APPENDIX I
Reductio ad Absurdum and the Square Root of Two
The original Pythagorean argument on the irrationality of the square root of 2 depended on a kind of argument called reductio ad absurdum, a reduction to absurdity: we assume the truth of a statement, follow its consequences and come upon a contradiction, thereby establishing its falsity. To take a modern example, consider the aphorism by the great twentieth-century physicist, Niels Bohr: ‘The opposite of every great idea is another great idea.’ If the statement were true, its consequences might be at least a little perilous. For example, consider the opposite of the Golden Rule, or proscriptions against lying or ‘Thou shalt not kill.’ So let us consider whether Bohr’s aphorism is itself a great idea. If so, then the converse statement, ‘The opposite of every great idea is not a great idea,’ must also be true. Then we have reached a reductio ad absurdum. If the converse statement is false, the aphorism need not detain us long, since it stands self-confessed as not a great idea.
We present a modern version of the proof of the irrationality of the square root of 2 using a reductio ad absurdum, and simple algebra rather than the exclusively geometrical proof discovered by the Pythagoreans. The style of argument, the mode of thinking, is at least as interesting as the conclusion:
Consider a square in which the sides are 1 unit long (1 centimeter, 1 inch, 1 light-year, it does not matter). The diagonal line BC divides the square into two triangles, each containing a right angle. In such right triangles, the Pythagorean theorem holds: 12+12=x2. But 12 + 12 = 1 + 1 = 2, so x2 = 2 and we write x = the square root of 2. We assume the square root of 2 is a rational number: The square root of 2 = p/q, where p and q are integers, whole numbers. They can be as big as we like and can stand for any integers we like. We can certainly require that they have no common factors. If we were to claim the square root of 2 = 14/10, for example, we would of course cancel out the factor 2 and write p = 7 and q = 5, not p = 14, q = 10. Any common factor in numerator or denominator would be canceled out before we start. There are an infinite number of p’s and q’s we can choose. From the square root of 2 = p/q, by squaring both sides of the equation, we find that 2 = p2/q2, or, by multiplying both sides of the equation by q2, we find
p2 = 2q2.
(Equation 1)
p2 is then some number multiplied by 2. Therefore p2 is an even number. But the square of any odd number is odd (12 = 1, 32 = 9, 52 = 25, 72 = 49, etc.). So p itself must be even, and we can write p = 2s, where s is some other integer. Substituting for p in Equation (1), we find
p2 = (2s)2 = 4s2 = 2q2
Dividing both sides of the last equality by 2, we find
q2 = 2S2
Therefore q2 is also an even number, and, by the same argument as we just used for p, it follows that q is even too. But if p and q are both even, both divisible by 2, then they have not been reduced to their lowest common factor, contradicting one of our assumptions. Reductio ad absurdum. But which assumption? The argument cannot be telling us that reduction to common factors is forbidden, that 14/10 is permitted and 7/5 is not. So the initial assumption must be wrong; p and q cannot be whole numbers; and the square root of 2 is irrational. In fact, the square root of 2 = 1.4142135 …
What a stunning and unexpected conclusion! How elegant the proof! But the Pythagoreans felt compelled to suppress this great discovery.
APPENDIX 2
The Five Pythagorean Solids
A regular polygon (Greek for ‘many-angled’) is a two-dimensional figure with some number, n, of equal sides. So n = 3 is an equilateral triangle, n = 4 is a square, n = 5 is a pentagon, and so on. A polyhedron (Greek for ‘many-sided’ ) is a three-dimensional figure, all of whose faces are polygons: a cube, for example, with 6 squares for faces. A simple polyhedron, or regular solid, is one with no holes in it. Fundamental to the work of the Pythagoreans and of Johannes Kepler was the fact that there can be 5 and only 5 regular solids. The easiest proof comes from a relationship discovered much later by Descartes and by Leonhard Euler which relates the number of faces, F, the number of edges, E, and the number of corners or vertices V of a regular solid:
V – E + F = 2
(Equation 2)
So for a cube, there are 6 faces (F = 6) and 8 vertices (V = 8), and 8 – E + 6 = 2, 14 – E = 2, and E = 12; Equation (2) predicts that the cube has 12 edges, as it does. A simple geometric proof of Equation (2) can be found in the book by Courant and Robbins in the Bibliography. From Equation (2) we can prove that there are only five regular solids.
Every edge of a regular solid is shared by the sides of two adjacent polygons. Think again of the cube, where every edge is a boundary between two squares. If we count up all the sides of all the faces of a polyhedron, nF, we will have counted every edge twice. So
nF = 2E
(Equation 3)
Let r represent how many edges meet at each vertex. For a cube, r = 3. Also, every edge connects two vertices. If we count up all the vertices, rV, we will similarly have counted every edge twice. So
rV = 2E
(Equation 4)
Substituting for V and F in Equation (2) from Equations (3) and (4), we find
If we divide both sides of this equation by 2E, we have
(Equation 5)
We know that n is 3 or more, since the simplest polygon is the triangle, with three sides. We also know that r is 3 or more, since at least 3 faces meet at a given vertex in a polyhedron. If both n and r were simultaneously more than 3, the left-hand side of Equation (5) would be less than 2/3 and the equation could not be satisfied for any positive value of E. Thus, by another reductio ad absurdum argument, either n = 3 and r is 3 or more, or r = 3 and n is 3 or more.
If n = 3, Equation (5) becomes
(1/3) + (1/r) = (1/2) + (1/E), or
(Equation 6)
So in this case r can equal 3, 4, or 5 only. (If E were 6 or more, the equation would be violated.) Now n = 3, r = 3 designates a solid in which 3 triangles meet at each vertex. By Equation (6) it has 6 edges; by Equation (3) it has 4 faces; by Equation (4) it has 4 vertices. Clearly it is the pyramid or tetrahedron; n = 3, r = 4 is a solid with 8 faces in which 4 triangles meet at each vertex, the octahedron; and n = 3, r = 5 represents a solid with 20 faces in which 5 triangles meet at each vertex, the icosahedron.
If r = 3, Equation (5) becomes
and by similar arguments n can equal 3, 4, or 5 only. n = 3 is the tetrahedron again; n = 4 is a solid whose faces are 6 squares, the cube; and n = 5 corresponds to a solid whose faces are 12 pentagons, the dodecahedron.
There are no other integer values of n and r possible, and therefore there are only 5 regular solids, a conclusion from abstract and beautiful mathematics that has had, as we have seen, the most profound impact on practical human affairs.