The economy of thought

[Hrvoje1]

As conciseness is one of main mathematical features, I would like to discuss one particular instance of it. Can someone please summarize in that context the usefulness of excluding number one from the set of prime numbers? As the definition of prime numbers would be more concise without it, ie if one was included, and in fact it was at the beginning, first great contributors to number theory who laid foundations to prime number theory considered it to be prime, exclusion was introduced later, without much change in the essence of the theory, so it must have payed off somehow in terms of development of shorter expressions of consequences of somewhat longer definition, and I would like to know all places where it showed to be the case. So, the definition is, for those who are really unfamiliar with the topic: prime numbers are natural numbers that are divisible by exactly two distinct divisors, by one and by themselves. The definition that would include one, would be like this: prime numbers are natural numbers that are divisible only by one and by themselves. Note that further shortening of the definition by omitting the crucial condition “only by” would be a blunder, since all natural numbers are divisible by one and by themselves, which would of course make the definition pointless.

I know there are already many answers online to the question “why 1 is not prime”, such as https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/ for example, but I didn’t find a satisfactory one.

[John Cuthber]

2 hours ago, Hrvoje1 said:

As conciseness is one of main mathematical features

Prove that before you go any further

[Bufofrog]

2 hours ago, Hrvoje1 said:

Can someone please summarize in that context the usefulness of excluding number one from the set of prime numbers?

Look at this article:  https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/

From the article: 

My mathematical training taught me that the good reason for 1 not being considered prime is the fundamental theorem of arithmetic, which states that every number can be written as a product of primes in exactly one way. If 1 were prime, we would lose that uniqueness. We could write 2 as 1×2, or 1×1×2, or 1⁵⁹⁴⁸²⁷×2. Excluding 1 from the primes smooths that out.

[Hrvoje1]

21 minutes ago, Bufofrog said:

Look at this article:  https://blogs.scientificamerican.com/roots-of-unity/why-isnt-1-a-prime-number/

From the article: 

My mathematical training taught me that the good reason for 1 not being considered prime is the fundamental theorem of arithmetic, which states that every number can be written as a product of primes in exactly one way. If 1 were prime, we would lose that uniqueness. We could write 2 as 1×2, or 1×1×2, or 1⁵⁹⁴⁸²⁷×2. Excluding 1 from the primes smooths that out.

That excerpt from the article is the main reason I posted it in the first place. And I knew that argument from my mathematical training too, ie before reading that article. But, if that is the only place where it shows to be useful, it is not a great economy, because you could have said there: “as a product of primes other than 1 in exactly one way”, and have smoother definition that does not exclude 1 in awkward way, right?

[studiot]

Definition

If a is a natural number then a isdivisible by 1 (the quotient bing a) and by a (the quotient being 1).
Any factor of a other than 1 or a is called a proper factor.
Any 'a' which has no proper factors is prime.

 

Eisenstein's theorem.

If p is a prime number, and a(x) = a₀ + a₁x + ...........a_nxⁿ  is any polynomial with integer coefficients such that

(i) a₀ is not divisible by p².
(ii) a₁, a₂....a_(n-1) are each divisible by p
(iii) a_n is not divisible by p

then (x) has no proper factor with integer coefficients.

This theorem would not hold if we allowed 1 to be prime.

 

 

 

 

[Hrvoje1]

Thanks studiot, at the moment, it seems to me that the summarization I was looking for, is rather trivial. Namely, nearly each theorem of the form that includes the statement “let p be prime” and further statement(s) that deal with divisibility of something by p, and I can imagine most of them do in some way, would most probably need to be rewritten so that the first statement states instead “let p be prime other than 1”, in order to maintain their point, right? And if not each, precisely which not?

Besides that, this definition you posted includes 1 to primes.

 

Edited November 29, 2020 by Hrvoje1
Typo

[wtf]

19 hours ago, Hrvoje1 said:

Can someone please summarize in that context the usefulness of excluding number one from the set of prime numbers?

I was going to explain that the reason 1 isn't prime is that 1 is a unit in the ring of integers. However, the SciAm article by  Evelyn Lamb that you linked already has a thorough discussion of this point. If it doesn't satisfy you then perhaps nothing will; because that's the deepest reason 1 isn't a prime. It's because it's a unit, meaning that it has a multiplicative inverse.

A ring is a system of numbers in which you can add and multiply, and there are additive inverses, but not necessarily multiplicative inverses. The integers are a ring because, for example, 5 doesn't have a multiplicative inverse. In the rationals 1/5 is the multiplicative inverse of 5 but 1/5 isn't an integer.

In a ring, an element that does happen to have a multiplicative inverse is called a unit. In the integers, 1 and -1 do happen to have multiplicative inverses. In fact in each case they are their own multiplicative inverse. It doesn't make sense to ask if a unit is a prime. for reasons that Ms. Lamb goes into. I suggest re-reading that part of the article. 

For reasons of simplicity for SciAm readers, she gives the more general definition of prime number but can't use the word ring or the term prime ideal, which is really what's going on here. When she says,  "Specifically, one important change was the development of sets of numbers beyond the integers that behave somewhat like integers." she is talking about rings. And the concept that defines what a prime is, is called a prime ideal.

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

@studiot, The article mentions Eisenstein's criterion, so that's related.

If you consider the multiples of 5 in the integers, namely the set {0, -5, 5, -10, 10, ...}, it has the property that if the product xy of two integers is in that set, then at least one of x and y must be in the set. 

If you consider the multiples of 6, namely {0, -6, 6, -12, 12, ...} it's possible for the product xy of two integers to be in the set yet neither of x and y is; for example, 3 and 4. 

Such sets are called ideals. The actual definition is more general and I'm simplifying too, but not as much as the SciAm article did. So the multiples of 5 are a prime ideal and the multiples of 6 aren't. 

Part of the definition of a prime ideal is that it can't be all of the original ring. But the set of multiples of 1 is all of the integers. So the ideal generated by 1 can't be a prime ideal. 

But again that's just because of the definition ... which is what you are unhappy about! So perhaps this will still not be satisfactory to you. 

The other reasons people gave here are good too. Unique factorization doesn't work with 1 as a prime unless you add phrases to qualify it. 

But really, if you didn't find the SciAm article satisfactory, I don't think there's going to be a better explanation. The real answer is that when we generalize the integers to rings, and we want to define what primes are, we do so in terms of prime ideals. And prime ideals don't include the multiples of 1 because the ideal generated by 1 (or any other unit) is the whole ring. And now we're into abstract algebra. But again, the SciAm article actually explains all that without using the words ring or prime ideal. I think you should just read the article again because it contains the answer you're looking for.

 

 

Edited November 29, 2020 by wtf

[Hrvoje1]

If we get back to fundamental theorem of arithmetic, and its unique factorization, it would be formally obstructed by admitting 1 as prime, but essentially not much, because everyone understands that (paraphrasing Lamb):

We don’t consider units to be either prime or composite because you can multiply by them without changing much

 

On the other hand, here:

5 hours ago, Hrvoje1 said:

and further statement(s) that deal with divisibility of something by p

maybe it was more precise to say that if these statements deal with indivisibility by p it would be problematic to maintain the point without excluding 1, because no number is indivisible by 1, which is similar argument to yours, of prime ideal.

Edited November 29, 2020 by Hrvoje1

[Hrvoje1]

...maybe it would be more precisely to say...

[studiot]

29 minutes ago, Hrvoje1 said:

...maybe it would be more precisely to say...

The explanation is not short and simple, however much we might like it to be so.

@wtf has offered a much more complete and modern answer than I did.  +1 for that.

It is not 'wrong' to consider 1 as a prime number, in fact some mathematicians still do so.

But there are consequences to taking this view.

For a thousand years or more everyone followed the ancient Greeks in allocating prime status to 1.

However when complex numbers began to be investigated and accepted some difficulties with desirable algebraic structures began to emerge.
This was subsequently generalised to the modern structure of abstract algebra including rings, ideals and so forth which was crystallised during the first half of the 20th century.
The results of this structure have proved very useful in laying down the rules for all sorts of later developments including wolfram alpha and other symbolic structures you can ask questions of, coding and cyphers, and much more background theory.
All at the 'cost' of excluding 1 from the primes.

Eisenstein is the only example in the real numbers (including integers) I could think of.

So the only way to get a full explanation is to study modern abstract algebra

Did you understand why this theorem would not be allowable in a modern structure ?

 

[Hrvoje1]

On 11/28/2020 at 4:20 PM, studiot said:

This theorem would not hold if we allowed 1 to be prime.

I am sorry, but this sounds a bit like a complete nonsense. Of course that it would still hold, you would just need to start differently, for example with:

“If p is a prime number greater than 1, and...”

6 hours ago, studiot said:

It is not 'wrong' to consider 1 as a prime number, in fact some mathematicians still do so

Precisely that is the reason why your previous claim is nonsense: it would be sad if a validity of a theorem could depend on personal preference regarding convention. Fortunately that is not the case in reality, because that would mean that the theorem does not hold for them.

6 hours ago, studiot said:

Did you understand why this theorem would not be allowable in a modern structure ?

No, I did not, why? I mean I studied its proof, just for fun and relaxation, but that did not answer your question.

 

[wtf]

3 hours ago, Hrvoje1 said:

Precisely that is the reason why your previous claim is nonsense: it would be sad if a validity of a theorem could depend on personal preference regarding convention. Fortunately that is not the case in reality, because that would mean that the theorem does not hold for them

Can you help me to understand your point? 

For example if I write 3 + 4 x 5 = ?, everyone agrees that the anseer is 23; because by convention we multiply first and then add. But that's just a convention. If we all agreed to add first then multiply, the answer would be 35. It's a purely arbitrary convention that could easily be different, as long as everyone agrees on what the convention is. As you know, the Internet is full of order-of-operations puzzles that are the result of poor math education regarding operator precedence.

If it makes you happy, 1 isn't a prime because that's the convention. We could make a different convention, regard 1 as prime, and adjust all the theeorems accordingly. It really makes no difference.

So what is your core issue or concern? It's just a convention. The reason for the convention can be motivated by higher algebra, but in the end it's still a convention and truly makes no substantive difference. It's similar to why we go on green lights and stop on red. At the dawn of the automobile age we could have adopted the opposite convention. It makes no difference as long as everyone agrees. 

Edited November 29, 2020 by wtf

[Hrvoje1]

On 11/29/2020 at 10:16 PM, wtf said:

Can you help me to understand your point? 

My point is that Eisenstein's criterion holds regardless of whether we consider 1 prime or not. You just have to express it a little bit differently, depending on your choice.

On 11/29/2020 at 10:16 PM, wtf said:

For example if I write 3 + 4 x 5 = ?, everyone agrees that the anseer is 23; because by convention we multiply first and then add. But that's just a convention. If we all agreed to add first then multiply, the answer would be 35. It's a purely arbitrary convention that could easily be different, as long as everyone agrees on what the convention is. As you know, the Internet is full of order-of-operations puzzles that are the result of poor math education regarding operator precedence.

I agree, validness that should not be arbitrary must not depend on something that is arbitrary, that was my point too. Is there some other point here?

On 11/29/2020 at 10:16 PM, wtf said:

We could make a different convention, regard 1 as prime, and adjust all the theeorems accordingly. It really makes no difference.

Yeah well, it obviously mattered something to those who established that convention, and their choice although arbitrary was not whimsical, I presume. And if it makes no difference to you, you shouldn’t have bothered to explain their reasons, which must be rational, and hence the subject of my interest.

If you include 1 once, in the definition of primes, and you have to exclude it hundreds of times afterwards, during the development of the number theory, in the premises of theorems, that makes no sense. That is why I asked how many such cases might be there, besides the fundamental theorem of arithmetic, and what is their common denominator. studiot showed up, and singled out just one, claiming later it is the only example in the real numbers he could think of. Now you mention all the theorems that should be adjusted accordingly, if the convention was changed. That does not necessarily contradict studiot, because of his constraint, but I guess the number is much bigger than just these two theorems.

[HallsofIvy]

If we allow 1 as a prime number, we would no longer have "unique prime factorization".

Now,  12= (2^2)(3).

If we allow 1 as a prime number,  12= (1)(2^2)(3)= (1^2)(2^2)(3)= (1^3)(2^2)(3)= (1^4)(2^2)(3)= ....

Edited December 1, 2020 by HallsofIvy

[Hrvoje1]

HallsofIvy, how did you manage to add so little new information to this discussion, almost nothing? I have searched for some new sources of information, for example:

https://math.stackexchange.com/questions/120/why-is-1-not-a-prime-number

Here user7530 says: Given how often "let p be an odd prime" shows up in theorems, sometimes I wonder if we'd be better off defining 2 as non-prime too

I read that as if 1 was prime, every theorem that now says “let p be an odd prime” would have to be adjusted to say “let p be a prime greater than 2”, and every one that now says “let p be a prime” would have to say “let p be a prime greater than 1”. That sounds to me like a compelling reason to exclude 1 by definition, even though this was not mentioned by Lamb at all.

 

Edited December 2, 2020 by Hrvoje1

[Hrvoje1]

As the discussion is not alive anymore, it is time for a small evaluation of performance of each participant. I will start with me, pointing out what I am satisfied with, and what I am less happy with. I made nice generalization of wtf’s example: validness of results of arithmetic operations is mandatory and therefore must not depend on conventions on how to write down arithmetic expressions, which are arbitrary. And it really does not depend, as long as conventions are strictly followed, that is another instance of general truth, which says that something that is not arbitrary, cannot depend on something that is. First instance of that truth was that validness of theorems cannot depend on definitions, which, however naturally chosen, are still conventions, and therefore, kind of arbitrary (not entirely, because noone wants unnatural choice). The interesting thing here in our case, is that perspective on which definition is more natural, rapidly changes with acquiring mathematical experience. Every newbie in elementary school wonders why is 1 excluded when presented for the first time with the definition of prime number, that is, if she or he is sufficiently perceptive, and sufficiently well informed, otherwise, one might leave school unaware of that fact. However, that wondering should quickly disappear if one is also presented with at least basics of number theory during school education, such as I was not. Because, if we get back to Eisenstein’s criterion for example, it is obvious that all three conditions regarding integer coefficients of polynomial are equally damaged by inclusion of 1, because not only there can be no case for (i) and (iii), if p is 1, but (ii) is also always the case, which is equally damaging to the point of theorem, and therefore possibility that p is 1 simply must be excluded. I immediately made nice generalization of studiot’s example, and correctly made conclusion that there is no reason to single out this theorem, regarding that matter. Because, there is a huge difference between conclusions: “every theorem that states...” and “just these two...”, or “just this one, which would even not hold if...?!”

Anyway, I am less proud of the fact that I tried to be more precise than it was possible, because, indivisibility is a kind of divisibility (a negated one), and, well, for a moment it seemed to me that condition (ii) is less damaged than the other two, which is not true.

I also think that I pointed out nicely the huge difference between these two explanations: “the deepest reason 1 isn't a prime is because it's a unit in the ring of integers” and “if it makes you happy, 1 isn't a prime because that's the convention, it really makes no difference, what is your problem?, it is just convention, and if you don’t understand what that means let me give you an example with traffic lights”. 

[Hrvoje1]

On 11/28/2020 at 4:20 PM, studiot said:

(i) a₀ is not divisible by p².

But it is divisible by p, and you left it out in condition (ii). Besides that, not only that this theorem holds, but it also doesn’t need any adjustments if 1 is considered prime. 1 is simply always skipped in the search for adequate prime that satisfies all three conditions, it doesn’t have to be excluded in the premise.

[Hrvoje1]

Since my generalization failed so miserably, I decided instead of that to exploit in constructive way information gathered here. So I considered two simple examples, x²+1 and 2x²+2x+1 and realized that although I couldn’t find a prime p that satisfies Eisenstein’s criterion for them, I also couldn’t factorize them, which I double checked here:

https://www.mathportal.org/calculators/polynomials-solvers/polynomial-factoring-calculator.php 

But everything is cool, because the theorem only states that if you can find it, then polynomial is irreducible, not the other way around. And moreover, you can still use the theorem indirectly by applying transformations that preserve reducibility, such as these for example:

x=y+1 => x²+1=y²+2y+2    for which p=2 satisfies the criterion

and reversing coefficients on that to achieve the second example. Proof that it works is here:

https://math.stackexchange.com/questions/1758745/prove-that-fx-is-irreducible-iff-its-reciprocal-polynomial-fx-is-irred 

Edited December 13, 2020 by Hrvoje1

[Hrvoje1]

OK, for those who didn’t quite get the concept of reversing coefficients, and what did mathguy actually mean when he said there: 

>>Observe also that p+q=n where p and q are the degrees of g and h respectively; then distribute the 1/xⁿ between g and h<<

let us consider another slightly different example, this time reducible:

2x²+3x+1=(x+1)(2x+1)  if we multiply both sides by 1/x² we get:

2+3x^-1+x^-2=(1+x^-1)(2+x^-1)  where we had 1+1=2 for p+q=n and 1/x² distributed between two factors, each received 1/x, now let y=x^-1 =>
2+3y+y²=(1+y)(2+y)  what we got is reciprocal polynomial to the original, that is obviously also reducible, and has reversed coefficients.

The logic is that if we started with irreducible one, such as y²+2y+2, which we know is irreducible by applying Eisenstein’s criterion, reversing coefficients gives 2x²+2x+1, the second example, which then also must be irreducible.