My Favorite One-Liners: Part 92

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

This is one of my favorite quote from Alice in Wonderland that I’ll use whenever discussing the difference between the ring axioms (integers are closed under addition, subtraction, and multiplication, but not division) and the field axioms (closed under division except for division by zero):

‘I only took the regular course [in school,’ said the Mock Turtle.]

‘What was that?’ inquired Alice.

‘Reeling and Writhing, of course, to begin with,’ the Mock Turtle replied; ‘and then the different branches of Arithmetic — Ambition, Distraction, Uglification, and Derision.’

My Favorite One-Liners: Part 18

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. This is a quip that I’ll use when a theoretical calculation can be easily confirmed with a calculator.

Sometimes I teach my students how people converted decimal expansions into fractions before there was a button on a calculator to do this for them. For example, to convert  x = 0.\overline{432} = 0.432432432\dots into a fraction, the first step (from the Bag of Tricks) is to multiply by 1000: How do we change this into a decimal? Let’s call this number x.

1000x = 432.432432\dots

x = 0.432432\dots

Notice that the decimal parts of both x and 1000x are the same. Subtracting, the decimal parts cancel, leaving

999x = 432

or

x = \displaystyle \frac{432}{999} = \displaystyle \frac{16}{37}

In my experience, most students — even senior math majors who have taken a few theorem-proof classes and hence are no dummies — are a little stunned when they see this procedure for the first time.

To make this more real and believable to them, I then tell them my one-liner: “I can see that no one believes me. OK, let’s try something that you will believe. Pop out your calculators. Then punch in 16 divided by 37.”

Indeed, my experience many students really do need this technological confirmation to be psychologically sure that it really did work. Then I’ll tease them that, by pulling out their calculators, I’m trying to speak my students’ language.

TI1637

See also my fuller post on this topic as well as the index for the entire series.

 

My Favorite One-Liners: Part 14

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. This quip is similar to the “bag of tricks” one-liner, and I’ll use this one if the “bag of tricks” line is starting to get a little dry.

Sometimes in math, there’s a step in a derivation that, to the novice, appears to make absolutely no sense. For example, to find the antiderivative of \sec x, the first step is far from obvious:

\displaystyle \int \sec x \, dx = \displaystyle \int \sec x \frac{\sec x + \tan x}{\sec x + \tan x} \, dx

While that’s certainly correct, it’s from from obvious to a student that this such a “simplification” is actually helpful.

To give a simpler example, to convert

x = 0.\overline{432} = 0.432432432\dots

into a decimal, the first step is to multiply x by 1000:

1000x = 432.432432\dots

Students often give skeptical, quizzical, and/or frustrated looks about this non-intuitive next step… they’re thinking, “How did you know to do that?” To lighten the mood, I’ll explain with a big smile that I’m clairvoyant… when I got my Ph.D., I walked across the stage, got my diploma, someone waved a magic wand at me, and poof! I became clairvoyant.

Clairvoyance is wonderful; I highly recommend it.

The joke, of course, is that the only reason that I multiplied by 1000 is that someone figured out that multiplying by 1000 at this juncture would actually be helpful. Subtracting x from 1000x, the decimal parts cancel, leaving

999x = 432

or

x = \displaystyle \frac{432}{999} = \displaystyle \frac{16}{37}.

In my experience, most students — even senior math majors who have taken a few theorem-proof classes and hence are no dummies — are a little stunned when they see this procedure for the first time. I learned this procedure when I was very young; however, in modern times, this procedure appears to be a dying art. I’m guessing that this algorithm is a dying art because of the ease and convenience of modern calculators. As always, I hold my students blameless for the things that they were simply not taught at a younger age, and part of my job is repairing these odd holes in their mathematical backgrounds so that they’ll have their best chance at becoming excellent high school math teachers.

For further reading, here’s my series on rational numbers and decimal expansions.

15-square puzzle

From the category “This Is Completely Useless”: here’s what a 15-square puzzle looks like when you arrange the tiles in order of how many factors they have.

puzzle15

My Mathematical Magic Show: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show the mathematical magic show that I’ll perform from time to time.

Part 1: Introduction.

Part 2a, Part 2b, and Part 2c: The 1089 trick.

Part 3a, Part 3b, and Part 3c: A geometric magic trick.

Part 4a: Part 4b, Part 4c, and Part 4d: A trick using binary numbers.

Part 5a, Part 5b, Part 5c, and Part 5d: A trick using the rule for checking if a number is a multiple of 9.

Part 7: The Fitch-Cheney card trick, which is perhaps the slickest mathematical card trick ever devised.

Part 8a, Part 8b, and Part 8c: A trick using Pascal’s triangle.

Part 9: Mentally computing n given n^5 if 10 \le n \le 99.

Part 6: The Grand Finale.

And, for the sake of completeness, here’s a recent picture of me just before I performed an abbreviated version of this show for UNT’s Preview Day for high school students thinking about enrolling at my university.

magician

 

Engaging students: Reducing fractions to lowest terms

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 Madison duPont. Her topic, from Pre-Algebra: reducing fractions to lowest terms.

green line

How can this topic be used in your students’ future courses in mathematics or science?

Reducing fractions to lowest terms can be applied to future mathematics topics such as ratios and proportions, and scientific topics such as chemistry or physics. Ratios can be represented as fractions and are not typically reduced to lowest terms because they represent relationships of two subjects using numbers. Being able to reduce these ratios can help students better identify the underlying relationship and apply this relationship to other aspects of the math problem, such as problems using unit price or map scales. Proportions relate to the concept of reducing fractions to lowest terms when using cross-multiplication. Having both sides of the proportion reduced to lowest terms makes the cross-multiplication much easier to compute and derive a final reduced answer. Chemistry uses fractions reduced to lowest terms with topics, like stoichiometry, that use potentially small and large numbers in several ratios that are multiplied together to obtain a final converted and reduced answer. Physics often uses ratio-like formulas and problems that are applied to real-world scenarios, which typically require fractions reduced to lowest terms because answers like miles per one hour are the goal. All of these topics use concepts of reducing fractions to lowest terms to more easily accomplish problems using a series of fractional computations, or to get an answer that is in terms of a single unit or most reduced so that it makes sense to real-world application.

 

 

green line

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

This topic extends previously learned topics such as concepts of unique prime factorizations, greatest common divisor, manipulating fractions, and multiplication facts. The concept of unique prime factorizations greatly aids students in finding the greatest common divisor, which is used to find the greatest factor of the value of both the numerator and denominator. Next, manipulation of fractions is used to properly divide the numerator and denominator by the greatest common divisor. This process of dividing both parts of the fraction utilizes multiplication facts as well to determine what the answer to the division problem on both the top and bottom of the fraction would be. These previously learned concepts are all subtle and important applications when reducing fractions to lowest terms.

 

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How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic? Note: It’s not enough to say “such-and-such is a great website”; you need to explain in some detail why it’s a great website.

 

 

This video reminded me of many students that I have tutored or encountered in classrooms that were determined that a calculator was all they needed when doing math. Applied to reducing fractions to lowest terms, this video is extremely relevant in displaying that technology cannot be the only source of intelligence when thinking mathematically. Reducing fractions with extremely large numbers or numbers that do not have well-known factors can seem exhausting or impossible. Punching several factors of the numerator and denominator into a calculator attempting to reduce numbers with each common factor, and then not being sure of whether the fraction appearing on their screen is truly in the most reduced form surely indicates the technology is not the only way of solving the problem. Many students hop on a procedural escalator when beginning varying types of problems (in addition to reducing fractions to lowest terms) using memorized steps, punching calculator buttons, feeling comfortable, until suddenly—there is a horribly unattractive fraction halting their progress. This is when using mathematical problem solving skills such as reducing the numerator and denominator by the greatest common divisor or checking to see that the numerator and denominator are relatively prime becomes pertinent. Using these conceptual skills can save someone that is stuck waiting for a calculator to do the work for them, or that has given up on finishing a problem because it seems impossible or difficult, from thinking they are incapable of working out a problem efficiently and successfully. This video highlights the importance of being capable of knowing when it is time to take the effort to climb the stairs to reach your destination.

 

References:

“Stuck on an Escalator” Video link:

https://www.youtube.com/watch?v=VrSUe_m19FY

found via Google video search

 

 

 

 

 

Lessons from teaching gifted elementary school students: Index (updated)

I’m doing 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 various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students. (This is updated from my previous index.)

Part 1: A surprising pattern in some consecutive perfect squares.

Part 2: Calculating 2 to a very large exponent.

Part 3a: Calculating 2 to an even larger exponent.

Part 3b: An analysis of just how large this number actually is.

Part 4a: The chance of winning at BINGO in only four turns.

Part 4b: Pedagogical thoughts on one step of the calculation.

Part 4c: A complicated follow-up question.

Part 5a: Exponentiation is multiplication as multiplication is to addition. So, multiplication is to addition as addition is to what? (I offered the answer of incrementation, but it was rejected: addition requires two inputs, while incrementation only requires one.)

Part 5b: Why there is no binary operation that completes the above analogy.

Part 5c: Knuth’s up-arrow notation for writing very big numbers.

Part 5d: Graham’s number, reputed to be the largest number ever to appear in a mathematical proof.

Part 6a: Calculating $(255/256)^x$.

Part 6b: Solving $(255/256)^x = 1/2$ without a calculator.

Part 7a: Estimating the size of a 1000-pound hailstone.

Part 7b: Estimating the size a 1000-pound hailstone.

Part 8a: Statement of an usually triangle summing problem.

Part 8b: Solution using binomial coefficients.

Part 8c: Rearranging the series.

Part 8d: Reindexing to further rearrange the series.

Part 8e: Rewriting using binomial coefficients again.

Part 8f: Finally obtaining the numerical answer.

Part 8g: Extracting the square root of the answer by hand.

Pizza Hut Pi Day Challenge (Part 8)

On March 14, 2016, Pizza Hut held a online math competition in honor of Pi Day, offering three questions posed by Princeton mathematician John H. Conway. As luck would have it, years ago, I had actually heard of the first question before from a colleague who had heard it from Conway himself:

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must be one of the following 10 numbers:

1 , 4 7 2 , 5 8 9 , 6 3 0,

7 , 4 1 2 , 5 8 9 , 6 3 0,

1 , 8 9 6 , 5 4 3 , 2 7 0,

9 , 8 1 6 , 5 4 3 , 2 7 0,

7 , 8 9 6 , 5 4 3 , 2 1 0,

9 , 8 7 6 , 5 4 3 , 2 1 0,

1 , 8 3 6 , 5 4 7 , 2 9 0,

3 , 8 1 6 , 5 4 7 , 2 9 0,

1 , 8 9 6 , 5 4 7 , 2 3 0,

9 , 8 1 6 , 5 4 7 , 2 3 0.

Up until now, I have used the divisibility rules to ensure that the property works for n = 1, 2, 3, 4, 5, 6, 8, 9, 10. But I haven’t used n = 7 yet.

Step 10. The number formed by the first seven digits must be a multiple of 7. There is a very complicated divisibility rule for checking to see if a number is a multiple of 7. However, at this point, it’s easiest to just divide by 7 and see what happens.

1 , 4 7 2 , 5 8 9 / 7 = 210,369.857\dots: not a multiple of 7.

7 , 4 1 2 , 5 8 9 / 7 = 1,058,941.285\dots: not a multiple of 7.

1 , 8 9 6 , 5 4 3 / 7 = 270,934.714\dots: not a multiple of 7.

9 , 8 1 6 , 5 4 3 / 7 = 1,402,363.285\dots: not a multiple of 7.

7 , 8 9 6 , 5 4 3 / 7 = 1,128,077.571\dots: not a multiple of 7.

9 , 8 7 6 , 5 4 3 / 7 = 1,410,934.714\dots: not a multiple of 7.

1 , 8 3 6 , 5 4 7 = 262,363.857\dots: not a multiple of 7.

3 , 8 1 6 , 5 4 7 / 7 = 545,221: a multiple of 7!!!

1 , 8 9 6 , 5 4 7 / 7 = 270,935.285\dots: not a multiple of 7.

9 , 8 1 6 , 5 4 7 / 7 = 1,402,363.857\dots: not a multiple of 7.

So, by inspection, only one of these works, yielding the answer to the puzzle:

3,816,547,290.

Pizza Hut Pi Day Challenge (Part 7)

On March 14, 2016, Pizza Hut held a online math competition in honor of Pi Day, offering three questions posed by Princeton mathematician John H. Conway. As luck would have it, years ago, I had actually heard of the first question before from a colleague who had heard it from Conway himself:

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must have one of the following four forms:

O , 4 O 2 , 5 8 1 , 6 O 0,

O , 4 O 2 , 5 8 9 , 6 O 0,

O , 8 O 6 , 5 4 3 , 2 O 0,

O , 8 O 6 , 5 4 7 , 2 O 0.

where O represents the remaining three odd digts. (The last digit is 0 and not an odd number.) There are 24 possible answers left.

Step 9. The number formed by the first three digits must be a multiple of 3. By the divisibility rules, this means that the sum of the first three digits must be a multiple of 3.

For the first form, that means that O + 4 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 3, 7, and 9. We can directly test this to see that it’s impossible:

3 + 4 + 7 = 14, not a multiple of 3.

3 + 4 + 9 = 16, not a multiple of 3.

7 + 4 + 9 = 20, not a multiple of 3.

For the second form, that means that O + 4 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 3, and 7. We can directly test this:

1 + 4 + 3 = 8, not a multiple of 3.

1 + 4 + 7 = 12, a multiple of 3.

3 + 4 + 7 = 14, not a multiple of 3.

Therefore, for the second form, the first three digits could be either 147 or 714.

For the third form, that means that O + 8 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 7, and 9. We can directly test this to see that it’s impossible:

1 + 8 + 7 = 16, not a multiple of 3.

1 + 8 + 9 = 18, a multiple of 3.

7 + 8 + 9 = 24, a multiple of 3.

Therefore, for the third form, the first three digits could be either 189, 981, 789, or 987.

For the fourth form, that means that O + 8 + O must be a multiple of 3, where O is chosen from the remaining odd digits, which are 1, 3, and 9. We can directly test this to see that it’s impossible:

1 + 8 + 3 = 12, a multiple of 3.

1 + 8 + 9 = 18, a multiple of 3.

3 + 8 + 9 = 20, not a multiple of 3.

Therefore, for the fourth form, the first three digits could be 183, 381, 189, or 981.

1 , 4 7 2 , 5 8 9 , 6 O 0,

7 , 4 1 2 , 5 8 9 , 6 O 0,

1 , 8 9 6 , 5 4 3 , 2 O 0,

9 , 8 1 6 , 5 4 3 , 2 O 0,

7 , 8 9 6 , 5 4 3 , 2 O 0,

9 , 8 7 6 , 5 4 3 , 2 O 0,

1 , 8 3 6 , 5 4 7 , 2 O 0,

3 , 8 1 6 , 5 4 7 , 2 O 0,

1 , 8 9 6 , 5 4 7 , 2 O 0,

9 , 8 1 6 , 5 4 7 , 2 O 0.

Indeed, for each of these, there is only one odd digit left, which means we automatically know what it has to be for each of these 10 answers by process of elimination:

1 , 4 7 2 , 5 8 9 , 6 3 0,

7 , 4 1 2 , 5 8 9 , 6 3 0,

1 , 8 9 6 , 5 4 3 , 2 7 0,

9 , 8 1 6 , 5 4 3 , 2 7 0,

7 , 8 9 6 , 5 4 3 , 2 1 0,

9 , 8 7 6 , 5 4 3 , 2 1 0,

1 , 8 3 6 , 5 4 7 , 2 9 0,

3 , 8 1 6 , 5 4 7 , 2 9 0,

1 , 8 9 6 , 5 4 7 , 2 3 0,

9 , 8 1 6 , 5 4 7 , 2 3 0.

So we’re down to 10 possible answers left.

In tomorrow’s post, I’ll finally find the answer.

Pizza Hut Pi Day Challenge (Part 6)

On March 14, 2016, Pizza Hut held a online math competition in honor of Pi Day, offering three questions posed by Princeton mathematician John H. Conway. As luck would have it, years ago, I had actually heard of the first question before from a colleague who had heard it from Conway himself:

I’m thinking of a ten-digit integer whose digits are all distinct. It happens that the number formed by the first n of them is divisible by n for each n from 1 to 10. What is my number?

I really like this problem because it’s looks really tough but only requires knowledge of elementary-school arithmetic. So far in this series, I described why the solution must have one of the following four forms:

O , 4 O 2 , 5 8 O , 6 O 0,

O , 6 O 2 , 5 8 O , 4 O 0,

O , 2 O 6 , 5 4 O , 8 O 0.

O , 8 O 6 , 5 4 O , 2 O 0.

where O is one of 1, 3, 7, and 9. (The last digit is 0 and not an odd number.) There are 96 possible answers left.

Step 8. The number formed by the first eight digits must be a multiple of 8. By the divisibility rules, this means that the number formed by the sixth, seventh, and eighth digits must be a multiple of 8.

For the first form, that means that 8O6 must be a multiple of 8. We can directly test this:

816/8 = 102: a multiple of 8.

836/8 = 104.5: not a multiple of 8.

876/8 = 109.5: not a multiple of 8.

896/8 = 112: a multiple of 8.

For the second form, that means that 8O4 must be a multiple of 8. This is impossible. Let O = 2n+1. Then

8O4 = 800 + 10 \times O + 4

= 800 + 10(2n+1) + 4

= 800 + 20n + 14

= 2(400 + 10n + 7).

We see that 7 is odd, and therefore 400 + 10n + 7 is not a multiple of 2. Therefore, 2(400 + 10n + 7) is not a multiple of 4 (let alone 8).

For the third form, that means that 4O8 must be a multiple of 8. This is also impossible. Let O = 2n+1. Then

4O8 = 400 + 10 \times O + 8

= 400 + 10(2n+1) + 8

= 400 + 20n + 18

= 2(200 + 10n + 9).

We see that 9 is odd, and therefore 200 + 10n + 9 is not a multiple of 2. Therefore, 2(200 + 10n + 9) is not a multiple of 4 (let alone 8).

For the fourth form, that means that 4O2 must be a multiple of 8. We can directly test this:

412/8 = 51.5: not a multiple of 8.

432/8 = 54: a multiple of 8.

472/8 = 59: a multiple of 8.

492/8 = 61.5: not a multiple of 8.

In other words, we’re down to

O , 4 O 2 , 5 8 1 , 6 O 0,

O , 4 O 2 , 5 8 9 , 6 O 0,

O , 8 O 6 , 5 4 3 , 2 O 0,

O , 8 O 6 , 5 4 7 , 2 O 0.

For each of these, there are 3! = 6 ways of choosing the remaining odd digits. Since there are four forms, there are 4 x 6= 24 possible answers left.

In tomorrow’s post, I’ll cut this number down to 10.