Timeline for First known proof of $\sqrt 2$ is irrational with prime factorization?
Current License: CC BY-SA 3.0
7 events
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| May 7, 2013 at 19:58 | comment | added | Lennart Meier | Proof 11 is also quite remarkable. @Ryan: While this is true, it requires something like unique factorization (more precisely: the Euclid lemma) if we want to do the same proof for bigger numbers. | |
| Sep 18, 2012 at 11:32 | comment | added | Manya | 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 (cut-the-knot.org/proofs/sq_root.shtml). If you haven't seen 8''' before, that one is quite nice. | |
| Sep 16, 2012 at 0:31 | comment | added | Ryan Reich | 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. | |
| Sep 15, 2012 at 14:56 | comment | added | Dick Palais | 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. | |
| Sep 15, 2012 at 13:00 | comment | added | Franz Lemmermeyer | I fail to see what this has to do with the question. | |
| Sep 15, 2012 at 4:58 | comment | added | Sidney Raffer |
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.
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| Sep 14, 2012 at 23:50 | history | answered | Dick Palais | CC BY-SA 3.0 |