
Why isn’t the fundamental theorem of arithmetic obvious? (2011) - ColinWright
https://gowers.wordpress.com/2011/11/13/why-isnt-the-fundamental-theorem-of-arithmetic-obvious/
======
Someone
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.

~~~
psyklic
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.

~~~
fenomas
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?

~~~
psyklic
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.

~~~
fenomas
> 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.

~~~
psyklic
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".

------
cousin_it
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)

------
evanpw
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.

~~~
cousin_it
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.

~~~
delazeur
> 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.

~~~
jdc
I generally agree, however I would like to know for what cases you think the
skill of solving unfamiliar problems is dangerous.

~~~
delazeur
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.

------
ball_of_lint
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.

~~~
CarolineW
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.

------
cousin_it
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.

~~~
thaumasiotes
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.

~~~
ColinWright

        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.

~~~
evanpw
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).

~~~
ColinWright

        ... 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.

------
nilkn
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.

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

~~~
Double_Cast
Do mathematicians actually have an answer for this? Or is it considered an
open question.

~~~
jackmaney
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)

------
vladsotirov
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)

------
jackmaney
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)

~~~
n4r9
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.

------
PepeGomez
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.

~~~
ColinWright

        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.

~~~
bluecalm
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.

~~~
ColinWright

        > 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.

~~~
bluecalm
>>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.

~~~
jerf
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.

~~~
ellyagg
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.

------
mcguire
That's not my question. My question is, why is it the fundamental theorem of
_arithmetic_?

~~~
twic
Yeah! I do arithmetic most days, and i very rarely have to prime factorise
anything!

------
playerqq
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.

~~~
Someone
_" 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.)

~~~
playerqq
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!

------
zwischenzug
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.

------
lacker
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.

------
ingenter
Why isn't 2+2==4 obvious?

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

~~~
pavelrub
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.

~~~
ingenter
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.

~~~
pavelrub
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.

~~~
thaumasiotes
> 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.

~~~
jessriedel
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.

------
illivah
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.

~~~
owenjonesuk
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)

~~~
illivah
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?

~~~
mrkgnao
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? ;)

------
NobleSir
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).

~~~
Chinjut
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.

------
pjdorrell
My own attempt to deal with this issue (2013):
[http://thinkinghard.com/blog/UniqueFactorisation.html](http://thinkinghard.com/blog/UniqueFactorisation.html)

------
snake_plissken
I agree 50%. But, it becomes pretty obvious once you realize the significance
of prime numbers, their applications and the scarcity of them.

------
pflats
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.

~~~
tel
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".

------
curiousgal
I've recently learned about Gödel numbering and I have to admit; it's my
favourite application of this theorem. Sorry RSA.

~~~
Smaug123
You don't actually need FTA to perform Gödel numbering. ASCII will do just as
well.

~~~
curiousgal
Yes. But Gödel originally used FTA, no?

~~~
Smaug123
He probably did. But for a programmer, ASCII is already something we're very
familiar with, so why not use that?

~~~
curiousgal
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.

~~~
kazinator
"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?

~~~
JadeNB
> "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.

------
trengrj
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.

~~~
rahimnathwani
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.

~~~
AimHere
> 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.

~~~
vinchuco
Your comment has destroyed the browsing experience on mobile :)

~~~
AimHere
Hah. Apologies for that! I thought the downvotes were because it was a lame
joke, not that I was destroying the site!

~~~
umanwizard
FWIW, I downvoted you because it's a lame joke.

------
hackuser
This is from 2011.

------
TopologyDough
"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.

~~~
ColinWright
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.

~~~
TopologyDough
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.

~~~
ColinWright

        > 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.

