THE OPEN-ENDED CONTEST PROBLEM AND SOLUTIONS

As promised, here is the open-ended, rubric-based, holistically scored, performance-assessed, student-constructed first problem from MathNotation's Third Contest:

1. A primitive Pythagorean triple is defined as an ordered triple of positive integers (a,b,c) in which a² + b² = c² and the greatest common factor (divisor) of a, b and c is 1. If (a,b,c) form such a triple, explain why c cannot be an even integer.

Comments
(a) The content here is number theory. Is some of this covered in your district's middle school curriculum or beyond? More importantly, at what point do students begin to formulate and write valid mathematical arguments?

(b) The immediate reaction of most students was that this seemed like a fairly simple problem. However, only a couple of teams scored any points. Perhaps the challenge here was the construction of a deductive argument, although as you will see below, there is one challenging part.

(c) There were two successful approaches used by the teams. Both involved indirect reasoning. Do your students begin to do these in middle school or are "proofs" first introduced in geometry?

(d) I allowed students to assume without proof the following:

(i) The general rules of parity of the sum of two integers
(ii) The square of a positive integer has the same parity as the integer

(e) Interestingly, none of the teams considered an algebraic approach to the one challenging case, i.e., demonstrating that the sum of the squares of two odd integers is not divisible by 4.

If a and b are odd, they can be represented as
a = 2m+1 and b = 2n+1, where m and n are integers.
Then a² + b² = (2m+1)² + (2n+1)² =
(4m² + 4m + 1) + (4n² + 4n + 1) =
4(m² + n²) + 4(m + n) + 2, which leaves a remainder of 2 when divided by 4.
BUT, if c is even, say c = 2k, then c² = 4k², which is divisible by 4.

(f) The two best solutions came from our first and second place teams, Chiles HS in FL and Hanover Park Middle School in CA. Both used the ideas of congruence modulo 4.

Here is the indirect method used by Chiles:

Let's assume that c can be an even integer. We'll prove by contradiction. An even integer can be summed in two ways:
1. with two even integers or
2. two odd integers
If it is the latter case, then looking at the residuals of modulo 4, the two odd integers summed will be equal to 2, but this is not the case as 2 is not a modulo of 4 residue. If it is the former case, then it does not satisfy the problem as then a, b, and c have common factor of 2. Therefore c must be an odd integer. Q.E.D.


Here is the indirect method used by Hanover Park:

Suppose, for the sake of contradiction, that there is a PPT (primitive Pythagorean Triple) s.t. c is even. Then c² ≡ 0 (mod 4).

We break this into cases based on the parity of a,b.

Case I: Both a and b are even; gcd(a,b,c) ≥ 2 because a,b,c are even, a contradiction.

Case 2: One of a and b is even. Then, a² + b² ≡ 0 + 1 ≡ 1
not ≡ 0 (mod 4), a contradiction.

Case 3: Both of a, b are odd. Then a² + b² ≡ 1 + 1 ≡ 2
not ≡ 0 (mod 4), a contradiction.
We have covered all cases for a, b with no valid cases. Thus, in a PPT, c cannot be even.

Both of these arguments represent a more sophisticated understanding of mathematics and the methods of proof. Clearly, these students are quite advanced and exceptional, however, I feel many middle school teachers begin early on to encourage their students to explain their thought processes both orally and in writing. Am I right? I would like to hear your thoughts on this...

0.99999.. equals 1: Oh no, not another 'Proof!'

For the remainder of this post, the statement 0.99999... = 1 will be denoted by S.

Over the course of my math education and my professional teaching career, S has occupied considerable time and provoked much thought on my part and reflection among my students, countless mathematicians and, now, the math blogosphere (see Polymathematic's famous series of posts!). Sane individuals (aka, non-mathematicians) remain skeptical about S, unwilling or unable to grasp the equality in the statement.
They argue: "0.99999... gets closer and closer to 1 but how can you say it EQUALS 1. There's always a gap!" Ah, the mystery of limits!

For many years now, I have been posting my 'proof' of S on various listservs, discussion groups (including MathShare, the one I moderate) and blogs. Here's the reaction I 've generally received: ____________________
That's right - silence. Because I like to put a positive spin on things, I take that to mean no has found a way to refute it! I've even occasionally heard a student say that this convinced her/him.

I don't want to bore the veterans out there who've heard and read all of the well-known arguments, most of which have 'holes' in them (or should I say, discontinuities!). Even using the basic formula for the sum of an infinite geometric series doesn't necessarily satisfy the Odd Thomases (sorry, I'm a Dean Koontz addict) who will continue to question the validity of the statement.

Any attempt to justify S necessarily requires (to paraphrase Liping Ma) a profound understanding of fundamental principles regarding the real number system and my argument is no different.

Enough already -- Here it is:

Non-Rigorous Explanation: If 0.9999... is less than one, then there must be a decimal between it and 1. But this is impossible!

Rigorous Explanation:

Step 1: Consider the sequence: 0.9, 0.99, 0.999,...
Since this is an increasing sequence of real numbers bounded above by 1, this sequence has a limit, L, namely its least upper bound. As many of you know, I am using the Completeness Axiom for the Reals (known by other names). An excellent reference for the axiomatic structure of the real number system can be found here.

This demonstrates that 0.99999... does exist (i.e., it is a real number). Thus,
0.99999... is the limit L of the above sequence. Verification of the existence of 0.99999... is what is often lacking in other demonstrations of S.

Step 2: L is either greater than 1, equal to 1 or less than 1. We need only consider the last 2 cases.

Step 3: Reasoning indirectly, assume that L<1. By the density property of the real numbers, there must exist at least one real, x, between L and 1. Since L is different from x, it must differ from it in some decimal place. The tenths place? No! Since x is less than 1 and greater than L, it must have 9 in the tenths place. The hundredths place? No, again for the same reason. Need I continue or do you see we've reached a contradiction? Therefore, our assumption that L is less than 1 is false. Thus, L = 1 or, equivalently, 0.99999... = 1. QED!

Ok, your turn! Feel free to critique the proof or present your own favorite argument for or against S. Also, would you consider using this type of argument when teaching this topic?

A Simple Geometry Proof or Circular Reasoning??

[Update: Partial solution now in the comments. Fascinating discussion taking place about this innocent-looking problem...]

Figure 1

A standardized test sample problem I saw this evening, got me to thinking, which is often very dangerous. I will pose this 'open-ended' problem in terms of Jake making an assertion and Jack trying to convince him he's wrong. Jake keeps arguing and so does Jack. Who will win the argument logically?

Jake shows Jack a piece of wood he cut out in the machine shop in the shape of a circular arc bounded by a chord (See Figure 1 above). Jake claimed that the arc was not a semicircle, and, in fact, he claimed it was shorter than a semicircle, i.e., segment AB was not a diameter and arc ACB was less than 180 degrees. Jack knew this was impossible and argued: "Don't you see, Jake, that O must be the center of the circle and that OA, OB and OC are radii!" Jake wasn't buying this since he measured everything precisely. He argued that just because they could be radii didn't prove they had to be!

Here's your challenge for today:

(a) Find at least THREE different ways to PROVE that Jake is wrong, i.e., AB, in Figure 1, must be a diameter and O is the center. [Note: Can one assume perpendicularity?]

Figure 2

(b) Assume, in Figure 2 above, that PQ is a diameter, O is the center, chord AB is parallel to PQ and radius OC ⊥ AB. Determine the length of segment CD, and in particular, show that CD < DB.

(c) Refer to Figure 2, but let's generalize by removing the numerical values. Prove, in general, that for any chord AB parallel to and above diameter PQ, CD < DB.
[Pls note: We are no longer requiring that the length of chord AB be half of the diameter!]