Pythagorean Theorem:
Some False Proofs

Even smart people make mistakes. Some mistakes are getting published and thus live for posterity to learn from. I'll list below some fallacious proofs of the Pythagorean theorem that I came across. Some times the errors are subtle and involve circular reasoning or fact misinterpretation. On occasion, a glaring error is committed in logic and leaves one wondering how it could have avoided being noticed by the authors and editors.

Proof 1

One such error appears in the proof X of the collection by B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.3, n. 6/7 (1896), 169-171.)

Suppose the theorem true. Then AB² = AC² + BC², BC² = CD² + BD², and AC² = AD² + CD². Combining the three we get

But CD² = AD·BD. Therefore,

From which

which is true. The supposition is true.

Critique

By the same token, assume 1 = 2. Then, by symmetry, 2 = 1. By Euclid's Second Common Notion, we may add the the two identities side by side: 3 = 3. Which is true, but does not make the assumption (1 = 2) even one bit less false.

As we know, falsity implies anything, truth in particular.

Proof 2

This proof is by E. S. Loomis (Am Math Monthly, v. 8, n. 11 (1901), 233.)

Let ABC be a right triangle whose sides are tangent to the circle O. Since CD = CF, BE = BF, and AE = AD = r = radius of circle, it is easily shown that

And if

(1)	a + 2r = b + c

then (1)² = (2):

(2)	a² + 4ra + 4r² = b² + 2bc + c².

Now if 4ra + 4r² = 2bc, then a² = b² + c². But 4ra + 4r² is greater than, equal to, or less than 2bc.

If 4ra + 4r² > or < 2bc, then a² + 4ra + 4r² > or < b² + 2bc + c²; i.e. a + 2r < or > b + c, which is absurd. Hence, 4ra + 4r² = 2bc and, therefore, a² = b² + c².

This proof is accompanied by an editors's Note:

Critique

The proof is incorrect, let alone indirect, and generally written in a style that would not be acceptable nowadays. The main problem, however, is the circularity of the argument: the proof uses the Pythagorean identity it set out to establish.

In a subsequent issue of the magazine (Am Math Monthly, v. 8, n. 12 (1901), 258), the editors admit the mishap and quote Prof. B. F. Yanney to the effect that "such reasoning as employed in the proof given by Dr. Loomis would make 4ar = b² + c² or 4r² = b² + c², or even r² = b² + c²."

A correct and also direct proof has been supplied by F. L Sawyer. Starting with (1),

(3)	2a + 2r = a + b + c.

Such that,

But evaluating the area of DABC in two different ways (see proof #42)

Hence,

(4)	2r(a + b + c) = 2bc.

By substituting (1) into (4) we obtain

Because of (2) we now get the desired a² = b² + c².

Remark

Note that (3) has been known to Liu Hui and on which is based dissection proof #45 by Douglas Rogers. In a private correspondence, Douglas also mentioned that Loomis included Sawyer's proof as the eighty-sixth algebraic proof in his compilation The Pythagorean Proposition without reference to his earlier erroneous effort, but claiming that he hit on it for himself before seeing the issue of The Monthly with Sawyer's proof. He also makes reference to Fourrey.

Proof 3

This proof appears as the geometric proof 22 in Loomis' collection. Loomis refers to a 1914 publication by J. Versluys which listed 96 proofs of the Pythagorean proposition. The proof at hand appears as Fig. 44, p. 43.

Critique

While the proof looks like many other proofs by dissection and rearrangement, this one is not complete in that the construction does not go through where one of the legs of the given triangle is essentially less than the other:

The original diagram does not provide a clue as to the handling of a more general case. (The diagrams above have been obtained with a dynamic illustration.)

Proof 4

This is what could be called a trigonometric proof. Several times it was suggested for the inclusion on the Pythagorean page until Eng. Marìa L. Mean, a retired professor at Havana University, pointed out that's the right place for the derivation is among other false proofs. So here it goes.

In a right-angled triangle ABC, let a, b be its legs, c its hypotenuse, and β the angle opposed to leg b. Then, by the definition of the trigonometric functions

As is well known, the sum of sin² β and cos² β is equal to 1:

which is a fundamental trigonometric identity. Consequently,

implying a² + b² = c².

Critique

The identity sin² β + cos² β = 1 is indeed fundamental in trigonometry. However, its derivation is based on the Pythagorean theorem, to start with. Thus, the trigonometric "proof" above may well serve as an example of the circular reasoning, a vicious circle as the latter is sometimes referred to.

A Later Remark

As it comes out sin² β + cos² β = 1 admits a proof independent of the Pythagorean theorem but based solely on the subtraction formulas for sine and cosine. Thus my critique of the above proof should be taken with the grain of salt. It nonetheless can be used in a polemique. Any one who promotes the above proof with no knowledge as to how to derive sin² β + cos² β = 1 with no recourse to the Pythagorean theorem makes a severe logical mistake.

Proof 5

A mistake of a higher order is sometimes committed by more advanced students of mathematics who went beyond trigonometry and ventured into the multidimensional geometry. In multidimensional spaces whose elements are vectors, one often defines what is known as the scalar product and then also an angle between two vectors. Say, for two vectors a and b, if the scalar product is denoted a·b, then the angle γ between the two is defined via the cosine function as in:

where ||a|| is the norm of vector a: ||a||² = a·a and similarly for ||b||.

For γ = 90°, it follows from the properties of the scalar product that

which, in a 2-dimensional case, is easily seen to express the common Pythagorean theorem.

Critique

The problem with this derivation is that the theory of vector spaces is entirely algebraic (or analytic, if you will). Vectors are defined as n-tuples of real numbers which is followed by the componentwise definition of vector operations. Within this framework, the vector Pythagorean identity above is indeed an easy consequence of the axioms and definitions. However, the relationship between the common geometry and the geometry of vector spaces is that of a model and an abstract theory. The above vector identity does not prove the Pythagorean theorem. It only shows that there is a tight relation between the model and the theory. It confirms this relation, perhaps offers an additional insight into the Pythagorean theorem, but does not prove it by any means.

Proof 6

A proof by John Molokach under the caption Calculus Proof for Pythagorean Theorem has been posted on the web. I reproduce the proof on a separate page.

References

1. E. S. Loomis, The Pythagorean Proposition, NCTM, 1968

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