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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)

#91
post #87

I know I'm wrong, but it feels like tis' in the definition of what a prime factor is, a prime factor being the fundamental indivisible integers. If so, considering that multiples of the same numbers are always the same, and all numbers that are prime are indivisible, the only way the conjecture could be false is if there were indivisible numbers that aren't prime. Definition inconsistency. That's why I'm not a mathem…

Actually, you're pretty much spot on. The word you want is "irreducible", rather than "indivisible". In general rings there are irreducible elements and prime elements, and they have different definitions. You're looking for a unique factorisation into irreducibles.

https://en.wikipedia.org/wiki/Irreducible_element https://en.wikipedia.org/wiki/Prime_element

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

#92
post #71

Earlier quoted context omitted.

2+2=4 is obvious. The axiomatic proofs are mostly a meaningless and boring exercise that mathematicians invented when they wanted to axiomatize everything. They have nothing to do with whether something is obvious or not. It isn't as if it was possible to doubt that 2+2=4 before the invention of the Peano axioms.

There's a lot of historical context you're missing if you think axiomatic proofs are meaningless. Around the late 1800s, a few contradictory proofs started popping up because people weren't being rigorous enough (the example I know involved proofs about pointwise vs uniformly continuous functions just being referred to as "continuous"). Then a few paradoxes were discovered (like Russel's paradox, the set of all sets…

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.

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

#93
post #90
post #87

I know I'm wrong, but it feels like tis' in the definition of what a prime factor is, a prime factor being the fundamental indivisible integers. If so, considering that multiples of the same numbers are always the same, and all numbers that are prime are indivisible, the only way the conjecture could be false is if there were indivisible numbers that aren't prime. Definition inconsistency. That's why I'm not a mathem…

It could also be false if there were two different pairs of prime numbers with the same product. See Answer 4 in the article.

Ah

I love it when my error is so simple that there isn't any question in my mind I missed something.

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

#94

Wouldn't multiple possible factorizations require numbers that both are and aren't divisible by certain numbers? If it's divisible by a number, the number must appear in its factorization and vice versa.

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 may reply.

Well, this is the non-obvious fact. If you don't already know for certain that prime factorizations are unique (and this is the claim in contention), then it's not obvious why X couldn't have multiple prime factorizations, some including p directly and some not including p directly even though p ends up dividing the overall product.

The non-obvious fact is that a prime can't divide a product unless it divides one of the factors. Why should that be true?

(It IS true of the integers, but it's NOT true of other very similar structures. The reason it's true of the integers, the special property that makes the whole thing tick, is, ultimately, that Euclid's algorithm shows us how the values not divisible by prime p are in fact invertible modulo p (and vice versa), so that the values not divisible by p are closed under multiplication. But that's not at all immediately obvious!)

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

#95
post #87

I know I'm wrong, but it feels like tis' in the definition of what a prime factor is, a prime factor being the fundamental indivisible integers. If so, considering that multiples of the same numbers are always the same, and all numbers that are prime are indivisible, the only way the conjecture could be false is if there were indivisible numbers that aren't prime. Definition inconsistency. That's why I'm not a mathem…

Actually, you're pretty much spot on. The word you want is "irreducible", rather than "indivisible". In general rings there are irreducible elements and prime elements, and they have different definitions. You're looking for a unique factorisation into irreducibles. https://en.wikipedia.org/wiki/Irreducible_element https://en.wikipedia.org/wiki/Prime_element

Thanks!

Trying to read that is like trying to learn an entire new language by reading a sentence in it. Way too many words to already understand what it's even defining (commutative ring, irreducible polynomials, UFDs, principle ideal, nozero prime ideal... and I'm not following why p divides ab in R... or even exactly what that means).

Searching youtube kahn academy and numberphile, but not turning up anything. exactly how high level is this?

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

#96
I'm going to be a little bit contrarian and disagree. I totally see where the author is coming from, but it comes across of mathematical self-aggrandizement.

I especially disagree with his interpretation of the layman's experience. They're not assuming the proof or begging the question; they have a secondary unrealized assumption that is not at all their fault.

The fundamental theorem of arithmetic is "obvious" because we grow up learning a system where if p|ab then p|a or p|b. As mathematicians, we know that Euclid's lemma is a requirement for a unique factoring domain, and we understand that the choice of set can affect whether the lemma is true.

For a non-mathematician, this is an inherent part of their conception of numbers, factoring, and their definition of the word prime.

Put yourself in the position of talking to someone who thinks prime factorization is "obvious". Where will things go wrong in their explanation? In answer 2: they have a completely deterministic way to prime factor, and it relies on Euclid's lemma.

So you ask them: "Well we know 6 is divisible by 2. What if we break it into 2 numbers that aren't divisible by 2?"

"Well then this wouldn't work. But that's not possible."

"But what if it was?"

"But that's not how numbers work."

"But what if I took each of those numbers and replaced them with two numbers and that I'm going to call one big number. And when I multiply those together I'm going to multiply the first number of each pair together, then multiply the second number of each pair together. Then I'll multiply the result of that second multiplication with -5, and add that final answer with the product of the first pair I did earlier."

"uh"

"If I do that, then do you believe that I could get two factors of 6 where neither pair of numbers is completely divisible by 2? Try (1,1) and (1,-1) and you'll see that it works: 1 x 1 = 1, 1 x -1 = -1. That -1 x -5 = 5, plus the original 1 gives me 6. See! It's not obvious."

"Yeah, uh, you didn't tell me I could make a new type of number and multiplication up on the spot."

"No no no, it's all mathematically sound. You see, those second numbers are just the coefficients of the square root of negative 5."

"Okay, so is there any rule I've learned that I can trust?"

"So I bet you think it's obvious that when you add two numbers, you can always get an answer..."

We are essentially telling kids that (American) football is a game where two groups of people line up, block, run formations and routes, give the ball to someone, and try to get it into the end zone for points.

Then, when it's your turn to go on offense, you drop back, throw a pass to your undefended wide receiver, and act amazed that the other team didn't even consider that you might throw the ball simply because you didn't forbid it.

The fundamental theorem of arithmetic was proven around 1,800 years before sqrt(-5) was even conceived of. Over two millennia before Ring Theory. Those proofs were correct for the systems in which they were written. They might even be obvious within that system. That that they are not generally true does not change things.

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

#97

Earlier quoted context omitted.

I didn't claim that FTA is the most practical way to perform Gödel numbering. I just said that it's my favorite application of it, in a general sense, not that I'd use it to do so. Preferences are subjective.

"Application" in mathematics usually means that something is used as a necessary component of a solution in another area. If Gödel numbering is an "application" of the FTA, that strongly suggests that you're saying that FTA somehow enables the possibility of Gödel numbering, or else that it is exploited somehow to endow that numbering with convenient properties (without loss of generality). Is that true?

Yes, exactly. The reason I said that is because Gödel himself used FTA to build his observation (Göodel numbering).

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

#98

Earlier quoted context omitted.

How are you getting a "theoretical pure black" without having a rigorous theory of color? If you're just going to call some things black and some things red, you'll find that a lot of things that people call black are lighter than a lot of other things that people call red.

I’ve already told you what to do if the term “pure black” causes problems. The color represented by #600000 on my monitor is always darker than the color represented by #FF0000, regardless of the existence of any theory of color. If you claim that some formal theory of color convinced you of that, then it would mean that a different theory of color could conceivably convince you otherwise. But this is clearly impossi…

But none of this is true. #600000 on your monitor might be lighter than #FF0000 on your monitor if they're compared at different times, or if part of your monitor happens to be in the shade. Or if I have a weird viewing angle to your monitor.

If you want to argue about what #600000 and #FF0000 should look like, you're back to a rigorous theory of color.

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

#99
post #94

Wouldn't multiple possible factorizations require numbers that both are and aren't divisible by certain numbers? If it's divisible by a number, the number must appear in its factorization and vice versa.

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.

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

#100

Earlier quoted context omitted.

> Why not? It seems at this point you are assuming something that is generally deduced as a consequence of the FTA. According to the OP, this result does not rely on the FTA -- he claims to derive it from the Euclidean Algorithm. makomk does the derivation in a sibling comment. > In particular, you have assumed that the product of the p_m is k (with appropriate exponents), and because c is in p_n we know that c|k, an…

>> Why not? It seems at this point >> you are assuming something that is >> generally deduced as a consequence >> of the FTA. > According to the OP, this result > does not rely on the FTA -- he > claims to derive it from the > Euclidean Algorithm. Yes. > I can't do that, so I'm taking > his word for it, but that doesn't > make the proof circular. But you should say that you are relying on this. As it is you are simpl…

> But you should say that you are relying on this.

I do say that. It's right there at the bottom of my comment.

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