The Physicist's World

A Proof of the Pythagorean Theorem

Introduction

The following proof of the Pythagorean theorem is an adaptation of a proof given in From Pythagoras to Einstein by K. O. Friedrichs, Chapter 1. The proof uses geometric intuition and approach, however, it relies on a modern algebraic interpretation of geometric objects. Figure 1 depicts the theorem to be proved. The quadrilaterals labeled A, B, and C in the figure denote areas and we want to show that A + B = C by areas. Using modern algebraic techniques, we know that the areas of A, B, and C are, respectively, a*a, b*b, and c*c, where a, b, and c denote lengths. It is common to refer to the square A, or the side a, however, one must remember that strictly speaking A denotes an area and is not a name for the quadrilateral and similarly, "a" denotes a length and is not a name for the side.

Traditional Euclidean geometry until the time of Descartes (c. 1630) identified geometric entities as magnitudes and referred to them by name, such as a list of names of corner points for a square. They did not allow "non-homogeneous" mixing of magnitudes of different dimensions. For example, magnitude a*a was a product of linear magnitudes, resulting in a linear magnitude which could never equal the magnitude of a planar object. Consequently, denoting the area of the square A by a*a would make no sense.

Assumptions

The following propositions are assumed in the proof below. A proof relies on commonly agreed upon truths and the most significant assumptions are usually stated. Typically many "obvious" assumptions are not stated.

  1. In a right triangle, one angle is a right angle and the other two angles sum to a right angle.
  2. Two right angles combine to form a straight line, and cannot combine to form any other angle.
  3. If three angles combine to a straight line and two of the three angles sum to a right angle, then the third angle must be a right angle. (Follows from assumptions 1 and 2)
  4. A quadrilateral is a rectangle if the corners are right angles.
  5. A rectangle is a square if the sides are all equal in length.
  6. The area of a rectangle with sides of length a and b is a*b.
  7. The area of a square with sides of length a is a*a
  8. The area of a triangle with base b and height a is (a*b)/2.

Proof of the Pythagorean Theorem

Figure 1 depicts the theorem to be proved. We must show that the area A plus the area B equals the area C.

  1. Construct the square in Figure 2. First arrange the squares A and B so that the adjoining corners of A and B form four right angles. Next complete the larger square by extending the outer edges. The larger square consists of the squares A and B plus two rectangles each with sides a and b. The total area of the larger square can be described in two ways.
    (1) (a+b)*(a+b)
    (2) a*a + b*b + 2*a*b

  2. Construct the square in Figure 3. Arrange four triangles each with sides a and b to adjoin at corners with the long leg (length b) to the left of the short leg and with the outer adjoined edges lined up straight to form a quadrilateral. The large quadrilateral is a square because the four corners are all right angles and the sides all have the same length (a+b). The enclosed area shown by the dotted lines forms a square with sides of length c. To see this, first consider the corners of the enclosed area which coincide with the points on the side of the larger square where the triangles adjoin to form three angles. The inner angle must be a right angle because the other two angles sum to a right angle and all three angles combine to a straight line (see assumptions 1,2,3 above). The sides of the enclosed area are evidently length c, the length of the hypotenuse of the enclosing triangles with sides a and b. The total area of the larger square in Figure 3 can be described in two ways.
    (3) (a+b)*(a+b)
    (4) c*c + 2*a*b

  3. The area formulas for figures 2 and 3 can are equal, because they are both (a+b)*(a+b) from (1) and (3) above. According to the alternate formulas (2) and (4) for areas of figures 2 and 3, we obtain the following equality of areas:
    (5) a*a + b*b + 2*a*b = c*c + 2*a*b
    (6) a*a + b*b = c*c (by subtracting 2*a*b from both sides of the equation)
    Equation (6) is the result to be proved.