evolution

The titles are Everyday Mathematics for grades K-5 and Math Thematics beyond that.

Superintendent Mike Mathes said the new math curriculum is recommended by the U.S. Department of Education.

“It’s not drill and practice, drill and practice,” Mathes said. “Not all (math) problems have the same path to a solution.”

Mathes said, for example, teaching how to determine the pitch of a roof can be taught in a calculus class or in a construction class.

District officials have said they expect some resistance from parents to the Everyday Math and Math Thematics materials.

“It’s a different way of doing math,” said Craig Carter, assistant director of curriculum. “But there is a lot of help online for students and parents.”

During the board’s February meeting, Carter said math scores have risen in every district that has adopted the curriculum. He added parental involvement is key to the curriculum’s online program.

The NCTM, of course, loves it too. I have a special place in my heart¹ for the NCTM.

A group calling themselves Mathematically Correct, with whose work I became familiar while a graduate student at Kansas State, gave the following grades to the curricula: 2nd grade got a C,², 5th grade got a C-,³ and 7th grade got a D.⁴

Let’s just say that in my short teaching career, I showed promise as an instructor of college-age and adult students, but that I was (ahem) not renowned for my ability to teach younger students in a grade-school classroom. I have a master’s degree in mathematics, so I feel qualified to make judgments as to whether or not one is covering a certain amount of it. So make of all this what you will.

1. near the CHUNK of ICE of my HATRED!!1!!1one! [↩]
2. ”Despite good coverage of some topics, it may be difficult to identify a situation where the use of this program is very appropriate. If expectations are high, then the program seems to be inappropriate due to the lack of support for the mastery of central topics. For situations with lower expectations, the program may contain too much attention to the higher-level topics and not enough attention to support success with addition and subtraction. Thus, it is difficult to recommend this program despite the circumstances.” [↩]
3. ”The program comes across with the flavor of a survey of some rather sophisticated areas of mathematics for fifth-grade students without support for the development of topics or student mastery of content. This unusual combination of features makes it difficult to imagine a fifth-grade circumstance where such a program could be recommended.” [↩]
4. ”This low rating reflects weakness in content, weakness in presentation, and weakness in student work as discussed immediately above and in each of the content topic reviews. It is not possible to recommend this book to anyone for any purpose.” [↩]

We talked about this before (here and here) but I do love to bore you with it.

Hot.

Interesting, and previously a topic of discussion on this site. The results cross “diverse backgrounds”, which is education-speak for racial and economic lines. From the blog of Scientific American:

Prior studies have shown that children as young as infants can judge simple mathematical relationships. When shown bunches of dots on a computer screen, for example, a preschool- or kindergarten-age child can tell that there are more dots combined in an image of 21 dots followed by one of 30 dots compared with a third 34-dot image.

A group including cognitive psychologist Elizabeth Spelke of Harvard University wondered if children could apply that ability, called nonsymbolic arithmetic, to Arabic numerals after learning to count but before they learned to add and subtract.

To find out, they gave several groups of children a laptop-based audiovisual test that asked whether one person had more or fewer candies or other objects than another person. The screen showed numbers to be added, such as 21 and 30, or subtracted, such as 64 and 13, followed by another number, such as 34, with which to compare the added or subtracted value.

The children answered correctly from 64 to 73 percent of the time, according to a report published online today by Nature. More affluent kids tested in the laboratory did better than their less well off peers tested in their classrooms, the group reports. The reason for the difference could be the testing environment, says Spelke, who adds that the important point is that kids from diverse backgrounds all showed the ability. “We never dreamed that you could simply give children the symbols and they will succeed,” she says.

Just imagine what we could teach them once we spent time developing that logic rather than spending it on math + “self-esteem building” exercises or cultural narratives.

I have confirmed that I did indeed pass the Society of Actuaries Course P exam. I received an “unofficial pass” at the time of the test, but the official results were posted today. I have not yet received my score, but for practical purposes that is purely a matter of pride.

In about two months it will be time to order my Course FM exam materials.

I’m studying for the first in the series of actuarial exams, so a topic that’s been top-of-mind for me lately has been probability theory. Scientific American has an accessible explanation of Bayes’s theorem, which relates to calculating conditional probabilities. Bayes’s theorem depends upon the Law of Total Probability.

The simplest example I can think of to illustrate the Law of Total Probability is the “Monty Hall” problem. It turns out that the old game show Let’s Make a Deal is the perfect illustration of the law. You know how it goes — Monty calls someone dressed up like Pippi Longstocking, a Hershey’s Kiss, or a galvanized-steel trash can up to the floor with him, and offers him/her a choice of what’s behind one of three doors. One of them has a valuable prize, and the other two have farm animals and boxes of Martha Gooch linguine.

The standard gambit is this: Ms Trash Can from Toledo chooses a door. Then Monty opens one of the other two doors — one without a prize, and offers her the chance to change her choice. The question is: how likely is Ms Trash Can to win if she switches?

There are two probability events in this situation. Let B represent the event of Ms Trash Can’s winning on her first choice. Either she will win a prize or not. Let A represent the event of Ms Trash Can’s winning on her choice after Monty’s offer. Again, she will win a prize or not. For some event X, let P(X) be the likelihood (or probability) that X will occur. This number will be between 0 (never) and 1 (certainty). Let X’ refer to the negative; i.e., X does not occur. By P(A|B) I mean the probability that A occurs, given that B has already occured.

What I want to know, in mathematical terms, is P(A). The Law of Total Probability says

I think even the laypersons among you can see where the name comes from. One must exhaust all the possible situations in which A occurs — both if Ms Trash Can wins or loses on her second choice. Let’s translate these symbols into the terms of our problem.

P(A|B) is the likelihood that Ms Trash Can will win on her second choice — given that she has won on her first. Well, if she won the first time she won’t make a second choice, making this likelihood 0.

P(B) is 1/3, since there are three doors and the prize is equally likely to be behind any one.

P(A|B’) is the likelihood that she will win on her second choice — given that she has chosen incorrectly the first time. She picked a door with no prize. Monty opened the other one with no prize. If she switches in this case, she is guaranteed a win, so this probability is 1.

P(B’) is the probability that she chose incorrectly the first time, which is clearly 2/3.

That means P(A) = P(A|B)P(B) + P(A|B’)P(B’) = 0 * 1/3 + 1 * 2/3 = 2/3, or a win in two out of three cases when she switches after Monty’s offer.

Seems counter-intuitive, no?

The article mentions a classic Bayes problem — the problem of a testable disease for which the test has a slight unreliability factor; i.e., the test shows positive for a certain percentage of non-afflicted patients and negative for a certain percentage of afflicted patients. Bayes’s theorem is used in conjunction with the Law of Total Probability to analyze such a situation.

Grigory Perelman turned down the Fields Medal, after having apparently proved the Poincare conjecture. Many think he’ll turn down the Clay Institute’s money — $1 million.

[inherited from: Seed Magazine.]

Some mathematicians think that Russian recluse Grigory Perelman has done just that. If so, that means the Clay Institute owes him a million dollars — which he has said he will refuse. It is also rumored that he will win the Fields Medal (mathematics’ equivalent to the Nobel Prize) when it is awarded later in the year.

So what is this conjecture? The original statement goes something like this: Any closed, simply-connected (= “having no holes”) three-dimensional space is homeomorphic (= “may be continuously stretched and bent”) to the equivalent of a sphere in four dimensions (the surface in four-dimensional coordinates where all points are equidistant from the origin).

Chop off a dimension and this is relatively easy to prove — Take some simply-connected “blob” in “two-space”, the “coordinate plane” from your high-school algebra class. It may clearly be continuously stretched and bent into the shape of a disk in two-space without creating holes. Then, one may identify it with the hollow sphere (and here I’m glossing over some serious machinery) in three-space. What this conjecture speaks of is this situation’s higher-dimensional analogue.

In fact, Perelman is believed to have proved something stronger — something known as the “geometrization conjecture”, which splits closed three-dimensional spaces (any closed three-dimensional space, even one with “holes”) into their connected parts and describes the geometry that holds within them.

This story — a bit old now; Perelman finished his work in 2002 — has much greater detail.

It’s time to begin down the path to real numbers, but we have some things to do first. Namely, we’ll construct the natural numbers using basic concepts related to sets. We shall assume the validity of the standard (Zermelo-Frankel with choice) set theory, which is known believed to be a consistent logical system as discussed previously. To enhance readability(!), I shall paper over some subtleties. I’ll assume that you know (at least intuitively) what a set is, and that you know what the union of two sets is (the collection of all elements in both sets; it helps to recall the Venn diagrams you did in school).

We are all apparently innately familiar with the natural numbers. So, we have some idea where we are going. Any model set we hope to build that does what we intend must accomplish the following:

1. Every member of the set must have a “successor”, i.e., given an element of the set, there must be a way to construct the elements that “come after”. These elements are also members of the set.
2. There is one element that is not the successor of any other element. We shall christen this element 0. (This is just a name, it need not correspond to what we know as zero. But, the only other number we know of that would work is 1.)
3. Elements with the same successor are equal.
4. Any subset of this model set which contains our 0, and given any element the subset also contains its successor, then this subset must in fact be the whole set. You may have heard this called the “induction principle”.

It has been proven that at least one set with these properties does in fact exist, assuming the validity of set theory.

Very well, let us plod on. Throughout, we denote the empty set ∅ and the phrase “defined to be equal to” by “:=”. Then, define:

• 0 := ∅
• 1 := {∅} — “a box containing an empty box”
• 2 := {∅, {∅}} = {0, 1}
• 3 := (∅, {∅}, {∅, {∅}}} = {0,1,2}

Let us define the successor S(a) of a element a to be a ∪ {a}. I claim that this construction satisfies the conditions outlined above (known as the Peano axioms).

All right — so how do you “add” two of these things together? Define an operation called “addition” and denoted by + as follows:

• a + 0 = a; and
• a + S(b) = S(a) + b

This is an example of a recursive definition; i.e. a definition that is applied repeatedly until one arrives at the first of these statements, and then unwound from there. For example, substitute 3 for a and 4 for b. One gets from the definition that 3 + 4 = 3 + S(3) = S(2) + S(3) = S(2) + S(S(2)) = S(S(1)) + S(S(S(1)), and so forth.

This operation we call “addition” is commutative; i.e., a + b = b + a for any choice of a or b. 0 is also what we call an “identity” element; by definition a + 0 = a for any a. Furthermore, 0 together with addition “generate” the entire set — remember, if the set contains an element, it also contains the element’s successor. (In jargon, this set with addition is a “free monoid on one generator”.)

You can define “multiplication” thusly: a * 0 = 0 and a * S(b) = (a * b) + a. This is also a recursive definition. You can test it if you like by substituting 3 for a and 2 for b. If you apply the definition correctly, you will get the expected result. (The set with multiplication is a “commutative monoid with identity 1.” The set with both operations together is a “commutative semi-ring”.)

With some effort it can be shown that these operations satisfy the expected properties of arithmetic, like the distributive law.

The mathematician Leopold Kronecker asserted that “the natural numbers are the work of God, all else is the work of man.” Here, we see that it is not required to assume their existence, although even infants have been shown to grasp the concept of “two” or “three”. They may be constructed from scratch out of a consistent logical system that does not assume their existence.

The question of whether this construction and the theory behind it expresses some fundamental concept inherent in our existence or is simply a word-and-symbol game is another matter entirely.

[References: Wikipedia, Kenneth Ross's accessible Elementary Analysis: The Theory of Calculus, which basically continues where I left off here. It had been a while since I thought about the ideas in this post! UPDATE: And thanks once again to David's fact-checking.]

In the comments on my post regarding the cardinality of the set of real numbers vs. that of the rationals, I have been caught in an act of laziness.

I originally wrote, after having proved that √2 was not rational:

We have demonstrated that there is a “real number” (whatever that is) that is not rational, so the set of “real numbers” is larger than the set of all rational numbers.

David catches me in an act of laziness:

You prove that the square root of 2 is not rational and then state that the reals must be of a higher cardinality than the rationals because of that (I reckon that you mean that by saying: “the set of ‘real numbers’ is larger than the set of all rational numbers”). But isn’t this argument of the same kind as ‘proving’ the cardinality of the integers bigger than that of the naturals by showing that 1 - 2 is not natural?

Indeed it is. What I had intended is to get those of you without mathematical training (which I assume is most of my readership) to make the intuitional leap to the fact that the reals contain the rationals, i.e., rational numbers are also real. That was not clear, and I should have made it so.

In fact, all I actually proved with any rigor is that the set of rational numbers and the set of real numbers are not equal; I demonstrated a real number that is not rational.

To prove that the set of rational numbers is indeed smaller in cardinality than the set of real numbers, one needs to do two things:

1. Prove that the rational numbers are countable, and hence may be placed into one-to-one correspondence with the natural numbers. This is done by mapping the fraction (lowest-terms or not) m/n to the triplet (m, n, a), where a is 0 or 1 depending on whether m/n is positive or negative. The set of all triplets so formed is clearly countable, being the product of two countable sets (the set of natural numbers, counted once for the m variable and again for n) with a finite set ({0,1}). It is then easy to show that this is a one-to-one correspondence;
2. Prove that the rationals are contained in the reals. This is intuitively obvious, but to prove it requires the construction of the reals and of the rationals, neither of which we’ve done; and
3. Prove that the reals are uncountable. Georg Cantor did this in two different ways, one of which is sheer genius in its simplicity: Convert each real number to its binary representation and apply Cantor’s diagonal argument.

This collection of arguments shows that the rationals are countable and that the reals are uncountable. We still have to show that the rationals are contained in the reals to complete our program, but we shall have to construct both of these sets to do it.

So, there is far more to it — but I attempted to paper it over with a mis-worded appeal to intuition. Mea culpa.