# Talk:Fundamental theorem of arithmetic

## Simple proof

This may be a simplification of the proof: Assume N=p1.p2...pn = q1.q2...qm is the smallest number with two prime factorizations. Assume q1 < p1. (p1-q1).p2...pn = q1.(q2...qm - p2...pn) is smaller than N, hence has unique factorization. q1 must then be a factor of p1-q1. So p1-q1=k.q1 for some integer k. Thus p1=(k+1)q1, which contradicts p1 being prime.

The simpler proof above (about uniqueness) is neat. Anyone willing to put that in the article (I'm no wiki-savvy)?

Except that the use of dot is uncluttered, consistent, and used by mathematicians as it is easier to write than cross. If you are worried by inconsistency with rest of wiki, then think of it as elegant variation, cf. Fowler Stan 23:49, 24 Feb 2004 (UTC)

## Joke: There are no Uninteresting Numbers

First, one must consider what numbers actually are interesting: 1 is unity, 2 is the smallest even and smallest prime number, 3 is the smallest odd prime number, 4 is the first whole number that is the square of another integer, 5 is significant in Biology (many vertebrates have some version or form of 5 digits), 6 is the first perfect number, 7 is a historically important number in many religions and mythologies, etc, etc....

Assume that one can eventually find and group all of the Uninteresting numbers. By the principly of Well Ordering, they can be arranged from smallest to largest. Let's look carefully at this group:

The first number that one comes across is the smallest of all the Uninteresting numbers known to exist in the universe. Well, that's interesting....

--Daydreamer302000 16:37, 28 February 2007 (UTC) PS: More thorough analysis of the Interesting Number Paradox

## Applications

The 5th and 6th lines of this paragraph are absolutely obscure

---

I second. Except I would venture to say that this entire section is poorly worded. I was unable to understand what was being said until I read it over and over and over again AND read some other references elsewhere.

I replaced now at some places in this paragraph 'factor' by 'divisor' (prime and non-prime) to perhaps diminish the confusions between a divisor and a prime factor. Hopefully this helps. Bob.v.R 17:07, 15 September 2005 (UTC)

## changeing the wording

How about changing the wording from

and there are no other such factorizations of 6936 or 1200 into prime numbers, except for reorderings of the above factors.

to

and there are no other possible factorizations of 6936 or 1200 into prime numbers, if we ignore the ordering of the factors.

## Paragraph sounds funny

Suppose there were a positive integer which can not be written as a product of primes. Then there must be a smallest such number: let's call it n. This number n cannot be 1, because of our convention above. It cannot be a prime number either, since any prime number is a product of a single prime, itself. So it must be a composite number. Thus n = ab where both a and b are positive integers smaller than n. >>>Since n was the smallest number for which the theorem fails, both a and b can be written as products of primes.<<< But then n = ab can be written as a product of primes as well, a contradiction.

There is something wrong with the paragraph above.

Since n is the smallest number which can not be written as a product of primes then both a nd b cannot be written as a product of primes. BUT didn't we say n is the smallest number which can not be written as a product of primes and the number n , a and b are positive integers. This means a is smaller than n and b is smaller than n. If a cannot be written as a product of primes and a is smaller than n then n CANNOT be the smallest number which can not be written as a product of primes. Hence a contradiction.

I prefer the first version. The first sentence of the second version jumps a step.

Charles Matthews 10:19, 29 Nov 2004 (UTC)

I think the original poster misunderstands the nature of the proof. It is simply not true that "n is smallest number which cannot be written as product of primes" implies "a and b cannot be written as a product of primes". In fact, it is just the opposite, by our choice of n. We are using well-ordering, and then showing that the existence of such a least n always leads to a contradiction. Revolver 20:33, 29 Nov 2004 (UTC)

## Proof by infinite descent

The unicity proof given as "by infinite descent" uses the division algorithm but not the Euclidean algorithm (much less its generalisation Bézout's lemma). In this way, it's more elementary (you need fewer facts about division, or if you prefer, you need not consider the consequences of the division algorithm in as much depth).

Who first came up with that argument?

Toby Bartels 09:35, 26 July 2006 (UTC)

## One

Might one say that the set of prime factors of one is the empty set? Thus, one is not prime because prime numbers are those where the set of factors has a cardinality of 1.

Several sources I've seen only state the Fundamental Theorem of Arithmetic for integers strictly greater than 1. Of course, I understand the argument about 1 being an empty product of primes, and I do admit that this convention makes the statement of the Fundamental Theorem of Arithmetic a lot cleaner. Nevertheless, I think it's worthwhile at least to mention that there are two prevalent conventions. In other words, I propose we leave in the explanation of 1 being the empty product, but that we also state that the result is often stated only for integers greater than 1. I don't mind making the change myself, but I wanted some kind of discussion first. For some odd reason, these small trivialities produce huge controversy and I don't want to step on anyone's toes. VectorPosse 06:32, 3 March 2007 (UTC)

The article as it is right now looks ok to me. The statement that 1 is the empty product is sufficient to avoid confusion. Some references indeed only state the Fundamental Theorem of Arithmetic for integers > 1 and just don't bother with 1. But there is no controversy as long as they do not claim that 1 is not representable as a product of primes. After all 0 would be the real exception. 85.0.100.1 09:18, 3 March 2007 (UTC)
I definitely see your point. I guess I'm just trying to answer the inevitable question that arises when a non-mathematically trained person (surely a person to whom this article ought to be directed) wonders why their textbook says n > 1 and this article says n = 1 too. I suppose they could click on Empty product, but I think this concept is a bit advanced for, say, a junior high student. Also, I'm concerned with the page Integer factorization. It makes no mention of empty product, so to find out why there's a difference between their textbook and the statement made on that page, they would have to click through Fundamental theorem of arithmetic and then on to Empty product. VectorPosse 10:49, 3 March 2007 (UTC)
This discussion started because someone claimed on the Integer factorization page that 1 does not have a prime factorization, which was incorrect. A link to the Empty product rule could possibly help to make this page clearer. Excluding the factorization of 1 from the Fundamental theorem of arithmetic would be a bad idea because it leads to other confusions. E.g., try computing GCDs by factoring the numbers first. What is the GCD of two integers that have no prime factor in common? How do you explain ${\displaystyle \gcd(1,n)=1}$? Junior high school is also the time when students are typically faced with ${\displaystyle a^{0}=1}$. While some might be confused by a claim that the empty product is 1 they should at least be able to understand that paradoxically looking definitions exist. 85.2.101.149 12:10, 3 March 2007 (UTC)
To make sure I'm being clear: I do not wish to remove n = 1 from any statement anywhere. I just want to acknowledge in the text that such a discrepancy does occur, and explain why n = 1 is okay despite what many textbooks say. (Or rather, what many textbooks fail to mention.) VectorPosse 12:20, 3 March 2007 (UTC)
I do think VectorPosse has got the right approach here. When there is disagreement among the sources we should document the disagreement not try to resolve it one way or the other. If might also be worth noting that 1 was consider prime a hundred years back, and that for most purposes it does not really matter which convention is chosen. --Salix alba (talk) 15:06, 3 March 2007 (UTC)
I mostly agree with VectorPosse too, though I'd give slightly different reasoning. We need to keep in mind that Wikipedia is a reference work, not a teaching tool. So I'm a little uncomfortable with trying to think of questions the reader might have. That line of thinking tends to lead to chatty, discursive writing not appropriate to an encyclopedia. But he's completely right that it's not our role to choose between commonly used conventions based on what we think is the more reasonable of them. Nor should we evade our responsibility to report the conventions by arguing that the n>1 sources don't explicitly say that it isn't also true for n equal to one. --Trovatore 17:25, 3 March 2007 (UTC)
While this debate is fascinating, the article contained a fatal flaw in the statement. I have fixed that, in a manner I hope is satisfactory and self-explanatory. (See paragraph three.) In the process, I added a reference to the incomparable Hardy & Wright, simultaneously introducing use of the lovely new {{Citation}} template and {{Harv}} references (which I find far superior to prior alternatives), per WP:CITET. Actually, I take advantage of the convenient {{Harvtxt}} template. Still, I have compromised, in that I strongly prefer Hardy & Wright's "positive integer" over the ambiguous "natural number", which my favorite authors (such as Mac Lane) begin at zero. (Since we already have Z+, why bother to have N without 0?) A tempting variation is Michael Artin's "every nonzero integer", with the form c×p1×⋯×pk, c = ±1, k ≥ 0; but this seems less common. --KSmrqT 00:39, 4 March 2007 (UTC)
If the theorem had been phrased as: "Every positive integer is the product of a unique multiset of primes", not only would the issue with the factorization of 1 have been skirted, but also the issue of the meaning of "unique product". After all, 6 = 2 × 3, but we also have that 6 = 3 × 2, so how unique is the product?  --LambiamTalk 06:31, 4 March 2007 (UTC)
Uniqueness is covered here by "Because multiplication is commutative, the order of factors is irrelevant." Hardy & Wright first introduce a standard form,
${\displaystyle n=p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}\qquad (a_{1}>0,a_{2}>0,\dots ,p_{1}
Theorem 2 (The fundamental theorem of arithmetic). The standard form of n is unique; apart from rearrangement of factors, n can be expressed as a product of primes in one way only.
However phrased, this requires less machinery than the introduction of multisets (also known as "bags"). I trust you concur with their elimination of an unnecessary distinction between composite and prime numbers, since both factor uniquely. --KSmrqT 08:36, 4 March 2007 (UTC)

## seems inconsistent

The 'fundamental theorem of arithmetic' states "every natural number greater than 1 can be written as a unique product of prime numbers". If 'product' implies multiplication, what are the factors of a prime number? The typical statement "a prime has no divisors other than 1 and itself", is contradictory because '1' is not considered prime, (if 1 is not prime then the 'ftoa' is not correct), and redundant because all natural numbers >1 are divisible by 1 and therefore this is not a qualifying condition. If the set M is all natural numbers>1, i.e. multiples of 1, then the primes are a subset P of M and can be defined as 'multiples of 1 only". Since 1 is not included in M it cannot be considered a prime.This gives a reason and not a convenience. The set M - P is C, the composites. The 'ftoa" could be restated "all composite natural numbers can be written as a unique product of prime numbers. phyti64.24.176.115 19:38, 19 March 2007 (UTC)

You are misinterpreting the word "product". A product need not have two or more numbers in it. Think of product as a list of numbers: if there is more than one number in the list, then you multiply them together. But if there's only one number, then the "product" consists of only that number. So a prime number can be written uniquely as a "list" of prime factors, namely the "list" containing just that prime number. This avoids questions about one. (See the discussion higher up on this page for the interpretation of the FTOA for the number one.) VectorPosse 00:55, 20 March 2007 (UTC)
You have to do better than just a bald statement here cob. We might be dysnumerate, but we ain't idiots. Of course, the idiots may well be the teachers who taught that "a product is the multiplication of two or more numbers." You need to produce a citation to back up your claim. Specifically, we are considering the integer "2", in the Fundamental Theorem. 203.161.102.82 (talk) 03:16, 18 February 2012 (UTC)
Yes, citations would be good—but this page is about primes, not about the definition of the word "product", so the citations don't belong here but elsewhere in the encyclopedia. I agree that the Product (mathematics) and Multiplication#Capital_Pi_notation could use some references. The books listed at Empty product might do the job. Jowa fan (talk) 12:50, 18 February 2012 (UTC)
I just ran across this problem myself. A product is the result of multiplication; multiplication, as defined by Wikipedia, as well as various other sources and dictionaries, is the "scaling of one number by another." Therefore, in order to be a product, it must be the result of the scaling of two numbers. Mind you that this Wikipedia belongs to everybody, not just the idiots or intellectuals. If we're talking about having a link to "Empty Product" having 0 factors and still being a multiplication, so 1 factor must also have a product, then it should be mentioned. This is Madness300 13:40, 30 June 2012 (UTC)
It's mentioned at Multiplication#Products_of_sequences. And I've just added a sentence to Product (mathematics). Does that clear things up? Jowa fan (talk) 01:43, 1 July 2012 (UTC)
That should clear things up for the average reader should they decide to go there. Thanks. This is Madness300 03:11, 2 July 2012 (UTC)

## Merge from Abnormal number?

Abnormal number has almost no content and I think it would fit better as a short comment in the proof section here. http://mathworld.wolfram.com/AbnormalNumber.html says "Hardy and Wright (1979) prove the fundamental theorem of arithmetic by showing that no abnormal numbers exist". PrimeHunter 15:16, 22 May 2007 (UTC)

## Proof by infinite descent

I rewrote some of this proof with the aim of greater clarity. (I had a little difficulty following the old version, probably because of my old age and exhausted brain cells:) ) I also thought that, since this is a purely number-theoretic result, it would be nice to offer a proof that doesn't require any fractions (as the original version did) and sticks purely to natural numbers. I hope that others find it easy to follow! This is my first venture into mathematical proofs on Wikipedia, so please forgive me if I've violated any policies or guidelines. Cheers.... Grover cleveland (talk) 05:19, 14 March 2008 (UTC)

## Some suggested changes

The article gives a very incomplete and somewhat misleading impression of the history of this result.

First: the sense in which Euclid "essentially" proved FTA should be spelled out: he proved Euclid's Lemma (p divides ab implies p divides a or p divides b). Assuming -- or easily proving -- the existence of factorizations into primes, FTA is easily seen to be equivalent to Euclid's Lemma.

Second: the induction proof presented here is not attributed to anyone. The way the article is written the only possible guess would be that it was due to Euclid, which is of course false (the notions of well-ordering and mathematical induction were not available to the ancient Greeks).

This raises the question: to whom is this induction proof (which I find very pretty) due? In the literature it seems to be most often attributed to Zermelo -- e.g. one can find several papers proving R a UFD implies R[t] a UFD by a similar argument (not by Gauss' Lemma) and the authors claim to be generalizing Zermelo. A look at the references on MathWorld shows that there is a 1934 paper by Zermelo on this, but also a 1933 paper by F.A. Lindemann and a 1928 paper by Hasse. That Zermelo (a famous German mathematician) was unaware of Hasse (a very famous German mathematician)'s paper published in Crelle (a leading German journal) six years before seems unlikely in the extreme, so some explanation of priority concerning this argument would be most desirable. Plclark (talk) 23:20, 16 June 2008 (UTC)Plclark

## Is this Inacurate?

<article quote>The fundamental therom of arithmetic states that every number greater than one can be written as a unique product of prime numbers."</article quote>

However prime numbers themselves cannot be written as a unique procuct of prime numbers. They are either the prodcut of 1 * themsleves (therefore not the product of primes) or simply themselves.

Am I missing something? 216.186.53.164 (talk) 19:50, 26 September 2008 (UTC)

When product refers to a product which doesn't necessarily have exactly two terms, it's usually allowed to have only one term (so the product is just the term). See for example Multiplication#Capital pi notation. PrimeHunter (talk) 20:31, 26 September 2008 (UTC)
I wouldn't have used such a trivial example to back up my case. Extending the short definition, "Capital pi notation" is a shorthand foreshadowing the use of calculation devices expressing the general case of multiplying sequential integers, as for example the factorial notation "n!" However, in real life one rarely -- if ever -- gets a situation where the factors of a number reduce to only one. In fact, the problem of the FToA as applied to prime numbers is the only mono-factorial example I have ever seen. So we need to see a reputable authority cited. 203.161.102.82 (talk) 03:36, 18 February 2012 (UTC)

## Why use a proof by infinite descent?

Just curious. It could easily be restated as a proof using the well-ordering property of Natural numbers (like the one above it), which I personally find more intuitive. It seems to me that it is inconsistent to use that property in one of the proofs, but not in the other, when it lends itself so easily in both.

It's not a problem; I'm just wondering if there's any real justification behind it. --72.177.97.222 (talk) 00:28, 8 February 2009 (UTC)

I see that the proof has been changed. Never mind, then. --72.177.97.222 (talk) 01:00, 10 February 2009 (UTC)

## Detailed proof of uniqueness removed

Which would be OK, except that I can't follow the new "proof" at all. Probably my senile old gray cells are failing me, but if someone could fill in the gaps and actually make the new proof a step-by-step affair then I would be happy. Otherwise I propose that the old proof be restored, perhaps in addition to the new. Cheers. Grover cleveland (talk) 07:18, 18 February 2009 (UTC)

The current proof does not start from first principles but appeals to what it calls "Euclid's algorithm: if p does not divide a then a and p are coprime so some linear combination ax + py = 1".
Now the result may well be derivable somehow from the proof of Euclid's algorithm, but the statement in italics is nowhere found within the current article on Euclid's algorithm. As best I can tell, nothing obviously equivalent to it is currently in that article: the words "coprime" and "prime" are completely absent from that article. I am therefore going to restore the original proof of uniqueness, since that spelled out all the steps. If anyone can fill in the gaps and make a pretty airtight connection between this article and the Euclid's algorithm so that no speculation is required, go ahead and remove it again. Grover cleveland (talk) 05:51, 2 March 2009 (UTC)
Well, consider the following proof: Let a and b be positive integers (we will prove that the GCD of a and b is of the form s*a + t*b). Let S be the set of all linear combinations of a and b (i.e S = {s*a + t*b | s, t are integers (possibly negative)}). Choose the smallest positive element in S; call this g. We assert that g is the GCD of a and b. Since g is in S, g is a liner combination of a and b. First of all, by "long division" we know that a = g*q + r for some q and r where either r is 0, or between 0 and g. But g = s*a + t*b (for some s and t) since g is in s. Therefore, a = a*q*s + t*b*q + r. In particular, r = a*(1 - q*s) + b*(-t*q). So that r must be 0 (otherwise, r would be an element of S, since it is positive as well as being a linear combination of a and b. If r was not 0, r is in between 0 and g which contradicts the unique property of g). Therefore, a = g*q and g | a. Similarly, g | b. If c divides a and b, then c divides a*s + b*t for all s and t (if c divides a and b, then c | a + b and any multiple of either summand). Since g is of this form, c divides g. Therefore, this last result, shows that g is the greatest common divisor of a and b. Q.E.D. --PST 13:20, 2 March 2009 (UTC)
Now actually, proving uniqueness in this theorem, does not use the above result. Suppose a = p1* ...... *pn = q1* ...... *qm. Then p1 divides the product of the qi's; simple induction shows that p1 divides one particular qj (note that if p is a prime and if p divides a*b, then p divides either a or b; if you don't know this already, I can prove it for you). But, qj is a prime so p1 = qj (we assume no redundancy so p1 is assumed to not equal 1) (note also that here is where uniqueness comes in because we are corresponding p1 with a term "deep" inside the qj's). From here, the qi's are simply the pj's up to rearrangement (of course you have to show that the case n < m or the case m > n are both impossible; concluding that n = m and so uniqueness holds up to rearrangement). --PST 13:20, 2 March 2009 (UTC)
Thanks! It appears that this statement is included in Wikipedia at Bézout's identity, so I'll rewrite the main proof a little to reference that. Grover cleveland (talk) 22:33, 13 March 2009 (UTC)

## Jordan Holder theorem

It may be interesting to note, in the article, that a simple proof of the Fundamental Theorem of Arithmetic may be obtained by applying the Jordan Holder theorem to cyclic groups (or even abelian groups). --PST 03:55, 21 April 2009 (UTC)

Of course, to apply this theorem, we would have to prove it first, so the proof of the Fundamental Theorem of Arithmetic is probably a better alternative. On the other hand, this application of a theorem of group theory allows one to see that essentially simple groups are a generalization of prime numbers; just as prime numbers are the core of all numbers, finite simple groups are the core of all finite groups. --PST 03:58, 21 April 2009 (UTC)

## "Truly Unique" Prime Factors?

I came to this discussion page's main article on the wings of a search that got drawn into the "Unique-Prime-Factorization Theorem" redirection, because I had just been dabbling with the concept of those numbers whose ultimate factors are all single examples of any primes (all those that aren't there exactly once are absent). (Y'know, 1 = (), 2=(2), 3=(3), not 4 because it is (22), having two '2's in there, but 5=(5), and 6=(2x3) and thus also contains no more than one of each prime it has... the number 903,210 is similarly valid because it =(2x3x5x7x11x17x23).)

I know this is probably not the place to start such a discussion, and especially request a new page, but if someone reading this knows of what I'm talking about (my personal maths education peaks at the applied/physics-orientated university level, but pure maths didn't get beyond that of Further Education) and knows what one might call this kind of number other than my original idea of "unique prime factors", it might be worth a link adding for the next interested party who wanders along with this same idea. This being a mere leisure-search for me I'm probably not going to find out anyway, never mind come back to add this in, myself.

(Also, as there appear to be 60 "TUPF"s below 100, 607 below 1000 and 607,925 of them below 1,000,000, by my hopefully correct calculations, any idea what the number that seems to be approximately 0.607... is that might well be emerging as some sort of constant/ratio? I couldn't find anything obviously relevent when I searched around for that.)

Anyway, forgive my intrusion with this thought. Quite possibly there's nothing intrinsically special (c.f. the "No Uninteresting Numbers" joke) about this set. Save that they're all factors of some hypothetical and inifinity-esque number which is the product of all possible primes... (Or should that instead be that the perhaps slightly less than 40% of natural numbers in the infinite set which are non-TUPFs are special because they aren't factors of this???) 82.132.248.232 (talk) 23:11, 22 January 2011 (UTC)

You are talking of square-free numbers. (And 1 is one of them). --Daniel5Ko (talk) 23:39, 3 June 2011 (UTC)

## Exclusion of one

There is no purpose in excluding 1 from the domain of the theorem. Today there are dozens or hundreds of examples of books and authors including the 1 as being uniquely factorizable, albeit sometimes with mentioning "empty products" in scare quotes.

I think the lead should formulate the theorem including 1 as a very normal positive integer (because it is, and everything works out just fine. Who cares about historical baggage? Or is there really more to it?). --Daniel5Ko (talk) 23:39, 3 June 2011 (UTC)

It's good to have an opening paragraph which is accessible to non-mathematicians. Mentioning or assuming the empty product would probably make it less accessible so I think it's OK to only bring it up later in the lead, as it is now. PrimeHunter (talk) 00:59, 4 June 2011 (UTC)
Hmm, maybe. But: Where's your data supporting the claim that the current state of the article is more well-suited for non-mathematicians (whoever that might be) ? --Daniel5Ko (talk) 23:42, 4 June 2011 (UTC)
I was only referring to the opening paragraph. I don't have verifiable data, just knowledge that some people don't know about the empty product and object to it - but there are also people objecting to a prime being a "product" of one prime factor and I certainly wouldn't remove that from the opening paragraph. PrimeHunter (talk) 02:05, 5 June 2011 (UTC)
I think, especially because of the apparent weirdness, both the unary and nullary product have their place in the lead (not necessarily explicitely mentionioned in the first sentence, but required later for justifying the claim made in it). Probably there is a greater learning effect for everyone prematurely jumping to the conclusion that the claim is wrong; everyone else doesn't care anyway. (I have no supporting data either, but it sounds about right! :) ) --Daniel5Ko (talk) 22:50, 6 June 2011 (UTC)

## nomenclature/history quesstion

Who coined the phrase "fundamental theorem"? I don't think it's Euclid or Euler or Gauss.

Virginia-American (talk) 16:36, 17 December 2011 (UTC)

## Alternate proof is really the same proof

It looks to me like the only difference is the wording used and the fact that the "alternate" proof assumes Euclid's lemma without proving it.

Since the proof of Euclid's lemma can be found in its own main article, I think it makes sense either to simply reference Euclid's lemma here (removing its proof from this page, and condensing the two uniqueness proofs into one since they're really the same), or to build up the entire proof of unique prime factorization from first principles, with separate sections for the division algorithm, Bézout's identity, etc.

18.189.112.192 (talk) 06:44, 22 December 2011 (UTC)

I agree, the alternate proof is the same as the main proof but skips steps (including Euclids lemma) to pull "Therefore,p_1q_1|s-p_1q_1" out of nowhere. Further the proofs concludes by stating nothing more than it's initial assumption (that s is the smallst integer with a non-unique factorization) and thus does not prove anything. I have removed the alternate proof. 213.246.84.162 (talk) 11:43, 1 April 2012 (UTC) Ali
Someone has now added a good alternate proof so this discussion is now mute. 213.246.117.133 (talk) 08:43, 14 May 2012 (UTC) Ali

I've rewritten the article in an atttempt to make it clearer and less redundant.

Virginia-American (talk) 19:27, 1 July 2012 (UTC)

## Invalid "proof"

I delted Shrohaneinstein's "proof" becaues it doesn't make sense: It started out assuming the product of primes could have two different valuse, rather than assuming that two different products of primes could have the same value. Virginia-American (talk) 12:54, 2 July 2012 (UTC)

## Von Mangoldt function

A new section Fundamental_theorem_of_arithmetic#Illustration was recently added, claiming that "the fundamental theorem of arithmetic is encoded by the von Mangoldt function." I deleted it, because it had no citations and I couldn't see how it related to the rest of the article. It's been restored with an edit summary saying "Watch lectures by Terence Tao and you will see that it is not original research." I'd like to see:

1. an expression for the (i,j)-th term of the matrix;
2. a clear explanation of how this matrix "encodes the fundamental theorem"—sure, the products of the rows give the natural numbers, but we also need to know why the entires of the matrix will always be primes (this may become obvious once point 1 is addressed), and there's the question of uniqueness;
3. a citation to a reliable source for this material.

Jowa fan (talk) 03:08, 25 September 2012 (UTC)

Under the assumption that Jowa fan's questions are answered, I have a further objection to this material: it appears to be a method of generating prime factorizations of numbers (well, I think that's what it's supposed to be doing, but actually the section doesn't say: it says that the Von Mangoldt function is the only function with "this property", but actually no property is stated). If so, this is maybe interesting but not about the Fundamental Theorem of Arithmetic at all; it might be on topic in e.g. the article prime factorization. --JBL (talk) 20:52, 25 September 2012 (UTC)
The material seems way off topic, even provided a source can be found. Sławomir Biały (talk) 22:57, 26 September 2012 (UTC)

Ok, delete it then. Matsgranvik (talk) 16:35, 27 September 2012 (UTC)

## why is this theorem so important?

As someone who is interested in mathematics, but far too dumb to be a mathematician, I would appreciate a brief and generally comprehensible statement at the beginning as to the importance of this theorem. _Why_ is it so important that it has the name "fundamental theorem of arithmetic"? 120.18.223.66 (talk) 10:38, 6 September 2013 (UTC)