MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:,, and of course

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

share|cite|improve this question
I'm not very good with tags, so please feel free to edit if there are better choices. Thanks. – Manya Sep 14 '12 at 8:19
Retagged as invited – Yemon Choi Sep 14 '12 at 8:31
Dear Manya, Did you see the book "The Square Root of 2" by Flannery, David? The link in Springer is: Also there is another link: I hope it will be helpful. – Shahrooz Janbaz Sep 14 '12 at 11:17
No, thanks for the reference. – Manya Sep 14 '12 at 15:39
up vote 7 down vote accepted

"The Discovery of Incommensurability" by Kurt von Fritz [ ] 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

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.

share|cite|improve this answer
Thanks for the information and the links. I wonder if Gauss himself could have proven it...? – Manya Sep 14 '12 at 9:55
that seems to be Barry Mazur's conclusion [added reference] – Carlo Beenakker Sep 15 '12 at 18:34
Thanks for the Mazur paper. It is very good! But is it so clear that Gauss applied the fundamental theorem to the $\sqrt 2$. Or am I missing something? – Manya Sep 18 '12 at 8:47
well, once you have proven the fundamental theorem, there does not seem to be anything left to prove regarding the irrationality of the square root of any number that is not a perfect square; I imagine asking Gauss why he doesn't care to publish that proof and receiving a gentle smile.. – Carlo Beenakker Sep 18 '12 at 10:31
Ok, I accept that :-) – Manya Sep 18 '12 at 11:22

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.

share|cite|improve this answer
Ok, thanks, that seems like an interesting book. – Manya Sep 18 '12 at 19:08

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.

share|cite|improve this answer
A variation on this: If $\sqrt{2}$ is a rational number, say $a/b$ with $a,b\in \mathbb{Z}$, then $\sqrt{2}-1$ is a rational between 0 and 1. Hence $(\sqrt{2}-1)^n\to 0$ as $n\to\infty$. But $(\sqrt{2}-1)^n$ has the form $u\frac{a}{b}+v$ for some $u,v\in\mathbb{Z}$, so $(\sqrt{2}-1)^n\ge1/b$. Contradiction. – SJR Sep 15 '12 at 4:58
I fail to see what this has to do with the question. – Franz Lemmermeyer Sep 15 '12 at 13:00
Your right Franz, it doesn't. It's just that there seems to be a belief that you NEED unique prime factorization to prove the irrationality of non-square integers, and when I first saw this (much more elementary) proof I found it an eye-opening experience. – Dick Palais Sep 15 '12 at 14:56
This is a nice proof, but I don't think the other standard proof (write $\sqrt{2} = a/b$ with $a,b$ not both even) requires unique factorization. It only requires one lemma about even and odd squares. – Ryan Reich Sep 16 '12 at 0:31
Yeah, I agree, this thread is off-topic, but there are of course other cool proofs out there. You can find a nice collection at the link I gave above ( If you haven't seen 8''' before, that one is quite nice. – Manya Sep 18 '12 at 11:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.