the isis problem and pre-service teachers’ ideas about proof brian greer, dirk de bock and wim van...

The Isis problem and pre-service teachers’ ideas about proof

Brian Greer, Dirk De Bock and Wim Van Dooren

The Isis problem

What rectangles with integral sides (in some unit) have area and perimeter numerically equal? Prove the result.

Solutions: “Isis rectangles”

4

4

Perimeter: 16

Area: 16

3

6

Perimeter: 18

Area: 18

The Egyptians relate that the death of Osiris occurred on the seventeenth (of the month), when the full moon is most obviously waning. Therefore the Pythagoreans call this day the "barricading" and they entirely abominate this number. For the number seventeen, intervening between the square number sixteen and the rectangular number eighteen, two numbers which alone of plane numbers have their perimeters equal to the areas enclosed by them, bars, discretes, and separates

them one from another... (Plutarch, quoted by Davis and Hersh, 1981)

Why is this problem interesting?

• It is very simple to prove the solutions; the interest lies in the variety of forms of argument that can be used in alternative approaches

• It is accessible to a wide age range of age (“developmental proof”) and technical mathematical knowledge

• It provides an instrument for probing students’ ideas about proof

• It is connected with deep principles regarding dimensionality

• It has fascinating extensions

Structure of this presentation

• Variety of approaches leading to proofs

• Experimental study with future teachers

• Extensions and pedagogical possibilities

Empirical approach

With squared paper, young children can explore the problem by just generating examples.

Even more simply, small children could be given tiles and strips and asked to find rectangles where the number of tiles equals the number of strips.

•A more systematic approach would be to construct a table showing “Area – Perimeter” .

length

width

1

2

3

4

5

6

7

1 2 3 4 5 6 7

-4 -4 -4 -4 -4 -4 -4-3 -4 -5 -6 -7 -8 -9

-5 -4 -3 -2 -1 0 1

-9 -4 1 6 11 16 21

-6 -4 -2 0 2 4 6-7 -4 -1 2 5 8 11

-8 -4 0 4 8 12 16

• Empirical investigations are good for generating conjectures.

• The more difficult part is moving from the conjecture, via systematization of the cases considered, to an argument that is the basis for a proof.

yy

1xx

Area increases by yPerimeter increases by 2Area – perimeter increases by y - 2

length

width

1

2

3

4

5

6

7

1 2 3 4 5 6 7

-4 -4 -4 -4 -4 -4 -4-3 -4 -5 -6 -7 -8 -9

-5 -4 -3 -2 -1 0 1

-9 -4 1 6 11 16 21

-6 -4 -2 0 2 4 6-7 -4 -1 2 5 8 11

-8 -4 0 4 8 12 16

The table:

• shows the solutions

• is rich in patterns

• provides the basis of a rigorous proof

• suggests the insight that xy increases faster than 2x + 2y, an example of a fundamental principle about dimensionality

Algebraic approaches

A person with a training in formal algebra is likely, as a matter of routine expertise, to write the equation:

xy = 2x + 2y

The question, then, is how to manipulate this expression to find the integer solutions.

There are two next steps based on routine expertise that might be taken.

Option 1.

“Express one variable as a function of the other”:

y = 2x/(x-2) = 2 + 4/(x-2)

One may know as routine expertise that this is the equation of a hyperbola (or you may have access to graphing software). From here, it is easy to see the solution and create a formal proof…

Option 2.

“Move everything to the left-hand side”:

xy - 2x - 2y = 0

Now think of “completing the rectangle” by analogy with “completing the square”, thus:

xy - 2x - 2y + 4 = 4

The expression on the left now factorizes:

(x - 2)(y - 2) = 4

Since x - 2 and y - 2 are whole numbers (if x and y are) then the only possibilities are 1 x 4, 2 x 2, 4 x 1, from which the result follows.

There are infinitely many other ways to rewrite xy = 2x + 2y

The “trick” is to find useful ways.

For example, it might seem of little sense to rewrite as:

yx + xy = 4x + 4y

And yet…

And yet …

If you flexibly think of the y in yx and the x in xy as coefficients rather than variables, it is clear from the equation:

yx + xy = 4x + 4y

that y and x cannot both be greater than 4 … and so to a proof

Or even ….

If you know about the harmonic mean, you will see that this says that the harmonic mean of x and y is 4. Hence, either they are both 4, or one is greater than 4 and the other less. A proof follows easily by checking cases.

xy = 2x + 2y

= 41/x + 1/y

2

Or ….

xy = 2x + 2y

1/x + 1/y = 1/2

1/x + 1/y = 1/2

This form of the equation brings us back to the Egyptians via unit fractions.

From the equation, either 1/x and 1/y each equal 1/4, or one is greater and the other less than 1/4. Again this quickly leads to a proof by considering the small number of cases.

If x = y, then it is easy to see that x = y = 4 is the only solution of 2x + 2y = xy.

If x ≠ y, assume that x < y. Then 2x + 2y < 4y, so if xy = 2x + 2y, xy < 4y, whence x < 4 (and so on).

1/x + 1/y = 1/2

xy = 2x + 2y

y = 2 + 4/(x - 2)

(x - 2)(y - 2) = 4

xy - 2x - 2y = 0

= 41/x + 1/y

2

yx + xy = 4x + 4y

Geometrical solutions

Idea: decompose a plane figure in triangles and/or squares that equally contribute to the area and to the perimeter of that plane figure.

2

2

2

2

22

2

2

x

y

And now for my favourite!

“thick perimeter”

Area = G + WPer. = G + 4If Area = Per.W = 4

x

y (x - 2)(y - 2)

Structure of this presentation

• Variety of approaches leading to proofs

• Experimental study with future teachers

• Extensions and pedagogical possibilities

Study with future teachers

• Group 1: Future middle school teachers in a class at a West Coast US urban university (N = 9).

• Group 2: Future lower high school teachers in two classes in Belgium (N = 23).

• Group 3: Future upper high school teachers in two classes in Belgium (N = 16) who already had a degree in mathematics or were expected shortly to complete it.

Part 1

Solve the problem, looking for more than one solution

Part 2

Study the five proofs, rank them in order of quality (not defined!) and comment on them

Group 1 Group 2 Group 3__________________________________________________

C P C P C PGraph 1 3Factorization 3Tiles 4Divisibilitya 1 1 4 2Exhaustionb 2 2 3Other 1 5__________________________________________________Total 0 1 4 3 19 5

a This refers to arguments based on 2x being divisible by x - 2, the natural number x being one dimension of the rectangleb This refers to arguments considering all the possibilities for 2x/(x - 2) being a positive integer

Watch this guy!

A Belgian student, Xander Verbeke, produced 5 proofs, all clearly and fully argued. Besides the factorization, tiles, and divisibility proofs, he produced two others.

Xander’s fourth proof

In the quadratic equation X2 - cX + 2c = 0 with roots x and y

xy = 2c and x + y = c

so xy and 2(x + y) are equal!

For x and y to be natural numbers, c2 - 8c must be a perfect square.

For what values of c is c2- 8c a perfect square?

Xander’s fifth proof

Let the sides be a and a + x, and see what happens …

(Can also let the sides be a and ka, or 2m.a and 2n.b where a and b are odd numbers…)

Proof evaluations

The students ranked the five proofs (factorization, tiles, unit fractions, graph, table) from worst (1) to best (5).

What is meant by “worst” and “best” was left open.

0

0,5

1

1,5

2

2,5

3

3,5

4

4,5

5

Factorization Unit fractions Tiles Table Graph

Group 1 Group 2 Group 3 Total

Comments on the proofs

• Preference of most students in Groups 2 and 3 for the algebraic proof by factorization

"In my view, the factorization proof is the best because it is the least intuitive. Every step is mathematical"

"The proof by factorization … is very tight, mathematically correct and easy to follow. Moreover, no tricks are used; the problem becomes easier by adding 4 on both sides".

• Rejection of experimentation

(a)Low ratings for table proof (in fact, rejection of it as a proof in some cases).

(b) Confusion between proof by exhaustion and “trial and error”.

“Proof by means of a table succeeds in this case because there are few possibilities that have to be considered. In general, an ‘enumeration’ is not a good technique. In fact, it is not a ‘nice’ proof”

• Ambivalent reactions to tiles proof

“Proof with tiles: This is a better proof because it is clear and from the beginning till the end, it is neatly reasoned. Nevertheless, I miss some equations and it is a rather intuitive proof” (Xander)

“This is a very nice proof: fast, it is not necessary to know “real” mathematics. On the other hand, it is very much focused on the concrete problem. It is ad hoc, not immediately generalizable to other problems”

• Emotional and aesthetic reactions

“Checking by trial and error which number can and which cannot [work] and I do not find this pleasant. It is indeed a proof, but it doesn’t look like it”.

“The factorization is very simple, clear and beautiful”.

“The proof with the tiles comes over as a little bit playful”

“The proof with unit fractions is far-fetched”

• Proofs that convince logically v proofs that illuminate

“The proof with the tiles is the most visual one: you are not only convinced about the truth of the judgment, you also get the feeling that you ‘see’ why it is so”.

Structure of this presentation

• Variety of approaches leading to proofs

• Experimental study with future teachers

• Extensions and pedagogical possibilities

Extensions

• To solids (cuboids, …), with area = volume

• To other plane figures: triangles, circles, polygons

Dimensionality

xy ‘grows’ faster than 2x + 2y

In particular, if x en y both double, then xy increases by factor 4, while 2x + 2y only doubles.

At a time when many scholars are bemoaning the decreasing attention to problem solving, proof,and proving in the school curriculum (e.g. Hanna, 2007), the problem also has many aspects thatsuggest that it would be a powerful teaching tool. These aspects include:

1. Showing how systematicity, and then proof, can emerge out of empirical explorations.

2. Providing practice in the flexible rewriting of algebraic equations.

3. Showing the connections between different forms of mathematical representations, as in thecase of the algebraic and geometric forms of (x – 2)(y – 2) = 4

4. Illustrating a wide range of forms of argument and proof, including proof by exhaustion, andreductio ad absurdum.

5. Demonstrating clearly the intimate relationship between proof, problem solving and representations(Davis, 1993).

6. Exemplifying the heuristic that Polya (1945) identified as “think of a similar problem.”

Examples include:(a)Walter (in Walter & Klamkin, 1986) reported a form of the

divisibility proof offered by two mathematics faculty who suggested that they may have thought of it because they had recently been teaching techniques of integration for functions such as 2x/(x – 2) in which division may be usefully applied.

(b) Greer (1993) suggested an analogy between completing the square (as in solving a quadratic equation) and “completing the rectangle” by adding 4 to xy – 2x – 2y.

(c) Xander’s fourth proof (see above) may have been inspired by the realization that the terms x + y and xy occur together as the coefficients of a quadratic expression with roots x and y.

7. Opening up discussion on problem generalization and problem posing.

8. Drawing attention to the long history of mathematics, as shown indeed by the name given to the problem.

………..