HPS 0628 Paradox

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Paradoxes of Impossibility: Incommensurability

John D. Norton
Department of History and Philosophy of Science
University of Pittsburgh
http://www.pitt.edu/~jdnorton


Paradoxes of impossibility arise when we have some task that appears to be quite achievable. However, try as we might, we can find no way to do it. And then with further analysis it turns out that the task set is provably impossible. We shall explore the earliest of these here, the ancient scandal of the incommensurability of the sides of an isosceles, right angle triangle. In a subsequent chapters, we will look at paradoxes of impossibility concerning classical geometric constructions and computation.

The Pythagoreans

One of the oldest of these paradoxes of impossibility and perhaps one of the oldest of provocations was a discovery attributed by tradition to the ancient Pythagoreans. They were a school of philosophers founded by Pythagoras of Samos in the 6th century BCE, at the inaugural moments of the Greek tradition in philosophy. The Pythagoreans were as much a school of philosophers as a cult. Their members were required to live an ascetic life governed by many rules. It is traditional to attribute to the Pythagoreans the results to be developed here. They are the Pythagorean theorem and the irrationality of the square root of two. The latter is attributed to the Pythagorean, Hippasus of Metapontum. However all these attributions are in doubt for lack of clear sources.

Nonetheless, the historical legends are engaging. Hippasus, it is said, discovered an impossibility so disturbing to the Pythagoreans that he was drowned at sea.

A numerical foundation for geometry

The Pythagoreans, by tradition, took as a fundamental principle of nature that "all is number." That principle worked well in music. Harmonious chords are produced by plucking stretched strings whose lengths stand in simple numerical ratios.


Image from Franchino Gaffurio, Theorica Musicae, 1492.

Whether the Pythagoreans actually proved the result as a theorem in some proof system seems less likely. The result really only can be properly called a theorem when such a proof is provided, such as is given in Euclid's Elements.

Similar whole number ratios initially seem able to express the facts of geometry. Here the Pythagorean theorem, as it later came to be known, is the key result. According to it:



The square of the length of the hypotenuse of a right angle triangle is equal to the sum of the squares of the lengths of the other two sides.

The celebrated application is through Pythagorean triples, such as (3, 4, 5), (5, 12, 13) and (8, 15, 17). They conform with the condition in the Pythagorean theorem since

32 + 42 = 52
52 + 122 = 132
82 + 152 = 172

It follows that the associated right angled triangles have sides that can be represented by whole numbers. One might imagine beads to be laid along each side. A 3-4-5 right angle triangle has sides of 3 beads, 4 beads and 5 beads.

It seems that whole numbers provide a foundation for geometry.

The square root of two

What upset this promising approach was the simplest of right angle triangles. It is an isosceles, right angle triangle, in which the right angle is enclosed by two equal sides. If we take each side to be of unit length, the hypotenuse must be of length √2, the square root of 2. Pythagoras' theorem tells us:

(√2)2 = 12 + 12

No whole numbers can deal with this simple case. √2 lies somewhere between 1 and 2. The obvious remedy is that we need to use many more numbers on each side. We come pretty close to capturing the geometric relations if we assign ten to the smaller sides and 14 to the hypotenuse.

We come close but not quite close enough:

102 + 102 = 200
142 = 196 ≠ 102 + 102
152 = 225

We press on. What about assigning 100 to the shorter sides and 141 to the hypotenuse? We are closer, but still not at the equality required by Pythagoras' theorem:

1002 + 1002 = 20,000
1412 = 19,881 ≠ 1002 + 1002
1422 = 20,164

We could continue the exercise. However as you all surely know, it is fated to fail There are no whole numbers that can be assigned to the sides and the hypotenuse so that Pythagoras' theorem can be satisfied.

Irrationality, Incommensurability

This is sometimes described by the assertion that the sides and the hypotenuse are "incommensurable." That means that there is no common measure for both. Here, we would think of the beads in the figure as unit measuring sticks. We might measure a smaller side in inches and get a whole number. Then we will not get a whole number for the hypotenuse. We might instead use a smaller unit of measure, say, a millionth of an inch. We would still not find whole number lengths for all the sides.

That is, there is a failure of ratios:

ratio length hypotenuse to length side ≠ ratio of whole numbers

To put this in a more familiar form:

The length of hypotenuse divided by the length of side is irrational. We now represent this fact by writing this ratio of lengths, the square root of 2, as a non-terminating, non-repeating decimal:

√2 =1.414213562373095048801688724209...

The ratios tried above are simply drawn from the first few digits of this decimal expansion.

14 to 10; 141 to 100; 1,414 to 1,000; 14,142 to 10,000; ...

By drawing more digits, we can come closer to the ratios required by the Pythagorean theorem. However no finite set of them can satisfy the theorem exactly.

This failure was a disaster for the Pythagorean program. It showed that a whole number foundation for geometry could not be found.

Traditionally, this failure led the Greek tradition to replace numbers as the foundation for geometry by geometrical entities. The revised approach takes points, lines, surfaces and volumes as the primitive notions that are posited as the foundation of geometry.

We can see this shift in a subtle reconception of the Pythagorean theorem. As stated above, it is a relation among numbers: the squares in the theorem are simply multiplications. "The square of the hypotenuse" refers to taking the number that measures the hypotentuse and multiplying it by itself.

The traditional Euclidean formulation of the Pythagorean theorem replaces this multiplication with a geometric construction: "the square ON the hypotenuse." and it is expressed in terms of the areas of these squares.

The square on the hypotenuse is equal [in area] to the sum of [the areas of] the squares on the other two sides.

(This is the diagram in Billingsley's 1482 first English translation of Euclid's Elements.)

Algebraic Proof
of the Irrationality of √2

The traditional demonstration of the irrationality of √2 resides is a short piece of algebra. It is a proof by reductio:

Assume that there are whole numbers p and q such that

√2= p/q

We assume moreover that we have eliminated all common factors from p and q. For example, we would not use the ratio 14/10 since the two numbers have a common factor 2 that can be cancelled: 14/10 = 5/7.

Squaring both sides we have 2 = p2/q2, so that

p2 = 2q2

It follows that p2 is an even number. This is only possible if p is itself even. (If p were odd, then p2 would also be odd.) That is, there is a whole number a such that

p2 = (2a)2 = 4a2

Combining the last two equations, we have

q2 = 2a2

It now follows by similar reasoning that q is also even. That is, both p and q are even and so share a common factor of 2.

This contradicts the assumption of no common factors. The reductio is complete.

There are no whole numbers p and q such that √2= p/q.

Geometric Proof
of the Irrationality of √2

The algebraic proof works and works well. However it is unsatisfying in that it makes no obvious connection with triangles and geometry. That geometric connection can be made in a variation of the proof.

The basis of the algebraic proof is that we suppose we have an irreducible ratio of whole numbers for √2, p/q, that is, p and q share no common factor. We then show that both p and q are even numbers, so there is another small ratio recovered by dividing both p and q by 2.

A geometric version of this proof shows how to construct a smaller right angle triangle that has this smaller-number ratio. The proof starts with an isosceles, right angle triangle, that is, one with two equal sides. It is assumed for purposes of reductio that the hypotenuse is of length p in some unit and the other two sides are each of length q.

The two numbers p and q are selected such that p has the smallest value possible; that is, if there are any common factors among p and q, they are factored out. We already have from the Pythagorean theorem that p is greater than q since

p = √2q > q.

We use the result from the algebraic proof that p must be an even number. That means that it is possible to bisect the hypotenuse into two equal lengths of p/2 and connect the midpoint to the right angle.

We now have two smaller right angled triangles. Each has a hypotenuse of length q and the lengths of the other two sides are p/2. We have a contradiction with the original assumption:

We assumed hypotenuse/side
      = p/q and p was the smallest whole number that can serve.
We infer that hypotenuse/side
      = q/(p/2), where q is less than p.

The reductio is complete and the original assumption is rejected.

The Shrinking Construction

While this completes the proof, a striking extension of it is possible. The construction above lets us build a smaller, right angle triangle within the larger one. We can iterate. With in this smaller right angle triangle, we can construct another still smaller right angle triangle; and so on indefinitely.

Each of these indefinitely nestled, right angle triangles has equal length sides enclosing the right angle; and with each two steps in the nestling, the hypotenuse is halved. As the halving continues, we will arrive at triangles with hypotenuses arbitrarily close to zero.

No matter how large the length p of the largest triangle is made, we will eventually come to a triangle whose hypotenuse cannot be measured even by a single unit.

Once again we have arrived at a contradiction with the initial assumption that the sides of an isosceles, right angled triangle

This construction may seem to be too powerful. It depends on the similarity of a right angle triangle with a smaller version of itself. Can this shrinking proof be used to rule out all right angle triangles, including the ones that do have whole number ratios like a 3-4-5 triangle? The proof cannot be applied to these cases. The proof requires more than self-similarity. It also requires that the whole number of the length of the sides is preserved as we move to a smaller triangle. That condition will fail in general. To begin, all hypotenuse lengths of Pythagorean triples are odd numbered. That means that we cannot bisect the hypotenuse into two equal, whole number lengths for triangles whose sides are Pythagorean triples.

To Ponder

In Cartesian geometry, we cover the Euclidean plane with an x-y coordinate system. Geometric structures arise as algebraic formulae. A straight line is "y=mx+b," for example. Has the Pythagorean goal of founding geometry on numbers been realized here?

The analyses above show that √2 is irrational. What about other numbers? How common is it for a whole number to have an irrational square root? Are there proofs for these other cases?*

The golden ratio φ is reputed to be the most pleasing proportion for a rectangle and, for this reason, appears often in classical art. Its value is roughly 1.618, but the exact value is irrational. It has a simple geometric definition: take a rectangle whose sides are in the golden ratio; remove a square; and the remaining rectangle is still a golden rectangle. This fact enables a "shrinking geometrical proof" of the irrationality of φ.

The algebraic expression of this geometrical fact is that (φ-1) = 1/φ. Solve for φ. (You will need the quadratic formula.)

*Hint: the algebraic proof generalizes if we replace 2 by any number N that has no whole number square root. The geometric shrinking proof works best for the case of a rectangle with the ratio of sides √N to 1, divided into N smaller rectangles of the same shape. All these proofs fail if N has a whole number square root. The first step of all the proofs is to show that, if √N = p/q, then p is a multiple of N. This step fails since, if the square root is a whole number, then we can write √N = p/1. Then p is not a multiple of N.

July 12, 2021

Copyright, John D. Norton