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Why isn’t the fundamental theorem of arithmetic obvious? (2011)

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Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#111

Post by ColinWright »

PepeGomez wrote: post #99
Chinjut wrote: post #94

Earlier quoted context omitted.

Who says "if it's divisible by a number, the number must appear in its factorization"? Why is that true? For example, 24 is divisible by 6, though 24 has the prime factorization 2 * 2 * 2 * 3, with no 6 to be found. "Right, but all the prime factors of 6 appear in the prime factorization of 24. If X is divisible by the prime p, then there is a unique prime factorization of X, which must include a factor of p", you ma…

No other prime is divisible by 3 than 3 itself and so on. And you can't ever get a number divisible by 3 by multiplying numbers that are not divisible by 3. (3n-1) * m mod 3 and (3n-2) * m mod 3 cycle predictably and never become zero unless m mod 3 = 0. The same principle should hold for every number.

No other prime is divisible by (1-sqrt(-5)) than (1-sqrt(-5)) itself, and yet you can multiply other numbers together and get a number divisible by (1-sqrt(-5)), namely 2 and 3 to get 6.

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#112

Post by GFK_of_xmaspast »

nilkn wrote: post #110

I know I'm going up against a brilliant mathematician and Fields medalist here, but I find this article to be unenlightening. It seems that Gowers has glossed over something about the integers that's built incredibly deeply into our intuition about them when he talks about Z[sqrt(-5)]: > These numbers have various properties in common with the integers: you can add them and multiply them, there are identities for bot…

Here's a stumper then: why do Z[sqrt(-1)] and Z[sqrt(-3)] have unique factorization but Z[sqrt(-5)] doesn't?

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#113

Post by nilkn »

GFK_of_xmaspast wrote: post #112
nilkn wrote: post #110

I know I'm going up against a brilliant mathematician and Fields medalist here, but I find this article to be unenlightening. It seems that Gowers has glossed over something about the integers that's built incredibly deeply into our intuition about them when he talks about Z[sqrt(-5)]: > These numbers have various properties in common with the integers: you can add them and multiply them, there are identities for bot…

Here's a stumper then: why do Z[sqrt(-1)] and Z[sqrt(-3)] have unique factorization but Z[sqrt(-5)] doesn't?

Did you mean to write Z[sqrt(-2)] instead of Z[sqrt(-3)]?

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#114

Post by jerf »

bluecalm wrote: post #83
ColinWright wrote: post #81

Earlier quoted context omitted.

> Now let's assume that our number A has > two factorings F1 and F2. Let's sort > them from the smallest to the biggest > divider. OK, I've done that. > Is it possible that F1 and F2 are > different at the first position? It > isn't as that would mean the same > number has different smallest prime > divider. So why is this false in Z[ sqrt(-5) ] ?? There we have: 6 = 2 x 3 6 = (1 - sqrt(-5)) x ((1 + sqrt(-5)) Now 6 h…

>>So why is this false in Z[ sqrt(-5) ] I don't know, I don't know what sqrt is, let alone sqrt for a negative number. It's like asking someone who made a nice geometric proof of Pitagoras theorem why it doesn't work on a 4 dimensional sphere for stuff which is kinda like triangles. It's different multiplication you are mentioning here. I can't do (1-sqrt(-5) x (1 + sqrt(-5) by putting some stones in a rectangle and…

You're approaching this with a hostile attitude, which is preventing you from understanding and/or addressing what other people are saying, and substituting (light) mockery for attempts to understand what others are saying. You're not going to learn anything or convince anybody of anything this way.

The point of using the ring with the sqrt in it is to conveniently demonstrate that the FTA is non-obvious. Since it is only being used for demonstration, and not as part of a proof, faking ignorance of sqrts and imaginary numbers is not helpful to you. There are many things in mathematics where the subtleties only came out later; heck, that's basically the entire history of set theory. Sets are also trivial if you come at them with an attitude of artificial ignorance like that.

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#115

Post by GFK_of_xmaspast »

nilkn wrote: post #113
GFK_of_xmaspast wrote: post #112

Earlier quoted context omitted.

Here's a stumper then: why do Z[sqrt(-1)] and Z[sqrt(-3)] have unique factorization but Z[sqrt(-5)] doesn't?

Did you mean to write Z[sqrt(-2)] instead of Z[sqrt(-3)]?

https://en.wikipedia.org/wiki/Eisenstein_integer

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#116

Post by jerf »

mcguire wrote: post #107
pavelrub wrote: post #92

Earlier quoted context omitted.

I think I wasn't clear enough. I never meant that axiomatic proofs in general are meaningless, only that axiomatic proofs of trivial arithmetic facts (1+1=2, 2+2=4...) are meaningless.

You have two choices here: 1. You assume "trivial arithmetic facts" as axioms. Result: You have an infinite number of axioms. (Whee!) The likelihood that you have snuck in non-trivial assumptions is pretty high, unless you are very strict about how you define "trivial" (which is probably as much work as just proving the trivial facts), and in that case, there's a high probability that some of your trivial facts are f…

From a mathematical point of view, one must be very careful, and we have abundant evidence of that.

However, we are fully justified in saying that if anybody came up with a mathematical system in which 2 + 2 != 4, we can dismiss it without having to do some sort of deep analysis of it. 2 + 2 = 4 is obvious. We can literally do it with 4 little objects right in front of us. If we can not accept that as obvious, we are hopelessly ignorant and have no reason to trust our fancy proofs, either. (Italicized to emphasize my main point.) If you can't trust that, you certainly can't trust the significantly more complicated number theory axioms do anything useful.

Note that 2 + 2 = 4 carries some implicit context when we say it without qualification, and subtly sliding in a context change is not a disproof. 2 + 2 = 1 modulo 3, but that's not what anybody means without qualification. Clearly we're operating on "the numbers I can hold in my hand" here, or some superset thereto. Note how I'm not even willing to say "the natural numbers" necessarily; it isn't obvious to me what some billion digit number added to some other billion digit number is. It's actually crucial to my point here that I'm not extending "obvious" out that far; I can only run an algorithm on that and trust the algorithm. But I'm just being disingenuous if I claim ignorance of 2 + 2. And being disingenuous like that tends to turn people off, and doesn't encourage them to try to learn more.

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#117

Post by kosievdmerwe »

hugh4 wrote: post #35
CarolineW wrote: post #30

Earlier quoted context omitted.

Of course they are more likely to believe their intuitions. They also believe that it makes no difference whether or not you swap doors in the Monty Hall problem, and don't believe that with only 23 people the odds of a shared birthday are more than 50%. To some extent, there is the problem. People trust their intuitions, and their intuitions are often wrong. That's why for some things we need proper proofs.

But proofs always come back to axioms, and on what basis do we accept axioms? That they sound intuitively right. So we've just kicked the problem upstairs a bit, we can't avoid using our intuition. Personally I'm more likely to believe 2 + 2 = 4, something I can easily check to my own satisfaction using four objects, than I am to believe the Axiom of Choice.

We still use our intuitions, but now everyone knows the starting set of assumptions.

As for the axioms in use, I think the big reasons they were chosen is: They lead to results we already wanted/proved to be true.

Another thing to keep in mind, not everyone works with the same sets of axioms. Which, as someone with a formalist[2] view on mathematics, I find interesting. For example, not everyone studying logic assumes the principle of the excluded middle[1]. One of the consequences of this is that you can no longer do proofs by contradiction.

The axiom of choice is another example of this where two groups of mathematicians accept it or not. I'm a formalist, so I don't have issues with this (as long as both sets of axioms are interesting and "intuitive"), other philosophies of maths might.

[1] https://en.wikipedia.org/wiki/Law_of_excluded_middle [2] https://en.wikipedia.org/wiki/Formalism_(philosophy_of_mathe...

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#119

Post by ColinWright »

evanpw wrote: post #118

Every one of the bad arguments that Gowers points out have been made in this thread full of smart people, even after everyone read the article. I think that's pretty good evidence that the theorem really isn't obvious.

What's the bet most of them have at best skimmed the article, and that many haven't even bothered to read further than the first few paragraphs, if at all.

Edit: typo

Re: Why isn’t the fundamental theorem of arithmetic obvious? (2011)

#120

Post by scott_s »

ColinWright wrote: post #86
bluecalm wrote: post #83

Earlier quoted context omitted.

>>So why is this false in Z[ sqrt(-5) ] I don't know, I don't know what sqrt is, let alone sqrt for a negative number. It's like asking someone who made a nice geometric proof of Pitagoras theorem why it doesn't work on a 4 dimensional sphere for stuff which is kinda like triangles. It's different multiplication you are mentioning here. I can't do (1-sqrt(-5) x (1 + sqrt(-5) by putting some stones in a rectangle and…

So what you are saying is that if you don't know what can go wrong then it's "obviously true." Let's try some other things. * If you draw a distorted circle in the plane then it's obviously true that it has an inside and an outside. * The inside is obviously contractable to a point, and the outside is obviously contractable to a plane with a hole in it. * In three dimensions if you have a distorted sphere then it obv…

Please eventually provide the answer, because I'm curious!
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