How to Find the Day of the Week for a Given Date

Pope Gregory VIII (1502-1585)
Pope Gregory VIII (1502-1585)

A calendar is a fundamental tool that we routinely use to plan and organize our lives. We rarely think about the many different types of calendars that mankind has used over the centuries. In this post I will discuss the difference between the Julian and Gregorian calendars, and present algorithms for finding the day of week for a given Julian or Gregorian calendar date. Hopefully teachers and parents will use this post to create a variety of interesting mathematical activities to give students practice with basic math operations, and develop the ability to maintain concentration by following the steps in a somewhat complicated algorithm. Answer keys can be easily created by accessing an online Day-of-Week calculator.

The Julian calendar was introduced in 46 BC by Julius Caesar, and Pope Gregory VIII introduced the Gregorian calendar in 1582. A steady shift in the date of Easter was the motivation for adopting the Gregorian calendar by the Catholic church. Other names for the Gregorian calendar are the Western calendar and Christian calendar. Countries such as Italy, France and Spain adopted the Gregorian calendar shortly after it was introduced, but England did not adopt the Gregorian calendar until 1752. Russia didn’t convert to the Gregorian calendar until 1918 because the Gregorian calendar had its origin in the Roman Catholic church. The Gregorian calendar is now the most internationally used civil calendar. If you do any type of historical research, it’s important to know whether a date is a Julian or Gregorian calendar date. Dates in time are not necessarily simple facts.

So what is the essential difference between the Julian and Gregorian calendars? Both calendars have leap years, but leap years occur slightly less often in the Gregorian calendar than in the Julian calendar. I will give an intuitive explanation of leap years in a calendar by considering Earth orbits having different periods.

  • In the Julian calendar, the mean length of a year equals 365.25 days = 365 ¼ days. To account for the extra ¼ day each year, an extra day is added to February in all years that are multiples of 4 such as 1628, 1752, 1800 and 1964.
  • Now suppose that the length of a year equals 365.2 days = 365 1/5 days. To account for the extra 1/5 day each year, the calendar would add an extra day in February in all years that are multiples of 5 such as 1775, 1800, 1945.and 2120.
  • Next suppose that the length of a year equals 365.5 days = 365 ½ days. To account for the extra ½ day each year, the calendar would add an extra day in February in all years that are multiples of 2 such as 1778, 1802, 1946 and 2136.
  • In the Gregorian calendar, the mean length of a year equals 365.2425 days. To account for the extra 0.2425 days each year, it was decided to add an extra day in February in the years that are multiples 4 or 400, but not multiples of 100. Hence the years 1600, 2000, 2400 and 2800 are leap years, but 1700, 1800, 1900 and 2100 are not leap years. All other years that are multiples of 4 such as 1776, 1812, 1936 and 2008 are leap years.

Before I can show you the Julian and Gregorian calendar algorithms for finding the day of week for given a date, I need to explain how the floor(x) function and binary mod operator works. The floor(x) function rounds the input value x down to the nearest integer. The p mod q operation gives us the remainder when integer p is divided by integer q ≠ 0. If p or q is negative, different programming languages will produce different mod operator results. In this post, it’s only necessary to deal with integer values of p and q such that p ≥ 0, and q > 0. The text box below illustrates how these functions work.

day_of_week_txtbx1a

The text box below describes Zeller’s Day-of-Week algorithm for Gregorian calendars. Experienced computer programmers or veteran spreadsheet users will find it easy to implement the algorithm in a computer program or spreadsheet. The algorithm should give correct results until about year 4900.

day_of_week_txtbx2a

The two text boxes below show how the Gregorian Day-of-Week algorithm can be applied. I find it interesting and fun to pick significant historical dates. Since England adopted the Gregorian calendar in 1753, the Gregorian calendar must have been adopted in the United States by 1800.

day_of_week_txtbx3a

day_of_week_txtbx4a

The text boxes below show how the Day-of-Week algorithm for Julian calendars. As previously mentioned, it’s important to know if a date is a Julian or Gregorian calendar date. In the two examples for Isaac Newton’s birthday, note that one date is a Julian calendar date and the other is a Gregorian calendar date. I’m sure Newton would have preferred that future historians report his Julian date of birth because it’s nicer for a Christian to say they were born on Christmas day 1642 rather than on January 4, 1643.

day_of_week_txtbx5

day_of_week_txtbx6

day_of_week_txtbx7

Orbit and Rotation of Planet Earth

earth

As we gaze across a beautiful valley or stare in awe at a distant mountain, it is easy to forget that we are on a spinning platform that is traveling on an elliptical orbit around the sun at an average speed of 66,600 miles per hour. I find this seemly unending journey truly amazing. In this post, I would like to take a look at some of the facts that mankind has learned about this journey.

Before Nicholas Copernicus (1474 – 1543), many people thought that the Sun, planets, and stars rotated about the Earth, and each planet in turn rotates on its own private circular arc. This complicated Earth centered view of nature became so entrenched that it became an article of faith in the Catholic Church. In fact, the Catholic Inquisition threated Galileo (1564 – 1642) with torture on the rack unless he publicly retracted his belief in the Sun centered circular orbit Copernican world system. Galileo publicly retracted his belief in the Copernican world view and was spared torture on the rack, but spent the remaining years of his life under house arrest.

Johannes Kepler (1571 – 1630) discovered three laws of planetary motion which is a relatively simple description of planetary motion. (You may find it helpful to read my post Demonstrating Dynamics in a Mathematical Model.) Kepler’s first law stated that the orbit of a planet around the Sun is an ellipse where the Sun is located at one of the two foci of the ellipse. An ellipse is a very special curve where every point P on the ellipse, the distance from P to one focus point plus the distance from P to the other focus point, is a constant. The diagram below shows an ellipse with foci at F1 and F2, length of major axis = 10 units, length of minor axis = 6 units, and center point at (0, 0). For very point P on an ellipse, the sum of the distances from point P to the two focus points equals the length of the major axis. As indicated in the diagram below, an ellipse can be drawn by first anchoring the endpoints of a length of string on a piece of paper or cardboard. Use a pencil to make the string taunt, and then trace the curve by keeping the string taunt as you move the pencil along the elliptical curve.

earth_orbit_fig1

To better understand planetary orbits, it’s necessary to understand what we mean by the eccentricity of an ellipse. If a = half the length of the major axis, and c = the distance from the center to a focus point, then the eccentricity e of the ellipse = c/a. Thus elliptical eccentricity e ranges from 0 to 1. If e = 0, the ellipse is a circle, and if e = 1, the ellipse degenerates to a line segment with foci at the endpoints of the major axis. (By definition, the eccentricity of a parabola equals 1, and the eccentricity of a hyperbola is greater than 1.) The two diagrams below show eccentricity values for five ellipses where the ellipse and foci have the same color. Note that eccentricity approaches 1 as the foci approach the endpoints of the major axis. The eccentricity of the Earth’s orbit = 0.0167086. This is the reason, I suspect, that Copernicus thought the Earth’s orbit was circular, not elliptical. Since half the length of the major axis of the Earth’s elliptical orbit equals 149.6 million km, it follows that the Sun is 0.0167086*149.6 million km = 2.4996 million km from the center of the Earth’s orbit.

earth_orbit_fig2

earth_orbit_txtbx1

The two diagrams below show an exaggerated oval shape of the Earth’s yearly orbit around the Sun; the purpose is to draw your attention to key time periods in a year. Orbital dates can vary slightly from year to year, and therefore the dates shown in the diagrams are approximate. The following points describe the key time periods in Earth’s orbit:

  • At the point of perihelion, the Earth is at its closest point of 147.1 million km from the Sun. In northern latitudes, the direction of the Earth’s polar axis is tilted away from the Sun, which results is less direct sunlight and cooler average temperatures.
  • At the point of aphelion, the Earth is at its farthest point of 152.1 million km from the Sun. In northern latitudes, the direction of the Earth’s polar axis is tilted towards the Sun, which results in more direct sunlight and warmer average temperatures.
  • The equinoxes and solstices divide a year into approximately four equal time periods or seasons. At the fall and spring equinoxes, the Earth’s polar axis is perpendicular to the plane of the Earth’s orbit which results in equal periods of daylight and darkness. At the summer and winter solstices, the Earth’s polar axis is tilted towards or away from the Sun which results the longest and shortest days of the year.
  • At the point of perihelion, the Earth reaches its fastest orbital speed of 109,080 km/hour.
  • At the point of aphelion, the Earth reaches its slowest orbital speed of 105,480 km/hour.
  • The average or mean orbital speed of the Earth equals 107,200 km/hour or 66,600 mph.
  • It takes the Earth 365.256 363 004 days to orbit the Sun. Because of the extra 0.256 363 004 days in a year, it’s necessary to add an extra day to our calendar every four years in February. To be more specific, leap years occur in years that are multiples of 4 or 400, but not multiples of 100. Hence the years 2000 and 2400 are leap years, but the years 1800, 1900, 2100, 2200 and 2300 are not leap years. All other years that are multiples of 4 such as 1868, 1936 and 2016 are leap years.
  • In the diagrams below, note that seasons in the northern and southern hemispheres occur at opposite times of the year.

orbit_northern

orbit_southern

Everyone knows that the Earth does a daily rotation about its polar axis. Here are a few facts about the Earth’s rotation.

  • The Earth rotates in about 24 hours with respect to the Sun and once every 23 hours, 56 minutes and 4 seconds with respect to the stars.
  • The Earth’s rate of rotation rate is slowing with time. Atomic clocks have demonstrated that a modern-day is about 1.7 milliseconds longer than a day in 1900. (I doubt that this fact will be reported in the national news any time soon.)
  • In the northern hemisphere, the Earth rotates east towards the Sun in the morning hours and away from the Sun in the west in the evening hours. This is the reason that the folks in New York see the Sun about 4 hours before the folks in California.
  • Technically speaking, there is no such thing as sunrise and sunset. The Sun only appears to rise and set in the sky because of the rotation of the Earth. Buckminster Fuller who was an American architect (geodesic domes) and systems theorist suggested that we should the terms sunsight and sunclipse because the terms sunrise and sunset do not accurately describe what we observe.
  • The Earth’s rate of rotation is not constant. The true solar day is about 10 seconds longer at the point of perihelion and 10 seconds shorter at the point of aphelion.
  • At the equator, the Earth’s linear speed of rotation is 465.1 m/s, 1,674.4 km/h or 1,040.4 mph. At higher latitudes, the linear rate of rotation is reduced by a factor of Cos(angle of latitude). Example: The Kennedy Space Center is located 28.59° North latitude and has a linear rotation rate of 1,674.4Cos(28.59°) = 1,470.23 km/h = 913.56 mph.

I will close this post about an epiphany I experienced many years ago. As I recall, it was about March of 1975 when my neighbor Chuck Beck invited me into his back yard to view Sun spot activity. Chuck had placed his expensive Celestron telescope with an attached power cord and lens filter on his picnic table. As I adjusted a knob on the Celestron in order to keep the Sun in view, I had the same physical sensation in my legs as if I was riding a merry-go-round. I thought to myself, “Johnson you really ARE on a moving and spinning platform in space!”

Why Division by Zero Can Lead to Absurd and Disastrous Results

USS Yorktown (CG-48)
USS Yorktown (CG-48)
In an earlier post, Why is Division by Zero is Forbidden, I discussed why division by zero is a mathematically meaningless operation. The purpose of this post is to show how attempted or actual division by zero can lead to absurd and disastrous results. I will also take a look at two rational polynomial functions to show how division by zero at point x = k can be interpreted. It’s important to understand why division by zero results in an undefined value, but it’s more important to be aware of actual or potential division by zero, and how to avoid, interpret, and deal with division by zero.

The first example is based on an article from Wikipedia that was brought to my attention by one of my students. On 21 September 1997, the U.S. Navy cruiser USS Yorktown (CG-48) was rendered dead in the water due to a failure in the ship’s propulsion system. A crew member entered a zero in a data base field causing attempted division by zero which in turn caused a buffer overflow in a computer network. As a result, all machines in the network were shut down until the problem was corrected. This undetected data entry error caused the crew to frantically work several hours to make the ship seaworthy. In light of standard programming practices, it’s difficult to understand why the possibility of a division by zero error was not considered by the network control software programming team. It’s not reasonable to assume that a human will always enter the correct value in a data base field; humans make mistakes.

The next example gives a flawed 9-step mathematical proof that 1 = 2. A class discussion to determine the flaw in the proof can be very interesting and enlightening. One only needs to understand the basic properties of real numbers and simple factoring to follow the logic of the proof. The fatal error in the proof occurs in step 8 because p – q = 0 which in turn leads to division by zero. It’s fun to see the initial reaction of a class when you announce that you will give a proof that 1 = 2. When presenting the proof, I have students give a justification for each step.

dividezero2fig1

In terms of a practical application, let’s see how to avoid the error message #DIV/0! in an Excel spreadsheet cell. It’s important that spreadsheet users are aware of the possibility of division by zero, and they know how to handle this type of error. Suppose the formula =C2/D2 is in some cell of an Excel spreadsheet such as cell E10. If cell D2 is blank or has a value equal to zero, the value of cell E10 will be set to #DIV/0!. This division by zero problem can be avoided by setting the value of cell E10 to blank on division by zero with the conditional statement =IF(D2=0,” “,C2/D2) in cell E10. Another way to avoid this problem is to set the contents of cell E10 to “Undefined” on division by zero with the conditional statement =IF(D2=0,”Undefined”, C2/D2) in cell E10.

Next consider the expression (x2 + x – 6) / (x + 3) = (x + 3)(x – 2) / (x + 3) and the graph of the function y = (x2 + x – 6) / (x + 3) shown below. Depending on the class and mathematical level of the students, there are a variety of interesting questions and points of discussion that can be raised. Remind students that the hole in the graph at (-3, -5) has an infinitely small diameter, and therefore can only be seen by a mathematically aware person or God.

  • Why is the expression (x2 + x – 6) / (x + 3) mathematically equivalent to x – 2 except when x = -3?
  • Why is the graph of the function y = (x2 + x – 6) / (x + 3) appear to be equal to the graph of y = x – 2?
  • What is the value of f(-3)?
  • What is the limit of f(x) as x approaches -3 or when x is arbitrarily close to -3?
  • Why is the function y = (x2 + x – 6) / (x + 3) continuous at every real number x except when x = -3?
  • Why does the graph of the function y = (x2 + x – 6) / (x + 3) contain a single hole at the point (-3, -5)?
  • Graph the function y = (x2 + x – 6) / (x + 3) on your favorite graphing utility. Most likely the graph is just the graph of the line y = x – 2 without any indication that the graph contains a hole? Why is this so?

dividezero2fig2

Lastly, consider the rational polynomial function f(x) = y = 3(x – 2)(x + 4) / ((x + 2)(x – 5)). The graph of this function and a table of key function values are shown below. Depending on the class and mathematical level of the students, there are a variety of interesting and important concepts that can be discussed. Teachers can use rational polynomial functions to give students a gentle and concrete introduction to important math concepts. In this example, students are confronted with division by zero when x = -2 or 5. Notice how the graph of f(x) behaves near x = -2 or x = 5. Warning! Initially, many students will not get the same results shown below because they failed to enclose (x + 2)(x – 5) in a pair of matching parentheses when they entered the expression in their graphing calculator. I often have to remind students that calculators obey the order of operation rules.

  • By only looking at the equation for y = f(x), how can you tell that f(2) = 0 and f(-4) = 0?
  • From a geometric point of view, what does f(2) = 0 tell us? What does f(-4) = 0 tell us?
  • Verify that f(0) = 2.4. From geometric point of view, what does this tell us?
  • By only looking at the equation for y = f(x), how can you tell that both f(-2) and f(5) are undefined?
  • As shown in the table below, pick values of x that are very close to -2 or 5, and then use a calculator to verify that the absolute value f(x) is a relatively large number.
  • When x is close to -2 or 5, the absolute value of (x + 2)(x – 5) is relatively small compared to the absolute value of 3(x – 2)(x + 4). Therefore |f(x)| is relatively large because dividing by smaller numbers gives us larger numbers.
  • Since f(-2) and f(5) are undefined, the lines x = -2 and x = 5 are vertical asymptotes or poles. This tells us that the graph of y = f(x) is literally ripped apart along the vertical asymptotes. In other words, the function y = f(x) is not continuous at x = -2 or x = 5. Continuous functions can be drawn without lifting the pencil from the sheet of paper.
  • As illustrated in the table below, pick values of x, negative or positive, that have very large absolute values, and then use a calculator to verify that f(x) is very close to 3. This tells us that the line y = 3 is a horizontal asymptote. In other words, as x approaches ±∞, f(x) approaches 3. You can verify that f(-1,000,000) = 2.999985 and f(1,000,000) = 3.000015.

dividezero2fig3

dividezero2fig4

I will close this post by mentioning that the free handout, Rapid Curve Sketching, explains how one can quickly sketch the graph of a polynomial or rational polynomial function when the equation of the function is completely factored. You can also download this handout by visiting our free instructional content page in the algebra and pre-calculus tab.

Linear Transformation Rule to Reflect over Oblique Line y = mx + b

reflect_over_line_headerThis post discusses how the linear transformation rule for reflecting a figure over an oblique line y = mx + b can be used to create learning opportunities for high school students at different grade levels. Initially one might think that this topic is too advanced for most high school students. In my opinion, it’s possible to create a variety of activities based on linear transformations that will give mathematical nourishment to high school students at many different grade levels. An understanding of the concept of using a linear transformation to change a 2D graphic object to another 2D graphic object can definitely benefit college linear algebra students and computer graphic programmers. You may find it helpful to read my post, Geometric Transformations to Practice Basic Skills and Introduce Fundamental Concepts.

The content of this post is based on my free handout, Reflection over Any Oblique Line. This handout covers the following topics and items:

  • The derivation of the liner transformation rule (p, q) → (r, s) that reflects a figure over the oblique line y = mx + b where both r and s are functions of p, q, m, and b.
  • The derivation of the linear transformation rule (p, q) → (r, s) that reflects a figure over the oblique line y = mx + b where both r and s are functions of p, q, b, and θ = Tan-1(m).
  • Given the specific equation of a line y = mx + b, show different ways of finding a linear transformation rule to reflect a preimage figure over the line y = mx + b.
  • Graphic images showing the reflection images of various polygons over different oblique lines.
  • A 13-step algorithm for the TI-84 graphing calculator to draw preimage and image polygons under a linear transformation.

The linear transformation rule (p, s) → (r, s) for reflecting a figure over the oblique line y = mx + b where r and s are functions of p, q, m, and b is given below. Finding the linear transformation rule given equation y = mx + b involves substituting the values for m and b in the formulas below, and then using basic operations with fractions to simplify each of the six coefficients of the linear transformation rule. What a sneaky way to get kids to practice operations with fractions.

reflect_over_line_fig1

As shown in the handout, Reflection over Any Oblique Line, the linear transformation rule for reflecting over the line y = -2x + 4 is (p, q) → (-3/5p – 4/5q + 16/5, -4/5p + 3/5q + 8/5). When reflecting over the line y = 3/5x – 4, the linear transformation rule is (p, q) → (8/17p + 15/17q + 60/17, 15/17p – 8/17q – 100/17). The graph below shows the reflection images of polygons over the lines y = -2x + 4 and y = 3/5x – 4.

reflect_over_line_fig2

The linear transformation rule (p, s) → (r, s) for reflecting a figure over the oblique line y = mx + b where r and s are functions of p, q, b, and θ = Tan-1(m) is shown below. Finding the linear transformation rule given the equation of the line of reflection equation y = mx + b involves using a calculator to find angle θ = Tan-1(m), and then calculating each of the six coefficients of the linear transformation rule.

reflect_over_line_fig3

The handout, Reflection over Any Oblique Line, shows the derivations of the linear transformation rules for lines of reflection y = √(3)x – 4 and y = -4/5x + 4.

  • Line y = √(3)x – 4: θ = Tan-1(√(3)) = 60° and b = -4. The corresponding linear transformation rule is (p, q) → (r, s) = (-0.5p + 0.866q + 3.464, 0.866p + 0.5q – 2).
  • Line y = -4/5x + 4: θ = Tan-1(-4/5) = -38.66° and b = 4. The corresponding linear transformation rule is (p, q) → (r, s) = (0.2195p – 0.9756q + 1.9512, -0.9765p – 0.2195q + 2.4390).

The graph below shows the reflection images of a polygon over the lines y = √(3)x – 4 and y = -4/5x + 4.

reflect_over_line_fig4

Suggestions for activities that teachers might consider:

  1. Give students a sheet of graph paper with the line of reflection and preimage polygon drawn. Also give them the equation of the line of reflection and the linear transformation rule corresponding to the equation of the line of reflection. Have students use the linear transformation rule to calculate the vertices of the image polygon, and then draw the image polygon. A completed assignment should include lists of the x-y coordinates of the vertices of the preimage and the image polygon.
  2. Give students a sheet of graph paper with the line of reflection and preimage polygon drawn. Have students find the equation of the line of reflection in slope-intercept format, and the linear transformation rule corresponding to the equation of the line of reflection. Depending on the level and interest of the students, allow students to calculate transformation rule coefficients in terms of parameters m and b or θ and b. Then have students use the linear transformation rule to calculate the vertices of the image polygon and draw the image polygon. A completed assignment should include lists of the x-y coordinates of the vertices of the preimage and the image polygon.
  3. Have upper level students derive the linear transformation rule with parameters m and b. As illustrated in the handout, Reflection over Any Oblique Line, explain and discuss the general strategy for deriving the linear transformation rule with parameter m and b. Derivation of the equations for a linear transformation only requires an understanding of concepts already encountered in their math classes. This is a great activity for promoting mathematical reasoning and presenting ideas in an organized manner to explain mathematical relationships.
  4. Have upper level trig students derive the linear transformation rule with parameters θ and b. Using the handout, Reflection over Any Oblique Line, as a guide, review and discuss the trig identities needed to convert the six coefficients of a linear transformation with parameters m and b to the coefficients of a linear transformation with parameters θ and b. Students now have a need to use trig identities.
  5. If students have access to the appropriate technology, have them do a project in which they graph the reflection image of a polygon over an oblique line y = mx + b. The completed project should include the following items: 1) A graph showing the line of reflection, preimage polygon, and image polygon. 2) The equation of the line of reflection. 3) A describtion of the linear transformation rule corresponding to the equation of the line of reflection. 4) A list of the x-y coordinates of the vertices of the preimage polygon. 5) A list of the x-y coordinates of the vertices of the image polygon. Students who have access to a TI-84 graphing calculator can use the 13-step algorithm given in the handout Reflection over Any Oblique Line.
  6. The handout, Reflection over Any Oblique Line, shows how linear transformation rules for reflections over lines can be expressed in terms of matrix multiplication. After showing students matrix multiplication based transformation rules, they better understand why matrix multiplication is done the way it is. Programmers use matrix multiplication to perform 2D and 3D transformations of objects on a computer screen. Computer video cards are optimized to perform millions of matrix multiplications per second.
  7. Recruit a team of computer programming geeks in your school to write a program that calculates the x-y coordinates of a reflection image of a figure over any oblique line y = mx + b, and then graph the line of reflection, the preimage figure, and the image figure. The set of features the program could offer is limited only by the ability and imagination of the programmers.

Geometric Transformations to Practice Basic Skills and Introduce Fundamental Concepts

IntGeoTrans_lead1This post shows how learning to apply geometric transformation rules to slide, reflect, rotate or resize a figure can benefit students. Geometric transformation graphing activities can help students learn important math concepts in the following ways:

  • Learn how to interpret a symbolic description of a geometric transformation rule to find the image point of a preimage point.
  • Learn what it means to reflect a figure over a line.
  • Learn what it means to slide or translate a figure.
  • Learn what it means to rotate a figure about a point.
  • Learn what it means to stretch or shrink a figure.
  • The idea that preimage and image points is equivalent to the idea of function inputs and outputs.
  • Worthwhile practice with the basic operations of signed numbers.
  • Worthwhile practice plotting points and drawing geometric transformation images.
  • A tacit introduction to the important mathematical concepts of function, inverse of a function and composition of functions.

This post does not discuss the general theory of affine transformations nor does it discuss the study of geometry from a geometric transformation point of view. For a general discussion of 2D matrix based geometric transformations, download my free handout Matrix Geometric Transformations or visit our free instructional content page.

The table below describes the geometric transformations considered in this post. Assume constants j and k are positive.

Transformation Operation
Reflect point (x,y) over the x-axis. (x,y) → (x,-y)
Reflect point (x,y) over the y-axis. (x,y) → (-x,y)
Reflect point (x,y) over the line y = x. (x,y) → (y,x)
Translate or slide point (x,y) right/left j units and up/down k units. (x,y) → (x ± j, y ± k)
Rotate point (x,y) 90° CW about (0,0). (x,y) → (y,-x)
Rotate point (x,y) 90° CCW about (0,0). (x,y) → (-y,x)
Rotate point (x,y) 180° about (0,0). (x,y) → (-x,-y)
Expand or contract point (x,y) by a factor of k from (0,0). (x,y) → (kx,ky)

The setup and parameters for a geometric transformation activity are shown below:

  1. Each student is provided a handout containing directions for the activity, an x-y coordinate axes with the graph of the preimage polygon drawn in black, a symbolic description of a geometric transformation, a table for the preimage points, and blank table for the image points to be filled in by the student. I prefer lattice point coordinate axes, but grid line coordinate axes are fine. The coordinate axes should be properly labeled and laid out so that it’s easy to plot points. It’s important that the coordinate axes be drawn with a 1:1 aspect ratio so that perpendicular lines appear to be perpendicular and graphs of circles appear to be circles, not ovals.
  2. Each student should have a ruler to aid in drawing graphs. Sloppy hand drawn graphs are not allowed. I have found that a 6 inch or 15 cm ruler works best.
  3. It is assumed that students can do the basic operations with signed numbers and plot points.
  4. No calculators allowed; strictly old school. Students may use scratch paper of course.
  5. The initial preimage point is arbitrary; just move from vertex to vertex around the polygon in either a clockwise or counterclockwise direction.
  6. To enhance the visual effect, allow colored ink pens or pencils to draw the image figures.

The tables and graphs below show the results of reflecting the same black preimage figure over the x-axis, the line x = -4 and the line y = x. Normally a geometric transformation graphing activity should have no more than two transformations to perform on a figure, but to conserve space three transformation images are graphed on the same x-y coordinate axes. The tables of x-y coordinates of the image points are color coded to match the color of the image polygon. In an actual transformation activity handout for students, the column of image point x-y coordinates is blank. The first two or three rows of the tables may show both preimage and image point coordinates to help students understand how the transformation rule works. The reflection over the line x = -4 is accomplished by chaining together transformations as follows:

  1. Slide the polygon right 4 units.
  2. Reflect the image over the y-axis.
  3. Slide the last image left 4 units.

IntGeoTrans_Fig1

IntGeoTrans_Fig2

IntGeoTrans_Fig3

The next demonstration involves a slide, rotation and size transformations of a preimage polygon drawn in black. The 90° counterclockwise rotation about (-2, 1) is accomplished by chaining together three transformations as follows:

  1. Slide the polygon right 2 units and down 1 unit.
  2. Rotate the image 90° counterclockwise about (0, 0).
  3. Slide the last image left 2 units and up 1 unit.

The image polygon drawn in green was obtained by chaining together a size transformation with expansion factor = 1.5 and a 180° rotation about (0, 0).

IntGeoTrans_Fig4

IntGeoTrans_Fig5

IntGeoTrans_Fig6

The graph below shows the decomposition of the geometric transformation (x, y) → (-y + 3 , x – 13) that rotates the black flag 90° counterclockwise about the point (8, – 5). Using the transformation mapping functions in the table above and the graphs of the 4 flags below, we can see that the geometric mapping function can be created as follows: (x, y) → (x – 8, y + 5) → (-(y + 5), x – 8) → (-(y + 5) + 8, x – 8 – 5) = (-y + 3, x – 13). Note that the transformations (x, y) → (x – 8, y +5) and (x, y) → (x + 8, y – 5) are inverse transformations.

IntGeoTrans_Fig7

Suggestions for fun follow up activities relating to geometric transformations:

  • After graphing the reflection image of a polygon over a line, have students fold the sheet of graph paper on the reflecting line, and then hold the sheet of graph paper up to the light to verify that preimage and image polygons are congruent.
  • After translating a polygon, use the theorem of Pythagoras to determine the magnitude of the slide and a protractor to determine the polar direction of the slide where the polar direction ranges from 0° to 360°.
  • Have students use a compass and protractor to verify that a polygon has been rotated a certain number of degrees about a point.
  • Let students be creative by having them make up their own transformation rule, and then use the rule to graph the image of a preimage polygon. Exceptional work can be posted in the classroom for all to enjoy.
  • Give students a graph similar to the graph above, and have them find a geometric transformation rule that maps a preimage to an image. You can let the preimage be any of the figures in the graph, and the image can be any of the other figures. You will be amazed to see that some students struggle to find a transformation rule that maps a given preimage figure onto itself, but this can be a great teaching opportunity!
  • Given the graphs of a preimage polygon and the image polygon under a size transformation, find the lengths of corresponding side pairs and verify that the ratios of the lengths of corresponding side pairs are equal.
  • Tell students that geometric transformations make it possible for game developers to create whose wonderful video games they love to play.

Here’s some exercises based on the examples in this post that you can give to your students:

Introduction to Geometric Transformations (student version)

Introduction to Geometric Transformations (teacher version)

Giving Students Meaningful Practice with Signed Numbers

IntEq_leadimgThe purpose of this post is to show how I have used simple equation graphing activities very early in a beginning algebra course to practice basic operations with signed numbers. Students not only practice applying the rules for adding, subtracting, multiplying and dividing signed numbers, they also learn how to plot x-y data pairs, draw graphs of equations, and get a glimpse of future course concepts which include the ideas that 1) equations and graphs of equations describe a relationship between two variables and 2) variables in linear relationships change at a constant or steady rate with respect to each other. After students have learned how to do the basic operations with signed numbers, there is no real reason to wait another two or three chapters in the book to start learning how to graph simple equations. They now have a reason to use what they have just learned.


You can download two free graphing activities by clicking the links below. These activities are examples of the approach presented in this post that you can directly use with your students or as a starting point for making your own activities.

Intro to Graphing Equations 1 (student version)

Intro to Graphing Equations 2 (student version)

🙂 The completed teacher versions of these activities are also available on our free instructional content page under the algebra and pre-calculus tab.


I will begin by describing the setup and parameters for a graphing activity.

  1. Each student is provided a handout containing directions for the activity, equations to be graphed, a table of equation input values, and x-y coordinate axes. I prefer lattice point coordinate axes, but grid line coordinate axes are fine. The coordinate axes should be properly labeled and laid out so that it’s easy to plot points.
  2. Each student should have a ruler to aid in drawing graphs. Sloppy hand drawn graphs are not allowed. I have found that a 6 inch or 15 cm ruler works best.
  3. No ink pens or calculators allowed; strictly old school. Student may use scratch paper of course.
  4. The input values for each equation are carefully selected so that most points in the relation are integer pairs, and the key features of the graph are graphed.
  5. Even if all values in a table are the same, students are required to write every value in the table; no down arrows to indicate that numeric values continue.

Shown below are sample equations and table setups that I have used to introduce graphing equations to my students. My handouts usually have two or three equations with tables located at the sides and/or bottom of a blank x-y coordinate axes. The tables presented in this post show the output values in red color text. In an actual graphing activity handout for students, the column of output values are blank. This is brand new stuff for beginners, and therefore I do a lot of coaching by explaining what the equations mean, and by doing a few of each type of problem. What is obvious to more advanced students is not so obvious to beginning students.

Tables A, B and C below are similar to tables that I have used to introduce graphing equations. These equations are so simple that some students initially find these equations somewhat difficult to understand; it’s true. As shown below, I like to give intuitive descriptions of the equations.

  • Table A: The value of the y-variable always equals the opposite of x.
  • Table B: No matter what x equals, the value of y always equals 7.
  • Table C: No matter what y equals, the value of x always equals -6.

IntEqTablesABC

The equations in tables D, E and F give students meaningful practice adding, subtracting and multiplying signed numbers. After filling in the tables, hopefully some students will notice a common pattern in the tables in that the y-variable goes up or down a certain amount whenever the x-variable goes up or down a certain amount. Depending on the class, I might ask students if they see a way to predict the steady rate of change between the variables from the equation. Later in the course, I use similar tables and corresponding graphs to show the relationship between the slopes of parallel lines and perpendicular lines. Intuitive descriptions of the equations D, E and F are shown below.

  • Table D: y always equals 2/3 of x plus 4.
  • Table E: y always equals 2/3 of x minus 2.
  • Table F: y always equals the opposite of 3/2 of x plus 1.

IntEqTablesDEF-1

The equations in tables G, H and I give students more meaningful practice adding, subtracting and multiplying signed numbers. The equation x + y = 4 requires students to think differently because the y-variable is not explicitly stated in terms of the x-variable. After filling in the table, hopefully students will notice patterns in the tables. Depending on the class, I might discuss the steady rate of change pattern in table G and the symmetry patterns in tables H and I. Intuitive descriptions of the equations G, H and I are shown below.

  • Table G: The sum of x and y must always equal 4.
  • Table H: y always equals x squared minus 8.
  • Table I: y equals the product of the quantities (x + 2) and (x – 4).

IntEqTablesGHI

After filling in the tables of equation x-y data pairs, I decrible how equations should be graphed as follows:

  1. For each x-y data pair that fits on the x-y coordinate axes provided, draw a heavy dot at location (x, y). The x-value indicates how many spaces to the left/right of the origin (0, 0) the point is. The y-value indicates how many spaces above/below the x-axis the point is.
  2. The points for equations A through G should fall on a straight line. If they don’t, correct the mistake in your table and replot the point.
  3. For each of the graphs A through G, use your ruler to draw a line segment through all of the equation points. At each endpoint of the line segment, draw an arrow to indicate that the graph continues forever in both directions.
  4. The graphs of equations H and I are named parabolas. If the points don’t fall on a smooth U shaped curve, correct the mistake in your table and replot the points.
  5. For graphs H and I, draw a smooth U shaped curve through the x-y data pairs. At each endpoint of the parabolic curve, draw an arrow to indicate that the graph continues forever in both directions.
  6. Remind students that they are only plotting a small sample of the infinitely many x-y real number data pairs that satisfy the equation, not just x-y integer pairs.

After drawing the graph of an equation, I show students what the graph of some equations should look like so that they can make corrections before turning in the graphing activity for grading. Shown below are the graphs of equations A through I. I cheated by using my equation graphing program.

IntEqGraph 1b

IntEqGraph 2b

IntEqGraph 3b


I will close this post by mentioning that readers can download blank x-y coordinate axes graphs in either lattice point or grid line format by clicking the links below. These graphs are in JPEG format which makes it easy to paste and resize them to create handouts, tests, presentations, etc.

Blank graph with grid lines (5×5 x-y axes scale)

Blank graph with grid lines (10×10 x-y axes scale)

Blank graph with grid lines (15×15 x-y axes scale)

Blank graph with lattice points (5×5 x-y axes scale)

Blank graph with lattice points (10×10 x-y axes scale)

Blank graph with lattice points (15×15 x-y axes scale)

🙂 These blank graphs are also available on our free instructional content page under the algebra and pre-calculus tab.

Getting Comfortable with Negative Exponents

Used green chalkboardThis post shows how I use a reciprocal pairs approach to introduce zero and negative integer exponents to my students. Several exercises that I have used to make students more comfortable using zero and negative integer exponents are also included. It seems to me that students learn to tolerate negative exponents, but they are not really comfortable using negative exponents.

I begin my intuitive discussion of negative exponents by creating a table of the first 4 positive integer powers of an integer such as 6 for example. Refer to the table below.

NegExpFig1

From this point, the discussion goes something like this:

  • What do we do to 1,296 to get 216? What do we do to 216 to get 36?
  • Some groups immediately notice that we divide by 6 to get the next number in the list, and other groups require a bit of coaching to see that we divide by 6 to get the next number.
  • In order to maintain the pattern, it’s clear that the fifth row of the table must be 60 = 1.
  • To get the remaining rows, I say, “Reduce the exponent by one and divide by whole number six to get the next fraction in the list.” (Some students see a rational number like 3/8 only as a single entity, not one whole number divided by another whole number. Also some groups need to be reminded how a fraction is divided by a whole number.)
  • As shown in the diagram, I bracket the reciprocal pairs of numbers in the list.
  • Now the important question; “What is the relationship between bracketed pairs of numbers?”
  • Some groups immediately notice the reciprocal pair relationship, and other groups need a little coaching to see the relationship. It takes a lot of coaching to get them to say out loud, “They are reciprocal pairs because all products of a pair of bracketed numbers equals 1.”
  • I point out that we could do the same process with any nonzero real number, but the math might be a little messy.

Shown below is a reciprocals pairs table for -2/5 and -5/2. If a teacher decides to show his/her class a reciprocal pairs table for two fractions, and fraction operation skills are fragile, I strongly advise that the implied multiplication and division of fractions in the table be explicitly demonstrated. It’s sufficient to demonstrate multiplication and division of fractions for one or two lines in the table.

NegExpFig2

After exploring one or two tables of reciprocal number pairs, I give my students an informal summary of what can be deduced from the table. Of course, the concepts that I discuss depends on the class I’m teaching. Base b equals any nonzero real number, p is an integer, and x and y are real numbers. Students should use their calculators to verify the specific examples.

Concept Example
b0 = 1 3.140 = 1
1 / (x/y) = y/x 1 / (-7/2) = -2/7
All real numbers have an implied exponent of one. π = π1
Changing the sign of the exponent of a number gives us the reciprocal of the number. 4-1 = 1/4
If integer n > 0, bn means repeated multiplication of b by itself n times. 43 = 4•4•4 = 64
If integer n > 0, b-n means repeated multiplication of 1/b by itself n times. 4-3 = (1/4)•(1/4)•(1/4) = 1/64
b-p = (1/b)p (1.2)-5 = (5/6)5
bp = (1/b)-p (-2.25)3 = (-1/2.25)-3
xy = y / (1/x) 0.008x = x/125
If x > 0, Log(x) = -Log(1/x) because logarithms are exponents. Log(5) = -Log(0.2) ≈ 0.699
The graphs of y = a(bx) and y = a(1/b)-x are equal. See graph below.
The graphs of y = a(bx) and y = a(b-x) are reflection images of each other over the y-axis because the graphs of y = f(x) and y = f(-x) are always mirror images over the y-axis. See graph below.

NegExpFig3

I will close this post by showing you some practice exercises that I have used to promote understanding of negative exponents, and to increase student comfort level in working with negative exponents. One may think that these exercises are nothing more than mental gymnastics, however, calculus students need to know how to rewrite expressions in terms of negative and fractional exponents. Readers can download my free handout, Properties of Exponents and Logarithms, by going to the algebra and pre-calculus tab in our instructional content page. This handout is filled with examples demonstrating the laws of exponents and logarithms.

NegExpFig4

NegExpFig5

Addition, Subtraction and Vectors

lead_imgEarly in a beginning algebra course, students are taught how to add and subtract signed numbers. Addition of signed numbers is taught by giving students a set of rules that they can follow to get the right answer. Some number line diagrams are thrown in to illustrate addition of signed numbers. Subtraction of signed numbers is usually taught by the “add the opposite” rule. This rule sounds plausible, however, it does not give students any insight into what subtraction is all about from a geometric point of view. You might ask, “Isn’t learning how to apply a set of rules to get the answer sufficient?” In my view, it’s not. As I stated several times in previous posts, whenever possible, students should understand a math concept from both an algebraic and geometric point of view. The primary purpose of this post to show how I have used vector diagrams to illustrate addition and subtraction of real numbers and 2D vectors. I will also discuss geometric interpretations of expressions and relationships such as (a + b)/2, |x – 4| < 5, and |x – 5| > 7.

Based on my own experience, most kids quickly learn how to apply the rules for adding and subtracting positive and negative numbers. When simplifying polynomial expressions, rational polynomial expressions, and polynomial long division, most mistakes occur at the step where a negative number is subtracted. Students are generally pretty good at adding signed numbers, but a little weak when it comes to subtracting signed numbers. This is why my favorite high school math teacher, Vivian Jones, said, “When subtracting a number, just change the sign of the number and add”. This is what I tell my college algebra students when they do polynomial long division. As taken from a developmental algebra textbook, the rules for adding and subtracting signed numbers are as follows:

Adding Two Real Numbers a and b

  • If a and b have the same sign, add their absolute values. Use the common sign of a and b as the sign of the answer.
  • If a and b have different signs, subtract their absolute values. The sign of the answer is the sign of the number that has the largest absolute value.

Subtracting Two Real Numbers a and b

  • ab = a + (-b)
  • In other words, ab equals a plus the opposite of b
Vector diagrams similar to diagrams A, B and C below are used to illustrate how the addition and subtraction rules for signed numbers work. The bottom half of diagram A shows that two negatives do not necessarily make a positive.

Diagrams B and C illustrate the subtraction rule for signed numbers. The problem with the subtraction rule and the corresponding vector diagram is that they don’t give us any insight into what subtraction is all about from a geometric point of view.

AddfigA

 

AddfigB

 

AddfigC

Diagrams D and E below illustrate the idea that subtracting two real numbers gives us the directed distance between two reals numbers. The concept of directed distance is fundamental in understanding subtraction from a geometric point of view. For real numbers a and b, the directed distance from b to a = a – b, and the directed distance from a to b = ba. The positive distance or just the distance from a to b = |a – b| = |b – a|. The distance concept is related to many concepts in mathematics such as the amount of change in a variable, how much a data value deviates from some fixed constant, margin of error, etc. Notice that the subtraction problems in diagrams D and E are the same problems in diagrams B and C. After comparing diagrams B and C with diagrams D and E, it becomes clear that diagrams D and E give us a much better way to understand subtraction from a geometric point of view.

AddfigD

AddfigE-2

We will now take a look at 2D vectors which are essential in understanding a variety of concepts in math and physics. We can treat 2D vectors as line segments that have the properties of length and direction. Any two vectors are equivalent if and only if they have the same length and direction. In this post, 2D vectors will be denoted by bold face capital letters, and a pair of vertical absolute value bars will denote the length or magnitude of a vector. The arrow at one tip of a segment indicates the direction the vector. Geometric rays have infinite length, and therefore 2D vectors are not geometric rays. In passing, I will mention that 2D vectors are just a special case of an abstract mathematical object named vector. If you want to stretch your mind, take a course in infinite dimensional vector spaces. The student and teacher versions of my free handout, Introduction to Vectors, can be downloaded by going to the trigonometry section of our instructional content page.

The graph below shows how two 2D vectors are added. All vectors drawn in the same color are equal to each other because they have the same length and direction. All 2D vectors can be represented by a pair of real numbers of the form < x, y > where x and y equal the x-component and y-components of the vector. Knowing the x-y components of a vector, it’s easy, at least for trig students, to calculate the vector’s magnitude and direction. Note the following as you study the graph:

  • A 2D vector can be expressed as the sum of its x-component and y-component. Example: Vector A = < -7, 0 > + < 0, 3 > = < -7, 3>.
  • To find the x-y components of a vector, start at the tail of the vector and count the number of spaces left/right and the number of spaces up/down to the head of the vector.
  • The x-y components of any two equivalent vectors are equal.
  • The tail of the second vector in a vector sum is located at the tip of the first vector.
  • Vector addition is commutative. In other words, it makes no difference in what order vectors are added.
  • Any two equivalent vector pairs will always result in equivalent vector sums.
  • The theorem of Pythagoras is used to calculate the length or magnitude of a vector.
  • The inverse tangent function, Tan-1(x), is used to calculate the direction of a vector.
  • Vectors A, B and S = A + B = B + A in the diagram have the following properties:
    • A = < -7, 3 >, |A| = √(58) units, and direction of A ≈ 156.800
    • B = < 3, 5 >, |B| = √(34) units, and direction of B ≈ 59.040
    • S = < -4, 8 >, |S| = √(80) units, and direction of S ≈ 116.570
    • |A| + |B| > |S|

Vectorfig1

The graph below shows how two 2D vectors are subtracted. Vector subtraction gives us a vector that represents a difference vector between the tips of the two vectors. Vector AB has its tail at the tip of B and its head at the tip of A.  Vector BA has its tail at the tip of A and its head at the tip of B. All vectors drawn in the same color are equal to each other because they have the same length and direction. Note the following as you study the graph:

  • The difference vector connects the tip of one vector to the tip of the other vector.
  • Vector subtraction is not commutative. Vectors AB and BA have the same length, but they point in opposite directions; they are vector opposites.
  • Vectors A, B, AB,  BA have the following properties:
    • A = < 6, 2 >, |A| = √(40) units, and the direction of A ≈ 156.800
    • B = < 2, 9 >, |B| = √(85) units, and the direction of B ≈ 59.040
    • AB = < 4, -7 >, |A – B| = √(65), and the direction of A – B ≈ 299.750
    • B – A = < -4, 7 >, |B – A| = √(65), and the direction of B – A ≈ 119.750
Vectorfig2

Once a student understands addition and subtraction from a geometric point of view, many math problems become much easier to solve. Consider the three routine math problems shown below.

Problem 1: Suppose the IQ score I of a person is in the normal range if the IQ score deviates from 100 by 10 points or less. What interval on a number line and inequality describes a normal IQ score?

Solution: The expression |I – 100| gives us the positive distance of the variable I from 100. Therefore the range of normal IQ scores is described by the inequality |I – 100| ≤10.

Prob1fig

Problem 2: A part will fail inspection if its diameter d deviates from 2.5 cm by more than 0.001 cm. What interval on a number line and inequality describes the rejection region?

Solution: A part will fail inspection if the positive distance from 2.5 to d is more than 0.001 cm. Therefore the rejection region can be described by the inequality |d – 2.5| > 0.001. Of course, we tacitly assume that there are practical restrictions on values of d.

Prob2fig

Problem 3:  The graph of the closed interval [0.84, 2.68] is shown below. Find the following:

  • Length of the interval
  • Coordinate of the midpoint M
  • Radius of the interval
  • Write an inequality that describes the interval.
Prob3fig

Solution: (Many of my elementary statistics students initially struggle with review problems like this.)

  • Length of interval = 2.68 – 0.84 = 1.84
  • Coordinate of midpoint M = (0.84 + 2.68)/2 = 1.76
  • Radius of the interval = 1.84/2 = 2.68 – 1.76 = 1.76 – 0.84 = 0.92
  • Inequality: |x – 1.76| ≤ 0.92

Miscellaneous facts I tell my students:

  • If you subtract a smaller number from a bigger number, the answer is positive.
  • If you subtract a bigger number from a smaller number, the answer is negative.
  • If you subtract a positive number from n, the answer is smaller than n.
  • If a positive influence is removed from your personal life, the quality of your personal life goes down.
  • If you subtract a negative number from n, the answer is bigger than n.
  • If you remove a negative influence from your person life, your personal life gets better.
  • If you add a negative number to n, the answer is smaller than n.
  • If a negative influence is introduced into your personal life, your personal life gets worse.
  • To find out how far apart two numbers are, subtract the numbers.
  • To find a number half way between two numbers, find the average of the numbers.

I will close this post with a true story about an epiphany I experienced early in my teaching career. The class was a regular high school geometry class. We were learning how to solve story problems involving complementary and supplementary angles. I could see that little Elmo (not his real name) was not getting the idea that if x is the measure of an acute angle, then 90 – x is the measure of the complement of the angle. So I asked Elmo a series of about 5 questions like: If an angle measures 200, what is the degree measure of the complement of the angle? Elmo got every one of my questions right. I then asked Elmo the following question: If an acute angle measures n degrees, what is the degree measure of the complement of the angle? All I got from Elmo was a blank stare. I’m thinking to myself, why doesn’t he get it? Then it hit me. When asked the degree measure of the complement of a 650, Elmo figured out how many degrees he needed to add to 650 to get 900. For me, this was an enlightening and humbling experience.

Division, Fractions, Proportional Parts, and Other Neat Stuff

division_leadAfter his freshman year in college, a former math student paid me a visit. Aaron (not his real name) indicated that his first year of college was a great experience. This was no surprise to me. I found Aaron to be a fairly perceptive individual who got along well with his high school classmates. Then Aaron said to me, “You know Mr. Johnson, after a year of college chemistry, I finally understand division.” He went on to explain how the concepts of division, fractions, decimals, ratios, rates, proportions and percent all fit together for him. In order to solve chemistry problems, Aaron was naturally motivated to learn and understand division and related concepts. Of course, I was surprised to hear Aaron make such a statement. He was an above average student who had attended a highly regarded local private school through junior high school. I then realized that if high school students like Aaron don’t really understand division, many of my students probably don’t really understand division either.

Based on numerous conversations with high school math teachers, in both public and private high schools, I have come to the conclusion that many high school students don’t really understand division and related concepts. It’s almost like a dirty little secret that high school math teachers don’t like to talk about in public. Many students have learned how to apply a set of rules to get the right answer, but they don’t really understand why a rule works or what the answer means. I see this all the time when I tutor developmental math students and above who are graduates of local area high schools.

Elementary, middle school, and high school teachers should NOT in any why interpret the above remarks to mean that they are not doing their job. In my view, teaching elementary and middle school students is far more difficult than teaching high school students. Of course, teaching high school math is not so easy either. If anyone thinks that they have a bullet proof method of teaching division and related concepts to a class of 30 typical middle school students, they should tell middle school teachers how to do it. I’m sure that middle school teachers would be ever grateful.

We constantly hear comments about how public education has deteriorated over the years. My now deceased father-in-law once complained to me that one of his office workers couldn’t figure out how many parts to ship to customers in proportion to the number of parts the customers ordered. (There were more parts ordered than parts in stock.). He said something like this, “Schools don’t teach kids the basics anymore. In my day, schools taught the fundamentals.” My father-in-law and many of his high school classmates learned basic math very well, but only 29% of the general population graduated from high school in the 1920’s. It’s my observation that today’s top students and athletes perform much better than the top students and athletes 40 or 50 years ago. As I see it, the perceived problem with modern education is due to the fact that public schools are attempting to reach a student population with a far greater range of abilities.

So what’s the purpose of this post? The purpose of this post is to share some of the methods that I have used in my classes to explain division and related concepts. Readers may find some of my examples and explanations to be ordinary, but other examples and explanations are different in that they are not found in textbooks. I invite readers to share some of their favorite methods that they use to explain division and related concepts to kids. I would love to do a post based on reader response. Readers should reply by sending an email to info@mathteachersresource.com. If your example has a graphic, please attach a file that contains the graphic. I will create the graphic for you if you give me a good description.

My first set of examples illustrates why the division rule for dividing two fractions gives us the correct answer. I assume that students already know how to apply the multiplication and division rules for fractions, and they know how to reduce a fraction. These examples reinforce the idea that dividing by a number is the same as multiplying by the reciprocal of the number.

  • Diagram A below illustrates the solution of following problem: If we have 12 pounds of hamburger and a meat loaf recipe that calls for 2 pounds of ground beef, how many meat loafs can we make? Solution: 12/2 = 12*(1/2) = 6 meat loafs.
division_figA


  • Diagram B below illustrates the solution of following problem: If we have 12 pounds of hamburger, how many 1/4 pound hamburger patties can we make? Solution: 12/(1/4) =12*(4/1) = 12*4 = 48 hamburger patties.
division_figB


  • Diagram C below illustrates the solution of following problem: If we have 12 pounds of hamburger, how many 3/4 pound hamburger patties can we make? Solution: 12/(3/4) = 12*(4/3) = 48/3= 16 hamburger patties. Alternatively, 12 pounds of hamburgers divided into 16 patties = 12/16 = 3/4 pounds per patty.
division_figC



When I introduce right triangle trigonometry to my students, I give them the sides of a right triangle, and then we find the trig ratios of the two acute angles α and β of the triangle. Students quickly notice that Sin(α) = Cos(β). I want my students to not only understand the definitions of trig ratios and understand that trig function values are ratios, but I also want them to see that trig function values are percentages in disguise. The diagram below shows a right triangle, accurately drawn to scale, with 150 and 750 acute angles. Knowing only the acute angles of the right triangle and a scientific calculator, we can deduce the facts below. When students study the diagram, they often say, “When you look at it this way, it makes more sense.”

  • Cos(150) = b/c ≈ 0.966. This tells us the length of the leg adjacent to a 150 angle is always about 96.6% of the length of the hypotenuse.
  • Sin(150) = a/c  ≈ 0.259. This tells us the length of the leg opposite a 150 angle is always about 25.9% of the length of the hypotenuse.
  • Tan(150) = a/b  ≈ 0.268. This tells us the length of the leg opposite a 150 angle is always about 26.8% of the length of the leg adjacent to a 150 angle.
  • Tan(750) = b/a ≈ 3.73. This tells us the length of the leg opposite a 750 angle is always about 373% of the length of the leg adjacent to a 750 angle.
division_trianglefig

In the next two problems, an intuitive approach is used to solve proportional parts problems. This would have made my father-in-law happy.

Problem 1: Find the three angles of a triangle if the angles are in a ratio of 3 : 4 : 5.

Solution: Refer to the diagram below. The problem boils down to dividing 1800 into a ratio of 3 : 4 : 5 parts. Because 3 + 4 + 5 = 12, we divide 180 into 12 equal parts with each part equal to 180/12 = 15. The smallest angle = 3 * 15 = 450. The next largest angle = 4* 15 = 600 and largest angle = 5 * 15 = 750. (I tell my students that I will only draw a diagram like the one below only one time because it takes too much time to draw the diagram. To my amazement, I once observed a student draw a diagram with well over 200 dots to solve a proportional parts homework problem. I’m not kidding.)

division_figD

Problem 2: Customers A, B, C, D, and E have placed orders for 5, 7, 10, 20, and 40 widgets respectively. Since the company has only 60 widgets in stock, it was decided to immediately fill the orders in proportion to the number of widgets ordered, and ship the remaining widgets when they become available. How many widgets should be shipped to each customer?

Solution: Because 5 + 7 + 10 + 20 + 40 = 82, we divide 60 into 82 equal parts with each part equal to 60/82 = 0.732.

  • Customer A: Ship 5 x 0.732 = 3.66 or 4 widgets
  • Customer B: Ship 7 x 0.732 = 5.124 or 5 widgets
  • Customer C: Ship 10 * 0.732 = 7.32 or 7 widgets
  • Customer D: Ship 20 x 0.732 = 14.64 or 15 widgets
  • Customer E: Ship the remaining 29 widgets

The classic “working together” problems are difficult for beginning algebra students to understand. Before I show students the standard algebraic technique for solving a working together problem, I show them how to use ratios to solve a working together problem. The problem below is a typical working together problem.

Problem: It would take homeowner Bob 7 days to roof his garage and professional roofer Clyde could roof Bob’s garage in 3 days. If Bob and Clyde cooperate with each other, how many days should it take them to roof Bob’s garage if they work together? Assume that a work day equals 8 hours. The text box below shows the solution of the problem. I suppose Bob and Clyde would work over time the second day so that they could finish the job in 2 days. (There is always someone who will make an initial guess of 5 days!)

division_txtbx1

Now let’s take a look at a couple of percent problems from a slightly different point of view.

Problem 1: A sofa that normally sells for $2,799.95 is on sale at 20% off. Local sales tax rate equals 9%. Find the sale price of the sofa and the cost of the sofa with sales tax. The solutions are given in the text box below.

division_txtbx2

Problem 2: Brad and his lovely wife Angelina dined at an upscale restaurant to celebrate their 20th wedding anniversary. The couple was fortunate to have a 15% discount coupon to help cover the $76.80 cost of their meal plus 9% sales tax. Since Brad was a high school math teacher, he thought that he had two options as to how he should apply the 15% discount. With option 1, he could take a 15% reduction of the cost of the meal plus sales tax. With option 2, he could first take a 15% discount on the cost of meal, and then pay sales tax on the discounted meal. With either option, Brad will leave a 20% tip. What option should Brad choose? (If Brad had stopped to really think about his options for a minute, he would have realized that it makes no difference what option he chooses.)

  • Option 1: Total cost = 0.85(1.09 x 76.80) = $71.16
  • Option 2: Total cost = 1.09(0.85 x 76.80) = $71.16
  • Total cost with tip = 1.20 x 71.16 = $85.39 or $85.00
I will close this post with some sample exercises that I have used to give students practice drawing valid conclusions and hopefully promote their mathematical reasoning ability. Each exercise involves a conditional relationship which is true or false depending on the values of the variables in the conditional statement. From a list of about 25 possible conclusions, student are to select all conclusions that are necessarily true. Given the long list of possible conclusions, most students don’t find these exercises so easy on first exposure. Each of the text boxes below shows a conditional statement and all valid conclusions which are contained in a list of possible conclusions.
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Why is the Product of Two Negative Numbers a Positive Number?

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Morris Kline (1908-1992)

When I ask most adults or high school students what is -5 times -10, I usually get the correct response of positive 50 or just 50. When I ask them if they can give me simple explanation or an example to show why the product or two negative numbers is a positive number, they can’t.  Students seem to remember the rule “two negatives make a positive”, but some forget that this rule applies to the product or quotient of two real numbers, but not to the sum or difference of two real numbers. I believe it’s fair to say that many lower level math textbooks and math courses treat this rule as just a simple mathematical fact of life, and rarely if ever give an intuitive explanation of why this rule is true.

The purpose of this post is to give examples of intuitive explanations as to why the product of two negative real numbers is a positive real number. It’s important that mathematics makes sense to students, and math should not be just a bunch of somewhat arbitrary rules that can be used to get the right answer. I have used and still use many of these examples when I explain why the product of two negative numbers is positive number. Some of these examples are probably familiar to many readers of this post. The first three examples are my favorites, and I use them exclusively. In all of the examples, it is assumed that students understand multiplication is repeated addition, and students have a basic understanding of the addition rules for positive and negative numbers. Of course, this is a big assumption.

Before I show you my examples, I will give you a brief description of the state of math education on the 1960’s and 1970’s in the United States. The Russian launch of the satellite Sputnik on October 4, 1957 caused a national panic which led to in complete revision of math and science curriculums. The revised math curriculum was called “new math” which was based on the advice of university research mathematicians and other professional mathematicians; not on the advice of wise and experienced math educators. New math placed emphasis on set theory, the fundamental properties of real numbers, functions, relations and the symbols of modern abstract mathematics. The new math approached mathematics from a more rigorous and abstract point of view as opposed to an intuitive and practical point of view. Students and parents found the new math strange and mystifying. Like most new waves in education, the new math was eventually replaced by another new wave. I clearly remember one nationally renowned math educator at a NCTM national convention in the 1980’s state (almost word for word), “Our research shows that a more rigorous abstract approach of teaching math only works with very bright students.” Of course, every experienced math teacher in the audience already knew this. The new math failed because you can’t make an abstraction (identify common core properties of different objects/systems) if you don’t have a base of knowledge and experience from which to make an abstraction. Since elementary, middle school, high school students, and normal adults don’t have this crucial base of knowledge and experience, it should be no surprise that the new math failed. By the early 1980’s, math education started to move in a new direction.

My first example is taken from the book Why Johnny Can’t Add: the Failure of the New Math written by Morris Kline (1908-1992) who was a scientist and professor of mathematics at NYU. One of Professor Kline’s core beliefs is that math concepts should be explained by using concrete examples that students can relate to. The example below is similar to the example Kline used to explain the product rule for positive and negative numbers.

Suppose homeowner Bob hires neighbor boy Bill to do general yard work at $10.00/hour. We have four situations to consider; two from Bob’s point of view and two from Bill’s point of view. Bill is gaining $10 every hour he works and Bob is losing $10 every hour every hour Bill works.

  • 4 hours in the future, Bill will be $40 richer. (+4 * +10 = +40)
  • 4 hours ago, Bill was $40 poorer. (-4 * (+10) = -40)
  • 4 hours in the future, Bob will be $40 poorer. (+4 * (-10) = -40)
  • 4 hours ago, Bob was $40 richer. (-4 * (-10) = +40)

The next example involves filming a person walking forward at the rate of 4 ft/sec for 10 seconds, and then filming the same person walking backwards at the rate of 4 ft/sec for 10 seconds. By running the two films forwards and backwards in a film projector, we have four cases to consider. Let film A show the person walking forward, and film B show the person walking backwards. (It’s fun to pace back and forth across the room to illustrate this example.)

  • If film A is run forward in the projector for 10 seconds, we will see the person walk forward 40 ft. (+4 * +10 = +40)
  • If film A is run backwards in the projector for 10 seconds, we will see the person walk backwards 40 ft. (4 * (-10)) = -40)
  • If film B is run forward in the projector for 10 seconds, we will see the person walk backwards 40 ft. (-4 * 10 = -40)
  • If film B is run backwards in the projector for 10 seconds, we will see the person walk forward 40 ft. (-4 * (-10)) = 40)

The next example is probably familiar to many readers. From my personal experience, a few students find the two previous examples somewhat confusing, but the pattern approach illustrated in the text box below seems to make the most sense to students. The first 4 rows of the table follow from the fact that the product of a positive number and a negative number is a negative number which makes perfect sense to most students. As the value of numbers in column A decrease by 1, the product of A and B gets bigger by 5. When A decreases from 0 to -1, I tell by students they can’t change horses in midstream; so the pattern must be maintained by increasing the product by 5 when A is decreased by 1.

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The next example hinges on the idea that multiplication is repeated addition under the following rules: (Rules are easier to understand if m and n are integers.)

  • If m is positive, then m * n equals n added to itself m times.
  • If m is negative, then m * n equals the opposite of n added to itself |m| times.

The text box below illustrates how these rules work then n = ±4 and m = ±6.

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My last example uses the properties of real numbers and mathematical reasoning to demonstrate  (-3)(-5) equals (3)(5) = 15. The demonstration hinges on the following properties of real numbers:

  • The distribute property
  • A negative number times a positive number is negative. (Previously established)
  • m + n = 0 if and only if m and n are opposites of each other.

Because this demonstration requires a higher level of mathematical maturity, I advise against showing this demonstration to younger learners.

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I will close this post with a discussion of the concept of positive and negative numbers by looking at two different number systems. You may be surprised to learn that in some number systems, the concept of positive and negative numbers does not exist. My post, A Simple Way to Introduce Complex Numbers, discusses the basics of complex numbers.

The set of real numbers:

  • For every real number x: x = 0, x < 0, or x > 0.
  • Every real number x not equal to zero has a unique opposite which is denoted by the symbol –x.
  • The opposite of a real number is the same as the additive inverse of a real number.
  • Real numbers x and y are opposites of each other if and only if x + y = 0.
  • If x > 0, then x is a positive number and –x is a negative number.
  • If x < 0, then x is a negative number and –x is a positive number.
  • The angular direction of all positive numbers is to the right or 00.
  • The angular direction of all negative numbers is to the left or 1800.
  • For all real numbers x and y, x*y = (-x)(-y). Note: This is true for any pair of real numbers. The expression (-x)*(-y) does not indicate we are multiplying two negative real numbers.
  • The symbol –x means the opposite of x; not negative x.

The set of complex numbers:

  • All complex number z can be expressed in the form z = a + bi where a and b are real numbers and i is the unit imaginary number such that i2 = -1.
  • Every complex number z not equal to zero has a unique opposite which is denoted by –z.
  • If z = a + bi, then –z = -a – bi.
  • The opposite of a complex number is the same as the additive inverse of a complex number.
  • Complex numbers w and z are opposites of each other if and only if w + z = 0.
  • The angular direction of complex number z can range from 00 to 3600.
  • In general, complex numbers are neither positive or negative because the angular direction of a complex number can range from 00 to 3600; not just 00 or 1800.
  • For all complex numbers w and z, w*z = (-w)(-z). Note: This is true for any pair of complex numbers. The expression (-w)*(-z) does not indicate we are multiplying two negative complex numbers because complex numbers in general don’t have a positive or negative property.
  • The symbol –z means the opposite of z; not negative z.