Why isn't the fundamental theorem of arithmetic obvious? (2011) - <AUTHOR> https://gowers.wordpress.com/2011/11/13/why-isnt-the-fundamental-theorem-of-arithmetic-obvious/ <comment><USER2> Let me try to explain with an example outside of mathematics:

            All swans are white.

For centuries (possibly millennia, as Juvenal thought it, too), that was obvious (in western Europe) to anyone studying nature. Then, Willem de Vlamingh returns from a journey to Australia with some dead black swans.

Now, there are various options. Some of them are:

            - You can drop your claim that all swans are white.
            - You state these aren't swans because they aren't white.

Both have merit, but even if you choose the latter, it is hard to keep claiming that it is _obvious_ that all swans are white.

This is somewhat similar. Smart men have studied numbers for centuries, and have thought hard about what it means to be a number. They came up with a set of properties and laws that they thought was necessary and sufficient for a set of objects to behave like numbers. The fundamental theorem of arithmetic was not in that list, as it was thought one could derive it from simpler laws, or, possibly, nobody even thought of doing that, as they thought it to be obvious (I do not know enough of the history of mathematics to know which is true)

Suddenly, somebody comes up with a set of objects that abides by those laws, but for that set, the fundamental theorem of arithmetic does _not_ hold.

Now, you can claim that set of objects isn't a set of numbers, but you can no longer claim that the fundamental theorem of arithmetic is obvious.

<comment><USER3> Gowers argues that even without considering generalizations of the reals, it is not "obvious". He argues that if the Theorem were "obvious", we should quickly be able to say that 23 x 1759 != 53 x 769, without multiplying them out.

<comment><USER4> Perhaps I'm dim but that seems like a ludicrous argument. The inequality _is_ obvious, if you know that all the numbers are prime, and it's not if you don't. Ergo, what's non-obvious is that the big numbers are prime, not that fundamental theorem is true.

Shouldn't the standard of proof should be something like whether it's obvious that:

        23 * 7 * 11  =/=  17 * 7 * 13

? In this case the inequality is obvious, to the extent that you can convince yourself of it without multiplying anything. Doesn't this imply that therefore the theorem is obvious?

<comment><USER3> It is only obvious if you take the theorem as a given (as mathematicians typically do). However, I bet if you are given a slightly different system, it would be difficult to say whether this is true: For some natural number N, do there exist n primes p1,..,pn such that p1..pn = N and no other m primes q1,..,qm where q1..qm = N ? As an example, Gowers gives the complex numbers, where it may intuitively appear to be true ... but it is not.

Gowers mentions that, to him, an obvious argument would be: 23 x 22 != 21 x 25, since 2 divides 23x22, but 2 does not divide 21x25.

<comment><USER4> > Gowers mentions that, to him, an obvious argument would be: 23 x 22 != 21 x > 25, since 2 divides 23x22, but 2 does not divide 21x25.

I don't follow the distinction - isn't the same true of my example, just swapping "2" for "11" or "23"?

I mean, given Gowers' standing I'm prepared to assume his premise is true, and I didn't follow the last argument (by comparison to other kinds of rings) at all so I'm guessing that it's the "real" reason why the theorem isn't obvious. But the preceding arguments made no sense to me at all.

<comment><USER3> The difference is that divisibility by 2 can be tested almost instantly. (Only the least significant digits of each factor matter.) However, testing each factor for divisibility is often as tedious as the original multiplication!

I think we're all at a loss for a good definition of "obvious".

</comment></comment></comment></comment></comment></comment><comment><USER5> Some more theorems that seem obvious but require non-trivial proofs:

[https://en.wikipedia.org/wiki/Intermediate_value_theorem](https://en.wikipedia.org/wiki/Intermediate_value_theorem)

[https://en.wikipedia.org/wiki/Jordan_curve_theorem](https://en.wikipedia.org/wiki/Jordan_curve_theorem)

[https://en.wikipedia.org/wiki/Isoperimetric_inequality](https://en.wikipedia.org/wiki/Isoperimetric_inequality)

[https://en.wikipedia.org/wiki/Hairy_ball_theorem](https://en.wikipedia.org/wiki/Hairy_ball_theorem)

[https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstei...](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem)

</comment><comment><USER6> 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.

<comment><USER5> I've kinda stopped believing that high intelligence always leads to high quality discussions. Every HN thread about math or physics has many misguided comments, coming from people who are probably very smart in their own fields. I've seen that on LessWrong too, really smart math/CS people talking about biology can get demolished by a second year biology student. Noticing my own cluelessness about a topic is a very subtle skill, it took me years to learn and I'm probably not there yet. Sometimes I even feel that math/CS education has damaged me in some ways, made me too arrogant, though obviously it gave me a huge advantage in other ways.

<comment><USER7> > I've seen that on LessWrong too, really smart math/CS people talking about > biology can get demolished by a second year biology student.

IMO, LessWrong and other communities based around critical thinking tend to either foster a sense of intellectual arrogance or attract people who already have that quality.

> Sometimes I even feel that math/CS education has damaged me in some ways, > made me too arrogant, though obviously it gave me a huge advantage in other > way.

A lot of education in science and engineering is based around teaching people to walk up to a problem they are completely unfamiliar with and make a reasonable attempt at solving it. That is an extremely useful skill in some cases, but very dangerous in others.

<comment><USER8> I generally agree, however I would like to know for what cases you think the skill of solving unfamiliar problems is dangerous.

<comment><USER7> It's dangerous when it leads a person to choose "I'm going to figure this out on my own" over "I'm going to find someone who can solve this" in cases where they really should have known to pick the latter.

</comment></comment></comment></comment></comment><comment><USER9> It actually becomes much easier to understand if you approach it from the other direction.

Some assumptions

\- A prime is only evenly divisible by itself and one.

\- One is not a prime.

\- A composite number is generated by multiplying two or more primes.

\- No number of primes can be multiplied together to make another prime. That would mean that the so called prime generated is composite.

Now, say we have a number n with two different prime factorizations, f1 and f2, where f1 includes a certain prime p, and f2 does not, and where n is evenly divisible by p, and n/p = m.

Now, if f1 and f2 are both prime factorizations of n, then consider what happens when we do this:

f1 = p * m = f2

m = f2 / p

f2 does not contain p, and we cannot construct a prime p from other primes or composites. This means that f2 cannot be evenly divisible by p, as that would require it to have a prime factor of p which it does not. Therefore f2 cannot be another prime factorization of n.

That may not be perfectly rigorous, but I'm certain the gist of it is true.

<comment><USER10> Something I learned a long time ago is this: if someone claims to have a proof, trying following the same proof in a different case where the conclusion is actually false, and see where their "proof" fails.

So let's try applying your reasoning to the example given in the linked article. There it shows that in the integers extended by sqrt(-5) we have 2x3=(1-sqrt(5))x(1+sqrt(5)). So using your specific reasoning:

        Let's take n=6

        f1 : 2 x 3
        f2 : (1-sqrt(5)) x (1+sqrt(5))
        p  : 2
        m  : 3

You now say:

        f2 does not contain p, ...

Correct.

        ... and we cannot construct a prime p
            from other primes or composites.

True.

        This means that f2 cannot be evenly
        divisible by p, as that would require
        it to have a prime factor of p which
        it does not.

But f2 _is_ evenly divisible by p, despite not including p in the list of primes being multiplied together to give n. So your line of logic fails at this point.

This is actually assuming (something equivalent to) the FTA. The example shows a case where f2 _is_ evenly divisible by p, so your deduction here is wrong.

It is subtle.

</comment></comment><comment><USER5> In math school we had a saying: "obvious means easy to prove". So the problem is about recognizing the difference between proofs and non-proofs. The hard but satisfying way to learn that difference is to start with axioms. Take some simple system of axioms that holds for Z, and try to prove the FTA from these axioms alone. Then check that the axioms aren't satisfied by Z[sqrt(-5)], or the even numbers, or some other ring where the FTA doesn't hold.

<comment><USER11> Eh. I read the commentary here and tried proving the fundamental theorem of arithmetic. Here goes:

Suppose some integer _k_ has two different prime factorizations -- it is the product of some set of _n_ primes raised to nonnegative integer powers, and also of some other set of _m_ primes raised to nonnegative integer powers. Call those sets p_n and p_m.

Observe that there is no prime number which is assigned a nonzero exponent by p_n but not p_m, and there is no prime number which is assigned a nonzero exponent by p_m but not p_n. If p_n assigned a positive exponent to any prime _c_ while p_m assigned _c_ a zero exponent, then the product of p_n would be congruent to 0 (mod _c_ ), but the product of p_m would not, and therefore the two products would not equal the same number. (And symmetrically.)

Therefore, p_m and p_n differ only in the nonzero exponents assigned to their various primes. Let _g_ be the set of primes in p_m and p_n with the minimum exponent assigned by either p_m or p_n, let p_M be p_m with all exponents reduced by the exponent assigned by _g_ , and let p_N be p_n with all exponents reduced by the exponent assigned by _g_. Observe that the exponent assigned to any prime is either 0 in each, or 0 in one of p_M or p_N and positive in the other. We can observe that, since p_m is not equal to p_n, one of p_M or p_N must assign a nonzero exponent to some prime.

Let γ, μ, and ν be the integer products of _g_ , p_M, and p_N. By hypothesis, γμ = γν, which means that μ = ν. But now we have two prime factorizations (p_M and p_N) of the same number (μ) for which one factorization assigns a positive exponent to some prime, and one factorization assigns an exponent of zero. By our earlier result we know that this is impossible.

\----

I needed a lot of symbols, and if I were formally typing this up I'd need more, but it didn't seem like a very difficult proof -- I spent more time typing up this comment than working out the proof. Reading the piece, I see that it is specifically called out:

> We'd be able to see instantly that 23 × 1759 ≠ 53 × 769 if we knew that a > product of two non-multiples of 23 was always a non-multiple of 23.

So I guess if you're comfortable with modular arithmetic, you can fairly consider this an obvious proof. It relies on another result about primes, but it's very common that one result makes another result easy, and a blanket disallowal of that approach leaves you saying that proving anything is as tricky and non-obvious as proving, _proving_ , that 2+2=4.

<comment><AUTHOR>
        If p_n assigned a positive exponent
        to any prime c while p_m assigned c
        a zero exponent, then the product of
        p_n would be congruent to 0 (mod c),
        but the product of p_m would not ...

Why not? It seems at this point you are assuming something that is generally deduced as a consequence of the FTA.

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, and hence we know that k=0 (mod c). So your claim here is false. It is actually assuming the FTA.

<comment><USER6> This is correct. To be fair, though, the standard terminology is confusing: calling a number only divisible by 1 and itself a "prime" already assumes the FTA.

In a more abstract setting, "p is prime" means that if p|ab, then p|a or p|b, and "irreducible" means only divisible by itself or a unit (in this case 1). The FTA corresponds to unique factorization into _irreducibles_ , and the fact that irreducible and prime are the same thing is a consequence of unique factorization. (In an integral domain, every prime is irreducible; in a unique factorization domain, the converse is also true).

<comment><AUTHOR>
        ... the standard terminology is confusing:
        calling a number only divisible by 1 and
        itself a "prime" already assumes the FTA.

Sort of, but not really. I'm not going to disagree with you, but make the following observation. People reading this article are likely to know about primes, and what you quote here is most likely the definition that they would be accustomed to. Introducing a new, technical term and then trying to describe the details of the difference would most likely derail the purpose, and abusing the terminology a little is perhaps justified, especially when it aligns with people's existing knowledge.

But you are correct, and the reason we have these terms is exactly to avoid some of the "intuitively obvious" misconceptions.

</comment></comment></comment></comment></comment><comment><USER12> 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 both addition and > multiplication, and every number has an additive inverse. And as with > integers, if you divide one by another, you don't always get a third, so the > notion of divisibility makes sense too. That means that we could if we > wanted try to define a notion of a "prime" number of the form a+b\sqrt{-5}.

He skipped over the fact that the positive integers are well-ordered by < (and in fact < also respects the arithmetic operations on the positive integers as well). I just can't take the comparison seriously without this being explicitly discussed, because the ordering of the integers is incredibly fundamental to human intuition about them.

<comment><USER13> Here's a stumper then: why do Z[sqrt(-1)] and Z[sqrt(-3)] have unique factorization but Z[sqrt(-5)] doesn't?

<comment><USER14> Do mathematicians actually have an answer for this? Or is it considered an open question.

<comment><USER15> There's a pretty good handle on it, in the form of the ideal class group[1].

[1]: [https://en.wikipedia.org/wiki/Ideal_class_group](https://en.wikipedia.org/wiki/Ideal_class_group)

</comment></comment></comment></comment><comment><USER16> This blog post illustrates well the infuriating tendency of academics to teach by calling people stupid. The substance of his answers are:

1\. " If you think it's obvious, then you're probably assuming what you need to prove" i.e. "If you think this, you're wrong."

2\. "Just because you've got a completely deterministic method for working out a prime factorization, that doesn't mean what you work out is the only prime factorization" i.e. "If you think doing it this way works, it doesn't."

3\. "Look, it just bloody well isn't obvious, OK?" i.e. "If you get frustrated that I keep telling you you're wrong, that's your fault for being wrong."

4\. "If it's so obvious that every number has a unique factorization, then why is the corresponding statement false in a similar context?" i.e. "I won't tell you what you did wrong, but instead I'll show you why your answer can't possibly be right."

It is unconscionable that he does not identify the actual fact people erroneously use without justification when thinking the fundamental theorem is obvious. FYI, that fact is

"If a number n can be written as the product of a fixed list of (not necessarily distinct) primes p_1, p_2, ..., p_n, then any prime p dividing n appears on this list."

(there is a minority of mathematicians who refer to _this_ fact as the fundamental theorem of arithmetic, not unique prime factorization; the obvious argument is going from the fact to unique prime factorization, proving the fact is the unobvious argument)

</comment><comment><USER15> A nice discussion, but a bit of a nitpick: in Z[sqrt(-5)], 2, 3, 1+sqrt(-5), and 1-sqrt(-5) are actually irreducible, not prime.

In an integral domain D, a nonzero element x is called irreducible if x is not a unit and whenever x = ab (for a, b in D), then one of a or b is a unit.

On the other hand, a nonunit element x in D is called prime if for all a, b in D, if x divides ab then x divides a or x divides b (by "x divides ab", I mean that there's some element--call it y--in D such that xy = ab).

In Z (or any unique factorization domain[1]), these concepts coincide. In Z[sqrt(-5)], however, there are irreducible elements that are not prime. In particular, 2 is irreducible in Z[sqrt(-5)], but it isn't prime, since 2 divides (1+sqrt(-5))*(1-sqrt(-5)), but 2 divides neither 1+sqrt(-5) nor 1-sqrt(-5).

[1]: [https://en.wikipedia.org/wiki/Unique_factorization_domain](https://en.wikipedia.org/wiki/Unique_factorization_domain)

<comment><USER17> This is brought up at the top of the comments section, along with Gowers' response and a neat comment from someone called Fabian:

>It seems it is the thoughtless combination of the definitions of "prime" and "irreducible" that makes the theorem appear obvious.

</comment></comment><comment><USER18> 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.

<comment><AUTHOR>
        Wouldn't multiple possible factorizations
        require numbers that both are and aren't
        divisible by certain numbers?

Yes.

        If it's divisible by a number, the number
        must appear in its factorization and vice
        versa.

That is what the FTA says, so you are saying that the FTA is true, but that's just saying that you believe it. The article is trying to point out why once you know enough about how arithmetic works,it's no longer obvious. In particular, there are fairly natural rings that look very similar to the integers, but where this is not true.

Taking the example in the linked article, it's true if we extend the integers by sqrt(-1), but it's not true if we extend the integers by sqrt(-5). In that ring we have:

       2 x 3 = (1 + sqrt(-5))x(1 - sqrt(-5))

So (1 + sqrt(-5)) appears in one factorisation of 6, but not in another. So in this case, factorisations are not unique.

<comment><USER19> Obvious is different than easy to prove. The concepts of multiplication, division, prime number and "divisible by" are much older than formal proofs and arbitrary sets of axioms.

Let's say I only now what multiplication is and that AxB = BxA and that prime number can't be written as AxB unless A or B are 1.

Now it's obvious that there are factorings of a number: you just divide it by smallest possible prime divider until you reach 1, that process obviously terminates every time and produces a finite factoring.

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.

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. We therefore divide our number by the prime divider at that first position and continue the process proving that dividers at all the positions must be the same or that F1 F2 are factorings of a different number. It is in fact obvious to someone who understands multiplication.

As to some points from the article:

>>it is obvious that 23\times 1759 is not the same number as 53\times 769.

It is, we sort them from the smallest to bigger prime in the factorization. If they were the same number that would mean the same number's smallest prime diviser is 23 and 53 at the same time.

You may want to argue that it's not obvious that to be divisible by a prime number it must appear in a factorization - here I just assume you understand it's obvious for anyone who knows what multiplication is and that the fact that it's not obvious to your system with your axioms is your problem altogether.

>>If it's so obvious that every number has a unique factorization, then why is the corresponding statement false in a similar context?

It's not a similar context, at least for non-mathematician. You are introducing some weird objects in the form of a + b*sqrt(-5). I may not even understand the concept of a complex number but I still can understand FTA. Btw, for that FTA isn't obvious because there is no order for those objects, it's not clear what is bigger or smaller than something else and which position given object is in if we sort them from the smallest.

I get it: FTA is hard to prove in your nice formal system with your nifty arbitrary chosen axioms and formal deduction rules. It's bloody obvious to the caveman who can do multiplication by putting stones in a rectangle though.

<comment><AUTHOR>
        > 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 has a "smallest" factor of 2, and a "smallest" factor of (1 - sqrt(-5)).

        > It is in fact obvious to someone who
        > understands multiplication.

So you are claiming that the author of the linked article, Prof Sir Tim Gowers, winner of the Fields Medal, Fellow of the Royal Society, doesn't understand multiplication? Might I instead suggest that you don't understand the things that might go wrong, and the subtleties that lurk underneath.

<comment><USER19> >>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 counting them. The concepts of primes, multiplication, divisor don't instantly make sense for those objects and I am not sure why you ask me to extend them. I am just claiming FTA is obvious for natural numbers and straightforward multiplication.

>> 6 = 2 x 3 >> 6 = (1 - sqrt(-5)) x ((1 + sqrt(-5))

I mentioned sorting from the smallest, can't do that with your sqrt thing. Another obvious thing with multiplication is that the more you multiply the more you get which isn't true for a + sqrt(-5).

>>So you are claiming that the author of the linked article, Prof Sir Tim Gowers, winner of the Fields Medal, Fellow of the Royal Society, doesn't understand multiplication?

No, I haven't claimed that. I think FTA is quite obvious to him but it's not obvious if you try formally define arithmetic using the smallest sensible subset of axioms. It is obvious is you just understand multiplication and can perform it by forming rectangles and then rectangles from rectangles (in case of 3 terms to multiply).

>>ight I instead suggest that you don't understand the things that might go wrong, and the subtleties that lurk underneath.

There are many subtleties in defining and proving things formally. That's a different point altogether.

<comment><USER20> 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.

<comment><USER21> But he clearly does understand, and some gentle mockery at the other side is called for when they play dumb.

I'm sure the fundamental theorem of arithmetic has a non-obvious proof. And perhaps that's precisely what a mathematician means every time they say "non- obvious". If that were all just made explicit to this general interest site in the first place, perhaps we'd have nothing to discuss.

But if we are trying to play coy here, 1+1=2 also requires a non-obvious proof to anyone not versed in formal methods. I looked a proof up:

[http://mathforum.org/library/drmath/view/51551.html](http://mathforum.org/library/drmath/view/51551.html)

I don't know how long it would take me, working alone, to come up with that proof. It's non-obvious because it took humans probably 100,000 years to come up with it, even though we've had the IQ to do it for a long time. I don't think we could agree on what constitutes a proof of it without some social aspect and convention, so non-obvious by means of proof.

But, we've been using and making predictions about the world using 1+1=2 for very much longer. That seems like a pretty worthwhile definition of obvious.

</comment></comment></comment></comment></comment></comment></comment><comment><USER22> That's not my question. My question is, why is it the fundamental theorem of _arithmetic_?

<comment><USER23> Yeah! I do arithmetic most days, and i very rarely have to prime factorise anything!

</comment></comment><comment><USER24> I haven't looked at any proofs for the FTA. Could somebody point out if I made any mistakes on the one I've arrived at?

[1] Proof by induction that if a positive integer has a prime factorization, then it is unique.

We're inducting over Z_N, where Z_N is the set of all positive integers with at least one known prime factorization using exactly N number of primes. Call this factorization Pn = p_1 * p_2 * ... p_n

For every case, split into proof by cases and contradiction: suppose there is an element Z in set Z_N had another factorization Fn.

\- If Fn has the same number of factors as Pn, divide both sides by p_i. Since Fn / p_i must be an integer, Fn must contain p_i, or else one of its factors f_i actually isn't prime by Euclid's Lemma.

\- If Fn has k more factors than Pn, then divide Pn by (f_1 * f_2 * ... * f_n). Then (f_n+1 * ... * f_n+k) = X for some composite integer X > 1\. Thus Fn = X * (f_1 * f_2 * ... * f_n) = (p_1 * p_2 * ... * p_n) = Pn. Since (p_1 * p_2 * ... * p_n)/Z must be an integer, Z must divide into one of the factors p_i by Euclid's lemma, which is impossible since they are prime by definition.

\- Similar argument to above if Pn has more factors.

[2] Proof by contradiction: Every positive integer Y greater than one has a prime factorization.

Suppose not. We know Y = Y * 1. So Y must be composite in order for it to not have a prime factorization. Hence, we know that Y = a * b, (a, b > 1). At least one of the two must be composite or else Y has a prime factorization, so let's say (a) is composite. Then a = c * d (c, d > 1). Then at least one of the two factors must be non prime or else we've found the prime factorization for Y. Repeat ad infinitum to show that if Y does not have a prime factorization, then it must be the product of an infinite number of composite factors, with every factor great than 1. Hence contradiction.

So every positive integer greater than 1 has a prime factorization, and it must be unique.

<comment><USER2> _" If Fn has the same number of factors as Pn, divide both sides by p_i. Since Fn / p_i must be an integer, Fn must contain p_i, or else one of its factors f_i actually isn't prime by Euclid's Lemma."_

You would have to prove that first. [https://en.m.wikipedia.org/wiki/Euclid%27s_lemma](https://en.m.wikipedia.org/wiki/Euclid%27s_lemma):

 _" This property is the key[4] in the proof of the fundamental theorem of arithmetic

[4] In general, to show that a domain is a unique factorization domain, it suffices to prove Euclid's lemma and ACCP."_

([https://en.m.wikipedia.org/wiki/Ascending_chain_condition_on...](https://en.m.wikipedia.org/wiki/Ascending_chain_condition_on_principal_ideals) looks more intimidating than Euclid's lemma to me, but may be easier to prove.)

<comment><USER24> Got it. I was trying to do a proof while avoiding abstract algebra as much as possible, given how rusty I am with it.

Thanks for the feedback, and additional things to take a look at!

</comment></comment></comment><comment><USER25> I'm surprised Wittgenstein hasn't been mentioned here. He thought and wrote extensively about the foundation of mathematics. In fact, the write-ups of a series of his lectures features discussions between him and Turing (and other luminaries) about what maths is about (and W tends to come over stronger on the subject):

[https://www.amazon.co.uk/Wittgensteins-Lectures- Foundations-...](https://www.amazon.co.uk/Wittgensteins-Lectures-Foundations- Mathematics-Cambridge/dp/0226904261)

It's great fun to read.

</comment><comment><USER26> I think a lot of people consider uniqueness of prime factorization to be obvious just because it's stated as true so frequently.

If you want to convince someone it's not obvious, I would ask them a simpler, weaker question:

 _Can you find four different prime numbers, a b c d, so that ab = cd? If not, why not? You may not cite unique-prime-factorization._

They will find themselves really wanting to cite unique prime factorization, and unable to prove it otherwise, which will convince people that it's not obvious.

</comment><comment><USER27> Why isn't 2+2==4 obvious?

[http://us.metamath.org/mpegif/mmset.html#trivia](http://us.metamath.org/mpegif/mmset.html#trivia)

<comment><USER28> 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.

<comment><USER27> I believe you're mistaken. There is value in axioms and axiomatic proofs: two different people will most definitively have a different notion of "obvious", and even have a different understanding of a mathematical problem. So a proof may be accepted by one person and rejected by another.

Given a set of axioms and proofs it's possible to mechanically check a proof. It's not quite possible to reliably check proofs otherwise.

<comment><USER28> I'm not saying anything about the ability to check proofs, or the value of axiomatic proofs in general, only that 2+2=4 specifically doesn't require an axiomatic proof in order to convince anybody that it is true. This is like saying that we need a rigorous theory of color in order to be convinced that black is darker than red. Mathematicians didn't axiomatize natural numbers in order to show that 1+1=2 or 2+2=4, or any other trivial arithmetical fact. They have never doubted it, and I don't know what "doubting 2+2=4" even means. In fact the entire process is reversed: they invented axioms that can form a formal basis for what we __already know to be true __. If Peano axioms proved that 2+2 = 6 - they wouldn 't be a valid axiomatization of the natural numbers. One cannot axiomatize the natural numbers without already assuming that all the basic arithmetic facts we know about them are true (or else he wouldn't be axiomatizing the natural numbers, but something else). Somebody who rejects 2+2=4 has a problem understanding human language, not proofs.

<comment><USER11> > This is like saying that we need a rigorous theory of color in order to be > convinced that black is darker than red.

You do, if you want to be right. The fact that you can get people to agree with you doesn't make you right, and red is frequently darker than black by some pretty normal definitions of "darker". Red and black are differentiated by the shape of their reflective spectrum, not the amplitude.

<comment><USER29> You guys are basically arguing over Moore's here-is-one-hand problem.

[https://en.wikipedia.org/wiki/Here_is_one_hand](https://en.wikipedia.org/wiki/Here_is_one_hand)

pavelrub's point is that you sometimes have less reason to believe the axioms of your formalization than their derived consequences. We have better reason to believe the intuitive idea that 2+2=4 than we do any putative axioms of arithmetic. If we derived that 2+2=5 from some particular axioms of arithmetic, we would conclude those axioms were wrong (or rather, were not the proper system for formalizing 2-plus-2-ness) rather than conclude that 2+2=5.

</comment></comment></comment></comment></comment></comment><comment><USER30> 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 mathematician.

<comment><USER31> 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/Irreducible_element) [https://en.wikipedia.org/wiki/Prime_element](https://en.wikipedia.org/wiki/Prime_element)

<comment><USER30> 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?

<comment><USER32> This is usually part of an introductory undergrad course in algebra. What you're looking for is called "ring theory".

Michael Artin's _Algebra_ is a good, concrete book with tons of motivation and zero prerequisites. Wanna try it? ;)

</comment></comment></comment></comment><comment><USER33> My favorite proofs of these basic number theory facts are in Apostols, Intro to Analytic Number Theory. At best the FTA using the proof there is only slightly less than obvious (of course obviousness probably varies widely based on mathematical background / maturity).

<comment><USER34> Sure, to a mathematician, these things become old hat, but note that Apostols spends several pages developing properties of the GCD that aren't generally considered "obvious"; in particular, he argues for Theorem 1.2 (that any two integers a and b have a common divisor of the form ax + by, which is in turn therefore divisible by all their other common divisors) by tracing out the steps to recursively compute this combination-as-GCD (i.e., the Euclidean algorithm), recast as an inductive proof. I don't think the layperson would find this obvious at all (in the way they might have claimed FTA to be obvious), or even be aware of this fact, though it plays a key role in building up to the eventual proof of the FTA.

</comment></comment><comment><USER35> My own attempt to deal with this issue (2013): [http://thinkinghard.com/blog/UniqueFactorisation.html](http://thinkinghard.com/blog/UniqueFactorisation.html)

</comment><comment><USER36> I agree 50%. But, it becomes pretty obvious once you realize the significance of prime numbers, their applications and the scarcity of them.

</comment><comment><USER37> 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.

<comment><USER38> I definitely get where you're coming from, but the introduction of Z[sqrt(-5)] was just there to illustrate that proof is hard. Even if you never invoke an exotic example it's hard to have proof that FTA holds.

The simplest way to get at that perhaps is to just increase the scope. It's obvious to you that 10 and 15 have unique factorizations. You can compute them and convince yourself that nothing else would work.

It's harder to do the same for 9^87654321-1 but you could in principle. Furthermore, you believe it would work (you could, e.g., program a computer to do it and wait a few hours? days? something like that) but only basically because you see no reason for your experience with small numbers to eventually stop.

But now let me tell you about the Ackermann function and ask you about A(1000, 1000). Let's be clear, this is _just_ a really large number, but now we're at a point where a computer couldn't write down the factorization in the lifetime of the universe. Are you _still_ sure it's obvious?

It might at this point become more clear that "I see no reason for my intuition about small numbers to stop" is weaker than "I know this always works".

</comment></comment><comment><USER39> I've recently learned about Gödel numbering and I have to admit; it's my favourite application of this theorem. Sorry RSA.

<comment><USER40> You don't actually need FTA to perform Gödel numbering. ASCII will do just as well.

<comment><USER39> Yes. But Gödel originally used FTA, no?

<comment><USER40> He probably did. But for a programmer, ASCII is already something we're very familiar with, so why not use that?

<comment><USER39> 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.

<comment><USER41> "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?

<comment><USER42> > "Application" in mathematics usually means that something is used as a > necessary component of a solution in another area.

I think that sentence is entirely true if you delete the word 'necessary', and otherwise entirely false. I think that almost everyone would agree that at the heart of modern cryptography is an application _par excellence_ of modular arithmetic, but I think that no-one would claim that cryptography _cannot_ be done without modular arithmetic.

</comment></comment></comment></comment></comment></comment></comment><comment><USER43> I remember reading a proof to the fundamental theorem of arithmetic (every number is composed of a unique multiple of primes) in a number theory book and really enjoying it. I would disagree with Gowers and say it is obvious intuitively, but I'd still argue it is worth writing a proof for.

<comment><USER44> Although it was ~30 years ago, I recall doing some school homework when we had just learned about prime numbers and factorisation. I remember trying to divide by random primes until I came across _a_ factorisation. It wasn't at all obvious to me (at ~10 years old) that prime factorisations were unique or that they could be found using a simple repetitive algorithm.

It _seems_ obvious now, but only because I've never come across an integer with more than one factorisation. If there were an article on HN tomorrow with the headline 'Integer with more than one prime factorisation found' I wouldn't be able to resist clicking the link.

<comment><USER45> > If there were an article on HN tomorrow with the headline 'Integer with more > than one prime factorisation found' I wouldn't be able to resist clicking > the link.

Curiously, I've only just found out that

      48016416432886585186892071037001629018831524915070361
      17449649760043615376581136847123881454516238486352419
      62687300988949648670959062041377941995335910356581948
      79838588416610716340382432762472099541373300228025778
      94213135434471675634979394732216151334015571089605667
      2861

has two distinct prime factorizations, thus providing a counterexample to Euclid's Fundamental Theorem of Arithmetic and showing that he made a mistake somewhere.

Unfortunately the truly marvellous lists of factors are too small to fit in a Hacker News comment, so you'll have to rediscover the details yourself.

<comment><USER46> Your comment has destroyed the browsing experience on mobile :)

<comment><USER45> Hah. Apologies for that! I thought the downvotes were because it was a lame joke, not that I was destroying the site!

<comment><USER47> FWIW, I downvoted you because it's a lame joke.

</comment></comment></comment></comment></comment></comment><comment><USER48> This is from 2011.

</comment><comment><USER49> "The fundamental theorem of arithmetic can be approximately interpreted 3 * 5 * 13 and 3 * 13 * 5" \- this is false. The fundamental theorem of arithmetic "can be more approximately interpreted as " 195 has one prime factorization and it only includes one 3, one 5, and one 13 and no other primes. Once we find one prime factorization we want to prove that 195 does not have a prime factorization that includes another 11 or other prime.

<comment><AUTHOR> It doesn't say that. What it says is:

        The fundamental theorem of arithmetic states that
        every positive integer can be factorized in one
        way as a product of prime numbers. This statement
        has to be appropriately interpreted: we count the
        factorizations 3x5x13 and 13x3x5 as the same, for
        instance.

That's not the same thing at all.

<comment><USER49> I miss quoted the original blog by replacing "has" for "can" and such. But the second sentence of what you quoted is misleading because that's not an correct interpretation of FTA.

<comment><AUTHOR>
        > But the second sentence of what you
        > quoted is misleading because that's
        > not an correct interpretation of FTA.

I wonder if this is a language thing. The second sentence says:

        ... we count the factorizations 3x5x13
        and 13x3x5 as the same, for instance.

This is giving an simple example of what the phrase "up to ordering" implies. The full statement of the FTA says that the factorisation is unique "up to ordering" and that's _exactly_ what the second sentence is saying.

So I still think you are mistaken, and I don't understand why you are disagreeing with what Gowers wrote. Perhaps you could give more detail about why you think he is wrong.</comment></comment></comment></comment>


Ask HN: Is it possible to make patient health records open source? - <AUTHOR> Hi there!
With Google and Amazon pushing into the healthcare space and collecting lots of available patient health records, I was wondering if it's possible to make that data open source so scientists and technical people all over the world can work on solutions, instead of only for-profit companies.
I've been wondering if it's necessary to build a company to do it - and your answers would greatly inform my decision and path to start one.
The following are the questions I've been pondering about - and would love your feedback:
- Nomenclature and data formats: how hard is it to develop a common language for many different, complex data formats with different naming conventions across countries? What exists already, what has been tried before?
- Are academic institutions (think Johns Hopkins University) allowed to share datasets publicly? I know the answer is 'Yes' since lots are already doing it, but what are specific (technical and legal) barriers to consider when doing this at scale and developing one platform?
- What are generally big challenges and problems for researchers in the medical sciences, or anyone else that professionally works with patient health data?
- I know data privacy is a big challenge - no secret. But does anyone have experience with specific technical and legal issues when it come to anonymisation, publishing data from secondary sources and their legal compliance? Please share! <comment><USER2> Have you heard of HIPAA? That means that every person would have to give permission to each data user

Anonymizing the data does not work, as you can infer and rematch against other databases. Even encrypted data is susceptible to reverse engineering

</comment><comment><USER3> Good luck. HIPAA will make that impossible.</comment>


Failure to Launch - <AUTHOR> https://medium.com/@jcahill84/failure-to-launch-ded04173ca10 <comment><USER2> I think it's alright to work on something and not launch it. Not everything is worthy of putting out into the world. And there is an opportunity cost.

<comment><AUTHOR> Agreed. This was really just written from my personal experience of being bummed out by never launching, then how awesome it felt to finally do it.</comment></comment>


EVE Online: Premeditated Murder - <AUTHOR> http://themittani.com/features/first-degree-awox-premeditated-murder?page=0%2C0

<comment><USER2> It'd be nice to have an explanation that covered the jargon for a beginner. E.g. why don't BDCI losses count? What even is a BDCI?</comment>


AI microchallenge: build a blackjack bot - <AUTHOR> https://www.kaggle.com/dansbecker/july-24-micro-challenge <comment><AUTHOR> Also interesting forum thread where users are sharing their approach and scores. So far the top performing bot has a 46% win rate: [https://www.kaggle.com/learn- forum/58735#post361651](https://www.kaggle.com/learn-forum/58735#post361651)

</comment><comment><USER2> It seems odd that the API they have defined only gives you the scores, rather than the actual cards.

</comment><comment><AUTHOR> Posted this earlier but with a link that hit a login wall, which is why I'm reposting.</comment>


A Type House Divided - <AUTHOR> http://nymag.com/news/features/jonathan-hoefler-tobias-frere-jones-2014-6/ <comment><USER2> I am partial to type design and I used to frequent Typophile and few other places where type designers hang out. It was always everyone's understanding that Hoefler and Frere-Jones were equal partners. H however has always been big on self-promotion and he generally had this aura of a cocky know-it-all, while FJ was more down to earth and approachable. FJ is 100% in the right here and H is just a greedy asshole. I lost all respect to him as I suspect did many in designer circles.

<comment><USER3> _It was always everyone 's understanding that Hoefler and Frere-Jones were equal partners_

The fact that the company was named "Hoefler & Frere-Jones" sure seems like it was designed to give that impression to the public, at least.

</comment></comment><comment><USER4> The original claim (filed on Jan. 16, 2014) can be found here: [https://iapps.courts.state.ny.us/fbem/DocumentDisplayServlet...](https://iapps.courts.state.ny.us/fbem/DocumentDisplayServlet?documentId=ydQwzs4EZjDkplRrBmkkwA==&system=prod)

Sad to see a (implied) partnership end with: "In the most profound treachery and sustained exploitation of friendship, trust and confidence, Hoefler accepted all of the benefits provided by Frere-Jones while repeatedly promising Frere-Jones that he would give him the agreed equity, only to refuse to do so when finally demanded."

Earlier discussion on the complaint and the fall-out of the partnership on Quartz (also Jan. 16): [http://qz.com/167993/frere-jones-is-suing-hoefler-for- his-ha...](http://qz.com/167993/frere-jones-is-suing-hoefler-for-his-half-of- the-worlds-preeminent-digital-type-foundry/)

<comment><USER5> Something similar happened to me when I was young and stupid and I still have a bad feeling about that decades later.

This should serve as a warning to anybody in a similar position: promises of shares are worth nothing, if you are in a vesting arrangement make sure you get the actual shares on the promised dates, otherwise you are being taken for a ride, if you are a co-founder and there is no vesting make sure you get your shares _right now_.

Particularly galling here is this paragraph from the linked text:

"Update (Jan. 17, 2:45pm ET): HFJ released a statement, set in the firm's Mercury typeface, denying Frere-Jones's claims and saying the company will henceforth be known as Hoefler & Co.: "

So, as long as you're working without a stake in the company but on vague promises your compensation is limited to your salary + having your name as an equal partner on the company letterhead but as soon as you so much as try to enforce your rights you're being removed as if you never existed. Classy.

I sincerely hope that this suit will be resolved amicably, and if it isn't that the firm will be named Frere-Jones Inc. when they're done.

Hoefler is going to have a hard time arguing that Frere-Jones was only in the masthead for decorative purposes, it certainly seems to spread the message that Frere-Jones is to be considered a full partner in the venture.

Beware of power asymmetry in relationships like these.

<comment><USER3> _Hoefler is going to have a hard time arguing that Frere-Jones was only in the masthead for decorative purposes, it certainly seems to spread the message that Frere-Jones is to be considered a full partner in the venture._

Their webpage as of a few months ago also listed him as a "Principal" [1], which isn't a precisely defined term, but usually implies something like a partner or co-owner.

[1] [https://web.archive.org/web/20131028065322/http://www.typogr...](https://web.archive.org/web/20131028065322/http://www.typography.com/about/biographies.php)

<comment><USER5> That might be worth saving somewhere as a pdf or as a notarized screenshot or something to that effect. The archive will remove pages if requested and this could come in very handy during the lawsuit.

</comment></comment></comment></comment><comment><USER6> Ah. Typography.

Please tell me I'm not the only one that expected some discussion of static vs. dynamic typing, etc. But was wondering why nymag would cover it.

<comment><USER7> The only reason I clicked was that I was wondering, "no, that couldn't be it."</comment></comment>


Microlending platform Zidisha (YC W14 nonprofit) bypasses third parties - <AUTHOR> http://www.startupdaily.net/2015/09/microlending-platform-zidisha-bypasses-third-parties-to-keep-interest-rates-down-for-people-in-impoverished-communities <comment><USER2> I wish the team at Zidisha nothing but success, but I'm skeptical that micro- lending is a practical solution for relieving poverty in the developing world.

If a viable opportunity for making small loans existed, then I would expect local financial institutions to fill that need. When those institutions do not exist, there is usually a reason, such as the lack of an adequate legal and financial system for recovering debts. Without the ability to enforce payment obligations, the line between lending and charity becomes rather slim.

It's also unrealistic to expect most people receiving micro-loans to operate effective businesses, given their limited educations and lack of direct exposure to Western entrepreneurial values. Perhaps their living conditions could be improved more efficiently by a non-profit that gave away the right to operate franchise businesses using a proven model?

I don't have the expertise to know what kind of business would be optimal for people with limited resources. But when I see the success of immigrant-run franchises within the United States, I can't help but imagine that the underlying principle could be harnessed to generate wealth for needy people overseas.

<comment><USER3> You are describing your "gut feeling", but allegedly micro-lending has been proven to work in theory and practice by actual economists.

On the other hand I am criticizing you for doing "armchair economics" without giving the references to support my previous statement :)

<comment><USER2> That's a fair point. But my view is informed by my (limited) review of the scholarly literature on microfinance. I was impressed by the economist David Roodman, who has done a lot of work on how the actual empirical evidence suggests that micro-loans are not effective at relieving poverty.

</comment></comment></comment><comment><USER4> A lot of the overhead from 'traditional' microfinance (the model popularized by Mohammed Yunus in the 70's-80s) comes from creating social/community-based incentives for repayment, referred to as social capital. This is considered the alternative to regular loans since in non-collateral lending the borrower loses nothing monetarily if they don't repay.

According to their site, the repayment rate for Zidisha is under 50%... thats pretty dismal in an industry where you normally consider a healthy MFI (microfinance institution) to be operating at high 90 percentiles of repayment.

<comment><USER5> Where do you get "under 50%"?

If you're looking at [https://www.zidisha.org/statistics](https://www.zidisha.org/statistics), you might be confused by

      Principal repaid: $607,262 (44.9% of amount disbursed)

This is for all loans made within the last year, many of which are very recent. You should add the line

      Principal held by borrowers repaying on time: $581,565 (43.0% of amount disbursed)

to get 87.9% either repaid or in good standing.

<comment><USER4> Ah, I stand corrected.</comment></comment></comment>


Here's why men have boneless es - <AUTHOR> http://arstechnica.com/science/2016/12/heres-why-men-have-boneless-penises/ <comment><USER2> > The average amount of time between penetration and ejaculation for men is > less than two minutes.

Sad news for women everywhere.

<comment><AUTHOR> I doubt it's news to them...</comment></comment>


ToTok, an Emirati messaging app, is the latest escalation of a digital arms race - <AUTHOR> https://www.nytimes.com/2019/12/22/us/politics/totok-app-uae.html <comment><USER2> > ToTok amounts to the latest escalation in a digital arms race among wealthy > authoritarian governments, interviews with current and former American > foreign officials and a forensic investigation showed. The governments are > pursuing more effective and convenient methods to spy on foreign > adversaries, criminal and terrorist networks, journalists and critics — > efforts that have ensnared people all over the world in their surveillance > nets.

So... um... how does the fact that the NSA monitors everything we do square with this? Is it only an arms race when other governments try to do it, too?

(A more low effort response would be "Me trying to uninstall apps that the US government monitors" and a GIF link to that scene in The Devil Wears Prada where she throws her cell phone into the fountain.)

<comment><USER3> > Is it only an arms race when other governments try to do it, too?

Yes, that's literally what defines an "arms race".

<comment><USER2> I guess my question is whether or not the New York Times considers the US to be a wealthy authoritarian government.

<comment><USER4> Wealthy, yes. If the US is considered authoritarian then the word starts to lose any useful meaning.

<comment><USER5> How could a country with the largest prison population in the world not fall into at least some definition of authoritarian?

<comment><USER6> The government is elected by the people, it doesn't fit that definition.

An authoritarian type of government is a government that is in power by force.

Democracy and slavary existed together. Having a prison population 1 in 125 people doesn't matter because the people decided to elect someone to make that decision.

<comment><USER7> > The government is elected by the people, it doesn't fit that definition.

Is it though? First rebuttal that comes to mind is that, in the US, one can legally be elected without the having the majority of the counter, valid, votes.

Another one would be that one can be elected without the approval of the majority of the people who _could_ vote.

<comment><USER8> >Is it though? First rebuttal that comes to mind is that, in the US, one can legally be elected without the having the majority of the counter, valid, votes.

Yes, spend a couple of moments reading up on the EC. The rules haven't changed in a long time. Even if you don't like it, that's only how the president is elected. Federal representatives for the state are elected by getting the most votes in a district.

>Another one would be that one can be elected without the approval of the majority of the people who could vote.

Every election system that allows more than two candidates or allows people to skip voting has this flaw.

<comment><USER9> >Every election system that allows more than two candidates or allows people to skip voting has this flaw.

This is completely false, there are plenty of voting systems out there that allow for more than 2 people to run, yet still require a majority to vote for you. Here's a playlist explaining several: [https://www.youtube.com/playlist?list=PLNCHVwtpeBY4mybPkHEnR...](https://www.youtube.com/playlist?list=PLNCHVwtpeBY4mybPkHEnRxSOb7FQ2vF9c)

</comment></comment></comment></comment></comment></comment></comment></comment></comment><comment><USER10>
       So instead of paying hackers to gain
       access to a target's phone — the going
       rate is up to $2.5 million for a hacking
       tool that can remotely access Android
       phones, according to recent price lists
        — ToTok gave the Emirati government a
       way to persuade millions of users to hand
       over their most personal information for
       free.

Yup.

<comment><USER11> I wonder why HN can't implement proper css for quotes that doesn't have scrollbars.

</comment></comment><comment><USER12> There's really not much in this, is there? Have we not concluded that

A) Modern governments will do anything they possibly can to infiltrate and track not just their own citizens but those of foreign states

and

B) The modern average consumer will place convenience above just about everything else in their lives, but especially any concerns about how their data is handled or its effect on their personal privacy

That the government in question in A, in this case, accomplished so much simply via marketing is a testament to B.

There's some chatter in the article pointing to the nationalities of those working in Dark Matter and Pax (Americans, Europeans, Asians and some Emirates), as though it too is consequential. How much of a step is it to make to work for the NSA as a contractor writing code to monitor people, or a US tech giant writing tracking and psycho-analysis tech via ads and other patterns, to doing the same for a company elsewhere for presumably a higher paycheck?

You could argue that some in the NSA are doing so out of patriotism and their new behaviour constitutes betrayal, but the same can't be said for former workers of Google or similar, and especially not Palantir.

I'm not sure what I'm expected to be outraged about here? The lack of something blatant points to why a clickbait headline may have been chosen for publishing.

<comment><USER13> [removed, wasn't actually relevant to article]

<comment><USER14> > The Chinese government, in general

...has nothing to do with the app from this article made by "Breej Holding" "most likely a front company affiliated with DarkMatter, an Abu Dhabi-based cyberintelligence and hacking firm where Emirati intelligence officials, former National Security Agency employees and former Israeli military intelligence operatives work."

<comment><USER13> You're right, I got confused about where I was and what I read! Previous comment over-written.

</comment></comment></comment></comment><comment><USER15> Did they hijack TikTok searches by calling it ToTok? Not a bad strategy to piggyback on another app's popularity...

<comment><USER16> My theory has always been, Chinese will start behaving like Americans in protecting their IP once they reach a similar level of reach and sophistication with their products. It's starting...

Then the shoe will be on the other foot. Not that a couple hundred years ago it wasn't the same.

<comment><USER17> I doubt this will happen unless there is motivation from the central government to crack down on it. There are many cultural reasons shameless copying is less stigmatized in China than in the west (not necessarily a negative thing).

<comment><USER18> Britain used industrial espionage to gather the technology to reach their industrial revolution. The USA did it in turn. (It used to be a death sentence for skilled technologists to emigrate.)

[https://www.allaboutlean.com/industrial-espionage-and- revolu...](https://www.allaboutlean.com/industrial-espionage-and-revolution/)

[https://www.history.com/news/industrial-revolution-spies- eur...](https://www.history.com/news/industrial-revolution-spies-europe)

</comment></comment></comment></comment><comment><USER19> Is there a term for the kind of phrasing used in this headline? It seems to be becoming more and more common.

<comment><USER20> Others have suggested "click-bait", which is probably the most likely term for a headline specifically. It does assume the reasoning though, that it's being done to drive views at the direction of marketing. Which granted is not an unreasonable assumption, even media with very strong editorial policies that will push back against most interference may consider "mere headlines" be a place they can sacrifice to marketing, it's not uncommon these days in many places to actually have a few different headlines created, served up at random to a sample of initial browsers, and then have whichever one gets the most clicks selected as the one that is displayed generally.

However if you want a more general term one might be "purple prose", which can apply not just to headlines but to any text in an article. It's a subjective judgement, but the idea is when there is text that is so excessively ornate, emotional, extravagant, etc that it actually breaks the flow of your attention and reading of the content itself. Basically, prose that stops serving to convey the content and starts drawing attention to itself for the sake of itself. It comes from a passage in one of the poetic works of Horace (Quintus Horatius Flaccus, from Rome).

Again, it is subjective, because sometimes content really does deserve heavy levels of verbage and different people have different tolerances. But when flowery prose gets in the way of comprehension rather than enhancing it it's a good candidate.

<comment><USER21> So it's settled. It's purple clickbait.

</comment></comment></comment><comment><USER22> Wired had an article on this too: see [https://www.wired.com/story/totok- alleged-emirati-spy-app/#i...](https://www.wired.com/story/totok-alleged- emirati-spy-app/#intcid=recommendations_wired-homepage-right-rail- popular_38fa3ffb-d229-4453-a2f0-226697dd6462_popular4-1)

</comment><comment><USER23> Interesting, looks like it's removed from the Play Store if i try to visit the link:

[https://play.google.com/store/apps/details?id=ai.totok.chat&...](https://play.google.com/store/apps/details?id=ai.totok.chat&hl=en_US)

Here's the WebCache:

[https://webcache.googleusercontent.com/search?q=cache:9_TRm1...](https://webcache.googleusercontent.com/search?q=cache:9_TRm1KhueIJ:https://play.google.com/store/apps/details%3Fid%3Dai.totok.chat%26hl%3Den_US+&cd=1&hl=en&ct=clnk&gl=us&client=firefox-b-1-d)

<comment><USER12> As per the article

>On Thursday, Google removed the app from its Play store after determining ToTok violated unspecified policies.

<comment><USER23> I didn't imply it wasn't did I? I was just pushing the cached version for those curious like me about the app itself and how popular it was (knowing first hand is better than he said she said). On the other hand, I wish Google and Apple would just disable downloading instead of getting rid of the whole darn page.

</comment></comment></comment><comment><USER24> >ToTok amounts to the latest escalation in a digital arms race among wealthy authoritarian governments

Why wealthy? You don't need much wealth to create a chat app.

<comment><USER25> You need the majority of your citizens to have enough money to buy smartphones to install your app. You need to care about making it look like there is a velvet glove on your surveillance.

<comment><USER26> Or you can just straight up arrest a citizen and ask why they spoke ill about the police on WeChat: [https://twitter.com/Snowden/status/1208469511051075585](https://twitter.com/Snowden/status/1208469511051075585)

<comment><USER27> Don't some European countries do the same with tweets? (eg [https://www.theverge.com/2016/3/24/11297128/matthew-doyle- ar...](https://www.theverge.com/2016/3/24/11297128/matthew-doyle-arrest- muslim-tweet-brussels))

They're both laws covering disturbing public order, which doesn't make them that different from China (because deciding what "disturbs" the public is a slippery slope)

</comment></comment></comment></comment><comment><USER8> How predictable, an expose of an Emirates spy tool and all of the top comments are "what about the US tho".

<comment><USER28> Comments on all platforms are easily ambushed from groups trying to push a specific point of view. Unfortunately, this place also suffers from this on political news.

</comment></comment><comment><USER29> Hypothetically, if people just got off their devices, wouldn't this entire digital arms race disappear? No phones. Not even computers except the bare necessity use.

</comment><comment><USER30> > But the service, ToTok, is actually a spying tool, according to American > officials familiar with a classified intelligence assessment [...] It is > used by the government of the United Arab Emirates to try to track every > conversation, movement, relationship, appointment, sound and image of those > who install it on their phones.

Only way to fight against terrorists and pedophiles I hear... [/sarcasm]

</comment><comment><USER31> Here's my problem:

Let's suppose that I, or anyone else that reads Hacker News (like many of the YCombinator companies) creates an app (as many of them have), and this app becomes very popular in foreign countries, and goes on to get tons of users in those foreign countries...

OK, now let's suppose that a foreign newspaper, like the New York times, but a foreign version of that -- all of a sudden prints an article like this one, alleging, but not showing a comprehensive proof (like the ones in mathematics or logic) that there are all kinds of things unspeakably wrong with the app...

Well... maybe all of those things are true, but maybe they aren't...

How would I, as the end-user of such an app -- know for sure?

If the article isn't true... think of the gigantic swath of economic damage that would be caused by an article like this...

And if the article is true... well, it could be argued that you've done a great thing, and saved countless users' privacy rights...

That's quite the dichotomy of potential outcomes, based on what the truth of the matter actually is...

So here's my question to the NYT: In this era of Fake News (which you are no doubt aware of!), how do I prove to myself (as someone who is 50/50, that is, a skeptic who is willing to believe, if the proper set of facts is presented to them) that what you're saying is true?

?

Here are two things for the NYT to think about:

"Semper necessitas probandi incumbit ei qui agit"

The burden of proof lays on he who makes the claim

and

"Extraordinary claims require extraordinary evidence" -Carl Sagan

[https://en.wikipedia.org/wiki/Sagan_standard](https://en.wikipedia.org/wiki/Sagan_standard)

Saying that this stuff is true because the Intelligence Community says that it's true -- is not unlike making an "Argument to/from Authority":

[https://en.wikipedia.org/wiki/Argument_from_authority](https://en.wikipedia.org/wiki/Argument_from_authority)

I don't know who the Intelligence Community is (much like I don't know who other abstract entities are like "China" or even the "U.S." are) I don't know who that is!

I do know who a guy named "John Smith" is, and I do know that if John Smith raises his right hand in a court of law and under oath promises to "tell the truth, the whole truth, and nothing but the truth, under penalty of perjury", then John Smith's claim, when he makes it under those circumstances, carries a whole lot more weight (to me) than the claims of any nameless, faceless, unaccountable organization -- by whatever name they go by...

Due Process (enshrined in our Constitution, which the media does not seem to give these days to all parties, despite the Constitution giving them their right of Free Speech) presumes innocence until proven guilty, in a court of law!

Now, all of that being said... I am willing to believe this article -- but I require a higher standard of proof...

<comment><USER32> > How would I, as the end-user of such an app -- know for sure?

Monitor network traffic, or maybe even decompile the binary. What information is getting sent out of the app? Does it seem reasonable? If not, you have cause for concern.

Unfortunately, modern platforms like iOS make this very difficult, frequently in the name of protecting security and privacy, ironically. _This_ is a very big problem.

<comment><USER31> I agree.

But to broaden your point, it could be argued that this problem affects every single application that uses the network... How do I know that Chrome, Firefox, IE, or any other U.S. made piece of software -- doesn't leak private information, one way or another? Intentionally or unintentionally?

I've thought about this long and hard (as someone who wanted to publish an app), and the only two things I can come up with are:

1) Open the source code; that is, make it open source, and subject to peer review.

2) Inform users that due to the complexity inherent in computers and software/hardware/firmware stacks in this day and age, there cannot be a guarantee that your private information will stay private; thus, the simplest solution is don't put any private information on the app / don't assume that it will stay private.

<comment><USER32> Investigate each layer one at a time.

Not every end user will be able to independently verify every piece of software on their machine—going fully open source wouldn't do that either. But these types of investigations should be _open_ to the general public so that peer review can take place.

</comment></comment></comment></comment><comment><USER33> I'm still mad that I got shadow banned. All I did was post ukulele videos.

<comment><USER34> Unless I'm shadowbanned with you I can assure you that you're not.

<comment><USER24> The only way to read guys who are shadowbanned is to enable "showdead" in your profile. You won't automatically read their comments even if you are shadowbanned yourself if you don't enable the showdead option.

<comment><USER34> I have showdead on, but swiley's post doesn't display as dead for me.

</comment></comment></comment></comment><comment><USER26> I'm glad there isn't another UAE based app that stores billions of messages on its servers, effectively in plaintext, including all group chats, all desktop client chats, and overwhelming majority of one-on-one chats that aren't opt-in secret chats. /s

"BuT TelEGram HasnT beEn ComPRomIsED iN tHe wILd haS it???"

<comment><USER35> You're making a valid point, but you could probably reword it without the sarcasm.

<comment><USER26> You're right. I apologize. I'm just absolutely terrified about the inevitable hack -- when intelligence establishment gets their hands on everyone's private messages on Telegram servers, or when some hacker leaks them.</comment></comment></comment>


Airbnb allegedly purged more than 1,000 New York listings to rig survey - <AUTHOR> http://www.theguardian.com/technology/2016/feb/10/airbnb-new-york-city-listings-purge-multiple-apartment-listings <comment><USER2> Honestly, it couldn't be easier to find listings that violate the law. This stuff isn't abstract, you can go look for yourself and see how pervasive short-term full apartment rentals are.

I just looked on the block where I live, for a one day stay starting next Monday. Six listings. One was legit, one was borderline (daytime office), three state that the guest will explicitly have the whole apartment to themselves, and with the final one you can infer that the host will not be present. Even giving the last one the benefit of the doubt, 50% illegal.

OK, let's look at somewhere that guests might not want to stay because it's far from tourist attractions, the Lower East Side. Random block, 4 listings. 2 legit (I'm giving one the benefit of the doubt), 2 illegal.

OK, let's go even further away, Fort Greene. Random block. 2 units, one of them is actually not legit. Damn, I thought I would be done here.

South Slope, harder to find blocks with units. Pick one near the park. 2 units, I'll give both of them the benefit of the doubt, because it's not listed one way or the other.

If the fines were high enough, the state of New York could make a healthy profit on paying inspectors to rent out units for one night and levying fines against violators.</comment>


Ask YC: Are talented web developers hard to find? - <AUTHOR>

A quick glance at TechCrunch's Job Board shows lots of employers looking for talented web developers. Their requirements usually include all the latest web 2.0 technologies like JQuery, YUI, Dojo, Prototype, Rails, plus PHP, SQL, etc etc...<p>I don't work in the industry, and am not sure what its actually like out there: Is it hard to find people who possess all of these skills? <comment><USER2> It's not hard to find people who claim to possess those skills. It is hard to find people who are proficient in them, and exceedingly difficult to find people who are very good at them.

If you're hiring, and you live in a metro area, and you don't know any hackers yourself; then I would suggest frequenting some user groups in your area to get to know some of the people there. People who go to user groups are far more likely to be good developers. In fact, anyone who does programming related stuff in their free time is likely going to be much stronger than someone who just does it for the check.

If you're not hiring and are instead thinking about getting into the industry, then there's a corollary somewhere in my second paragraph for you.

<comment><USER3> I'll agree with this. If you're hiring or looking for a job - do attend user groups, found on Meetup.com or the site for your local networking site for your city, if there is one (I know for Boston there's MarksGuide.com and myeventguru.)

</comment></comment><comment><USER4> It's always difficult to hire people based on a long shopping list of very specific skills. What're the chances that someone who is in exact match is in your area AND looking for work AND wants to work for you? Especially when the skills overlap... Why would someone use Rails and PHP? I bet you could find plenty who had one or the other, etc.

</comment><comment><USER5> It's not too hard to find people who actually possess those skills, but it's very difficult to hire them: They don't generally apply to you, you apply to them.

<comment><USER4> LOL yeah, it was like that in the late 90s too. Then NASDAQ imploded and all the "rockstars" got a reality check.

<comment><USER2> Define the "Reality check" you're talking about?

When the dotcom burst happened, I got a big fat pay raise to guarantee I'd stick around at the firm I was at.

[Edit: I realize this wasn't terribly clear.

Those of us who knew what we were doing at the burst generally didn't hit hard times; but the developers who were in the industry because they saw it as an easy paycheck are the ones who got hurt. In that way, the burst was good for our industry - it helped separate the wheat from the chaff.]</comment></comment></comment>


Building Security in Maturity Model 6 - <AUTHOR> http://www.csoonline.com/article/2995452/security-awareness/security-information-sharing-gets-even-bigger-with-bsimm6.html <comment><AUTHOR> Some science was done on a large amount of organizations and the state of their security.</comment>


Python implementation of algorithms from Russell & Norvig's Artificial Intelligence book - <AUTHOR> http://code.google.com/p/aima-python/

<comment><USER2> Ah, the fond memories.

The professor I had for my AI course taught 100% via the Socratic method. It was one of the most challenging courses to prepare for because you had absolutely no idea just how deep you would personally be challenged to discuss for the next class. Really taught me how to be prepared thoroughly before an event.

Sadly, this caused us to go so in-depth into the material that we didn't get as far as I would have liked. But the course clearly helped ground me in ideas we explored more in later studies.

The good pseudocode in that book always struck me as fairly easy to reproduce in Python-- it's great to see some of that captured.

</comment><comment><USER3> Holy Dijkstra! That's some great practice :-P

Edit: not that Norvig needs it. How does he have the free time for this?

<comment><USER4> You probably don't need that much free time each day. Doing something constructive for even half an hour a day can really add up over weeks and months.

<comment><USER5> Programming for just half an hour every day is like having sex for just one minute every hour.

<comment><USER4> Not really - I've managed to write a basic raytracer within these constraints. But I doubt one could make a baby with your analogy :)

</comment></comment></comment></comment><comment><USER6> Here's a page with a number of implementations: <http://aima.cs.berkeley.edu/code.html></comment>


Twitter vs. Facebook - <AUTHOR> http://gigaom.com/2010/04/13/twitter-vs-facebook/

<comment><USER2> I can't find anything to really disagree with in this article, but I feel like it gives Twitter too much credit or Facebook too little.

What Twitter has done so far is A) implement a cool and novel idea and B) cross the chasm with it to mainstream adoption. Those things are no small achievement. Of course the Twitter guys should be proud of that and I don't mean to disparage that in any way. It puts them in very good company.

However Facebook has gone an extra several orders of magnitude to bring them into the league of the Amazons and Googles. Two things Facebook has done are A) develop a business leadership that can reinvent the whole product every 6 months and actually continue to grow wildly as a result of these radical tactical shifts and (even more impressively) B) an engineering team that can keep up with this and deal with scalability profiles that make Google look like child's play all the while pushing code to production on a daily basis as if they were some kind of garage startup.

Twitter on the other hand took _years_ to build a small team that was capable of scaling the existing minimalistic application, and what do they have now? A simple application and API that is somewhat mainstream, but also with only 1% of the hooks and stickiness that Facebook has established.

I don't even particularly like Facebook, I think it's a waste of time and that Mark Zuckerberg probably would have no moral qualms about reducing the entire world productivity down to a trickle by means of viral Zynga-style brain-crap. However I have to admit that there's always some hook that brings me back into Facebook (oh I gotta see so-and-so's photos or I'll just use Facebook Connect to log into this site because it's easy).

For Twitter to get there they need to figure out how to get past the geeky niche users, the vapid celebrities, and the huckster marketeers to find something wildly compelling to people. 140-characters is a nice idea, but it's not the kind of idea that's easy to make essential to people. Twitter has a long and steep road ahead and I fully expect an undervalued exit at some point rather than an IPO or massive profitability.

</comment><comment><USER3> I'm still desperately trying to figure out where exactly the Twitter platform is supposed to end. Are we supposed to believe that a platform these days is an all encompassing primarily closed ecosystem like Facebook (which I suppose is good from a business standpoint) or something more fast and loose like Twitter up until this new direction?

</comment><comment><USER4> Tumblr is better.

</comment><comment><USER5> All this twitter and facebook nonsense must be explainable using cuils.</comment>


EFI Secure Boot for Linux - <AUTHOR> http://www.rodsbooks.com/efi-bootloaders/secureboot.html <comment><USER2> Good article. I think Ubuntu uses Debian's shim. The package is called something *-signed-shim... I've tried self signing modules before, sometimes the "blue screen" won't accept my password (not a numlock or length or chatacter problem) and one time it did but I had to go true root which I don't want. IMO, chances of boot sector malware invading a personal computer is low, and even lower under linux given that I verfiy all keys and checksums before booting an iso. Can it be caught _on-the-fly_?</comment>