# Please move the deer crossing

Nearly all of the posts on this blog lie somewhere in the union (and often in the intersection) of mathematics and education, discussing ways of deepening content knowledge and imparting that knowledge to students.

This is not one of those posts.

However, if I ever need to lighten the mood with my students, this never fails to get a laugh.

And, in case if you’re wondering if the above phone call was a fake, here’s the rest of the story:

# Full lesson plan: magic squares

Over the summer, I occasionally teach a small summer math class for my daughter and her friends around my dining room table. Mostly to preserve the memory for future years… and to provide a resource to my friends who wonder what their children are learning… I’ll write up the best of these lesson plans in full detail.

This was perhaps my favorite: fostering algebraic thinking through the use of 3×3 magic squares, which have the property that the numbers in every row, column, and diagonal have the same sum.

This lesson plan is written in a 5E format — engage, explore, explain, elaborate, evaluate — which promotes inquiry-based learning and fosters student engagement.

Magic Squares Lesson Plan

Post Assessment 1

Post Assessment 2

Vocabulary Sheet

Magic Squares Examples

# Local universities help train high school teachers

At UNT, we take very seriously our mission to improve both the quality and quantity of secondary teachers of science and mathematics. Part of this mission includes providing professional development for current high school teachers, including our alumni and the mentor teachers who provide the essential field experiences that form the backbone of our teacher education program. We had a nice write-up about these efforts in the local newspaper this week.

Local universities help train high school teachers,” by Jenna Duncan (Denton Record-Chronicle, July 16, 2013. Published online July 15, 2013.)

# Why does 0! = 1? (Part 2)

This common question arises because $0!$ does not fit the usual definition for $n!$. Recall that, for positive integers, we have

$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$

$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$

$3! = 3 \cdot 2 \cdot 1 = 6$

$2! = 2 \cdot 1 = 2$

$1! = 1$

In Part 1 of this series, I discussed descending down this lines with repeated division to define $0!$.

Here’s a second way of explaining why $0!=1$ that may or may not be as convincing as the first explanation. Let’s count the number of “words” that can made using each of the three letters A, B, and C exactly once. Ignoring that most of these don’t appear in the dictionary, there are six possible words:

ABC, ACB, BAC, BCA, CAB, CBA

With two letters, there are only two possible words: AB and BA.

With four letters, there are 24 possible words:

DABC, DACB, DBAC, DBCA, DCAB, DCBA.

Evidently, there are $4!$ different words using four letters, $3!$ different words using three letters, and $2!$ different words using two letters.

Why does this happen? Let’s examine the case of four letters. First, there are $4$ different possible choices for the first letter in the word. Next, the second letter can be anything but the first letter, so there are $3$ different possibilities for the second letter. Then there are $2$ remaining possibilities for the third letter, leaving $1$ possibility for the last.

In summary, there are $4 \cdot 3 \cdot 2 \cdot 1$, or $4!$, different possible words. The same logic applies for words formed from three letters or any other number of letters.

What if there are 0 letters? Then there is only 1 possibility: not making any words. So it’s reasonable to define $0!=1$.

It turns out that there’s a natural way to define $x!$ for all complex numbers $x$ that are not negative integers. For example, there’s a reasonable way to define $\left( \frac{1}{2} \right)!$, $\left(- \frac{7}{3} \right)!$ and even $(1+2i)!$. I’ll probably discuss this in a future post.

# Why does 0! = 1? (Part 1)

This common question arises because $0!$ does not fit the usual definition for $n!$. Recall that, for positive integers, we have

$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$

$4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$

$3! = 3 \cdot 2 \cdot 1 = 6$

$2! = 2 \cdot 1 = 2$

$1! = 1$

Going from the bottom line to the top, we see that start at $1$, and then multiply by $2$, then multiply by $3$, then multiply by $4$, then multiply by $5$. To get $6!$, we multiply the top line by $6$:

$6! = 6 \cdot 5! = 6 \cdot 120 = 720$.

Because they’re formed by successive multiplications, the factorials get large very, very quickly. I still remember, years ago, writing lesson plans while listening to the game show Wheel of Fortune. After the contestant solved the final puzzle, Pat Sajak happily announced, “You’ve just won $40,320 in cash and prizes.” My instantaneous reaction: “Ah… that’s $8!$.” Then I planted a firm facepalm for having factorials as my first reaction. (Perhaps not surprisingly, I was still single when this happened.) Back to $0!$. We can also work downward as well as upward through successive division. In other words, $5!$ divided by $5$ is equal to $4!$. $4!$ divided by $4$ is equal to $3!$. $3!$ divided by $3$ is equal to $2!$. $2!$ divided by $2$ is equal to $1!$. Clearly, there’s one more possible step: dividing by $1$. And so we define $0!$ to be equal to $1!$ divided by $1$, or $0! = \displaystyle \frac{1!}{1} = 1$. Notice that there’s a natural way to take another step because division by 0 is not permissible. So we can define $0!$, but we can’t define $(-1)!, (-2)!, \dots$. In Part 2, I’ll present a second way of approaching this question. # Engaging students: Solving exponential equations In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place. I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course). This student submission comes from my former student Jesse Faltys. Her topic: solving exponential equations. APPLICATIONS: What interesting (i.e., uncontrived) word problems using this topic can your students do now? Once your students have learned how to solve exponential equations, they can solve many different kinds of applied problems like population growth, bacterial decay, and even investment earning interest rate. (Examples Found: http://www.education.com/study-help/article/pre-calculus-help-log-expo-applications/) Examples 1. How long will it take for$1000 to grow to $1500 if it earns 8% annual interest, compounded monthly? $A = P \left( 1 + \displaystyle \frac{r}{n} \right)^{nt}$ • $A (t) = 1500$, $P = 1000$, $r = 0.08$, and $n = 12$. • We do not know $t$. • We will solve this equation for $t$ and will round up to the nearest month. • In five years and one month, the investment will grow to about$1500.

2. A school district estimates that its student population will grow about 5% per year for the next 15 years.  How long will it take the student population to grow from the current 8000 students to 12,000?

• We will solve for t in the equation $12,000 = 8000 e^{0.05t}$.

$12,000 = 8000 e^{0.05t}$

$1.5 = e^{0.05t}$

$0.05t = \ln 1.5$

$t = \displaystyle \frac{\ln 1.5}{0.05} \approx 8.1$

• The population is expected to reach 12,000 in about 8 years.

3. At 2:00 a culture contained 3000 bacteria.  They are growing at the rate of 150% per hour.  When will there be 5400 bacteria in the culture?

• A growth rate of 150% per hour means that $r = 1.5$ and that $t$ is measured in hours.

$5400 = 3000 e^{1.5t}$

$1.8 = e^{1.5t}$

$1.5t = \ln 1.8$

$t = \displaystyle \frac{\ln 1.8}{1.5} \approx 0.39$

• At about 2:24 ($0.39 \times 60 = 23.4$ minutes) there will be 5400 bacteria.

A note from me: this last example is used in doctor’s offices all over the country. If a patient complains of a sore throat, a swab is applied to the back of the throat to extract a few bacteria. Bacteria are of course very small and cannot be seen. The bacteria are then swabbed to a petri dish and then placed into an incubator, where the bacteria grow overnight. The next morning, there are so many bacteria on the petri dish that they can be plainly seen. Furthermore, the shapes and clusters that are formed are used to determine what type of bacteria are present.

CURRICULUM — How does this topic extend what your students should have learned in previous courses?

The students need to have a good understanding of the properties of exponents and logarithms to be able to solve exponential equations.  By using properties of exponents, they should know that if both sides of the equations are powers of the same base then one could solve for x.  As we all know, not all exponential equations can be expressed in terms of a common base.  For these equations, properties of logarithms are used to derive a solution.  The students should have a good understanding of the relationship between logarithms and exponents.  Logs are the inverses of exponentials.  This understanding will allow the student to be able to solve real applications by converting back and forth between the exponent and log form.  That is why it is extremely important that a great review lesson is provided before jumping into solving exponential equations. The students will be in trouble if they have not successfully completed a lesson on these properties.

TechnologyHow can technology be used to effectively engage students with this topic?

1. Khan Academy provides a video titled “Word Problem Solving – Exponential Growth and Decay” which shows an example of a radioactive substance decay rate. The instructor on the video goes through how to organize the information from the world problem, evaluate in a table, and then solve an exponential equation. For our listening learners, this reiterates to the student the steps in how to solve exponential equations.

2. Math warehouse is an amazing website that allows the students to interact by providing probing questions to make sure they are on the right train of thought.

For example, the question is $9^x = 27^2$ and the student must solve for $x$.  The first “hint” the website provides is “look at the bases.  Rewrite them as a common base” and then the website shows them the work.  The student will continue hitting the “next” button until all steps are complete. This is allowing the visual learners to see how to write out each step to successfully complete the problem.

# The “Don’t Suck” theory of improving graduation rates

From the Chronicle of Higher Education:

Maintaining an up-to-date list of available tutors, calculating financial aid accurately, placing students in the right classes, picking up garbage, and maintaining elevators aren’t “best” practices. They are “minimally competent” practices. Nobody is ever going to publish a research study finding a causal link between \$125,000-per-photocopier contracts, Caribbean cruises, and graduation rates.

But I’m quite sure that these things are much more important to helping students graduate than the presence or absence of specific retention programs. They all go to the basic competence and quality of the institution. Well-run universities that have student-focused organizational cultures and are properly accountable to outside regulatory bodies simply don’t behave this way. Well-run universities are also much more successful in helping student earn degrees. It’s unreasonable to think that a university like Chicago State, which enrolls many part-time, low-income, and academically diverse students, will have a 100 percent graduation rate. But based on the research and examples cited in the article, it’s reasonable to expect that CSU could graduate 1 in 2 students, as opposed to 1 in 10.

# A great algebra question. (Or is it?)

I absolutely love the following algebra question:

Mrs. Ortiz made a batch of cookies for Carlos, Maria, Tina, and Joe. The children shared the cookies equally and finished them all right away.

Then Mrs. Ortiz made another batch of cookies, twice as big as the first. When she took the cookies off the cookie sheet, 6 of them crumbled, so she didn’t serve them to the children. She gave the children the rest of the cookies.

Just then, Mr. Ortiz came home and ate 2 cookies from the children’s tray. Each of the children ate 3 more cookies along with a glass of milk. They were stuffed, so they decided to leave the last 4 cookies on the tray.

1. How many cookies were in the first batch?

2. How many cookies did each of the children eat?

The reason I love this algebra question is that it wasn’t an algebra question. It was a question that was posed to upper elementary students. (Here are the Google results for this question; most of the results are brain-teaser type questions for students ranging from 4th grade to 6th grade.)

As a math person, my first instinct probably would be to let $x$ represent the number of cookies that each child ate on the first day and then set up an equation for $x$ based on the information from the second day. There may be other algebraic ways of solving this problem that are just as natural (or even better than my approach.)

So try to think about this problem from the perspective of a child who hasn’t learned algebra yet.  How would you even start tackling a complex problem like this if you didn’t know you could introduce an $x$ someplace?

I encourage you to take a few minutes and try to solve this problem as a 4th or 5th grader might try to solve it.

While this problem doesn’t require the use of algebra, it does require the use of algebraic thinking. That’s what I love about this problem: even a 9-year-old child can be reasonably expected to think through a solution to this problem, even if the methods that they might choose may not be those chosen by students with more mathematical training.

My observation is that math majors in college — even those that have good teaching instincts and want to teach in high schools after graduating — have a difficult time thinking that far back in time. Of course, putting themselves in the place of students who have not learned algebra yet is a good exercise for anyone who wants to teach algebra. So that’s a major reason that I love this problem; it’s a good vehicle for forcing college students who are highly trained in mathematics to think once again like a pre-algebra student.