# Area of a Circle: Index

I’m using the Twelve Days of Christmas (with a week-long head start) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on the formula for the area of a circle.

Part 1: Why the circumference function $C(r) = 2 \pi r$ is the derivative of the area function $A(r) = \pi r^2$.

Part 2: Finding the area of a circle via integration by trigonometric substitution.

Part 3: Finding the area of a circle via a double integral.

Part 4: Justifying the formula $A(r) = \pi r^2$ to geometry students by slicing a circle into pieces and rearranging the pieces as a parallelogram (approximately).

# Common Core, Subtraction, and the Open Number Line: Index

While the implementation of the Common Core has left much to be desired (understatement of the day), I do endorse — whether it’s done through Common Core or not — the fostering of deeper conceptual understanding when teaching mathematics to elementary school students. I have plenty of opinions on teaching for conceptual understanding, Common Core mathematics, and (where the Common Core has utterly failed) assessing for conceptual understanding:

Division 1: A discussion about the usefulness of unorthodox ways of teaching long division.

Division 2: A continuation of the above discussion.

Subtraction 1: Introducing a viral picture about the Common Core, and its easy solution.

Subtraction 2: The pedagogical rationale for using an open number line (even though I personally do not endorse this technique as superior to other ways of teaching subtraction).

Subtraction 3: The abject failure of current developmentally inappropriate ways of assessing the depth of a student’s mathematical knowledge.

Subtraction 4: The importance of engaging parents when unorthodox methods are used to teach mathematics to children.

# Hands on SET

Every so often, I’ll publicize through this blog an interesting article that I’ve found in the mathematics or mathematics education literature that can be freely distributed to the general public. Today, I’d like to highlight “Hands-on SET®,” by Hannah Gordon, Rebecca Gordon, and Elizabeth McMahon. Here’s the abstract:

SET® is a fun, fast-paced game that contains a surprising amount of mathematics. We will look in particular at hands-on activities in combinatorics and probability, finite geometry, and linear algebra for students at various levels. We also include a fun extension to the game that illustrates some of the power of thinking mathematically about the game.

Full reference: Hannah Gordon, Rebecca Gordon & Elizabeth McMahon (2013) Hands-on SET®, PRIMUS: Problems, Resources, and Issues in Mathematics Undergraduate Studies, 23:7, 646-658, DOI: 10.1080/10511970.2013.764368

# Angular size

Source: http://www.xkcd.com/1276/

# Lessons from teaching gifted elementary school students (Part 4c)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

What is the chance of winning a game of BINGO after only four turns?

When my class posed this question, I was a little concerned that the getting the answer might be beyond the current abilities of a gifted elementary student. As discussed over the past couple of posts, for a non-standard BINGO game with 44 numbers, the answer is

$\displaystyle 4 \times \frac{4}{44} \times \frac{3}{43} \times \frac{2}{42} \times \frac{1}{41} = \displaystyle \frac{4}{135,751}$

What’s that in decimal?

With these gifted students, I encourage thinking as much as possible without a calculator… and they wanted me to provide the answer to this one in like fashion. For my class, this actually did serve a purpose by illustrating a really complicated long division problem so they could reminded about the number of leading zeroes in such a problem.

Gritting my teeth, I started on the answer:

At this point, I was asked the other question that I had anticipated but utterly dreaded… motivated by child-like curiosity mixed perhaps with a touch of sadism:

How long do we have before the digits start repeating?

My stomach immediately started churning.

I told the class that I’d have to figure this one later. But I told them that the answer would definitely be less than 135,751 times. My class was surprised that I could even provide this level of (extremely) modest upper limit on the answer. After some prompting, my class saw the reasoning for this answer: there are only 135,751 possible remainders after performing the subtraction step in the division algorithm, and so a remainder has to be repeated after 135,751 steps. Therefore, the digits will start repeating in 135,751 steps or less.

What I knew — but probably couldn’t explain to these elementary-school students, and so I had to work this out for myself and then get back to them with the answer — is that the length of the repeating block $n$ is the least integer so that

$135751 \mid 10^n - 1$

which is another way of saying that we’ve used the division algorithm enough times so that a remainder repeats. Written in the language of group theory, $n$ is the least integer that satisfies

$10^n \equiv 1 \mod 135751$

(A caveat:this rule works because neither 2 nor 5 is a factor of 135,751… otherwise, those factors would have to be taken out first.)

Some elementary group theory can now be used to guess the value of $n$. Let $G$ be the multiplicative group of integers modulo $135,751$ which are relatively prime which. The order of this group is denoted by $\phi(135751)$, called the Euler totient function. In general, if $m = p_1^{a_1} p_2^{a_2} \dots p_r^{a_r}$ is the prime factorization of $m$, then

$\phi(m) = n \left( \displaystyle 1 - \frac{1}{p_1} \right) \left( \displaystyle 1 - \frac{1}{p_2} \right) \dots \left( \displaystyle 1 - \frac{1}{p_r} \right)$

For the case at hand, the prime factorization of $135,751$ can be recovered by examining the product of the fractions near the top of this post:

$135751 = 7 \times 11 \times 41 \times 43$

Therefore,

$\phi(135,751) = 6 \times 10 \times 40 \times 42 = 100,800$

Next, there’s a theorem from group theory that says that the order $n$ of an element of a group must be a factor of the order of the group. In other words, the number $n$ that we’re seeking must be a factor of $100,800$. This is easy to factor:

$100,800 = 2^6 \times 3^2 \times 5^2 \times 7$

Therefore, the number $n$ has the form

$n = 2^a 3^b 5^c 7^d$,

where $0 \le a \le 6$, $0 \le b \le 2$, $0 \le c \le 2$, and $0 \le d \le 1$ are integers.

So, to summarize, we can say definitively that $n$ is at most $100,800$, and that were have narrowed down the possible values of $n$ to only $7 \times 3 \times 3 \times 2 = 126$ possibilities (the product of one more than all of the exponents). So that’s a definite improvement and reduction from my original answer of $135,751$ possibilities.

At this point, there’s nothing left to do except test all 126 possibilities. Unfortunately, there’s no shortcut to this; it has to be done by trial and error. Thankfully, this can be done with Mathematica:

The final line shows that the least such value of $n$ is 210. Therefore, the decimal will repeat after 210 digits. So here are the first 210 digits of $\displaystyle \frac{4}{135,751}$ (courtesy of Mathematica):

0.000029465712959757202525211600651192256410634175807176374391348866675015285338597874048809953517837805983013016478699972744215512224587664\
179269397647162820163387378361853688002298325610861061796966504850792996…

# Lessons from teaching gifted elementary school students (Part 4b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

What is the chance of winning a game of BINGO after only four turns?

When my class posed this question, I was a little concerned that the getting the answer might be beyond the current abilities of a gifted elementary student.  Still, what I love about this question is that it gave me a way to teach my class some techniques of probabilistic reasoning that probably would not occur in a traditional elementary school setting.

As discussed yesterday, for a non-standard BINGO game with 44 numbers, the answer is

$\displaystyle 4 \times \frac{4}{44} \times \frac{3}{43} \times \frac{2}{42} \times \frac{1}{41}$

For a standard BINGO board with 75 numbers, the denominators are instead 75, 74, 73, and 72.

Now, for the next challenge: getting my students to simplify this product. I’m always mystified when college students blindly multiply numerators and denominators together without bothering to attempt to cancel common factors. Fortunately, this class already understands how to simplify fractions, and so the next step was easy:

$\displaystyle 4 \times \frac{1}{11} \times \frac{3}{43} \times \frac{1}{21} \times \frac{1}{41}$

So I was ready for the next step: cancelling 3 from the numerator and denominator. To my surprise, this was a major stumbling block. I tried probing around to prod them to perform this cancellation, but no luck. Eventually, I guessed the issue that my class was facing: they were familiar with the mechanics of both adding and multiplying fractions and also with writing fractions in lowest terms, but they weren’t yet comfortable enough with fractions to cancel 3 from the numerator of one fraction and the denominator of a different fraction.

So, toward this end, I asked my class if it was OK to shuffle a couple of the numerators and rewrite this product as

$\displaystyle 4 \times \frac{1}{11} \times \frac{1}{43} \times \frac{3}{21} \times \frac{1}{41}$

It took a moment, but then they agreed that this was OK because the order of multiplication doesn’t matter, even volunteering the word commutative to explain their reasoning. (I’m going to try to remember this technique for future reference as a way to get students new to fractions more comfortable with similar cancellations.) Once they got past this conceptual barrier, it was straightforward to continue the simplification:

$\displaystyle 4 \times \frac{1}{11} \times \frac{1}{43} \times \frac{1}{7} \times \frac{1}{41}$

$= \displaystyle \frac{4}{135,751}$

So I explained that if a game of BINGO took one minute, we could play round the clock for 135,751 minutes (about 96 days) and expect to win in the minimal number of turns only four times. Not very likely at all. (Though I didn’t discuss this with my class, the answer is even smaller with a standard BINGO game with 75 numbers: you’d expect to win only once every 211 days.)

# Lessons from teaching gifted elementary school students (Part 4a)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

What is the chance of winning a game of BINGO after only four turns?

I leave a thought bubble in case you’d like to think this. One way of answering this question appears after the bubble.

When my class posed this question, I was a little concerned that my class was simply not ready to understand the solution (described below), as it takes more than a little work to get at the answer. Still, what I love about this question is that it gave me a way to teach my class some techniques of probabilistic reasoning that probably would not occur in a traditional elementary school setting. Also, I was reminded that even these gifted students might need a little help with simplifying the answer. So let me discuss how I helped these young students discover the answer. I found the ensuing discussion especially enlightening, and so I’m dividing this discussion into several posts.

Here’s a non-standard BINGO board:

Using the free space in the middle, there are four ways of winning the game in four moves:

• Horizontally (11-12-13-14)
• Vertically (3-8-17-22)
• Diagonally (1-7-18-24)
• Diagonally (5-9-16-20)

A standard BINGO board has 75 possible numbers (B 1-15, I 16-30, N 31-45, G 46-60, O 61-75). However, the board that I was using with my class (which was being used for pedagogical purposes) only had 44 possibilities. So the solution below assumes these 44 possibilities; the answer for a standard BINGO board is obvious.

My class quickly decided to start by solving the problem for the horizontal case. I began by asking for the chance that the first number will be on the middle row; after some thought, the class correctly answered $\displaystyle \frac{4}{44}$.

Next, I asked the chance that the next number would also be on the middle row. To my surprise, this wasn’t automatic for my young but gifted students. They felt that they didn’t know where the first number was, and so they felt like they couldn’t know the chance for the second number. To get them over this conceptual barrier (or so I thought), I asked them to pretend that the first number was 11. Then what would be the odds that the next number fell on the middle row? After some discussion, the class agreed that the answer was $\displaystyle \frac{3}{43}$.

Once that barrier was cleared, then the class saw that the next two fractions were $\displaystyle \frac{2}{42}$ and $\displaystyle \frac{1}{41}$. I then explained that, to get the answer for the four consecutive numbers on the middle row, these fractions have to be multiplied:

$\displaystyle \frac{4}{44} \times \frac{3}{43} \times \frac{2}{42} \times \frac{1}{41}$

I didn’t justify why the fractions had to be multiplied; my class just accepted this as the way to combined the fractions to get the answer for all four events happening at once.

Then I asked about the other three possibilities — the middle column and the two diagonals. The class quickly agreed that the answer should be the same for these other possibilities, and so the final answer should just be four times larger:

$\displaystyle 4 \times \frac{4}{44} \times \frac{3}{43} \times \frac{2}{42} \times \frac{1}{41}$

At this point, I was ready to go on, but then a student asked something like the following:

Shouldn’t the answer be $\displaystyle \frac{1}{44} \times \frac{1}{43} \times \frac{1}{42} \times \frac{1}{41}$? I mean, we chose 11 to be the first number so that we can figure out the chance for the second number, and the chance that the first number is 11 is $\displaystyle \frac{1}{41}$.

Oops. While trying to clear one conceptual hurdle (getting the answer of $\displaystyle \frac{3}{43}$ for the second number), I had inadvertently introduced a second hurdle by making my class wonder if the first number had to be a specific number.

I began by trying to explain that the first number really didn’t have to be 11 after all, but that only seemed to re-introduce the original barrier. Finally, I found an answer that my class found convincing: Yes, the chance that the first number is 11, the second number is 12, the third number is 13, and fourth number is 14 is indeed $\displaystyle \frac{1}{44} \times \frac{1}{43} \times \frac{1}{42} \times \frac{1}{41}$. But there are other ways that all the numbers could land on the middle row:

• The first number could be 12, the second number could be 11, the third number could be 13, and the fourth number could be 14.
• Quickly, light dawned, and my class began volunteering other orderings by which all the numbers land in the middle row.
• We then enumerated the number of ways that this could happen, and we found that the answer was indeed 24.
• I then tied the knot by noting that $4 \times 3 \times 2 \times 1 = 24$, and so gives another explanation for the numerators in the answer.

Having found the answer, it was now time to simplify the answer. More on this in tomorrow’s post.

# Lessons from teaching gifted elementary school students (Part 3b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

Suppose

$A \times A = B$

$B \times B \times B = C$

$C \times C \times C \times C= D$

If the pattern goes on, and if $A = 2$, what is $Z$?

In yesterday’s post, we found that the answer was

$Z =2^{26!} = 10^{26! \log_{10} 2} \approx 10^{1.214 \times 10^{26}}$,

a number with approximately $1.214 \times 10^{26}$ digits.

How can we express this number in scientific notation? We need to actually compute the integer and decimal parts of $26! \log_{10} 2$, and most calculators are not capable of making this computation.

Fortunately, Mathematica is able to do this. We find that

$Z \approx 10^{121,402,826,794,262,735,225,162,069.4418253767}$

$\approx 10^{0.4418253767} \times 10^{121,402,826,794,262,735,225,162,069}$

$\approx 2.765829324 \times 10^{121,402,826,794,262,735,225,162,069}$

Here’s the Mathematica syntax to justify this calculation. In Mathematica, $\hbox{Log}$ means natural logarithm:

Again, just how big is this number? As discussed yesterday, it would take about 12.14 quadrillion sheets of paper to print out all of the digits of this number, assuming that $Z$ was printed in a microscopic font that uses 100,000 characters per line and 100,000 lines per page. Since 250 sheets of paper is about an inch thick, the volume of the 12.14 quadrillion sheets of paper would be

$1.214 \times 10^{16} \times 8.5 \times 11 \times \displaystyle \frac{1}{250} \hbox{in}^3 \approx 1.129 \times 10^{17} \hbox{in}^3$

By comparison, assuming that the Earth is a sphere with radius 4000 miles, the surface area of the world is

$4 \pi (4000 \times 5280 \times 12) \hbox{in}^2 \approx 8.072 \times 10^{17} \hbox{in}^2$.

Dividing, all of this paper would cover the entire world with a layer of paper about $0.14$ inches thick, or about 35 sheets deep. In other words, the whole planet would look something like the top of my desk.

What if we didn’t want to print out the answer but just store the answer in a computer’s memory? When written in binary, the number $2^{26!}$ requires…

$26!$ bits of memory, or…

about $4.03 \times 10^{26}$ bits of memory, or…

about 5.04 \times 10^{25} bytes of memory, or ...

latex 5.04 \times 10^{13}$terabytes of memory, or… about 50.4 trillion terabytes of memory. Suppose that this information is stored on 3-terabyte external hard drives, so that about $50.4/3 = 16.8$ trillion of them are required. The factory specs say that each hard drive measures $129 \hbox{mm} \times 42 \hbox{mm} \times 167 \hbox{mm}$. So the total volume of the hard drives would be $1.52 \times 10^{19} \hbox{mm}^3$, or $15.2 \hbox{km}^3$. By way of comparison, the most voluminous building in the world, the Boeing Everett Factory (used for making airplanes), has a volume of only $0.0133 \hbox{km}^3$. So it would take about 1136 of these buildings to hold all of the necessary hard drives. The cost of all of these hard drives, at$100 each, would be about $1.680 quadrillion. So it’d be considerably cheaper to print this out on paper, which would be about one-seventh the price at$242 trillion.

Of course, a lot of this storage space would be quite repetitive since $2^{26!}$, in binary, would be a one followed by $26!$ zeroes.

# Lessons from teaching gifted elementary school students (Part 3a)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received:

Suppose

$A \times A = B$

$B \times B \times B = C$

$C \times C \times C \times C= D$

If the pattern goes on, and if $A = 2$, what is $Z$?

I leave a thought bubble in case you’d like to think this. (This is significantly more complicated to do mentally than the question posed in yesterday’s post.) One way of answering this question appears after the bubble.

Let’s calculate the first few terms to try to find a pattern:

$B = 2 \times 2 = 2^2$

$C = 2^2 \times 2^2 \times 2^2 = 2^6$

$D = 2^6 \times 2^6 \times 2^6 \times 2^6 = 2^{24}$

etc.

Written another way,

$A = 2^1 = 2^{1!}$

$B = 2^{2!}$

$C = 2^{3!}$

$D = 2^{4!}$

Naturally, elementary school students have no prior knowledge of the factorial function. That said, there’s absolutely no reason why a gifted elementary school student can’t know about the factorial function, as it only consists of repeated multiplication.

Continuing the pattern, we see that $Z = 2^{26!}$. Using a calculator, we find $Z \approx 2^{4.032014611 \times 10^{26}}$.

If you try plugging that number into your calculator, you’ll probably get an error. Fortunately, we can use logarithms to approximate the answer. Since $2 = 10^{\log_{10} 2}$, we have

$Z = \left( 10^{\log_{10} 2} \right)^{4.032014611 \times 10^{26}} = 10^{4.032014611 \times 10^{26} \log_{10} 2}$

Plugging into a calculator, we find that

$Z \approx 10^{1.214028268 \times 10^{26}} = 10^{121.4028628 \times 10^{24}}$

We conclude that the answer has more than 121 septillion digits.

How big is this number? if $Z$ were printed using a microscopic font that placed 100,000 digits on a single line and 100,000 lines on a page, it would take 12.14 quadrillion pieces of paper to write down the answer (6.07 quadrillion if printed double-sided). If a case with 2500 sheets of paper costs $100, the cost of the paper would be$484 trillion ($242 trillion if double-sided), dwarfing the size of the US national debt (at least for now). Indeed, the United States government takes in about$3 trillion in revenue per year. At that rate, it would take the country about 160 years to raise enough money to pay for the paper (80 years if double-sided).

And that doesn’t even count the cost of the ink or the printers that would be worn out by printing the answer!