

13. $\frac{29}{6}$

15. $15$

17. $5$

Exercise $\PageIndex{5}$

The formula $d = | b − a |$ gives the distance between any two points on a number line. Determine the distance between the given numbers on a number line.

$10$ and $15$
$6$ and $22$
$0$ and $12$
$−8$ and $0$
$−5$ and $−25$
$−12$ and $−3$

Answer

1. 5 units

3. 12 units

5. 20 units

Exercise $\PageIndex{6}$

Determine the reciprocal of the following.

$\frac{1}{3}$
$\frac{2}{5}$
$−\frac{3}{4}$
$−12$
$a$ where $a ≠ 0$
$\frac{1}{a}$
$\frac{a}{b}$ where $a ≠ 0$
$\frac{1}{ab}$

Answer

1. $3$

3. $−\frac{4}{3}$

5. $\frac{1}{a}$

7. $\frac{b}{a}$

Exercise $\PageIndex{7}$

Perform the operations.

$−4 (−5) ÷ 2$
$(−15) (−3) ÷ (−9)$
$−22 ÷ (−11) (−2)$
$50 ÷ (−25) (−4)$
$\frac{2}{3} (−\frac{9}{10})$
$−\frac{5}{8} (−\frac{16}{25})$
$\frac{7}{6} (−\frac{6}{7})$
$−\frac{15}{9} (\frac{9}{5})$
$\frac{4}{5} (−\frac{2}{5}) ÷ \frac{16}{25}$
$(−\frac{9}{2}) (−\frac{3}{2}) ÷ \frac{27}{16}$
$\frac{8}{5} ÷ \frac{5}{2} ⋅ \frac{15}{40}$
$\frac{3}{16} ÷ \frac{5}{8} ⋅ \frac{1}{2}$
Find the product of $12$ and $7$.
Find the product of $−\frac{2}{3}$ and $12$.
Find the quotient of $−36$ and $12$.
Find the quotient of $−\frac{3}{4}$ and $9$.
Subtract $10$ from the sum of $8$ and $−5$.
Subtract $−2$ from the sum of $−5$ and $−3$.
Joe earns $$18.00$ per hour and “time and a half” for every hour he works over $40$ hours. What is his pay for $45$ hours of work this week?
Billy purchased $12$ bottles of water at $$0.75$ per bottle, $5$ pounds of assorted candy at $$4.50$ per pound, and $15$ packages of microwave popcorn costing $$0.50$ each for his party. What was his total bill?
James and Mary carpooled home from college for the Thanksgiving holiday. They shared the driving, but Mary drove twice as far as James. If Mary drove for $210$ miles, then how many miles was the entire trip?
A $6 \frac{3}{4}$ foot plank is to be cut into $3$ pieces of equal length. What will be the length of each piece?
A student earned $72, 78, 84,$ and $90$ points on her first four algebra exams. What was her average test score? (Recall that the average is calculated by adding all the values in a set and dividing that result by the number of elements in the set.)
The coldest temperature on Earth, $−129$°F, was recorded in $1983$ at Vostok Station, Antarctica. The hottest temperature on Earth, $136$°F, was recorded in $1922$ at Al’ Aziziyah, Libya. Calculate the temperature range on Earth.

Answer

1. $10$

3. $−4$

5. $−\frac{3}{5}$

7. $−1$

9. $−\frac{1}{2}$

11. $\frac{6}{25}$

13. $84$

15. $−3$

17. $−7$

19. $$855$

21. $315$ miles

23. $81$ points

Exercise $\PageIndex{8}$

Perform the operations.

$7 − \{3 − [−6 − (10)]\}$
$− (9 − 12) − [6 − (−8 − 3)]$
$\frac{1}{2} \{5 − (10 − 3)\}$
$\frac{2}{3} \{−6 + (6 − 9)\}$
$5 \{2 [3 (4 − \frac{3}{2} )]\}$
$\frac{1}{2} \{−6 [− (\frac{1}{2} − \frac{5}{3})]\}$
$\frac { 5 - | 5 - ( - 6 ) | } { | - 5 | - | - 3 | }$
$\frac { | 9 - 12 | - ( - 3 ) } { | - 16 | - 3 ( 4 ) }$
$\frac { - | - 5 - ( - 7 ) | - ( - 2 ) } { | - 2 | + | - 3 | }$
$\frac { 1 - | 9 - ( 3 - 4 ) | } { - | - 2 | + ( - 8 - ( - 10 ) ) }$

Answer

1. $−1$2

3. $−1$

5. $75$

7. $−3$

9. $0$

Exercise $\PageIndex{9}$

Perform the operations.

$12^{2}$
$(−12)^{2}$
$−12^{2}$
$−(−12)^{2}$
$−5^{4}$
$(−5)^{4}$
$(−\frac{1}{2})^{3}$
$−(−\frac{1}{2})^{3}$
$−(−\frac{3}{4})^{2}$
$−(−\frac{5}{2})^{3}$
$(−1)^{22}$
$(−1)^{13}$
$−(−1)^{12}$
$−(−1)^{5}$
$−10^{2}$
$−10^{4}$

Answer

1. $144$

3. $−144$

5. $−625$

7. $−\frac{1}{8}$

9. $−\frac{9}{16}$

11. $1$

13. $−1$

15. $−100$

Exercise $\PageIndex{10}$

Simplify.

$5 − 3 (4 − 3^{2})$
$8 − 5 (3 − 3^{2})$
$(−5)^{2} + 3 (2 − 4^{2})$
$6 − 2 (−5^{2} + 4 ⋅ 7)$
$5 − 3 [3 (2 − 3^{2}) + (−3)^{2}]$
$10 − 5 [(2 − 5)^{2} − 3]$
$[5^{2} − 3^{2} ] − [2 − (5 + (−4)^{2} )]$
$−7^{2} − [ (2 − 7)^{2} − (−8)^{2} ]$
$\frac{3}{16} ÷ (\frac{5}{12} −\frac{1}{2} +\frac{2}{3}) ⋅ 4$
$6 \cdot \left[ \left( \frac { 2 } { 3 } \right) ^ { 2 } - \left( \frac { 1 } { 2 } \right) ^ { 2 } \right] \div ( - 2 ) ^ { 2 }$
$\frac { 3 - 2 \cdot 5 + 4 } { 2 ^ { 2 } - 3 ^ { 2 } }$
$\frac { \left( 3 + ( - 2 ) ^ { 2 } \right) \cdot 4 - 3 } { - 4 ^ { 2 } + 1 }$
$\frac { - 5 ^ { 2 } + ( - 3 ) ^ { 2 } \cdot 2 - 3 } { 8 ^ { 2 } + 6 ( - 10 ) }$
$\frac { ( - 4 ) ^ { 2 } + ( - 3 ) ^ { 3 } } { - 9 ^ { 2 } - \left( - 12 + 2 ^ { 2 } \right) * 10 }$
$−5^{2} − 2 |−5| $
$−2^{4} + 6 | 2^{4} − 5^{2} |$
$− (4− | 7^{2} − 8^{2} |)$
$−3 (5 − 2 |−6|)$
$(−3)^{2}− | −2 + (−3)^{3} | − 4^{2}$
$−5^{2} − 2 | 3^{3} − 2^{4} | − (−2)^{5}$
$5 ⋅ |−5| − (2 − |−7|)^{3}$
$10^{2} + 2 ( |−5|^{3} − 6^{3})$
$\frac{2}{3} − | \frac{1}{2} − (−\frac{4}{3})^{2} |$
$−24 | \frac{10}{3} − \frac{1}{2} ÷ \frac{1}{5} |$
Calculate the sum of the squares of the first three consecutive positive odd integers.
Calculate the sum of the squares of the first three consecutive positive even integers.
What is $6$ subtracted from the sum of the squares of $5$ and $8$?
What is $5$ subtracted from the sum of the cubes of $2$ and $3$?

Answer

1. $20$

3. $−17$

5. $41$

7. $35$

9. $\frac{9}{7}$

11. $\frac{3}{5}$

13. $−\frac{5}{2}$

15. $−35$

17. $11$

19. $−36$

21. $150$

23. $−\frac{11}{18}$

25. $35$

27. $83$

Exercise $\PageIndex{11}$

What is PEMDAS and what is it missing?
Does $0$ have a reciprocal? Explain.
Explain why we need a common denominator in order to add or subtract fractions.
Explain why $(−10)^{4}$ is positive and $−10^{4}$ is negative.

Answer

1. Answer may vary

3. Answer may vary

Footnotes

⁵³The result of adding.

⁵⁴The result of subtracting.

⁵⁵Given any real number $a, a + 0 = 0 + a = a$.

⁵⁶Given any real number $a, a + (−a) = (−a) + a = 0$.

⁵⁷Given real numbers $a, b$ and $c, (a + b) + c = a + (b + c)$.

⁵⁸Given real numbers $a$ and $b$, $a + b = b + a$.

⁵⁹A denominator that is shared by more than one fraction.

⁶⁰The least common multiple of a set of denominators.

⁶¹The result of multiplying.

⁶²The result of dividing.

⁶³Given any real number $a, a ⋅ 0 = 0 ⋅ a = 0 .$

⁶⁴Given any real number $a, a ⋅ 1 = 1 ⋅ a = a .$

⁶⁵Given any real numbers $a, b$ and $c, (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) . $

⁶⁶Given any real numbers $a$ and $b, a ⋅ b = b ⋅ a.$

⁶⁷Two real numbers whose product is $1$.

⁶⁸Parentheses, brackets, braces, and the fraction bar are the common symbols used to group expressions and mathematical operations within a computation.

⁶⁹The compact notation $a^{n}$ used when a factor $a$ is repeated $n$ times.

⁷⁰The factor $a$ in the exponential notation $a^{n}$.

⁷¹The positive integer $n$ in the exponential notation $a^{n}$ that indicates the number of times the base is used as a factor.

⁷²The result when the exponent of any real number is $2$.

⁷³The result when the exponent of any real number is $3$.

Search

Text Color

Text Size

Margin Size

Font Type