Skip to main content
Mathematics LibreTexts

3.1.1: Preparation 3.1

  • Page ID
    148737
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

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

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

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

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

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

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

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

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

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

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

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

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

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

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

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

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

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

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

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

    There was no Exercise for Unit 2.9. However, the content presented here will prepare you for Collaboration 3.1.

    Equivalent Fractions

    Four circles showing the following:
1st circle: Shown as one whole
2nd circle: Shown as two halves (one half is highlighted)
3rd circle: Shown as four quarters (two quarters are highlighted)
4th circle: Shown as eight eighths (four eighths are highlighted)

The same part of an object: 1/2 = 2/4 = 4/8

    The same part of an object: 1/2 = 2/4 = 4/8

    Two fractions are equivalent if they have the same value or represent the same part of an object.

    For example, the figure shows that \(\dfrac{1}{2}\), \(\dfrac{2}{4}\), and \(\dfrac{4}{8}\) all represent the same part of a whole. They are equivalent fractions.

    Recall that the denominator of a fraction represents the number of parts into which the whole has been divided. The numerator represents a count of the number of parts.
    So, \(\dfrac{4}{8}\) means that the whole is divided into eight equal parts, and four of these parts are counted.

    (1) Write another fraction that are equivalent to \(\dfrac{1}{2}\), other than \(\dfrac{2}{4}\) or \(\dfrac{4}{8}\).

    Simplest Form of a Fraction

    The fraction \(\dfrac{50}{100}\) is equivalent to \(\dfrac{1}{2}\). Note that you can write

    \[\dfrac{50}{100} = \dfrac{1 \cdot 50}{2 \cdot 50} = \dfrac{1}{2} \cdot 1\nonumber \]

    The above calculation shows that both 50 and 100 can be written as a number times 50. You say that 50 and 100 have a “common factor of 50.” Another way to think of this is that the number “1” (written as \(\dfrac{50}{50}\)) is embedded in the fraction \(\dfrac{50}{100}\).

    The number “1” is a special number in mathematics because if you multiply any number by 1, you get a result that is equivalent to the original.

    Simplify and Simplest Form

    The word, simplify means to rewrite a fraction in an equivalent form with smaller numbers. The simplest form means that the fraction is written using the smallest possible whole numbers. In general, answers should always be given in simplest form unless the question specifically calls for a different form

    For example:

    By dividing \(\dfrac{50}{50}\) to get 1, you simplify \(\dfrac{50}{100}\) to \(\dfrac{1}{2}\).

    Caution! It is common to say that you “cancel” the 50 from the numerator (top number) and denominator (bottom number) and write the fraction in “reduced form.” This language is misleading for two reasons. While numbers can be “canceled” by dividing by 1, it is better to think about the operation (add, subtract, multiply, divide) so you can understand why it works.

    Second, the value of the simpler fraction is the same as the original fraction, but the word “reduced” implies that the “reduced fraction” represents a smaller quantity. It can be confusing, so the terminology simplest form makes more sense.

    (2) Write each fraction in its simplest form.

    (a) \(\dfrac{24}{36}\)

    (b) \(\dfrac{9}{12}\)

    (c) \(\dfrac{20}{30}\)

    (d) \(\dfrac{35}{28}\)

    (3) Compare the two fractions to determine if they are equivalent.

    (a) \(\dfrac{3}{6}\), \(\dfrac{6}{16}\)

    (b) \(\dfrac{3}{4}\), \(\dfrac{6}{9}\)

    (c) \(\dfrac{10}{8}\), \(\dfrac{5}{4}\)

    (d) \(\dfrac{10}{24}\), \(\dfrac{5}{8}\)

    Multiplying Fractions

    The fact that common factors in the denominator and numerator of a number can be divided to make 1 can be used to make multiplying fractions easier. Consider the following multiplication problem.

    Image showing two-thirds times seven-eighths.

    If you see that there is a common factor of 2 in the numerator and denominator before multiplying, you can divide the common factors first. This makes the multiplication easier because you have smaller numbers to work with and the simplification is complete.

    Image showing multiplication procedure for two-thirds times seven-eighths.

    This is an important concept when working with ratios with units. You will learn more about this below.

    Dividing Fractions

    The division of fractions is not the most intuitive of procedures. So, instead of asking you to memorize the division rule, let’s start by looking at a few examples.

    Suppose you have $48 to spend on going to the movies during a month. How many tickets can you buy in a month? A movie ticket costs $8. One way to think about this is that you want to know how many groups of $8 there are in $48, or 48 ÷ 8.

    In the same way, suppose you had $10 to spend on downloading songs for a 1/2 dollar. (For the sake of the mathematics, you are going to express “a half of a dollar” as a fraction instead of as a decimal.) This means you want to know how many 1/2 dollars there are in $10. Your common sense probably tells you that the answer is 20 because every one dollar has two halves. So you multiplied 102. Look at this written as a calculation:

    One over two and two over one are called reciprocals.

    Division is the same as multiplying by the reciprocal. Here are more examples:

    Four divided by a half is the same as four times 2 over one.

    Perform the calculations indicated in Questions 4–7. Write your answer as a whole number or a fraction in reduced form. Note: If your answer is an improper fraction (the numerator is bigger than the denominator), leave it that way. Don't convert it to a decimal or a mixed number. Improper fractions are easier to work with and easier to type into a computer.

    (4) \(10 \cdot \dfrac{2}{5} =\)

    (5) \(8 \div \dfrac{3}{4} =\)

    (6) \(\dfrac{7}{8} \div \dfrac{5}{8} =\)

    (7) \(\dfrac{7}{8} \cdot \dfrac{8}{5} =\)

    (8) Jarrod is helping with his daughter’s school assembly. Every student will receive a decorative ribbon to wear. The ribbons have to be 2/3 of a foot long. A local store donates 30 feet of ribbon. How many decorative ribbons can Jarrod make?

    (9) Lorinda has started to think about saving for retirement. She reads a recommendation that says she should save at least 3/10 of her income because she is over 40 years old. Lorinda makes $42,000 a year. How much should she save in one year according to this recommendation?

    Conversion Factors

    A fraction that is a ratio of quantities can be equivalent to 1, even when the numerator and denominator are not the same number. However, it is necessary that the numerator and denominator represent equivalent quantities. For example, the following fractions are all forms of 1:

    16 ounces over 1 pound. 1 mile over 5280 feet. 60 minutes over 1 hour.

    These types of ratios are sometimes called conversion factors because they can be used to convert between units.

    (10) Complete the following fractions to make a conversion factor. Look up the conversion factor in a reference book or on the Internet if you do not know it.

    (a) 1 yard over ? feet.

    (b) ? fluid ounces over 1 gallon.

    (c) ? inches over 1 yard.

    Converting for Speed

    A rate, like 35 miles per hour, can be expressed as a fraction: \(\dfrac{35\;miles}{1\;hour}\). You can use the conversion factor of 60 minutes per 1 hour to convert the rate to the units of miles per 1 minute. The example below shows how to set up a multiplication problem with the rate and the conversion factor to convert miles per hour to miles per minute.

    35 miles over 1 hour times 1 hour over 60 minutes.

    Notice that the conversion factor is written so that the units of “hours” are in the numerator. This is because you want the “hours” to divide out in the same way that common factors divided out in the multiplication problems above. This leaves the units of miles/minute as shown below (on the right).

    Multiplication of Fractions

    Unit Conversion

    \(\dfrac{2}{5}\times\dfrac{15}{7}\rightarrow\dfrac{2}{3}\times\dfrac{5\times3}{7}\rightarrow\dfrac{6}{7}\)

    35 miles over 1 hour times 1 hour over 60 minutes equals 35 miles over 60 minutes which equals 0.58 miles over 1 minute.

    Both \(\dfrac{35\;miles}{hour}\) and \(\dfrac{0.58\;mile}{minute}\) are unit rates. Unit rates are ratios that have 1 unit in the denominator. Here are some other examples of unit rates:

    • \(\dfrac{35,000\;people}{square\;mile}\)
    • \(\dfrac{2,500\;calories}{day}\)
    • \(\dfrac{56,661\;views}{YouTube\;Video}\)

    You will use this concept in Collaboration 3.1.

    This assignment included a large number of vocabulary words that are important. Make sure you fully understand each term listed below:

    • Equivalent fraction
    • Simplify, simplest form
    • Common factor
    • Reciprocal
    • Unit rate
    • Conversion factor

    After Preparation 3.1 (survey)

    You should be able to do the following things for the next collaboration. Rate how confident you are on a scale of 1–5 (1 = confident and 5 = very confident).

    Before beginning Collaboration 3.1, you should understand the concepts and demonstrate the skills listed below.

    Skill or Concept: I can …

    Rating from 1 to 5

    multiply two fractions.

     

    divide two fractions.

     

    understand that a fraction can be simplified by dividing common factors in the numerator and denominator (simplify fractions).

     

    understand that multiplying by 1 doesn’t change a value.

     

    be familiar with basic units of measure of length (feet, miles) and time (seconds, hours, minutes).

     

    This page titled 3.1.1: Preparation 3.1 is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Carnegie Math Pathways (WestEd) .

    • Was this article helpful?