Math Bite: Irrationality of √m (1999)
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Math Bite: Irrationality of √m (1999)
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Re: Math Bite: Irrationality of √m (1999)
#2That proof is really great! It's always nice to see alternatives to well-known proofs. This demonstrates that you can always tackle problems from different angles.
However, I slightly disagree with the introduction of the proof:
| The really interesting thing about this proof is that it doesn't use divisibility, just mathematical induction in its "Z is well-ordered" form.
That's quite a bold statement. It would mean that this proof should generalize to other well-ordered sets that do not provide any notion of divisibility.
However, I don't see that. (Does anyone else see how this proof could work on such a generalized setting?)
Instead, it seems that this proof introduces the concept of "divisibility" through the backdoor, by calculating with fractions and making use of the basic fraction laws, which are based on divisibility. In particular, the equation
1 / (sqrt(m) - n) = (sqrt(m) + n) / (m - n^2)
Nevertheless, great proof!
Re: Math Bite: Irrationality of √m (1999)
#3Discovering wonderful sites like this is why I read Hacker News!
Re: Math Bite: Irrationality of √m (1999)
#4The print is too small and the document cannot be scaled, only pixellated. This makes it quite unusable for me. Too bad, math deserves better.
Re: Math Bite: Irrationality of √m (1999)
#5[deleted]
Re: Math Bite: Irrationality of √m (1999)
#6Discovering wonderful sites like this is why I read Hacker News!
Yes, such wonderful sites that must put annoying bouncing "Click here to see more" and "Join our newsletter" things around everything.
By the way, this particular note appears to be taken from "Biscuits of Number Theory". Compare with this scan from Google Books:
https://books.google.si/books?id=_g5TvMCJQB4C&pg=PA109&lpg=P...
Re: Math Bite: Irrationality of √m (1999)
#7That proof is really great! It's always nice to see alternatives to well-known proofs. This demonstrates that you can always tackle problems from different angles. However, I slightly disagree with the introduction of the proof: | The really interesting thing about this proof is that it doesn't use divisibility, just mathematical induction in its "Z is well-ordered" form. That's quite a bold statement. It would mean…
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Re: Math Bite: Irrationality of √m (1999)
#8That proof is really great! It's always nice to see alternatives to well-known proofs. This demonstrates that you can always tackle problems from different angles. However, I slightly disagree with the introduction of the proof: | The really interesting thing about this proof is that it doesn't use divisibility, just mathematical induction in its "Z is well-ordered" form. That's quite a bold statement. It would mean…
By "doesn't use divisibility" I assume the author means that the point of contradiction doesn't rest on the "divisibility properties" of the integer, not that division is never used. In particular, this proof doesn't rely on the fundamental theorem of arithmetic.
Re: Math Bite: Irrationality of √m (1999)
#9That proof is really great! It's always nice to see alternatives to well-known proofs. This demonstrates that you can always tackle problems from different angles. However, I slightly disagree with the introduction of the proof: | The really interesting thing about this proof is that it doesn't use divisibility, just mathematical induction in its "Z is well-ordered" form. That's quite a bold statement. It would mean…
By "doesn't use divisibility" I assume the author means that the point of contradiction doesn't rest on the "divisibility properties" of the integer, not that division is never used. In particular, this proof doesn't rely on the fundamental theorem of arithmetic.
> I assume the author means that the point of contradiction doesn't rest on the "divisibility properties" of the integer
I see what the author meant by that. I just think it was stated in a slightly exaggerated way. The "divisibility properties" of the integers are still used. That part has just has been moved to another corner of the proof, by transforming fractional equations.
One of the comments in the article is quite interesting here:
| Theodor Estermann proved the irrationality of 2√ without relying on the prime factorization of m.
I believe that this statement is more correct. The "traditional" proof uses not just divisibility, but prime factorization, which is quite a strong property. And that is something the alternative proof doesn't make use of.
Maybe the introduction should have been stated that way.
Re: Math Bite: Irrationality of √m (1999)
#10I don't understand. The end of the proof says r/p is a fraction, presumably because r must be an integer. But why must r be an integer?
There seems to be an implicit assumption that m is an integer, but the explicit assumptions only give the much weaker statement that "m is not a perfect square".