MathNotations

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...

Posted by Dave Marain at 6:04 AM 3 comments

Labels: indirect proof, math contest problems, MathNotations Contest, number theory, open-ended, proof

There is still time to register for the upcoming MathNotations Third Online Math Team Contest, which should be administered on one of the days from Mon October 12th through Fri October 16th in a 45-minute time period.

Registration could not be easier this time around. Just email me at dmarain "at" "gamil dot com" and include your full name, title, name and full address of your school (indicate if Middle or Secondary School).

Be sure to include THIRD MATHNOTATIONS ONLINE CONTEST in the subject/title of the email. I will accept registrations up to Fri October 9th (exceptions can always be made!).

• Your school can field up to two teams with from two to six members on each. (A team of one requires special approval).
  
• Schools can be from anywhere on our planet and we encourage homeschooling teams as well.
  
• The contest includes topics from 2nd year algebra (including sequences, series), geometry, number theory and middle school math. I did not include any advanced math topics this time around, so don't worry about trig or logs.
  
• Questions may be multi-part and at least one is open-ended requiring careful justification (see example below).
  
• Few teams are expected to be able to finish all questions in the time allotted. Teams generally need to divide up the labor in order to have the best chance of completing the test.
• Calculators are permitted (no restrictions) but no computer mathematical software like Mathematica can be used.
  
• Computers can be used (no internet access) to type solutions in Microsoft Word. Answers and solutions can also be written by hand and scanned (preferred). A pdf file is also fine.
  

The following is a sample of the open-ended "proof-type" questions on the test:

Explain why each of the following statements is true. Justify your reasoning carefully using algebra as needed.

The square of an odd integer leaves a remainder of 1 when divided by
(a) 2
(b) 4
(c) 8

I may post a sample solution to this or you can include this in your comments to this post.

Posted by Dave Marain at 8:44 AM 0 comments

Labels: math contest, MathNotations Contest, open-ended, proof

HAPPY HOLIDAYS!

The following is a series of apparently straightforward arithmetic problems for middle schoolers. However, the objective is to have students justify their reasoning beyond "guess and test" methods. Proving there is only one solution or none requires more careful logic using algebra as needed. Students will need some basic algebra for the "proofs." For the younger student, modify these questions to have them find the squares in questions 1,2,3 and 5. Take this as far as you wish...


In the following, square refers to the square of an integer. Justify your reasoning or prove each of the following.

(1) There is only one square which is 1 more than a prime.

(2) There is only one square which is 4 more than a prime.

(3) There is only one square which is 9 more than a prime.

(4) There is no square which is 16 more than a prime.

(5) There is only one square which is 25 more than a prime.

(6) Can one generalize this or not??

Click Read More for selected answers, solutions...



Selected Answers, Solutions

(2) If n² is 4 more than some prime, p, then we can write
p = n² - 4 = (n-2)(n+2). Since p is prime, the smaller factor must be 1, so
n-2 = 1 or n = 3. Thus, there is only one square, 9, which is 4 more than the prime, 5.

(4) p = n² - 16 = (n-4)(n+4). There n would have to equal 5, n² would equal 25 but 25 - 16 = 9 is not prime.

(6) If there were a general rule would that mean we'd have a formula for primes?

Your thoughts about these questions...

...Read more

Posted by Dave Marain at 8:04 AM 4 comments

Labels: algebra, middle school, more, number theory, proof

Ok, so most normal people are not thinking about the significance of the digit '8' in 2008 the day before the Super Bowl. Sorry, but in this post there will be no predictions about the score, no 'over-unders', no boxes, no betting at all. You do have to admit that this is a great time for lovers of mathematics. People are actually interested in mathematical odds and chances of all kinds of weird number combinations occurring in the score on Sunday night. However, this post will focus instead on the number 8, the units' digit in 2008. The Super Bowl comments above will no doubt soon become outdated but the mathematics below will live on! Who knows, maybe the number 8 will turn out to have special significance on Feb 3, 2008? Remember, I said that here before the game!!

2008 is a special number for so many reasons, being divisible by 4 of course: Leap Year, Prez Election year, Summer Olympics and much more. In fact, 2008 is divisible not only by 4 but also by 8 itself. In the good ol' days, some students were even taught the divisibility rules for 2, 4 and 8:
Divisible by 2: If the 'last' digit is divisible by 2 (of course!)
Divisible by 4: If the number formed by the last TWO digits is divisible by 4
Divisible by 8: If the number formed by the last three digits is divisible by 8.

Let's demonstrate this for 2008:
2008 us divisible by 2 because 8 is divisible by 2
2008 is divisible by 4 because '08' is divisible by 4
2008 is divisible by 8 because '008' is divisible by 8

A little weird with those zeros and not particularly interesting, right? Anyone care to guess a rule for divisibility by 16? Interesting, but none of this is the issue for today....

BACKGROUND FOR PROBLEM/INVESTIGATION/ACTIVITY
Today, we are are interested in the squaresof numbers and their remainders when divided by 8. Notice that 4² is divisible by 8 but 6² is not. So we cannot say that the square of any even number is divisible by 8. What about the squares of odd numbers when divided by 8?
1² leaves a remainder of 1 when divided by 8
3² leaves a remainder of 1 when divided by 8
5² leaves a remainder of 1 when divided by 8
7² leaves a remainder of 1 when divided by 8

What is going on here? That's for your crack investigative team to decipher.

TARGET AUDIENCE: Our readers of course; Middle schoolers through algebra

PROBLEM/INVESTIGATION FOR READERS/STUDENTS
1. Discover, state and prove a general rule for the remainder when the square of an even number is divided by 8.
2. Discover, state and prove a general rule for the remainder when the square of an odd number is divided by 8.

Comments:
(1) These are well-known relationships and not very difficult questions. Just something to extend thinking about divisibility, remainders and the use of algebra to deduce and prove generalizations. Prealgebra students may be able to explain their findings without algebra!
(2) 'Discovering' or stating the rule for question (2) is transparent from the examples above. Instructors may prefer 'data-gathering' and making a table first. That is, have students develop a table for the squares of the first 10 positive integers and their remainders when divided by 8. Proving the result for the squares of odd integers is more challenging, even algebraically. Most will see the remainder when dividing by 4, but 8 is slightly trickier.
(3) Those who are more comfortable with congruences and modular arithmetic can approach these questions another way.

Posted by Dave Marain at 6:28 AM 4 comments

Labels: algebra, divisibility, investigations, proof, remainders

The results below are well-known but, as usual, I am offering an investigation for the classroom that has many objectives:

(1) Digit properties of multiples of 9 (and their 'proofs')
(2) Review place-value and algebraic representation
(3) Investigate patterns based on data collection
(4) Develop inference and conjecture
(5) Introduce students to algebraic proof
(6) And, of course, practice for those open-ended questions we've all come to know and love...

Children are often fascinated by the discoveries they can make regarding 2- and 3-digit numbers. At some point in middle school all students should either discover on their own or be introduced to the remarkable properties of the number 9 in our base 10 number system. The investigation below will explore some of this.

Students are also fascinated by the results of taking a 2- or 3-digit number and reversing its digits. With or without calculators, students like to see how these numbers are related, particularly when they are added or subtracted. In this activity, they will have the opportunity to discover some of these properties and use basic algebra to explain why they work. Perhaps, this will also lead to questions about palindromes, but that's for another day...

The questions below are designed for middle schoolers through Algebra 1. The proofs require some basic algebra, so you can make those parts optional for the prealgebra group. For this group, having them state their conjectures and suggesting possible explanations are more than enough.

STUDENT/READER ACTIVITY/INVESTIGATION

(1) List all of the 2-digit multiples of 9. What do you notice about the sum of their digits?
(2) Using the fact that any 2-digit number can be represented algebraically as 10a+b, show/justify/explain/demonstrate/prove the following:

If a 2 -digit number is a multiple of 9, so is the sum of its digits AND
if the sum of the digits of a 2-digit number is divisible by 9, then the number is a multiple of 9.

(3) If you made sense of (2), why stop with 2-digit numbers! State and prove a similar result for 3- and 4-digit numbers!

Now for reversals:

(4) To be a mathematical researcher, one needs to do what the scientific researcher does. Collect lots of data first, then make conjectures and PROVE them! Choose at least 5 different 2-digit numbers, in addition to the examples below, and complete the table.

Number..........Reversal............Sum..........Difference (Larger-Smaller)
41.....................14.......................55...............27
33....................33......................66................0
72....................27......................99................45
Your turn - do this FIVE more times.

(5) Make conjectures about the how the sum and difference are related to the digits of the original number. Using the algebraic representation 10a+b for any 2-digit number, PROVE your conjectures (or disprove them!).

(6) 72 and 27 are not only reversals. They are are also both multiples of 9. Does this have to be true for any 2-digit multiple of 9? Explain! Further, is there a special property for the sum of the number and its reversal in this case. Make sure you verify conjectures for several cases before attempting to prove it.

(7) Make a similar table for 3-digit numbers. Is there an obvious relationship for the sum of the number and its reversal this time? The difference? Make conjectures and PROVE them!

If you feel this activity is useful, please comment, share it and rate it below. Enjoy!

Posted by Dave Marain at 7:33 AM 3 comments

Labels: investigations, middle school, nine, number theory, patterns, proof

[You may also want to look at the preview of the interview with Alec Klein, author of A Class Apart, to be hosted on MathNotations. Alec has agreed to answer my questions about Stuyvesant HS in NYC, other specialized schools and gifted education.]


Please comply with the 'Proper Attribution' statement that now appears in the sidebar.

Anyone recall a detailed investigation I posted in Feb '07 about numbers which can or cannot be written as a sum of 2 or more consecutive positive integers? You will probably want to quickly review that for this discussion. That investigation was implemented in a 9th grade prealgebra class with students who had struggled with math for a long time. They worked for the entire period (and into the next class as well) and expressed satisfaction and a sense of accomplishment. One student, KC, even found a way to express those numbers which were unsummable!

Today, we will take this question of which numbers are unsummable to a higher level. The challenge for my readers and for students is to use methods from Algebra 2 and basic number theory (primes, factors) to prove a conjecture made by one of my former students.

Quick background:
Students were asked to investigate those positive integers which can be written as a sum of 2 or more consecutive positive integers. I started them off with examples like
3 = 1+2; 5 = 2+3; 6 = 1+2+3, etc.
They worked in pairs and completed a table up to 36 over a couple of days. Most quickly realized that every odd positive integer starting with 3 could be represented but not every even positive integer. Working in pairs helped students to catch common arithmetic/logic errors and the results were reviewed after every 10 numbers or so in order to insure that all students had accurate results to work with.

Here's your challenge for today:
Rather than demonstrate which numbers can be represented as such a sum, your mission is to prove the following:
Powers of 2 are unsummable, i.e., a power of 2 can never be represented as a sum of 2 or more consecutive positive integers.
Notes:
(1) The whole notion of algebraic proof may be new for some students, so this may require some demonstration first.
(2) Students will need to know the formula for the sum of an arithmetic series, so this challenge would be appropriate after learning or reviewing that. The instructor however could develop that formula earlier on or simply provide the formula.
(3) Some understanding of the Fundamental Theorem of Arithmetic is needed here, i.e., every positive integer is either prime or can be written as a product of primes in a unique fashion. This theorem often goes unmentioned or taken for granted in middle school - time to bring it back?
(4) Some readers may find a way to prove this without using the algebraic formula mentioned above. Share that as well!

Enjoy...

Posted by Dave Marain at 5:45 AM 25 comments

Labels: algebra 2, factors, powers of 2, proof, sums of consecutive positive integers

Two of the opposite vertices of square PQRS have coordinates P(-1,-1) and R(4,2). (a) SAT-Type: Find the area of PQRS.
SAT Level of difficulty: 4-5 (i.e., moderately difficult to difficult).
Note: For standardized tests, in particular, students are encouraged to learn the special formula for the area of a square in terms of its diagonal.


Now for a more significant challenge that can be used to extend and enrich. Students can work individually or in small groups:

(b) Explain why there is only one possible square with the given pair of opposite vertices. Use theorems to justify your reasoning. Would this also be true if a pair of consecutive vertices were given? How many rectangles, in general, are determined if 2 vertices are given (opposite or consecutive)?

(c) Determine the coordinates of Q and S, the other pair of vertices of the square.
Note: There are many many approaches here. Students often get hung up on the distance formula leading to messy algebra (with 2 variables). There are simpler coordinate methods. You may want to provide a toolkit for students here: Graph paper, Geometer's Sketchpad, etc. Students who estimate or 'guess' the coordinates must verify (PROVE) that these vertices do in fact form a square. Students who quickly 'solve' the problem should be encouraged to find more than one method. This is the only way they will expand their thinking!

And now another coordinate problem that can be solved by a variety of methods...

In right triangle PQR, with right angle at Q, the coordinates of the vertices are:
P(-p,0)
Q(0,8)
R(r,0)
Determine the value of the area of the triangle. Assume p and r are positive.

Notes: Students should again be encouraged to try both synthetic (Euclidean) and analytical (algebraic, coordinates) methods.

General Comment: Students often forget how powerful slopes can be when solving geometry problems by coordinate methods!

Posted by Dave Marain at 4:59 PM 2 comments

Labels: coordinate problems, geometry, investigations, proof, SAT-type problems, squares

Update as of 6-2-07: All solutions to the problem below are now posted in the Comments section. Also, read Eric Jablow's astute comments and his challenge to students!

The following problem is well-known, however it is an excellent exercise for middle schoolers and as a challenge for older students as well.
Target Audience: Grades: 6-12
Prerequisite skills: Basic understanding of fractions and simple operations
Develops: Logical thinking, Systematic Counting/Listing, Fraction Concepts, Structure of Mathematical Proof
Recommended Classroom Organization for this activity: Students working in groups up to 4.
Online Resource: Here's one of the best sites on Egyptian Fractions I have found on the web. The problem is discussed but not solved!

Introduction for student:
A unit fraction is defined as 1/n, where n is a positive integer greater than 1.
The number 1 can be written as a sum of 3 unit fractions in 3 ways:
1 = 1/2 + 1/3 + 1/6
1 = 1/2 + 1/4 + 1/4
1 = 1/3 + 1/3 + 1/3
No other ways (other than rearrangement) can be found using the following reasoning:
The largest of the 3 fractions could not be less than 1/3. Why?
If 1/3 is the largest fraction, then the larger of the remaining fractions could not be less than 1/3. Why?

Here's your challenge:

There are 14 ways to write 1 as a sum of four unit fractions of the form:
1 = 1/a + 1/b + 1/c + 1/d, where a ≤ b ≤ c ≤ d.
Make a list of all of these ways. Have fun! Uh, no calculators please!

Posted by Dave Marain at 6:10 AM 5 comments

Labels: egyptian fractions, fractions, investigations, middle school, proof, unit fractions

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

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?]






(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!]

Posted by Dave Marain at 11:21 PM 12 comments

Labels: circles, geometry, indirect proof, investigations, proof

Pi day is over, but it seems fitting to continue exploring. Archimedes did more than develop an approximation procedure for pi ! There are many excellent websites that explain the following in greater detail and discuss many more of Archimedes' theorems about parabolas and tangents. I attempted to draw a diagram using Draw in Word. It's crude but you'll get the idea. The object is to share this extraordinary piece of history of mathematics and have your students finish the proof that a light ray from the focus that strikes a parabolic surface is reflected in a ray that is parallel to the axis of the parabola. This is equally interesting in reverse: External light rays and other forms of electromagnetic radiation that are parallel to a parabola's axis are reflected to the focus, very useful for radar and other 'collection' devices.
Considering that Archimedes' proofs used only geometric properties makes his work even more astounding (now of course we can use coordinate geometry, calculus, etc.). This type of investigation is usually deferred to College Geometry courses, but I believe we can deliver it to motivated geometry, 2nd year algebra or precalculus students. If nothing else, it makes for a wonderful long-term project!

Ok, here goes...

In the diagram below, I've gone out of my way to make the reflecting ray NOT look parallel to the axis, even though we're trying to prove it is. This is to help students avoid assuming collinearity, when, in fact, that needs to be proved!

The two angles marked X are equal by a reflection principle (angle of incidence equals...). The two angles marked Y are equal because it can be proved that the tangent line at P is the perpendicular bisector of segment FP', where F is the focus and P' is the foot of the perpendicular from P to the directrix. I chose not to derive Archimedes' very subtle argument, but it is worth studying the proof. The proof starts by constructing the perpendicular bisector and showing that this line passes through P but no other point of the parabola, thus it is tangent. Alex Bogomolny's excellent and in-depth treatment (with java applets) of this topic (on cut-the-knot) is very worthwhile reading.

The student is being asked to prove that the reflecting ray is parallel to the axis. This is equivalent to showing that the line containing PP' and the reflecting ray are one and the same. The argument is straightforward, but students may want to continue learning more about the genius of Archimedes.

[Good luck copying this diagram (jpg). Some of you may find errors in my argument or an extremely simply argument for the parallelism, so pls share!!]

Posted by Dave Marain at 10:02 AM 0 comments

Labels: archimedes, geometry, logic, parabolas, pi, proof, tangents

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