I recently went through this process myself when I got a particular mathematical ant in my pants. I am currently reading a novel that made a mention of the Pythagorean Theorem in passing. The character in the novel had such natural mathematical talents that the Theorem made immediate intuitive sense to her. Reflecting on this savant-like brilliance, I tested my own understanding of the Theorem. Almost everyone who has studied geometry remembers Pythagoras's most famous equation, "A squared plus B squared equals C squared." If you were a good enough math student, you also remember that this equation describes a natural relationship between the three-sides of any triangle that has one corner at an angle of 90-degrees a.k.a. a right triangle. My problem was I couldn't remember how to prove the Pythagorean Theorem. This seems especially silly to me because Pythagoras originally demonstrated his proof about 2500 years ago.
In mathematics it is never enough to simply trust that an equation or rule works, you must be able to prove it. You can't just demonstrate it inductively either. To prove Pythagoras it would not be acceptable to show that 3^2 + 4^2 = 5^2 (feel free to pause for a second and check that arithmetic if you want) and a triangle with sides measuring 3, 4 and 5 units would have a right angle. Mathematicians don't take a rule that applies to a handful of examples and say, "I assume it works like that all the time." An effective proof takes established rules and applies them to a generic example so as to create a universally applicable statement.
Lacking the ability to leave questions unanswered, I took my ignorance as an opportunity to test my mathematical thinking. I also decided to write down my process so that I could "unpack" the logical process necessary to prove something in math. Because I have blogged about education topics in the past, I though some segment of my readers might take an interest in unraveling the bundle of thoughts involved. By making my geometric contemplations explicitly clear and simple, I hope to help fellow educators find ways to help their students think through math problems.
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| An illustrative triangle |
I began by noting that for any triangle the angles of the corners must add up to exactly 180 degrees. This makes intuitive sense, because you can imagine a triangle as a path that takes three turns to point you in the opposite direction. As we all know, a reversal of direction is 180 degrees -or pi radians if you prefer.
Next I recalled the equation for calculating the area of a triangle, Area= 1/2*base*height. You can see how this works with a simple thought experiment. If you take any triangle and add a conjoined copy like a reflection across one of the sides, you have quadrilateral (i.e. a 4 sided shape) with an area equal to the triangle's base time's the triangle's height. Naturally the area of the original triangle is half the area of this new quadrilateral or 1/2*base*height.
The last fragment of knowledge I drew on was a little more advanced. I knew that there was a relationship between the size of an angle at the corner of a triangle the length of the side opposite to it. This is the basis for those trigonometric functions that probably traumatized most people in geometry class, sine, cosine, and tangent. This makes intuitive sense when dealing with a right triangle. The largest side, called the hypotenuse, is always opposite of the 90 degree angle (which must be, by definition, the largest angle in the triangle).
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| An example of a 45-45-90 triangle |
This is useful because it establishes that in a right triangle with two equal angles of 45 degrees each, a 45-45-90 triangle, the two sides other than the hypotenuse must be equal because they are opposite equal angles.
The reason I focused on the example of the 45-45-90 triangle is because it has some special properties. In the same way we could imagine a quadrilateral made out of two identical triangles, we can create one square out of four identical 45-45-90 triangles. Just picture the hypotenuse as one side of the square with the four 90 degree angles joined in the middles.
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| One square which contains four 45-45-90 triangles |
We also know that the total area of the square is the same as the combined area of the the four triangles. As discussed above the area of a triangle is given by the equation area = 1/2*base*height. You multiply the area of one of those triangles by 4 and you get this:
4 * area = 4 * (1/2*base*height) = 2*base*height = total area of square
Since the base and the height, which we can call side A and side B, are both the same length we can simplify the above equation for the square's area.
2*base*height = 2*side*side = 2 (side^2) = side^2 + side^2
We can substitute side A or side B in for either side length in this equation. That allows us to rewrite the area of the square as Area = A^2 + B^2 = C^2. This proves the Pythagorean Theorem for all 45-45-90 triangles.
I started with proving the Theorem for 45-45-90 triangles, because I knew they had special properties that made them easier to work with. Unfortunately you can't create a simple square out of every right triangle. Still, knowing that I had worked the proof for one class of triangles, gave me an insight for working a broader proof. In order to prove Pythagoras correct for all triangles, I had to find a more sophisticated method that I could apply to any type of right triangle. I was able to work from the foundation of my earlier exercise by searching for ways to create squares using the sides of triangles.
My breakthrough came when I realized that any right triangle can be used to define a square area. (This was important because, as I would learn later, earlier generations had used this as the basis for other proofs of Pythagoras.) You can form a square by arranging four identical copies of so that their right angles form the corners of the square.
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| The square defined by the combines sides of four copies of a triangle |
Area = (A+B) * (A+B) = A^2 + 2AB + B^2
When you arrange the triangles in this manner, the hypotenuse of each triangle forms the side of another smaller square with the square discussed above. Since all the sides of this square have the length of the hypotenuse, which we call C, you can define the area of this square as C^2. The area of the square with sides of A+B can be redefined as the area of the hypotenuse square plus the area of the four triangles. Each of the triangles has an area of 1/2*A*B.
Area = C^2 + 4(1/2*A*B) = C^2 + 2AB
We know from our earlier definition of the larger square's area that the area is equal to A^2 + 2AB + B^2. Therefore
Area = A^2 + 2AB + B^2 = C^2 + 2AB
Because the two quantities are already equal, we can perform any mathematical operation to both sides and they will still be equal. In this case we can subtract the quantity 2AB from both sides because in exists in both expression. This leaves us with A^2 + B^2 = C^2. This proves the Pythagorean Theorem for all right triangles.
Just for clarity, I do not recommend any math teacher tries to explain the Pythagorean Theorem using my meandering thought process. Instead I hope teachers of all kinds appreciate that though my method may not seem perfect it achieved the desired result. The purpose of unpacking is often to find the twists and kinks in a person's reasoning, but it can also demonstrate that differing methods of thinking can be acceptable. Students need to know that there is no one right way to go about thinking. Every teacher should encourage students to unpack their thinking, so they can see that -however they go about reasoning- they can always find ways to sharpen their thinking. You can always improver your problem solving even if it means solving a problem 2500 years too late.
