First known proof of $\sqrt 2$ is irrational with prime factorization? - MathOverflow

First known proof of $\sqrt 2$ is irrational with prime factorization?

2012-09-14T08:18:18Z

Do any of you happen to know the history of the standard prime factorization proof of $\sqrt 2$ is irrational? I know this theorem was known to Aristotle, and that the Fundamental Theorem of Arithmetic, on which the proof rests, is found already in Euclid, but I've not been able to track down the origin of this particular proof.

These sites I know about: http://www.cut-the-knot.org/proofs/sq_root.shtml, http://www.math.ufl.edu/~rcrew/texts/pythagoras.html, and of course http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

But any other references, online or in paper form, would be greatly appreciated!

Answer by Carlo Beenakker for First known proof of $\sqrt 2$ is irrational with prime factorization?

2012-09-14T09:13:15Z

"The Discovery of Incommensurability" by Kurt von Fritz [ http://www.jstor.org/stable/1969021 ] indicates that the early Greek mathematicians did not explicitly use the Fundamental Theorem to prove the irrationality of √2. The proof known to Aristotle ("the diagonal of the square is incommensurate with the side, because odd numbers are equal to evens if it is supposed to be commensurate") uses a restricted version of the Fundamental Theorem, as explained in http://en.wikipedia.org/wiki/Quadratic_irrational

Apparently, the explicit use of the Fundamental Theorem to prove the irrationality of √2 is post-Gauss. This is argued convincingly by Barry Mazur:

This fundamental theorem of arithmetic has a peculiar history. It is not trivial, and any of its proofs take work, and, indeed, are interesting in themselves. But it is nowhere stated in the ancient literature. It was used, implicitly, by the early modern mathematicians, Euler included, without anyone noticing that it actually required some verification, until Gauss finally realized the need for stating it explicitly, and proving it.

Answer by Barry Cipra for First known proof of $\sqrt 2$ is irrational with prime factorization?

2012-09-14T12:14:43Z

From the Wikipedia entry for Wilbur Knorr, on one of his books:

The Evolution of the Euclidean Elements: A Study of the Theory of Incommensurable Magnitudes and Its Significance for Early Greek Geometry (Dordrecht: D. Reidel Publishing Co., 1975).

This work incorporates Knorr's Ph.D. thesis. It traces the early history of irrational numbers from their first discovery (in Thebes between 430 and 410 BC, Knorr speculates), through the work of Theodorus of Cyrene, who showed the irrationality of the square roots of the integers up to 17, and Theodorus' student Theaetetus, who showed that all non-square integers have irrational square roots. Knorr reconstructs an argument based on Pythagorean triples and parity that matches the story in Plato's Theaetetus of Theodorus' difficulties with the number 17, and shows that switching from parity to a different dichotomy in terms of whether a number is square or not was the key to Theaetetus' success.

Answer by Dick Palais for First known proof of $\sqrt 2$ is irrational with prime factorization?

2012-09-14T23:50:04Z

Here is a proof based on the well-ordering of the positive integers rather than the FTA:

To begin with, we observe that if $\sqrt{2}$ is rational, then there is some positive integer q such that q × $\sqrt{2}$ is an integer. Since the positive integers are well ordered, we may suppose that q is the smallest such number. We next observe that since 1 < $\sqrt{2}$ < 2, then $\sqrt{2}$ – 1 < 1, and consequently q × ($\sqrt{2}$ – 1) = (q × $\sqrt{2}$ – q ) is less than q. Let us call this new number r, and observe that it too is a positive integer. But we now have r × $\sqrt{2}$ is also an integer, since r × $\sqrt{2}$ = (q ×$\sqrt{2}$ – q ) × $\sqrt{2}$ = (2q – q × $\sqrt{2}$). In short, r is a positive integer less than q and r × $\sqrt{2}$ is an integer. But we said that q was the smallest positive integer with this property, and so we have a contradiction.