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Comment Re:Poor QA (Score 1) 626

(1/10)^n for integer n is irrational in base 2 and the truncation was unavoidable.

Whether or not a number is irrational does not depend on the base. The number (1/10)^n is rational in any base. By irrational, maybe you meant "finite decimal expansion"?

Unrelated: The article starts with the example 599999999999999 - 599999999999998 = 0 in Google. Fortunately some software gives the correct result by default.

Comment Re:For anybody who's curious... (Score 3, Insightful) 94

That's a good explanation. I have to emphasize though, that they actually found all the congruent numbers up to a trillion only under the completely unproven hypothesis that the Birch and Swinnerton-Dyer conjecture is true. It's entirely possible that this conjecture is false, and some of the numbers they found are actually not congruent numbers. However, part of the conjecture is known (by work of Coates and Wiles -- the same Wiles who proved Fermat's Last Theorem), so we do know that all numbers they didn't list are definitely not congruent numbers.

Comment Re:Hard Drive? (Score 2, Interesting) 94

I own the 128GB RAM, etc., computer that the second group did the computation on. I have a Sun X4550 24TB disk array (ZFS) connected to it, but I only allocated a few terabytes of space for a scratch disk. They were well into the calculation when I found out what they were up to (I was initially annoyed, since they were saturating the network). I think they were just being polite to me and the other users by not using a lot more disk.

Comment Re:"outlined in detail" != "here's some pseudo cod (Score 1) 94

I asked them before this came out, and they said they didn't want to post their code on the press release in order to avoid being slashdotted. Seriously. I think the code is certainly available upon request, and will be made available later when the hoopla dies down. Much of it is in FLINT, which is part of Sage.

Comment Re:Why? (Score 2, Informative) 94

It is an *open problem* to show that there exists algorithm at all to decide whether a given integer N is a congruent number. Full stop. It's not a question of speed, or even skipping previous integers. We simply don't even know that it is possible to decide whether or not integers are congruent numbers. However, if the Birch and Swinnerton-Dyer conjecture is true (which we don't know), then there is an algorithm.

Software

Submission + - Open Source Software Brings Transparency to Math (uwnews.org)

William Stein writes: "The free open source mathematics program Sage that was recently mentioned on slashdot just won first prize in the scientific software category of Les Trophees du Libre, an international competition for free software. Soon Sage will face off against the major software companies in San Diego at the American Mathematical Society Meeting."

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