By Andrew Freda, posted August 31, 2015 —

A mathematician is an animal which turns coffee into theorems.—attributed to Paul Erdõs

What does it mean to prove something? This is a question that I ask my Geometry students often and in different contexts. Early in the year, we work through Euclid’s Proposition 1 from Book 1 of The Elements (see, Geometry and Euclid). As rigorous as that exercise seemed at the time, students are stunned to discover that Euclid falls short of modern standards for a mathematical proof. Specifically, he uses the intersection of two circles, but he gives no postulate that circles must intersect and that they must intersect in a point. I tell my students that I think Euclid is wonderful and rigorous, but my feelings and attitude can’t change what we mean by a mathematical proof. I heartily recommend showing the wonderful documentary on the proof of Fermat’s last theorem by Andrew Wiles; this is the single most vivid example of what the life of working mathematicians is like and what it means to prove something mathematically.

A SEQUENCE OF SQUARES (ONE PROBLEM WITH MANY TYPES OF PROOFS)

For a large square, with side length 4 units and successive internal squares, each made by connecting the midpoints of the outer square, show that the sum of the areas of all the squares is

16 + 8 + 4 + 2 + 1 + 1/2 + 1/4 + … = 32.

A Numeric Approach

• The largest square has sides of length 4, so its area is 16.

• The second-largest square has each side equal to the hypotenuse of the green triangle. Because the green triangle has sides of length 2, 2, and 2√2, the area of the square is (2√2)² = 8.

• The third-largest square is bordered by four isosceles right triangles with sides of length √2, √2, and 2, so its area is 2² = 4.

• The fourth-largest square, bordered by isosceles right triangles with sides of length 1, 1, and √2, has area (√2)² = 2.

• The fifth-largest square, bordered by isosceles right triangles with sides √2/2, √2/2, and 1, has area 1² = 1.

These values lead to the series above, in which the integers sum to 31. We should point out to students that we must find the area of each square; we have no proof that the next square’s area will be half that of the previous one. Taking for granted that the pattern will hold brings us to the next problem—how to find the value of the infinite geometric series given by the fractions. Once again, a pattern quickly emerges that the sum will be the fraction (2ⁿ – 1)/2ⁿ, which gets closer and closer to 1, but can we prove that it equals 1?

An Algebraic Approach

Let the largest square have side length a; then the sides for the green triangle (and the other three isosceles right triangles whose hypotenuses create the next smaller square) are given by a/2, a/2, and a√2/2. The square’s area equals (a√2/2)² = (2a²/4) = (1/2)a², which shows that each subsequent square has half the area of the previous square. Students should see that this generalization allows for an answer with the largest square of any length you wish. Factoring a² from the total of the areas leads us to the series above. To find the sum of the geometric series, let S be the sum and then subtract one-half S. The only term that will “survive” the subtraction is the first term, 1/2, so (1/2)S = 1/2 gives the value of S as one unit.

A Geometric Approach

Show that eight green triangles will create the largest square and show that four green triangles will create the second largest square. It follows that each successive square has half the area of the previous square, leading us to the series above. Again, students should note that this generalized proof does not rely on a numeric pattern. We must be careful and rigorous about proving the relation between the green triangle and the largest and next-largest squares.

The following example teaches us to be careful with diagrams.

THE MISSING SQUARE PARADOX (OR, BE CAREFUL WITH DIAGRAMS)

In the diagram, the same-colored polygons are congruent.

You may be tempted to surmise (incorrectly) that the top triangle is 13 squares wide and 5 squares high, so its area is (1/2) • 13 • 5 = 32.5, whereas the bottom triangle with the same width and height has one “missing” square, so its area is 31.5.

Why the missing square? The key is to see that the red and blue triangles have hypotenuses with different slopes—3/8 and 2/5, respectively. So we cannot form a straight line, a “true hypotenuse,” by having the red triangle touch the blue triangle.

This is a valuable paradox to share with our students. It shows them the importance of evaluating a diagram rather than taking it at face value. It reinforces the level of rigor that is needed for a mathematical proof. And it is great fun!

I recommend a YouTube video in which a chocolate bar is broken in a similar manner to create an infinite supply of chocolate squares—a wonderful dream for your chocolate-loving students.

ANDREW FREDA, afreda@deerfield.edu, teaches at Deerfield Academy in Deerfield, Massachusetts. He has spoken and written on a variety topics, including the importance of visual representations in algebra, teaching Geometry through stand-alone “units,” ways to understand probability through the use of a dice game, the use of fractals in the high school curriculum, and the role of CAS in the mathematics classroom.