Functions that commute (Part 2)

Math With Bad Drawings had a nice post with pedagogical thoughts on the tendency of students to commute two functions that don’t commute:

The author’s proposed remedies:

Teach the distributive law more carefully. Draw pictures. Work examples. Talk about “bags.” Make sure they understand the meaning behind this symbolism.

Teach function notation much more carefully. Give them the chance to practice it. Think like Dan Meyer and seek activities that create the intellectual need for function notation.

Keep stamping out the “everything is linear” error when it crops up. Like the common cold, it’ll probably never be entirely eradicated, but good mathematical hygiene should reduce its prevalence.

I agree with all three points. Concerning the third point, here’s an earlier post of mine concerning these kinds of mistakes (and others), with a one-liner I’ll use to try to get students to remember not to make these kinds of mistakes:

I wish I could remember the speaker’s name, but I heard the following one-liner at a state mathematics conference many years ago, and I’ve used it to great effect in my classes ever since. Whenever I present a property where two functions commute, I’ll say, “In other words, the order of operations does not matter. This is a big deal, because, in real life, the order of operations usually is important. For example, this morning, you probably got dressed and then went outside. The order was important.”
If that fails, then I’ll cite Finding Nemo, trying to minimize frustration by keeping the mood light.

And if that fails, I’ll cite The Princess Bride. One of the most common student mistakes with logarithms is thinking that
$\log_b(x+y) = \log_b x + \log_b y$ .

When I first started my career, I referred to this as the Third Classic Blunder. The first classic blunder, of course, is getting into a major land war in Asia. The second classic blunder is getting into a battle of wits with a Sicilian when death is on the line. And the third classic blunder is thinking that $\log_b(x+y)$ somehow simplfies as $\log_b x + \log_b y$ .

Sadly, as the years pass, fewer and fewer students immediately get the cultural reference. On the bright side, it’s also an opportunity to introduce a new generation to one of the great cinematic masterpieces of all time.

High School Teachers’ Problem Solving Activities to Review and Extend Their Mathematical and Didactical Knowledge

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 Manuel Santos-Trigo & Fernando Barrera-Mora (2011) High School Teachers’ Problem Solving Activities to Review and Extend Their Mathematical and Didactical Knowledge, PRIMUS: Problems, Resources, and Issues in Mathematics Undergraduate Studies, 21:8, 699-718, DOI: 10.1080/10511971003600965

Here’s the abstract:

The study documents the extent to which high school teachers reflect on their need to revise and extend their mathematical and practicing knowledge. In this context, teachers worked on a set of tasks as a part of an inquiring community that promoted the use of different computational tools in problem solving approaches. Results indicated that the teachers recognized that the use of the Cabri-Geometry software to construct dynamic representations of the problems became useful, not only to make sense of the problems statement, but also to identify and explore a set of mathematical relations. In addition, the use of other tools like hand-held calculators and spreadsheets offered them the opportunity to examine, contrast, and extend visual and graphic results to algebraic approaches.

The full article can be found here: http://dx.doi.org/10.1080/10511971003600965

The worst math education video on YouTube

We now have a winner for the worst math education video on YouTube:

My personal favorite part is demonstrating that 140*9 is a multiple of 9 by casting out nines.

Why is this so awful? There are two essential ideas that make this work:

Humans have chosen a convention that there are 360 degrees in a circle. There’s nothing particularly magical about 360; that’s just the number that humanity has chosen for measuring angles with degrees. Notice that 360 happens to be a multiple of 9.
In base-10 arithmetic, one can check an integer is divisible by 9 by checking if the sum of the digits is a multiple of 9.

The first part of the video shows that, when 360 degrees is successively bisected, the digits of the resulting angle still sum to 9. That’s because dividing by 2 is the same as multiplying by 5 and then dividing by 10. Dividing by 10 is unimportant for the purpose of adding digits, so the only operation that’s important is multiplying by 5. And of course, if a multiple of 9 is multiplied by 5, the product is still a multiple of 9.

Notice that’s important that the angles are successively bisected. If the angles were trisected instead, this would fail (360/3 = 120, which is not a multiple of 9.)

The second part of the video notes that the sum of the angles in a convex polygon is a multiple of 9. That’s because the sum of the angles is (in degrees) $180(n-2)$ , which of course is a multiple of 9. Furthermore, this formula is a consequence of the human convention of choosing 360 degrees to measure a complete rotation. From this number, the measure of a straight angle is 180 degrees. From this, the sum of the angles in a triangle is determined to be 180 degrees, and from this the sum of the angles in a convex polygon is found to be $180(n-2)$ degrees. All this to say, there’s nothing mystical about this. The second part of the video is a logical consequence of choosing 360 degrees for measuring circles.

The third part of the video is utter nonsense.

Lessons from teaching gifted elementary school students (Part 5d)

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.

A bright young student of mine noticed that multiplication is repeated addition:

$x \cdot y = x + x + x \dots + x$ ,

Also, exponentiation is repeated addition:

$x \uparrow y = x^y = x \cdot x \cdot x \dots \cdot x$ ,

The notation $x \uparrow y$ is unorthodox, but it leads to the natural extensions

$x \upuparrows y = x \uparrow x \uparrow x \dots \uparrow x$ ,

$x \uparrow^3 y = x \upuparrows x \upuparrows x \dots \upuparrows x$ ,

and so. I’ll refer the interested reader to Wikipedia and Mathworld (and references therein) for more information about Knuth’s up-arrow notation. As we saw in yesterday’s post, these numbers get very, very large… and very, very quickly.

When I was in elementary school myself, I remember reading in the 1980 Guiness Book of World Records about Graham’s number, which was reported to be the largest number ever used in a serious mathematical proof. Obviously, it’s not the largest number — there is no such thing — but the largest number that actually had some known usefulness. And this number is only expressible using Knuth’s up-arrow notation. (Again, see Wikipedia and Mathworld for details.)

From Mathworld, here’s a description of the problem that Graham’s number solves:

Stated colloquially, [consider] every possible committee from some number of people $n$ and enumerating every pair of committees. Now assign each pair of committees to one of two groups, and find $N$ , the smallest $n$ that will guarantee that there are four committees in which all pairs fall in the same group and all the people belong to an even number of committees.

In 1971, Graham and Fairchild proved that there is a solution $N$ , and that $N \le F(F(F(F(F(F(F(12)))))))$ , where

$F(n) = 2 \uparrow^n 3$ .

For context, $2 \uparrow^4 3$ is absolutely enormous. In yesterday’s post, I showed that $2 \uparrow^3 = 65,536$ . Therefore,

$2 \uparrow^4 3 = 2 \uparrow^3 (2 \uparrow^3 2)$

$= 2 \uparrow^3 65,536$

$= 2 \upuparrows 2 \upuparrows 2 \upuparrows \dots \upuparrows 2$ ,

repeated 65,536 times.

That’s just $2 \uparrow^4 3$ . Now try to imagine $F(12) = 2 \uparrow^{12} 3$ . That’s a lot of arrows.

Now try to imagine $F(F(12)) = 2 \uparrow^{F(12)} 3$ , which is even more arrows.

Now try to imagine $F(F(F(F(F(F(F(12)))))))$ . I bet you can’t. (I sure can’t.)

Graham and Fairchild also helpfully showed that $N \ge 6$ . So somewhere between 6 and Graham’s number lies the true value of $N$ .

A postscript: according to Wikipedia, things have improved somewhat since 1971. The best currently known bounds for $N$ are

$13 \le N \le 2 \uparrow^3 6$ .

Lessons from teaching gifted elementary school students (Part 5c)

A bright young student of mine noticed that multiplication is repeated addition:

$x \cdot y = x + x + x \dots + x$ ,

Also, exponentiation is repeated addition:

$x^y = x \cdot x \cdot x \dots \cdot x$ ,

So, my student asked, why can’t we define an operation that’s repeated exponentiation? Like all good explorers, my student claimed naming rights for this new operation and called it $x \hbox{~playa~} y$ :

$x \hbox{~playa~} y = x^{x^{x^{\dots}}}$

For example,

$4 \hbox{~playa~} 3 = 4^{4^4} = 4^{256} \approx 1.34 \times 10^{154}$

Even with small numbers for $x$ and $y$ , $x \hbox{~playa~} y$ gets very large.

Unfortunately for my student, someone came up with this notion already, and it’s called Knuth’s up-arrow notation. I’ll give some description here and refer the interested reader to Wikipedia and Mathworld (and references therein) for more information. Surprisingly, this notion has only become commonplace since 1976 — within my own lifetime.

Let’s define $x \uparrow y$ to be ordinary exponentiation:

$x \uparrow y = x \cdot x \cdot x \dots \cdot x$ .

Let’s now define $x \upuparrows y$ to be the up-arrow operation repeated $y$ times:

$x \upuparrows y = x \uparrow x \uparrow x \dots \uparrow x$ .

In this expression, the order of operations is taken to be right to left.

Numbers constructed by $\upuparrows$ get very, very big and very, very quickly. For example:

$2 \upuparrows 2 = 2 \uparrow 2 = 2^2 = 4$ .

Next,

$2 \upuparrows 3 = 2 \uparrow (2 \uparrow 2)$

$= 2 \uparrow (2^2)$

$= 2 \uparrow 4$

$= 2^4$

$= 16$

Next,

$2 \upuparrows 4 = 2 \uparrow (2 \uparrow (2 \uparrow 2))$

$= 2 \uparrow 16$

$= 2^{16}$

$= 65,536$

Next,

$2 \upuparrows 5 = 2 \uparrow (2 \uparrow (2 \uparrow (2 \uparrow 2)))$

$= 2 \uparrow 65,536$

$= 2^{65,536}$

$\approx 2.0035 \times 10^{19,728}$

Next,

$2 \upuparrows 6 = 2 \uparrow (2 \uparrow (2 \uparrow (2 \uparrow (2 \uparrow 2))))$

$= 2 \uparrow 2^{65,536}$

$= 2^{2^{65,536}}$

$\approx 10^{6.031 \times 10^{19,727}}$

We see that $2 \upuparrows 6$ is already far larger than a googolplex (or $10^{10^{100}}$ ), which is often (and erroneously) held as the gold standard for very large numbers.

I’ll refer the interested reader to a previous post in this series for a description of how logarithms can be used to write something like $2^{65,536}$ in ordinary scientific notation.

Knuth’s up-arrow notation can be further generalized:

$x \uparrow^3 y = x \upuparrows x \upuparrows x \dots \upuparrows x$ ,

repeated $y$ times. The numbers $x \uparrow^4 y$ , $x \uparrow^5 y$ , etc., are defined similarly.

These numbers truly become large quickly. For example,

$2 \uparrow^3 2 = 2 \upuparrows 2 = 4$ , from above.

Next,

$2 \uparrow^3 3 = 2 \upuparrows (2 \upuparrows 2)$

$= 2 \upuparrows 4$

$= 65,536$ , from above

Next,

$2 \uparrow^3 4 = 2 \upuparrows (2 \upuparrows (2 \upuparrows 2))$

$= 2 \upuparrows 65,536$

$= 2 \uparrow 2 \uparrow 2 \uparrow \dots \uparrow 2$ ,

where there are 65,536 repeated 2’s on this last line. It’d be nearly impossible to write this number in scientific notation, and we’ve only reached $2 \uparrow^3 4$ .

Lessons from teaching gifted elementary school students (Part 5b)

Here’s a question I once received (though I probably changed the exact wording somewhat):

Exponentiation is to multiplication as multiplication is to addition. In other words,

$x^y = x \cdot x \cdot x \dots \cdot x$ ,

$x \cdot y = x + x + x \dots + x$ ,

where the operation is repeated $y$ times.

So, multiplication is to addition as addition is to what?

My kneejerk answer was that there was no answer… while exponents can be thought of as repeated multiplication and multiplication can be thought of as repeated addition, addition can’t be thought of as some other thing being repeated. But it took me a few minutes before I could develop of proof that could be understood by my bright young questioner.

Suppose $y = 1$ . Then the expressions above become

$x^1 = x$

and

$x \cdot 1 = x$

However, we know full well that

$x + 1 \ne x$ .

Therefore, there can’t be an operation analogous to addition as addition is to multiplication or as multiplication is to exponentiation.

Lessons from teaching gifted elementary school students (Part 5a)

Here’s a question I once received (though I probably changed the exact wording somewhat):

Exponentiation is to multiplication as multiplication is to addition. In other words,

$x^y = x \cdot x \cdot x \dots \cdot x$ ,

$x \cdot y = x + x + x \dots + x$ ,

where the operation is repeated $y$ times.

So, multiplication is to addition as addition is to what?

Which then naturally led to my student’s next question, which I was dreading:

Can you prove that?

This led to another kneejerk reaction, but I kept this one quiet: “Aw, nuts.”

I suggested that $x + y$ can be thought of as starting with $x$ and then adding $1$ repeatedly $y$ times, but my bright student wouldn’t hear of this. After all, in the repeated renderings of $x^y$ and $x \cdot y$ , there’s no notion of starting with a number and then doing something with a different number $y$ times.

So I had to put my thinking cap on, and I’m embarrassed to say that it took me a good five minutes before I came up with a logically correct answer that, in my opinion, could be understand by the bright young student who asked the question.

I’ll reveal that answer in tomorrow’s post. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own.

15-year-old catches math error at the Boston Museum of Science

See the full article here: http://www.boston.com/news/local/massachusetts/2015/07/06/year-old-catches-math-error-the-museum-science/awhREamdn1KRg7nz2gGyPO/story.html?p1=Must_Reads

Is 2i less than 3i? (Part 4: Two other attempted inequalities)

In yesterday’s post, I demonstrated that there is no subset $\mathbb{C}^+ \subset \mathbb{C}$ of the complex numbers which satisfies the following four axioms:

If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1+z_2 \in \mathbb{C}^+$
If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1 z_2 \in \mathbb{C}^+$ .
For every $z \ne 0$ , either $z \in \mathbb{C}^+$ or $-z \in \mathbb{C}^+$ , but not both.
$0 \notin \mathbb{C}^+$

However, it’s instructive (and fun) to try to construct such a set. Yesterday I showed that the following subset satisfies three of the four axioms:

$\mathbb{C}^+ = \{x + iy : x > 0 \qquad \hbox{or} \qquad x = 0, y > 0\}$

Apostol’s calculus suggests two other subsets to try:

$\mathbb{C}^+ = \{x + iy : x^2 + y^2 > 0 \}$

and

$\mathbb{C}^+ = \{x + iy : x > y\}$

Neither of these sets work either, but I won’t spoil the fun for you by giving you the proofs. I leave a thought bubble if you’d like to try to figure out which of the four axioms are satisfied by these two notions of “positive” complex numbers.

Is 2i less than 3i? (Part 3: An inequality that almost works)

In yesterday’s post, I demonstrated that there is no subset $\mathbb{C}^+ \subset \mathbb{C}$ of the complex numbers which satisfies the following four axioms:

If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1+z_2 \in \mathbb{C}^+$
If $z_1, z_2 \in \mathbb{C}^+$ , then $z_1 z_2 \in \mathbb{C}^+$ .
For every $z \ne 0$ , either $z \in \mathbb{C}^+$ or $-z \in \mathbb{C}^+$ , but not both.
$0 \notin \mathbb{C}^+$

However, it’s instructive (and fun) to try to construct such a set. One way of attempting this is defining

$\mathbb{C}^+ = \{x + iy : x > 0 \qquad \hbox{or} \qquad x = 0, y > 0\}$

This set $\mathbb{C}^+$ leads to the lexicographic ordering of the complex numbers: if $z_1 = a_1 + i b_1$ and $z_2 = a_2 + i b_2$ , where $a_1, a_2, b_1, b_2 \in \mathbb{R}$ , we say that $z_1 \prec z_2$ if

$a_1 < a_2 \qquad \hbox{or} \qquad a_1 = a_2, b_1 < b_2$

I used the symbol $z_1 \prec z_2$ because, as we’ll see, $\prec$ satisfies some but not all of the usual properties of an inequality. This ordering is sometimes called the “dictionary” order because the numbers are ordered like the words in a dictionary… the real parts are compared first, and then (if that’s a tie) the imaginary parts are compared. See Wikipedia and Mathworld for more information.

In any case, defining $\mathbb{C}^+$ in this way satisfies three of the four order axioms.

Suppose . It’s straightforward to show that . Let and , where . Then , and so .
- Case 1: If $a_1 + a_2 > 0$ , then clearly $z_1 + z_2 \in \mathbb{C}^+$ .
- Case 2: If $a_1 + a_2 = 0$ , that’s only possible if $a_1 = 0$ and $a_2 = 0$ . But since $z_1, z_2 \in \mathbb{C}^+$ , that means that $b_1 > 0$ and $b_2 > 0$ . Therefore, $b_1 + b_2 > 0$ . Since $a_1 + a_2 = 0$ , we again conclude that $z_1 + z_2 \in \mathbb{C}^+$ .
Suppose , where . Then or . We now show that, no matter what, or , but not both.
- Case 1: If $a > 0$ , then $-a < 0$ , and so $z \in \mathbb{C}^+$ but $-z \notin \mathbb{C}^+$ .
- Case 2: If $a < 0$ , then $-a > 0$ , and so $-z \in \mathbb{C}^+$ but $z \notin \mathbb{C}^+$ .
- Case 3: If , then since . Also, if , then , so that and .
  - Subcase 3A: If $b > 0$ , then $-b < 0$ , and so $z \in \mathbb{C}^+$ but $-z \notin \mathbb{C}^+$ .
  - Subcase 3B: If $b < 0$ , then $-b > 0$ , and so $-z \in \mathbb{C}^+$ but $z \notin \mathbb{C}^+$ .
By definition, $0 = 0 + 0i \notin \mathbb{C}^+$ .

However, the fourth property fails. By definition, $i = 0 + 1i \in \mathbb{C}^+$ . However, $i \cdot i = -1 + 0i \notin \mathbb{C}^+$ .

Because this definition of $\mathbb{C}^+$ satisfies three of the four order axioms, the relation $\prec$ satisfies some but not all of the theorems stated in the first post of this series. For example, if $z_1 \prec z_2$ and $z_2 \prec z_3$ , then $z_1 \prec z_3$ . Also, if $z_1 \prec z_2$ and $w_1 \prec w_2$ , then $z_1 + w_1 \prec z_2 + w_2$ .