1.5: Mathematical Notation and the Factor-Label Method

Learning Objectives

• To be introduced to the concept of mathematics as symbolic language for understanding and problem solving
• To alter a pre-existing mathematical model to fit the perspective of a problem
• To use the factor-label method to identify whether an equation is set up correctly in a numerical calculation
• To use the factor-label method to facilitate the conversion of units

In "Introduction to Mathematical Thinking" Professor Keith Devlin describes two types of categories of math skills. The former category consists of the rote use of formulas and calculations to solve what is traditionally called a math problem. These problems are identified by their abstraction from reality and are valuable for teaching how to understand mathematical relationships. The latter category consists of modeling a problem with mathematics and then using the now-visible relationships to more precisely analyze the problem.

Modeling a problem with mathematics and then using the now-visible relationships to more precisely analyze the problem.

Modeling with Symbols and Precise Language  

 Problem-solving starts with symbols. In a school environment, these will often be the words you read which describe the problem. In chemistry, they may be illustrations of atoms or chemical formulas.

Definition: Symbol

A mark or character used as a conventional representation of an object, function, or process, e.g. the letter or letters standing for a chemical element or a character in musical notation.

A lone symbol is not necessarily helpful. Problems exist because of the impact parts of a system have on the whole system. One must understand the relationship between the symbols to understand the system. Symbolic notations, like in music, let a person express complex ideas (like sound), while not actually being those things.

Figure \(\PageIndex{i}\): The treachery of images (This is not a pipe). Rene Magritte, 1928-1929; Brussels, Belgium. (Fair Use)

Spoken and written languages are symbolic notations, but often rely on living and experience and context. Consider the unfortunate statement, "One American dies of aggressive cancer almost every hour." What does it mean? Given the order (i.e. relationship) between the words (i.e. symbols), a singular American undergoes an hourly cycle of rebirth and death. However, the context of reality leads most to interpret the statement as it should have been expressed, "Almost every hour, an American dies of aggressive cancer." The two statements seem interchangeable, particularly to native-speakers, due to a familiarity with how the world works. Yet, one cannot rely on familiarity when approach novel problems and unfamiliar circumstances. Mathematics provides a more precise language because its relationships are also represent by symbols.

Newton's third law of motion states, "For every action there is an equal and opposite reaction." The more precise definition is, "When one object exerts a force on a second object, the second object exerts an equal force on the first object." Modeled mathematically \[F_1=F_2\]

Force itself is a symbol representing the relationship between the mass of an object and its acceleration:

\begin{array}{l}F=ma\end{array}

Mathematics then allows for the substitution of F with what it is equal to, ma, and the resulting equation:

\begin{array}{l}m_1a_1=m_2a_2\end{array}

Starting with Newton's third law of motion, the resulting equation 1.5.3 shows an inverse relationship between mass and acceleration. If the first object has greater mass than the second object, then the second object must have a greater acceleration than the first object. Equations can also be manipulated, giving a change in perspective, by changing what is being solved for; or in mathematical terms whether one is solving for f(x) in terms of y or f(y) in terms of x.

Example \(\PageIndex{1}\)

Solve for y given x = 2y and interpret both equations.

Solution

\[x\ =\ 2y\\\frac{x}{2}=y\ \text{or}\ y=\frac{x}{2}\]

The first equation states x is always twice twice the value as y. The second equation states y is always half the value of x.

The Factor-Label Method

As described in the previous section, all measurements involve a description of what is being measured. The descriptions, or units, become part of the equation. Labels factor and expand as if they were numbers.

Factoring and Expansion

Factoring, or finding factors, results in the multiplication of a simpler equation. Expansion is its opposite.

To factor the equation 4x+2, begin by identifying the common factor of the terms being added together. In this case, it is 2. Divide both terms, 4x and 2, by the common factor, 2. Place the resulting equation in parentheses and multiple the result by the common factor.

\[\frac{4x}{2}=2x\ \text{and}\ \frac{2}{2}=1\\2x+1\ \text{is the resulting equation}\\2\left(2x+1\right)\ \text{is the factored solution}\]

To expand the equation 3(2x-5), multiply the terms inside the parentheses by the outside value.

\[3\times2x=6x\ \text{and}\ 3\times5=15\\6x-15\ \text{is the resulting equation and the expanded solution}\]

The factor-label method takes advantage of units following these mathematical rules to understand how properties interact and to convert between units. What is meant by, "how properties interact"? Consider the following problem. If you have three apples and are given two more apples, how many apples do you have? The obvious answer is five apples, but look at how this can be approached through factoring and expansion:

\begin{array}{l}3\ \text{apples}\ +2\ \text{apples}\ \ \\\text{apples}\left(3+2\right)\\\text{apples}\left(5\right)\\\text{5 apples}\end{array}

What about if you have three apples and are given two oranges?

\begin{array}{l}3\ \text{apples}\ +2\ \text{oranges}\ \ \end{array}

Apples and oranges cannot be factored. However, apples are a fruit and oranges are a fruit. Substitution can be applied.

\begin{array}{l}3\ \text{apples}\ +2\ \text{oranges}\\\text{apple = fruit and orange = fruit}\\3\ \text{fruit}\ +2\ \text{fruit}\\\text{fruit}\left(3+2\right)\\\text{fruit}\left(5\right)\\5\ \text{fruit}\end{array}

While the previous examples are a lot of work for what are simple problems, the goal is to understand the process so it can be applied to more complex problems.

Factor-Label Method and Unit Conversion

The factor-label method is most often used to convert between units. Unit conversion effectively describes the same physical reality from different perspective. A chunk of gold has a mass, a volume, a color, a shape; however, it is still the same chunk regardless of which property is being focused on. 

This can be explained with proportions and fractions. A numerator and denominator reduce to one when they are the same.

\[\frac{x}{x}=1\]

A value remaining unchanged when multiplied by one is a common math fact. Three times one remains itself, three. However, it is more precise to say that anything over itself achieves unity and anything multiplied by unity remains unchanged. Proportions are an example of unity. A meter has 100 centimeters:

\[1\ m\ =\ 100\ cm\]

Simple algebra means that the left side of the equation can be reduced to one by dividing by 1 m. However, any operation performed on the left side must be performed on the right side

\[\frac{1\ m}{1\ m}\ =\ \frac{100\ cm}{1\ m}\]

Completing the operations ends with

\[1\ =\ \frac{100\ cm}{1\ m}\]

Yet, how can 1 equal 100? Equation 1.5.13 shows unity, that 100 cm and 1 m are the same physically if not numerically. The incongruity of the values, however, results in a proportion which can be used to convert between the physical units in the numerator and physical units in the denominator. What happens, then, when they are used together in multiplication along with an object's measured length?

\[x\ \text{meter}\times\frac{100\ \text{centimeter}}{1\ \text{meter}}\\\frac{x\ \text{meter}\times100\ \text{centimeter}}{1\ \text{meter}}\\\frac{\text{meter}}{\text{meter}}\times\frac{x\times100\ \text{centimeter}}{1}\\100x\ \text{centimeters}\]

The calculation results in a value 100 times greater than what was initially measured, but the object's length has not physically become 100 times greater. A unity was created by taking two numerically different, but physically the same, measurements and using them as a conversion factor to change the object's measured length from meters to centimeters. When the factor-label method is used in this manner, it is called unit conversion. Unit conversion is the most common application of the factor-label method and the two terms are often used interchangeably. Dimensional analysis is another synonym as the process is often applied to analyzing different dimensions of a physical system.

Definition: Conversion Factor

A fraction in which the numerator and denominator are two different measured units which describe the same physical system.

Definition: Unit Conversion

Multiplying a measured value by a conversion factor in order to calculate the measured value in a different unit, or to calculate a proportional physical property.

 

Math With Fractions

Addition and Subtraction

With a common denominator

\[\frac{a}{x}+\frac{b}{x}=\frac{a+b}{x}\]

Without a common denominator

\[\frac{a}{x}+\frac{b}{y}\\\frac{y}{y}\times\frac{a}{x}+\frac{b}{y}\times\frac{x}{x}\\\frac{ay}{xy}+\frac{bx}{xy}=\frac{ay+bx}{xy}\]

Multiplication

\[\frac{x}{a}\times\frac{y}{b}=\frac{xy}{ab}\]

Canceling Out

\[x\times\frac{y}{x}=\frac{xy}{x}\\\frac{xy}{x}=\frac{x}{x}\times\frac{y}{1}\\\frac{x}{x}\times\frac{y}{1}=1\times\frac{y}{1}\\1\times\frac{y}{1}=y\]

Division

Non-fraction divided by a fraction

\[\frac{x}{\frac{a}{b}}=x\times\frac{b}{a}=\frac{xb}{a}\\\text{or}\\x\div\frac{a}{b}=x\times\frac{b}{a}=\frac{xb}{a}\]

Fraction divided by a fraction

\[\frac{\frac{x}{y}}{\frac{a}{b}}=\frac{x}{y}\times\frac{b}{a}=\frac{xb}{ya}\\\text{or}\\\frac{x}{y}\div\frac{a}{b}=\frac{x}{y}\times\frac{b}{a}=\frac{xb}{ya}\]

 

Proportions as Conversion Factors

If you have every planned a party, you have used conversion factors. The amount of drinks and food you will need depends on the number of people you expect. For example, if you are planning a Friday night party and expect 30 people you might estimate you need to go out and buy 120 bottles of sodas and 10 large pizza's. How did you arrive at these numbers? The following indicates the factor-label method solution to your party problem:

\[(30 \; \cancel{humans}) \times \left( \dfrac{\text{4 sodas}}{1 \; \cancel{human}} \right) = 120 \; \text{sodas} \label{Eq1} \]

\[(30 \; \cancel{humans}) \times \left( \dfrac{\text{0.333 pizzas}}{1 \; \cancel{human}} \right) = 10 \; \text{pizzas} \label{Eq2} \]

Notice that the units that canceled out are lined out and only the desired units are left (discussed more below). Finally, in going to buy the soda, you perform another unit conversion: should you buy the sodas in six-packs or in cases?

\[(120\; { sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3} \]

\[(120\; {sodas}) \times \left( \dfrac{\text{1 case }}{24\; {sodas}} \right) = 5 \; \text{cases} \label{Eq4} \]

Realizing that carrying around 20 six packs is a real headache, you get 5 cases of soda instead.

In this party problem, we have used the factor-label in two different ways:

• In the first application (Equations \(\ref{Eq1}\) and Equation \(\ref{Eq2}\)), the factor-label was used to calculate how much soda is needed need proportional to the number of people. This is based on knowing: (1) how much soda we need for one person and (2) how many people we expect; likewise for the pizza.
• In the second application (Equations \(\ref{Eq3}\) and \(\ref{Eq4}\)), dimensional analysis was used to convert units (i.e. from individual sodas to the equivalent amount of six packs or cases)

Using the Factor-Label Method to Convert Units

Consider the conversion in Equation \(\ref{Eq3}\):

\[(120\; {sodas}) \times \left( \dfrac{\text{1 six pack}}{6\; {sodas}} \right) = 20 \; \text{six packs} \label{Eq3a} \]

If we ignore the numbers for a moment, and just look at the units, we have:

\[\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) \nonumber \]

The units are treated like values for numerical analyses (i.e. any number divided by itself is 1). Therefore:

\[\text{soda} \times \left(\dfrac{\text{six pack}}{\text{sodas}}\right) = \cancel{\text{soda}} \times \left(\dfrac{\text{six pack}}{\cancel{\text{sodas}}}\right) \nonumber \]

So, size of the answer is six, but the unit is now describing packs instead of individual sodas, while the physical quantity of soda remains unchanged.

How can we use the factor-label method to be sure we have set up our equation correctly? Consider the following alternative way to set up the above unit conversion analysis:

\[ 120 \cancel{\text{soda}} \times \left(\dfrac{\text{6 sodas}}{\cancel{\text{six pack}}}\right) = 720 \; \dfrac{\text{sodas}^2}{\text{1 six pack}} \nonumber \]

• While it is correct that there are 6 sodas in one six pack, the above equation yields a value of 720 with units of sodas²/six pack.
• These rather bizarre units indicate that the equation has been setup incorrectly (and as a consequence you will have a ton of extra soda at the party).

The unit you want to cancel out (the unit which is not part of the answer) should be in the denominator. The unit you want your answer to be in should be in the numerator.

Using the Factor-Label Method in Calculations

In the above case it was relatively straightforward keeping track of units during the calculation. What if the calculation involves powers, etc? For example, the equation relating kinetic energy to mass and velocity is:

\[E_{kinetics} = \dfrac{1}{2} \text{mass} \times \text{velocity}^2 \label{KE} \]

An example of units of mass is kilograms (kg) and velocity might be in meters/second (m/s). What are the dimensions of \(E_{kinetic}\)?

\[(kg) \times \left( \dfrac{m}{s} \right)^2 = \dfrac{kg \; m^2}{s^2} \nonumber \]

The \(\frac{1}{2}\) factor in Equation \ref{KE} is neglected since pure numbers have no units. Since the velocity is squared in Equation \ref{KE}, the dimensions associated with the numerical value of the velocity are also squared. We can double check this by knowing the the Joule (\(J\)) is a measure of energy, and as a composite unit can be decomposed thusly:

\[1\; J = kg \dfrac{m^2}{s^2} \nonumber \]

Example \(\PageIndex{1}\)

Imagine that a chemist wants to measure out 0.214 mL of benzene, but lacks the equipment to accurately measure such a small volume. The chemist, however, is equipped with an analytical balance capable of measuring to \(\pm 0.0001 \;g\). Looking in a reference table, the chemist learns the density of benzene (\(\rho=0.8765 \;g/mL\)). How many grams of benzene should the chemist use?

Solution

\[0.214 \; \cancel{mL} \left( \dfrac{0.8765\; g}{1\;\cancel{mL}}\right)= 0.187571\; g \nonumber \]

Notice that the mL are being divided by mL, an equivalent unit. We can cancel these our, which results with the 0.187571 g. However, this is not our final answer, since this result has too many significant figures and must be rounded down to three significant digits. This is because 0.214 mL has three significant digits and the conversion factor had four significant digits. Since 5 is greater than or equal to 5, we must round the preceding 7 up to 8.

Hence, the chemist should weigh out 0.188 g of benzene to have 0.214 mL of benzene.

Example \(\PageIndex{2}\)

To illustrate the use of dimensional analysis to solve energy problems, let us calculate the kinetic energy in joules of a 320 g object traveling at 123 cm/s.

Solution

To obtain an answer in joules, we must convert grams to kilograms and centimeters to meters. Using Equation \ref{KE}, the calculation may be set up as follows:

\[ \begin{align*} KE&=\dfrac{1}{2}mv^2=\dfrac{1}{2}(g) \left(\dfrac{kg}{g}\right) \left[\left(\dfrac{cm}{s}\right)\left(\dfrac{m}{cm}\right) \right]^2 \\[4pt] &= (\cancel{g})\left(\dfrac{kg}{\cancel{g}}\right) \left(\dfrac{\cancel{m^2}}{s^2}\right) \left(\dfrac{m^2}{\cancel{cm^2}}\right) = \dfrac{kg⋅m^2}{s^2} \\[4pt] &=\dfrac{1}{2}320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) \left[\left(\dfrac{123\;\cancel{cm}}{1 \;s}\right) \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) \right]^2=\dfrac{0.320\; kg}{2}\left[\dfrac{123 m}{s(100)}\right]^2 \\[4pt] &=\dfrac{1}{2} 0.320\; kg \left[ \dfrac{(123)^2 m^2}{s^2(100)^2} \right]= 0.242 \dfrac{kg⋅m^2}{s^2} = 0.242\; J \end{align*} \nonumber \]

Alternatively, the conversions may be carried out in a stepwise manner:

Step 1: convert \(g\) to \(kg\)

\[320\; \cancel{g} \left( \dfrac{1\; kg}{1000\;\cancel{g}}\right) = 0.320 \; kg \nonumber \]

Step 2: convert \(cm\) to \(m\)

\[123\;\cancel{cm} \left(\dfrac{1 \;m}{100\; \cancel{cm}}\right) = 1.23\ m \nonumber \]

Now the natural units for calculating joules is used to get final results

\[ \begin{align*} KE &=\dfrac{1}{2} 0.320\; kg \left(1.23 \;ms\right)^2 \\[4pt] &=\dfrac{1}{2} 0.320\; kg \left(1.513 \dfrac{m^2}{s^2}\right)= 0.242\; \dfrac{kg⋅m^2}{s^2}= 0.242\; J \end{align*} \nonumber \]

Of course, steps 1 and 2 can be done in the opposite order with no effect on the final results. However, this second method involves an additional step.

Example \(\PageIndex{3}\)

Now suppose you wish to report the number of kilocalories of energy contained in a 7.00 oz piece of chocolate in units of kilojoules per gram.

Solution

To obtain an answer in kilojoules, we must convert 7.00 oz to grams and kilocalories to kilojoules. Food reported to contain a value in Calories actually contains that same value in kilocalories. If the chocolate wrapper lists the caloric content as 120 Calories, the chocolate contains 120 kcal of energy. If we choose to use multiple steps to obtain our answer, we can begin with the conversion of kilocalories to kilojoules:

\[120 \cancel{kcal} \left(\dfrac{1000 \;\cancel{cal}}{\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \cancel{J}}\right)= 502\; kJ \nonumber \]

We next convert the 7.00 oz of chocolate to grams:

\[7.00\;\cancel{oz} \left(\dfrac{28.35\; g}{1\; \cancel{oz}}\right)= 199\; g \nonumber \]

The number of kilojoules per gram is therefore

\[\dfrac{ 502 \;kJ}{199\; g}= 2.52\; kJ/g \nonumber \]

Alternatively, we could solve the problem in one step with all the conversions included:

\[\left(\dfrac{120\; \cancel{kcal}}{7.00\; \cancel{oz}}\right)\left(\dfrac{1000 \;\cancel{cal}}{1 \;\cancel{kcal}}\right)\left(\dfrac{4.184 \;\cancel{J}}{1 \; \cancel{cal}}\right)\left(\dfrac{1 \;kJ}{1000 \;\cancel{J}}\right)\left(\dfrac{1 \;\cancel{oz}}{28.35\; g}\right)= 2.53 \; kJ/g \nonumber \]

The discrepancy between the two answers is attributable to rounding to the correct number of significant figures for each step when carrying out the calculation in a stepwise manner. Recall that all digits in the calculator should be carried forward when carrying out a calculation using multiple steps. In this problem, we first converted kilocalories to kilojoules and then converted ounces to grams.

Converting Between Units: https://youtu.be/wSaOh48k8Wg

 

Summary

The factor-label method is used in numerical calculations, and in converting units. It can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected). Units are treated similarly to the associated numerical values, i.e., if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc.

Exercise \(\PageIndex{1}\)

1. A bus can hold 25 passengers.
   1. Make an equality to describe the relationship between a bus and the number of passengers it can hold.
   2. What conversion factor would you use if you knew the number of passengers, but needed to calculate the number of buses needed to carry them?
   3. What conversion factor would you use if you knew the number of buses, but needed to calculate the number of passengers they could carry?
2. A student uses the following calculation to determine the mass of 3 cubic centimeters of gold; knowing that a cubic centimeter of gold has a mass of 19.3 grams.
   \[3\times\frac{1}{19.3}\]
   Is the student going to get the correct answer? Why or why not?

Answer

1. Answers
   1. 1 bus = 25 passengers
   2. \(\frac{1\ bus}{25\ \ passengers}\)
   3. \(\frac{25\ passengers}{1\ \ bus}\)
2. The student will not get the correct answer. If they had included units, which they should, then the student would see that their final unit would be \(\frac{cm^3\times cm^3}{g}\)

 

Contributors and Attributions

• Mark Tye (Diablo Valley College)

• Mike Blaber (Florida State University)